Jee Mains 2019 All Papers With Solutions
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FIITJEE Solutions to JEE(Main)-2019 Test Date: 9th January 2019 (First Shift)
PHYSICS, CHEMISTRY & MATHEMATICS Paper - 1 Time Allotted: 3 Hours
Maximum Marks: 360
Please read the instructions carefully. You are allotted 5 minutes specifically for this purpose.
Important Instructions: 1.
The test is of 3 hours duration.
2.
This Test Paper consists of 90 questions. The maximum marks are 360.
3.
There are three parts in the question paper A, B, C consisting of Physics, Chemistry and Mathematics having 30 questions in each part of equal weightage. Each question is allotted 4 (four) marks for correct response.
4.
Out of the four options given for each question, only one option is the correct answer.
5.
For each incorrect response 1 mark i.e. ¼ (one-fourth) marks of the total marks allotted to the question will be deducted from the total score. No deduction from the total score, however, will be made if no response is indicated for an item in the Answer Box.
6.
Candidates will be awarded marks as stated above in instruction No.3 for correct response of each question. One mark will be deducted for indicating incorrect response of each question. No deduction from the total score will be made if no response is indicated for an item in the answer box.
7.
There is only one correct response for each question. Marked up more than one response in any question will be treated as wrong response and marked up for wrong response will be deducted accordingly as per instruction 6 above.
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JEE-MAIN-2019 (9th Jan-First Shift)-PCM-2
PART – A (PHYSICS) 1.
A convex lens is put 10 cm from a light source and it makes a sharp image on a screen, kept 10 cm from the lens. Now a glass block (refractive index 1.5) of 1.5 cm thickness is placed in contact with the light source. To get the sharp image again, the screen is shifted by a distance d. Then d is: (A) 1.1 cm away fro the lens (B) 0 (C) 0.55 cm towards the lens (D) 0.55 cm away from the lens
2.
A resistance is shown in the figure. Its value and tolerance are given respectively by: (A) 270 , 10% (B) 27 k, 10% (C) 27 k, 20% (D) 270 , 5%
3.
Drift speed of electrons, when 1.5 A of current flows in a copper wire of cross section 5 mm2, is v. If the electron density in copper is 9 × 1028 /m3 the value of v in mm/s is close to (Take charge of electron to be = 1.6 × 10–19 C) (A) 0.02 (B) 3 (C) 2 (D) 0.2
4.
An L-shaped object, made of thin rods of uniform mass density, is suspended with a string as shown in figure. If AB = BC, and the angle made by AB with downward vertical is , then: 1 1 (A) tan (B) tan 2 2 3 2 1 (C) tan (D) tan 3 3
5.
6.
ˆ , where K is a constant. The general A particle is moving with a velocity v K(yiˆ xj) equation for its path is: (A) y = x2 + constant (B) y2 = x + constant 2 2 (C) y = x + constant (D) xy = constant
A mixture of 2 moles of helium gas (atomic mass = 4 u), and 1 mole of argon gas (atomic mass = 40 u) is kept at 300 K in a container. The ratio of their rms speeds Vrms (helium) , is close to: Vrms (arg on) (A) 3.16 (C) 0.45
(B) 0.32 (D) 2.24
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JEE-MAIN-2019 (9th Jan-First Shift)-PCM-3
7.
A current loop, having two circular arcs joined by two radial lines is shown in the figure. It carries a current of 10 A. The magnetic field at point O will be close to: (A) 1.0 × 10–7 T (B) 1.5 × 10–7 T (C) 1.5 × 10–5 T (D) 1.0 × 10–5 T
8.
A block of mass m, lying on a smooth horizontal surface, is attached to a spring (of negligible mass) of spring constant k. The other end of the spring is fixed, as shown in the figure. The block is initially at rest in a equilibrium position. If now the block is pulled with a constant force F, the maximum speed of the block is 2F F (A) (B) mk mk F F (C) (D) mk mk
9.
For a uniformly charged ring of radius R, the electric field on its axis has the largest magnitude at a distance h from its centre. Then value of h is: R R (A) (B) 5 2 (C) R (D) R 2
10.
Two coherent sources produce waves of different intensities which interfere. After interference, the ratio of the maximum intensity to the minimum intensity is 16. The intensity of the waves are in the ratio: (A) 16 : 9 (B) 25 : 9 (C) 4 : 1 (D) 5 : 3
11.
Surface of certain metal is first illuminated with light of wavelength 1 = 350 nm and then, the light of wavelength 2 = 540 nm. It is found that the maximum speed of the photo electrons in the two cases differ by a factor of 2. The work function of the metal (in eV) is 1240 close to: (Energy of photon = eV) (in nm) (A) 1.8 (B) 2.5 (C) 5.6 (D) 1.4
12.
Temperature difference of 120°C is maintained between two ends of a uniform rod AB of length 2L. Another bent rod PQ, of same cross-section as AB 3L and length , is connected across AB (See 2 figure). In steady state, temperature difference between P and Q will be close to: (A) 45°C (B) 75°C (C) 60°C (D) 35°C FIITJEE Ltd., FIITJEE House, 29-A, Kalu Sarai, Sarvapriya Vihar, New Delhi -110016, Ph 46106000, 26569493, Fax 26513942 website: www.fiitjee.com.
JEE-MAIN-2019 (9th Jan-First Shift)-PCM-4 13.
A gas can be taken from A to B via two different processes ACB and ADB. When path ACB is used 60 J of heat flows into the system and 30 J of work is done by the system. If path ADB is used work down by the system is 10 J. the heat flow into the system in path ADB is (A) 40 J (B) 80 J (C) 100 J (D) 20 J
14.
Consider a tank made of glass (refractive index 1.5) with a thick bottom. It is filled with a liquid of refractive index . A student finds that, irrespective of what the incident angle I (see figure) is for a beam of light entering the liquid, the light reflected from the liquid glass interface is never completely polarized. For this to happen, the minimum value of is 5 3 (A) (B) 3 5 5 4 (C) (D) 3 3
15.
Mobility of electron in a semiconductor is defined as the ratio of their drift velocity to the applied electric field. If, for an n-type semiconductor, the density of electrons is 1019 m–3 and their mobility is 1.6 m2/(V.s) then the resistivity of the semiconductor(since it is an ntype semiconductor contribution of holes is ignored) is close to: (A) 2 m (B) 4 m (C) 0.4 m (D) 0.2 m
16.
A plane electromagnetic wave of frequency 50 MHz travels in free space along the positive x-direction. At a particular point in space and time, E 6.3ˆj V / m. . The corresponding magnetic field B, at that point will be ˆ ˆ (A) 18.9 × 10–8 kT (B) 2.1 × 10–8 kT ˆ ˆ (C) 6.3 × 10–8 kT (D) 18.9 × 108 kT
17.
Three charges +Q, q, +Q are placed respectively, at distance, 0, d/2 and d from the origin, on the x-axis. If the net force experienced by +Q, placed at x = 0, is zero, then value of q is (A) –Q/4 (B) +Q/2 (C) +Q/4 (D) –Q/2
18.
A copper wire is stretched to make it 0.5% longer. The percentage change in its electric resistance if its volume remains unchanged is (A) 2.0% (B) 2.5% (C) 1.0% (D) 0.5%
19.
A sample of radioactive material A, that has an activity of 10 mCi (1 Ci = 3.7 × 1010 decays/s), has twice the number of nuclei as another sample of different radioactive material B which has an activity of 20 mCi. The correct choices for half-lives of A and B would then be respectively: (A) 5 days and 10 days (B) 10 days and 40 days (C) 20 days and 5 days (D) 20 days and 10 days
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JEE-MAIN-2019 (9th Jan-First Shift)-PCM-5 20.
A heavy ball of mass M is suspended from the ceiling of car by a light string of mass m (m << M). When the car is at rest, the speed of transverse waves in the string is 60 ms–1. When the car has acceleration a, the wave-speed increases to 60.5 ms–1. The value of a, in terms of gravitational acceleration g is closest to: g g (A) (B) 30 5 g g (C) (D) 10 20
21.
A conducting circular loop made of a thin wire, has area 3.5 × 10–3 m2 and resistance 10 . It is placed perpendicular to a time dependent magnetic field B(t) = (0.4T) sin (50 t). The field is uniform in space. Then the net charge flowing through the loop during t = 0 s and t = 10 ms is close to: (A) 14 mC (B) 7 mC (C) 21 mC (D) 6 mC
22.
An infinitely long current carrying wire and a small current carrying loop are in the plane of the paper as shown. the radius of the loop is a and distance of its centre from the wire is d (d >> a). If the loop applies a force F on the wire then: a (A) F = 0 (B) F d a2 (C) F 3 d
a (D) F d
2
23.
A block of mass 10 kg is kept on a rough inclined plane as shown in the figure. A force of 3 N is applied on the block. The coefficient of static friction between the plane and the block is 0.6. What should be the minimum value of force P, such that the block does not move downward? (take g = 10 ms–2) (A) 32 N (B) 18 N (C) 23 N (D) 25 N
24.
A parallel plate capacitor is made of two square plates of side ‘a’, separated by a distance d(d <
K 0 a 2 ln K d(K 1)
K0 a2 ln K d 1 K0 a2 (D) 2 d
(C)
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JEE-MAIN-2019 (9th Jan-First Shift)-PCM-6 25.
A rod of length L at room temperature and uniform area of cross section A, is made of a metal having coefficient of linear expansion /°C. It is observed that an external compressive force F, is applied on each of its ends, prevents any change in the length of the rod, when it temperature rises by TK. Young’s modulus, Y, for this metal is F F (A) (B) A T A( T 273) F 2F (C) (D) 2A T A T
26.
A bar magnet is demagnetized by inserting it inside a solenoid of length 0.2 m, 100 turns, and carrying a current of 5.2 A. The coercivity of the bar magnet is (A) 285 A/m (B) 2600 A/m (C) 520 A/m (D) 1200 A/m
27.
Three blocks A, B and C are lying on a smooth horizontal surface, as shown in the figure. A and B have equal masses, m while C has mass M. Block A is given an initial speed towards B due to which it collides with B perfectly inelastically. The combined mass collides with C, also perfectly inelastically. The combined mass 5 collides with C, also perfectly inelastically if th of the 6 initial kinetic energy is lost in whole process. What is value of M/m? (A) 5 (B) 2 (C) 4 (D) 3
28.
When the switch S, in the circuit shown, is closed, then the value of current i will be (A) 3 A (B) 5 A (C) 4 A (D) 2 A
29.
If the angular momentum of a planet of mass m, moving a round the Sun in a circular orbit its L, about the center of the Sun, its areal velocity is: L 4L (A) (B) m m L 2L (C) (D) 2m m
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JEE-MAIN-2019 (9th Jan-First Shift)-PCM-7
30.
m are connected at the two ends of a 2 massless rigid rod of length . The rod is suspended by a thin
Two masses m and
wire of torsional constant k at the centre of mass of the rodmass system (see figure). Because of torsional constant k, the restoring torque is = k for angular displacement . If the rod is rotated by 0 and released, the tension in it when it passes through its mean position will be 3k02 2k02 (A) (B) 2 k0 k 0 2 (C) (D) 2
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JEE-MAIN-2019 (9th Jan-First Shift)-PCM-8
PART –B (CHEMISTRY) 31.
Two complexes [Cr(H2O)6]Cl3 (A) and [Cr(NH3)6]Cl3 (B) are violet and yellow coloured respectively. The incorrect statement regarding them is (A) 0 values of (A) and (B) are calculated from the energies of violet and yellow light, respectively. (B) both are paramagnetic with three unpaired electrons (C) both absorbs energies corresponding to their complementary colours (D) 0 value for (A) is less than that of (B)
32.
The correct decreasing order for acidic strength is (A) NO2CH2COOH rel="nofollow"> FCH2COOH > CNCH2COOH > ClCH2COOH (B) FCH2COOH > NCCH2COOH > NO2CH2COOH > ClCH2COOH (C) CNCH2COOH > O2NCH2COOH > FCH2COOH > ClCH2COOH (D) NO2CH2COOH > NCCH2COOH > FCH2COOH > ClCH2COOH
33.
The major product of the following reaction is 1 AlH iBu2 R C N ? 2 H2O (A) RCOOH (B) RCONH2 (C) RCHO (D) RCH2NH2
34.
The highest value of the calculated spin-only magnetic moment (in BM) among all the transition metal complexes is (A) 5.92 (B) 6.93 (C) 3.87 (D) 4.90
35.
0.5 moles of gas A and x moles of gas B exert a pressure of 200 Pa in a container of volume 10 m3 at 1000 K. Given R is the gas constant in JK–1 mol–1, x is 2R 2R (A) (B) 4 R 4 R 4 R 4 R (C) (D) 2R 2R
36.
The one that is extensively used as a piezoelectric material is (A) tridymite (B) amorphous silica (C) quartz (D) mica
37.
Correct statements among a to d regarding silicones are (i) they are polymers with hydrophobic character (ii) they are biocompatible (iii) in general, they have high thermal stability and low dielectric strength (iv) usually they are resistant to oxidation and used as greases. (A) (i), (ii), (iii) and (iv) (B) (i), (ii) and (iii) (C) (i) and (ii) only (D) (i), (ii) and (iv)
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JEE-MAIN-2019 (9th Jan-First Shift)-PCM-9 38.
The major product of the following reaction is
(A)
(B)
(C)
(D)
39.
In general, the properties that decrease and increase down a group in the periodic table respectively are (A) atomic radius and electronegativity (B) electron gain enthalpy and electronegativity (C) electronegativity and atomic radius (D) electronegativity and electron gain enthalpy
40.
A solution of sodium sulphate contains 92 g of Na+ ions per kilogram of water. The Molality of Na+ ions in that solution in mol kg–1 is (A) 12 (B) 4 (C) 8 (D) 16
41.
The correct match between Item-I and Item-II is: Item – I Item –II (drug) (test) (a) Chloroxylenol (p) Carbylamine test (b) Norethindrone (q) Sodium hydrogen carbonate test (c) Sulpha pyridine (r) Ferric chloride test (d) Penicillin (s) Bayer’s test (A) a r, b p, c s, d q (B) a q, b s, c p, d r (C) a r, b s, c p, d q (D) a q, b p, c s, d r
42.
A water sample has ppm level concentration of the following metals: Fe = 0.2, Mn = 5.0, Cu = 3.0, Zn = 5.0. The metal that makes the water sample unsuitable for drinking is (A) Cu (B) Mn (C) Fe (D) Zn
43.
The anodic half-cell of lead-acid battery is recharged using electricity of 0.05 Faraday. The amount of PbSO4 electrolyzed in g during the process is: (Molar mass of PbSO4 = 303 g mol–1) (A) 22.8 (B) 15.2 (C) 7.6 (D) 11.4 FIITJEE Ltd., FIITJEE House, 29-A, Kalu Sarai, Sarvapriya Vihar, New Delhi -110016, Ph 46106000, 26569493, Fax 26513942 website: www.fiitjee.com.
JEE-MAIN-2019 (9th Jan-First Shift)-PCM-10 44.
Which one of the following statements regarding Henry’s law is not correct? (A) Higher the value of KH at a given pressure, higher is the solubility of the gas in the liquids (B) Different gases have different KH(Henry’s law constant) values at the same temperature (C) The partial pressure of the gas in vapour phase is proportional to the mole fraction of the gas in the solution (D) The value of KH increases with increase of temperature and KH is function of the nature of the gas.
45.
The following results were obtained during kinetic studies of the reaction 2A B products Experiment
[A] [B] Initial rate of reaction (in mol L–1) (in mol L–1) (in mol L–1 min-1) I 0.10 0.20 6.93 10–3 II 0.10 0.25 6.93 10–3 III 0.20 0.30 1.386 10–2 The time(in minutes) required to consume half of A is (A) 5 (B) 10 (C) 1 (D) 100 46.
Major product of the following reaction is
(A)
(B)
(C)
(D)
47.
The alkaline earth metal nitrate that does not crystallise with water molecules is (A) Mg(NO3)2 (B) Sr(NO3)2 (C) Ca(NO3)2 (D) Ba(NO3)2
48.
20 mL of 0.1 M H2SO4 is added to 30 mL of 0.2 M NH4OH solution. The pH of the resultant mixture is [pkb of NH4OH = 4.7] (A) 5.2 (B) 9.0 (C) 5.0 (D) 9.4
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JEE-MAIN-2019 (9th Jan-First Shift)-PCM-11 49.
Adsorption of a gas follows Freundlich adsorption isotherm. In the given plot, x is the x mass of gas adsorbed on mass m of the adsorbent at pressure p. is proportional to m
(A) p2 (C) p1/2
(B) p1/4 (D) p
50.
Which amongst the following is the strongest acid? (A) CHBr3 (B) CHI3 (C) CH(CN)3 (D) CHCl3
51.
The ore that contains both iron and copper is (A) copper pyrites (B) malachite (C) dolomite (D) azurite
52.
For emission line of atomic hydrogen from n1= 8 to nf = n, the plot of wave number 1 against 2 will be(The Rydberg constant, RH is in wave number unit) n (A) Linear with intercept - RH (B) Non linear (C) Linear with slope RH (D) Linear with slope – RH
53.
54.
The isotopes of hydrogen are (A) tritium and protium only (C) protium, deuterium and tritium
According to molecular orbital theory, which of the following is true with respect to Li2 and Li2 ? (A) Li2 is unstable and Li2 is stable (C) Both are stable
55.
(B) protium and deuterium only (D) deuterium and tritium only
(B) Li2 is stable and Li2 is unstable (D) Both are unstable
The major product of the following reaction is:
(A)
(B)
(C)
(D)
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JEE-MAIN-2019 (9th Jan-First Shift)-PCM-12 56.
Aluminium is usually found in +3 oxidation state. In contrast, thallium exists in +1 and +3 oxidation states. This is due to (A) inert pair effect (B) diagonal relationship (C) lattice effect (D) lanthanoid contraction
57.
The increasing order of pKa of following amino acids in aqueous solution is Gly, Asp, Lys, Arg (A) Asp < Gly < Arg < Lys (B) Gly < Asp < Arg < Lys (C) Asp < Gly < Lys < Arg (D) Arg < Lys < Gly < Asp
58.
Consider the reversible isothermal expansion of an ideal gas in a closed system at two different temperatures T1 & T2 (T1 < T2). The correct graphical depiction of the dependence of work done(w) on the final volume(v).
59.
(A)
(B)
(C)
(D)
Arrange the following amine in the decreasing order of basicity:
(A) I > II > III (C) III > I > I 60.
(B) III > I > II (D) I > III > II
The compounds A and B in the following reaction are, respectively
(A) A = Benzyl alcohol, B = Benzyl cyanide (B) A = Benzyl chloride, B = Benzyl cyanide (C) A = Benzyl alcohol, B = Benzyl isocyanide (D) A = Benzyl chloride, B = Benzyl isocyanide
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JEE-MAIN-2019 (9th Jan-First Shift)-PCM-13
PART–C (MATHEMATICS)
61.
The value of
cos x
3
dx is:
0
4 3 4 (D) 3
(A) 0 (C)
(B)
2 3
62.
The maximum volume (in cu.m) of the right circular cone having slant height 3 m is (A) 6 (B) 3 3 4 (C) (D) 2 3 3
63.
For x 2 n 1, n N (the set of natural numbers), the integral
x.
2 sin x
(A) loge (C)
64.
1 sin 2 x
2
2
1 sec 2 x 2 1 c 2
x 2 1 1 loge sec 2 c 2 2
(B)
1 loge sec x 2 1 c 2
x2 1 (D) loge sec c 2
If y y x is solution of the differential equation x 1 y is equal to 2 7 (A) 64 49 (C) 16
65.
dx is 1
2 sin x 2 1 sin 2 x 2 1
dy 2y x 2 satisfying y 1 1, then dx
1 4 13 (D) 16 (B)
Axis of a parabola lies along x – axis. If its vertex and focus are at distance 2 and 4 respectively from the origin, on the positive x – axis then which of the following points does not lie on it? (A) 5, 2 6 (B) (8, 6)
(C) 6, 4 2
(D) (4, –4)
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JEE-MAIN-2019 (9th Jan-First Shift)-PCM-14 x2 y2 . If the eccentricity of the hyperbola 1 is greater than 2, 2 cos2 sin2 then the length of its latus rectum lies in the interval: 3 (A) 3, (B) , 2 2 3 (C) 2, 3 (D) 1, 2
66.
Let 0
67.
For x R 0, 1 , let f1 x
1 1 , f2 x 1 x and f3 x be three given functions. If x 1 x a function, J (x) satisfies f2oJof1 x f3 x then J x is equal to: 1 f3 x x (D) f1 x
(A) f3 x
(B)
(C) f2 x 68.
2 Let a ˆi ˆj, b ˆi ˆj kˆ and c be a vector such that a c b 0 and a.c 4, then c is
equal to: 19 (A) 2
(B) 9
(C) 8 69.
70.
71.
(D)
If a, b and c be three distinct numbers in G.P. and a b c x b then x can not be (A) –2 (B) –3 (C) 4 (D) 2
3 2 3 If cos1 cos1 , x then x is equal to: 4 3x 4x 2 145 145 (A) (B) 12 10 146 145 (C) (D) 12 11 Equation of a common tangent to the circle, x 2 y 2 6x 0 and the parabola, y 2 4x , is: (A) 2 3y 12x 1 (B) 3y x 3 (C) 2 3y x 12
72.
17 2
(D)
3y 3x 1
The system of linear equation x y z 2, 2x 3y 2z 5, 2x 3y a 2 1 z a 1 then (A) is inconsistent when a = 4
(B) has a unique solution for a 3
(C) has infinitely many solutions for a = 4
(D) inconsistent when a 3
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JEE-MAIN-2019 (9th Jan-First Shift)-PCM-15
73.
If the fractional part of the number (A) 6 (C) 4
2403 k is , then k is equal to: 15 15 (B) 8 (D) 14
74
The equation of line passing through (–4, 1, 3), parallel to the plane x 2y z 5 0 and x 1 y 3 z 2 intersecting the line is: 3 2 1 x 4 y 3 z 1 x 4 y 3 z 1 (A) (B) 2 1 4 1 1 3 x 4 y 3 z 1 x 4 y 3 z 1 (C) (D) 3 1 1 1 1 1
75.
Consider the set of all lines px qy r 0 such that 3p 2q 4r 0 . Which one of the following statements is true? 3 1 (A) The lines are concurrent at the point , . 4 2 (B) Each line passes through the origin. (C) The lines are all parallel (D) The lines are not concurrent
76.
lim
1 1 y4 2
y 0
y4
(A) exists and equals
(C) exists and equals
1
1
(B) exists and equals
4 2
2 2
1
2 1
(D) does not exist
2 2
77.
The plane through the intersection of the plane x y z 1 and 2x 3y z 4 0 and parallel to y – axis also pass through the point: (A) (–3, 0, –1) (B) (–3, 1, 1) (C) (3, 3, –1) (D) (3, 2, 1)
78.
If denotes the acute angle between the curves, y 10 x 2 and y 2 x 2 at a point of their intersection, then tan is equal to 4 9 7 (C) 17
(A)
8 15 8 (D) 17
(B)
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JEE-MAIN-2019 (9th Jan-First Shift)-PCM-16
79.
80.
cos sin If A , then the matrix A 50 when , is equal to 12 sin cos 3 1 1 3 2 2 2 2 (A) (B) 1 3 1 3 2 2 2 2 3 1 1 3 2 2 2 2 (C) (D) 1 3 3 1 2 2 2 2
If the Boolean expression p q ~ pq is equivalent to p q , where , , , then the ordered pair , is :
81.
82.
(A) ,
(B) ,
(C) ,
(D) ,
5 students of a class have an average height 150 cm and variance 18 cm2. A new student, whose height is 156 cm, joined them. The variance (in cm2) of the height of these six students is: (A) 16 (B) 22 (C) 20 (D) 18 4 2 For any , , the expression 3 sin cos 6 sin cos 4 sin6 equals: 4 2 (A) 13 4 cos2 6 sin2 cos2 (B) 13 4 cos6
(C) 13 4 cos2 6cos4
(D) 13 4 cos4 2 sin2 cos2
83.
The area (in sq. units) bounded by the parabola y x 2 1, the tangent at the point (2, 3) to it and the y – axis is: 8 32 (A) (B) 3 3 53 14 (C) (D) 3 3
84.
Let a1,a2 ,.....,a30 be an A.P., S ai and T a 2i1 . If a5 27 a and S 2T 75 ,
30
15
i 1
then a10 is equal to (A) 52 (C) 47
i 1
(B) 57 (D) 42
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JEE-MAIN-2019 (9th Jan-First Shift)-PCM-17
85.
86.
if x 1 5, a bx, if 1 x 3 Let f :R R be a function defined as f x . Then f is: b 5x, if 3 x 5 30, if x5 (A) continuous if a 5 and b 5 (B) continuous if a 5 and b 10 (C) continuous if a 0 and b 5 (D) not continuous for any values of a and b
3 2i sin Let A 0 , : purely imaginary . Then the sum of the elements in A 2 1 2i sin is: 5 (A) (B) 6 2 3 (C) (D) 4 3
87.
Consider a class of 5 girls and 7 boys. The number of different teams consisting of 2 girls and 3 boys that can be formed from this class, if there are two specific boys A and B, who refuse to be the members of the same team, is: (A) 500 (B) 200 (C) 300 (D) 350
88.
Let and be two roots of the equation x 2 2x 2 0 , then 15 15 is equal to: (A) –256 (B) 512 (C) –512 (D) 256
89.
Three circles of radii a, b, c ( a < b < c ) touch each other externally. If they have x – axis as a common tangent, then: 1 1 1 1 1 1 (A) (B) a b c b a c (C) a, b, c are in A.P.
90.
(D)
a, b, c are in A.P.
Two cards are drawn successively with replacement from a well shuffled deck of 52 cards. Let X denote the random variable of number of aces obtained in the two drawn cards. Then P X 1 P X 2 equals: 49 169 24 (C) 169
(A)
52 169 25 (D) 169
(B)
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JEE-MAIN-2019 (9th Jan-First Shift)-PCM-18
JEE (Main) – 2019 ANSWERS PART A – PHYSICS 1.
D
2.
B
3.
A
4.
B
5.
C
6.
A
7.
D
8.
D
9.
B
10.
B
11.
A
12.
A
13.
A
14.
B
15.
B
16.
B
17.
D
18.
C
19.
B
20.
C
21.
BONUS
22.
D
23.
A
24.
A
25.
A
26.
B
27.
C
28.
B
29.
C
30.
C
PART B – CHEMISTRY 31.
A
32.
D
33.
C
34.
A
35.
D
36.
C
37.
D
38.
A
39.
C
40.
C
41.
C
42.
B
43.
C
44.
A
45.
B
46.
D
47.
D
48.
B
49.
C
50.
C
51.
A
52.
D
53.
C
54.
B
55.
A
56.
A
57.
C
58.
A
59.
B
60.
D
PART C – MATHEMATICS 61.
A
62.
D
63.
D
64.
C
65.
B
66.
A
67.
A
68.
A
69.
D
70.
A
71.
B
72.
D
73.
B
74.
C
75.
A
76.
A
77.
D
78.
B
79.
C
80.
A
81.
C
82.
B
83.
A
84.
A
85.
D
86.
D
87.
C
88.
A
89.
A
90.
D
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JEE-MAIN-2019 (9th Jan-First Shift)-PCM-19
HINTS AND SOLUTIONS PART A – PHYSICS 1.
If u = –10 cm v = +10 cm f = 5 cm 1 2 Glass plate shift = t 1 1.5 1 = 0.5 cm 3 So, new u = 10 – 0.5 = 9.5 cm 1 1 1 v u f 1 1 1 v 9.5 5 After solving we get, 47.5 47.5 2.5 v= . Hence, shift 10 0.55 cm 4.5 4.5 4.5
2.
Fact based.
3.
i = ne AVd 1.5 = 9 × 1028 × 1.6 × 10–19 × 5 × Vd Vd = 0.02
4.
Lets considered mass of each rod is m for stable equilibrium the torque about point O should be zero. Torque balance about O a a mg sin mg cos a sin 2 2 1 tan = 3 1 tan1 3
O a
mg
a mg
5.
6.
dx dy y ; x dt dt dx y dy x y2 = x 2 + c (VRMS )He MAr (VRMS )Ar MHe
40 10 = 3.16 4
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JEE-MAIN-2019 (9th Jan-First Shift)-PCM-20
7.
Magnetic field at centre of an area subtending angle at the centre
0I . 4r
1 1 B 0 10 2 2 5 10 4 3 10 4 = B 104 105 10 5 30 3 8.
K
When Vmax acceleration = 0 F x K
m
F
Apply work energy theorem W sp + W F = K.E.
1 1 F2 F 2 1 2 Kx 2 F.x K.E. ; K 2 mumax 2 2 K K 2 F2 1 2 mumax 2K 2 9.
;
F Vmax mK
Electric field kQx E 2 (x R2 )3/2 dE For maxima 0 dx R After solving we get, x 2
10.
I1 I2 16 ; I I 1 1 2 3 I1 5 I2
11.
I1 I2 I1 I2
4 1
I1 25 I2 9
1240 (KE)I 4x …(1) 350 1240 (KE)II x …(2) 540 (1) – (2) 1240 1240 3.542 2.296 = 3x 350 540 1.246 = 3x ; x = 0.41 = 2.296 – 0.41 = 1.886
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JEE-MAIN-2019 (9th Jan-First Shift)-PCM-21
12.
TA TB
T1 T2 3R 3 120 = 45 8R 5 8 5
T1
R/2 A
B R/2
R
T2
R/4
R/4 R
13.
As temperature at point A and C is same. Internal energy change will be same. Q – W = Q – W 60 – 30 = Q – 10 Q = 40 J
14.
Sin 90° = sin 1 Sin sin = 1.5 sin r tan = 1.5 1.5 tan 3 1 Sin 2 9 4
16.
17.
Use I = neAvd and
r
vd E
E C B E 6.3 10 27 B = 2.1 × 1019 C 3 108 kq d 2
18.
92 = 9 + 42 3 5
15.
Completely polarized
2
k 2 2
3d 2 4Q 4q 9 Q q 9
A 2 2 R A V R
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JEE-MAIN-2019 (9th Jan-First Shift)-PCM-22
dR 2d R dR Hence, 1%. R 19.
Activity = (number of atoms) 10 = A NA …(1) 20 = B NB …(2) NA = 2NB …(3) A 1 Solving we get, B 4
20.
v T/ M
g
g2 a2 60.5 g 60
2
1 a2 1 using by binomial approximation. 1 2 2 g 60 g a 30 1
21.
B(t) = 0.4 Sin 50 10 2
50 = 0.4 Sin = 0.4 100 Q Q q = R R 0.4 3.5 10 3 = = 140 mC 10 No option matching. 22.
Force on one pole m oI m = pole F 2 d2 x 2 strength Total force = 2F sin 2 o Im x oI mx = 2 2 2 2 [d2 a2 ] 2 d a d a = m2x = M = I a2 oIa 2 Total force = 2(d2 a2 )
oIa 2 2d2
F m
x
x
d x Current wire
–m F
d a
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JEE-MAIN-2019 (9th Jan-First Shift)-PCM-23 23.
For equilibrium of the block net force should be zero. Hence we can write. mg sin + 3 = P + friction mg sin + 3 = P + mg cos . After solving, we get, P = 32 N.
24.
Let’s consider a strip of thickness dx at a distance of x from the left end as shown in the figure. y d x a d y x a
C1 d
K y C2 x
dx a
0 adx k adx ; C2 0 (d y) y C1C2 k0 adx Ceq C1 C2 kd (1 k)y Now integrating it from 0 to a C1
a
0
25.
k0 adx d kd (1 k) x a
k0a2 nk d(k 1)
L = L T L Strain = T L
Y
;
F A T
26.
B = oH ; oni = oH 100 5.2 H 0.2 H = 2600 A/m
27.
Apply LMC (Linear Momentum Conservation) mv = (2m + M)v mv v 2m M Initial energy 1 mv 2 2 Final energy
1 mv (2m M) 2 2m M
2
Initial kinetic energy – Final kinetic energy = After solving, we get,
5 of initial kinetic energy. 6
M 4. m
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JEE-MAIN-2019 (9th Jan-First Shift)-PCM-24 28.
i3 + i2 = i1 20 v 10 v v 2 4 2 v = 10 V 10 i1 2 = 5 amp.
29.
Based on Kepler’s law. dA L dt 2m
30.
v
20V 20
10V 4
2
k ; = 0 × I T m2 3 k 2 T m 2 0 where I m 3 I 3 2 k = 0
m
L/3
2L/3
m/2
PART B – CHEMISTRY 31.
0 is calculated from the energies of absorbed radiation not from emitted radiation (complementary colour).
32.
Acid strength in this case varies directly with the electron withdrawing power of the groups attached to the -carbon of CH3COOH. The order of electron withdrawing tendency is NO2 > CN > F > Cl
33. H2 O R C N Al H R C N Al H RC N Al RCHO H2N Al
34.
Maximum number of unpaired electron of metal or metal ion in complexes = n = 5
s n n 2 35 5.916 5.92 35.
PV = nRT 200 10 = (0.5 + x)R 1000 4R On solving x 2R
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JEE-MAIN-2019 (9th Jan-First Shift)-PCM-25 36.
Materials those produce electric current when they are put under mechanical stress are called piezoelectric materials.
37.
Silicones are polymers and hydrophobic due to presence of alkyl groups. They are used as greases as some of them are cyclic. Br
38.
Br Br2
EtOH OEt Br
39.
40.
On moving down a group, electronegativity decreases and atomic radius increases for representative elements.
w 1000 Molality of Na 2 ( Na2SO4 contains two Na+ ions) W M 92 1000 2 8 23 1000
41
42.
The permissible level in ppm unit is Fe = 0.2 Mn = 0.05 Cu = 3 Zn = 5 Mn is higher
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JEE-MAIN-2019 (9th Jan-First Shift)-PCM-26
43.
Pb s SO 24 PbSO4 2 e
For 2F current passed, PbSO4 deposited = 303 g/mol 0.05 303 7.6 g 2 For 0.05 F: PbSO4 deposited = 44.
Gases having higher KH value are less soluble.
45.
R = k[A]x [B]y 6.93 10–3 = k (0.1)x (0.2)y (i) 6.93 10–3 = k (0.1)x (0.25)y (ii) 1.386 10–2 = k (0.2)x (0.3)y (iii) y = 0 (from (i) & (ii)), zero order w.r.t. B x = 1 (from (i) & (iii)) First order wrt A 6.93 10–3 = k (0.1) k = 6.93 10–3 min-1 Cl
46.
Cl Cl
O
NH2
H2N
NH NH2
O
O
HCl
O
Et3N HCl Et3 NHCl Cl
O Free radical polymerization
NH
n
NH2 O
47.
Due to larger size of Ba2+ ion, Ba(NO3)2 can not hold water molecules during crystallization.
48.
H2SO4 2NH4OH NH4 2 SO 4 2H2O 2 m.m
6 m.m 2 m.m
pOH 4.7 log
2 m.m
4 5 2
pH = 14 – 5 = 9
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JEE-MAIN-2019 (9th Jan-First Shift)-PCM-27
49.
x KP1/n m x 1 or log logK logP m n 1 2 1 Slope n 4 2 x P1/2 m
50.
CH CN 3 C CN 3 H
Negative charge of the conjugate base C CN 3 in extensively delocalized through the C N group. 51.
Copper pyrites is CuFeS2 Malachite: CuCO3.Cu(OH)2 Azurite : 2Cu(CO)3.Cu(OH)2 Dolomite: CaCO3.MgCO3
52.
For emission line nf < ni 1 1 1 1 RZ 2 2 2 R 2 2 n 8 ni nf 1 1 or, RH 2 64 n R R H 2H 64 n 1 R RH 2 H n 64 y = mx + c Slope = - RH
53.
The isotopes are: 1 2 3 1H, 1 H and 1 H P,D,T
54.
Both Li2 and Li2 have same bond order. But the number of antibonding electrons is less in Li2 than in Li2
55.
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JEE-MAIN-2019 (9th Jan-First Shift)-PCM-28 56.
The outermost electron configuration of Tl is 6s26p1. The 6s electrons are strongly attracted towards the nucleus due to its more penetrating power and deshielding of d and f-electrons. Hence 6s electrons do not participate in bonding.
57.
58.
Let the gas is expanded from V1 to V at T1 and from V2 to V at T2 At T1 V |W 1| = nRT1 n nRT nV nV1 V1 Similarly at T2 |W 2| = nRT2(nV - nV2) W 1 = nRT1 nV – nRT1 nV1 W 2 = nRT2nV – nRT2nV2 Slope of W 2 > Slope of W 1 As nRT2 > nRT1(T2 > T1) The intercept of W 2 is more negative than that of W 1 because V2 > V1.
59.
In(III), nitrogen atom undergoes sp3, in(I) sp2 hybridization. In(II), the lone pair participate in resonance. CH2Cl
60. HCHO HCl
CH2NC
AgCN
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JEE-MAIN-2019 (9th Jan-First Shift)-PCM-29
PART C – MATHEMATICS 2
cos x
61.
3
cos x
3
0
2
dx
3
2 cos x dx 0
2
3
2 cos x dx 0
2 4 2 (By wallis formula) 3 3 62.
3 r 2 h2 9
Volume of cone is
1 2 r h 3
1 h 9 h2 3 dv 1 9 3h2 0 dh 3 9 3h2 0 h2 3, h 3 1 V 6 3 2 3 3 V
2 sin x
1 1 cos x
h
r
1
2 sin x 2 1 1 cos x 2 1 63.
x
2
2
x 2 1 sin 2 dx x x2 1 cos 2
x 2 1 x tan dx 2 x2 1 Let t 2xdx 2dt 2 tan t dt n sec t c x2 1 n sec c 2
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JEE-MAIN-2019 (9th Jan-First Shift)-PCM-30
64.
dy 2y x 2 dx dy 2 yx dx x
x
This is linear differential equation in
dy dx
2
dx Integrating factor e x x2 Solution of differential equation is yx 2 x 3 dx x4 c 4 Curve passes through (1, 1) 3 then c 4 4 x 3 yx 2 4 1 Put x 2 yx 2
4
1 3 2 1 y 4 4 49 y 16
65.
Vertex is (2, 0) a=2 Any general point on given parabola can be taken as 2 2t 2 , 4t t R .
(8, 6) does not lie on this. 66.
x2 y2 1 cos2 sin2 e 2 (given) sin2 4 cos2 1 tan2 4 tan2 3 , 3 2 sin2 2 tan sin Latus rectum 2 cos e2 4 1
for , , 2 tan sin is increasing function 3 2 Hence latus rectum 3, FIITJEE Ltd., FIITJEE House, 29-A, Kalu Sarai, Sarvapriya Vihar, New Delhi -110016, Ph 46106000, 26569493, Fax 26513942 website: www.fiitjee.com.
JEE-MAIN-2019 (9th Jan-First Shift)-PCM-31 67.
x R 0,1
1 1 , f2 x 1 x, f3 x x 1 x Given f2 J f1 x f3 x f1 x
1 J f1 x f3 x
J f1 x 1 f3 x 1
1 1 x
x x 1 1 x 1 J x x 1 1 1 x 1 J x f3 x 1 x J f1 x
68.
ˆ c xiˆ yjˆ zkˆ a ˆi ˆj, b ˆi ˆj k, ac b 0 ˆi ˆj kˆ
1 1 0 ˆi ˆj kˆ 0 x
y
z
ˆi z ˆj z kˆ y x
1 z 0 z 1,
Also x y 1, and a.c 4 x y 4
x
3 5 , y 2 2
2 9 25 38 19 c x 2 y 2 z2 1 4 4 4 2 69.
70.
a ar ar 2 xar 1 r2 r 1 since a 0 so x; 1 r x r r 1 r , 2 2, x , 1 3, r 3 2 3 cos1 cos1 x 4 3x 4x 2 2 3 4 9 cos 1 1 2 1 3x 4x 9x 16x 2 2
1 9x 2 4 16x 2 9 2x 2 12x 2
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JEE-MAIN-2019 (9th Jan-First Shift)-PCM-32
6 9x 2 4 16x 2 9 Square both side 36 144x 4 81x 2 64x 2 36 144x 4 145x 2 145x 2 145 x4 x ,0 144 12 3 145 x hence x 4 12 ty x t 2
71.
3 t2
3
1 t2
t 3 3y x 3 By applying Crammer’s Rule 1 1 1
72.
D 2 3
2 2
2 3 a 1
3 a2 1 6 2 a 2 1 4 2
2
a 1 2 a 3 If a 3 system has unique solution If a 3 2x 3y 2z 3 1 Hence system is inconsistent for a 3 x y z 1 2x 3y 2z 1
100
73.
2403 23.2400 8. 1 15 15 15 15
8
100
2
C0 100C1 15 100C2 15 ....... 15
8 2 8 100 C1 15 100C2 15 ...... 15 Remainder is 8.
74.
x 4 y 3 z 1 a b c L x 2y z 5 0 x 1 y 3 z 2 L intersects 3 2 1 Let the line L be
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JEE-MAIN-2019 (9th Jan-First Shift)-PCM-33 3
0
1
a
b
c 0
3 2 1 2a 0b 6c 0 Also a 2b c 0 a b c 3 1 1 x 4 y 3 z 1 L is 3 1 1
px qy r 0
75.
3p 2q px qy 0 4 3 2 p x q y 0 4 4 3 1 x and y . 4 2 76.
n
1 x So,
1 nx when x 0
1 y4 1 2
lim y 0
y4 2
y4 2 2 y4
y4 2 1 1 8 2 1 4 y 8 4 2
77.
Equation of required plane is x y z 1 2x 3y z 4 0
1 2 x 1 3 y 1 0 since given plane is parallel to y – axis 3 1 0
1 3
Hence equation of plane is x 4z 7 0 78.
y x 2 2 and y 10 x 2 Meet at 2, 6 m1 4 and m2 4 8 tan 15
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JEE-MAIN-2019 (9th Jan-First Shift)-PCM-34
79.
cos sin A sin cos cos sin cos sin A2 sin cos sin cos cos 2 sin 2 A2 sin 2 cos 2 By using symmetry cos 50 sin 50 A 50 sin 50 cos 50
At
A 50
80.
12
25 25 3 sin cos sin cos 6 6 6 6 2 25 25 1 sin cos sin cos 6 6 6 6 2
1 2 3 2
Check all option repeatedly (i) A B ~ A B A B ~ A B A B A B i is correct
(ii) A B ~ A B A ~ A B f B f
(iii) A B ~ A B B (iv) A B ~ A B B A ~ A B f f only (1) is correct
81.
Let 5 students are x1, x 2 , x 3 , x 4 , x 5 Given x 2 i
x 5
x x 5
2
5 i
750
150
………..(1)
i 1
x 18 5
2 i
2
150 18 5
xi2 22500 18 5
x
2 i
112590
…………(2)
i 1
Height of new student = 156 (Let x6) Then x1 x 2 x 3 x 4 x5 x 6 750 156
x new
x1 x 2 x 3 x 4 x 5 x 6 906 151 6 6
……….(3)
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JEE-MAIN-2019 (9th Jan-First Shift)-PCM-35
x Variance (new)
2
i
x new
2
6 from equation (2) and (3) 2 112590 156 2 variance (new) 151 22821 22801 20 6 82.
2
3 1 sin2 6 1 sin 2 4 sin6
3 1 2 sin 2 sin2 2 6 6 sin 2 4 sin6 2
6
9 3 sin 2 4 sin
9 12 sin2 cos2 4 1 cos2
3
9 12 1 cos2 cos2 4 1 3 cos2 3 cos4 cos6 2
4
2
4
6
13 12cos 12cos 12 cos 12cos 4cos 13 4 cos6 3
83.
Area
3
xdy xdy
5
1
3
3
y5 y 1dy 4 1 5
y2 5y 2 4
3
3
2 3/2 y 1 3
(2, 3)
3
1
–1
5
9 25 2 15 2 25 16 8 4 3 3
30
84.
–5
15
S ai , T a2i1 , a5 27, S 2T 75 i 1
i 1
Let ai a i 1 D
S a1 a2 a3 .................. a30 T a1 a3 a5 ............... a29 2T 2a1 2a3 2a5 .................. 2a29 S 2T a2 a1 a 4 a3 a 6 a5 .............. a30 a29 75 = 15D But S 2T 75 15D 75 D 5 Now a5 27 a 4D 27 a 27 20 a 7 now a10 a 9D 7 45 52
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JEE-MAIN-2019 (9th Jan-First Shift)-PCM-36 85.
For x = 1 R.H.L = a + b L.H.L = 5 So to be continuous at x = 1 a+b=5 …………(i) for x = 3 R.H.L. = b + 15 L.H.L = a + 3b b + 15 = a + 3b a + 2b = 15 ………….(ii) for x = 5 R.H.L = 30 L.H.L = b + 25 b + 25 = 30 b = 5. From equation (ii) a = 10 but a 10 and b 5 does not satisfied equation (i) So f (x) is discontinuous for a R and b R
86.
z
3 2i sin 1 2i sin 1 2i sin 1 2i sin
3 4 sin 8i sin 2
z
1 4 sin2 For purely imaginary real part should be zero. i.e. 3 4 sin2 0 . 3 2 2 2 2 , , , Sum of all values is 3 3 3 3 3 3 3
i.e. sin
87.
Number of ways = Total number of ways without restriction – When two specific boys are in 7 C3 5C2 350 If team without any restriction, total number of ways of forming team is two specific boys B1, B2 are in same team then total number of ways of forming team 5 5 equals to C1 C2 50 ways total ways = 350 – 50 = 300 ways
88.
x 2 2x 2 0 x 1 1
2
3 i 4
x 1 i 2 e
15 , 15
15
2
3 2cos 15. 4
1 28 2 256 2
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JEE-MAIN-2019 (9th Jan-First Shift)-PCM-37 C1
89.
C3
Length of direct common tangent for circle C1 and C2 is AB
a b
2
a b
2
For C2 and C3 Length of direct common tangent is BC
a c
2
a c
b
2
For C1 and C3 Length of direct common tangent is AC
b c
2
b c
A
C2 a B
c C
2
AB BC AC
a b
2
2
a b
b c
2
b c
a c
2
2
a c
2
ab ac bc 1 1 1 c b a 90.
4 48 24 2 52 52 169 4 4 1 P x 2 52 52 169 25 P x 1 P x 2 169 P x 1
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FIITJEE Solutions to JEE(Main)-2019 Test Date: 9th January 2019 (Second Shift)
PHYSICS, CHEMISTRY & MATHEMATICS Paper - 1 Time Allotted: 3 Hours
Maximum Marks: 360
Please read the instructions carefully. You are allotted 5 minutes specifically for this purpose.
Important Instructions: 1.
The test is of 3 hours duration.
2.
This Test Paper consists of 90 questions. The maximum marks are 360.
3.
There are three parts in the question paper A, B, C consisting of Physics, Chemistry and Mathematics having 30 questions in each part of equal weightage. Each question is allotted 4 (four) marks for correct response.
4.
Out of the four options given for each question, only one option is the correct answer.
5.
For each incorrect response 1 mark i.e. ¼ (one-fourth) marks of the total marks allotted to the question will be deducted from the total score. No deduction from the total score, however, will be made if no response is indicated for an item in the Answer Box.
6.
Candidates will be awarded marks as stated above in instruction No.3 for correct response of each question. One mark will be deducted for indicating incorrect response of each question. No deduction from the total score will be made if no response is indicated for an item in the answer box.
7.
There is only one correct response for each question. Marked up more than one response in any question will be treated as wrong response and marked up for wrong response will be deducted accordingly as per instruction 6 above.
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JEE-MAIN-2019 (9th Jan-Second Shift)-PCM-2
PART –A (PHYSICS) 1.
At a given instant, say t = 0, two radioactive substances A and B have equal activates. R R The ratio B of their activities. The ratio B of their activates after time t itself decays RA RA with time t as e–3t. If the half-life of A is n2, the half-life of B is: (A) 4n2 (C)
n2 4
(B)
n2 2
(D) 2n2
2.
A power transmission line feeds input power at 2300 V to a step down transformer with its primary windings having 4000 turns. The output power is delivered at 230 V by the transformer. If the current in the primary of the transformer is 5A and its efficiency is 90%, the output current would be: (A) 50 A (B) 45 A (C) 35 A (D) 25 A
3.
The energy associated with electric field is (UE) and with magnetic field is (UB) for an electromagnetic wave in free space. Then: U (A) UE B (B) UE > UB 2 (C) UE < UB (D) UE = UB
4.
A force acts on a 2 kg object so that its position is given as a function of time as x = 3t2 + 5. What is the work done by this force in first 5 seconds? (A) 850 J (B) 950 J (C) 875 J (D) 900 J
5.
A particle having the same charge as of electron moves in a circular path of radius 0.5 cm under the influence of a magnetic field of 0.5 T. If an electric field of 100 V/m makes it to move in a straight path, then the mass of the particle is (given charge of electron = 1.6 × 10–19 C) (A) 9.1 × 10–31 kg (B) 1.6 × 10–27 kg –19 (C) 1.6 × 10 kg (D) 2.0 × 10–24 kg
6.
Two point charges q1( 10 C) and q2 ( 25 C) are placed on the x-axis at x = 1 m and x = 4 m respectively. The electric field (in V/m) at a point y = 3 m on y-axis is, 1 9 10 9 Nm2C2 take 4 0 2 ˆ 10 (A) (63iˆ 27i) (B) ( 63iˆ 27iˆ ) 10 2 ˆ 10 2 (C) (81iˆ 81i)
(D) ( 81iˆ 81iˆ ) 10 2
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JEE-MAIN-2019 (9th Jan-Second Shift)-PCM-3 7.
Expression for time in terms of G (universal gravitational constant), h (Planck constant) and c (speed of light) is proportional to: (A) (C)
hc 5 G Gh c5
(B) (D)
c3 Gh Gh c3
8.
Ge and Si diodes start conducting at 0.3 V and 0.7 V respectively. In the following figure if Ge diode connection are reversed, the value of Vo changes by: (assume that the Ge diode has large breakdown voltage) (A) 0.8 V (B) 0.6 V (C) 0.2 V (D) 0.4 V
9.
The top of a water tank is open to air and its water level is maintained. It is giving out 0.74 m3 water per minute through a circular opening of 2 cm radius in its wall. The depth of the centre of the opening from the level of water in the tank is close to: (A) 6.0 m (B) 4.8 m (C) 9.6 m (D) 2.9 m
10.
The energy required to take a satellite to a height ‘h’ above Earth surface (radius of Earth = 6.4 × 103 km) is E1 and kinetic energy required for the satellite to be in a circular orbit at this height is E2. The value of h for which E1 and E2 are equal is (A) 1.6 × 103 km (B) 3.2 × 103 km 3 (C) 6.4 × 10 km (D) 1.28 × 104 km
11.
Two Carnot engines A and B are operated in series. The first one, A, receives heat at T1(= 600 K) and rejects to a reservoir at temperature T2. The second engine B receives heat rejected by the first engine and, in turns, rejects to a heat reservoir at T3 (=400 K). Calculate the temperature T2 if the work outputs of the two engines are equal: (A) 600 K (B) 400 K (C) 300 K (D) 500 K
12.
A series AC circuit containing an inductor (20 mH), a capacitor (120 F) and a resistor (60 ) is driven by an AC source of 24 V/50 Hz. The energy dissipated in the circuit in 60 s is: (A) 5.65 × 102 J (B) 2.26 × 103 J (C) 5.17 × 102 J (D) 3.39 × 103 J
13.
A particle is executing simple harmonic motion (SHM) of amplitude A, along the x-axis, about x = 0. When its potential energy (PE) equals kinetic energy (KE), the position of the particle will be A A (A) (B) 2 2 2 A (C) (D) A 2 FIITJEE Ltd., FIITJEE House, 29-A, Kalu Sarai, Sarvapriya Vihar, New Delhi -110016, Ph 46106000, 26569493, Fax 26513942 website: www.fiitjee.com.
JEE-MAIN-2019 (9th Jan-Second Shift)-PCM-4 14.
A mass of 10 kg is suspended vertically by a rope from the roof. When a horizontal force is applied on the rope at some point, the rope deviated at an angle of 45° at the roof point. If the suspended mass is at equilibrium, the magnitude of the force applied is (g = 10 ms–2) (A) 200 N (B) 140 N (C) 70 N (D) 100 N
15.
A 15 g mass of nitrogen gas is enclosed in a vessel at a temperature 27°C. Amount of heat transferred to the gas, so that rms velocity of molecules is doubled, is about: [Take R = 8.3 J/K mole] (A) 0.9 kJ (B) 6 kJ (C) 10 kJ (D) 14 kJ
16.
In a Young’s double slit experiment, the slits are placed 0.320 mm apart. Light of wavelength = 500 nm is incident on the slits. The total number of bright fringes that are observed in the angular range –30° 30° is: (A) 640 (B) 320 (C) 321 (D) 641
17.
Two plane mirrors are inclined to each other such that a ray of light incident to the first mirror (M1) and parallel to the second mirror (M2) is finally reflected from the second mirror (M2) parallel to the first mirror (M1). The angle between the two mirrors will be: (A) 45° (B) 60° (C) 75° (D) 90°
18.
A rod of length 50 cm is pivoted at one end. It is raised such that if makes an angle of 30° fro the horizontal as shown and released from rest. Its angular speed when it passes through the horizontal (in rad s–1) will be (g = 10 ms–2). 30 (A) (B) 30 2 20 30 (C) (D) 2 2
19.
A carbon resistance has a following colour code. What is the value of the resistance? (A) 530 k 5% (B) 5.3 M 5% (C) 6.4 M 5% (D) 64 k 10%
20.
One of the two identical conducing wires of length L is bent in the form of a circular loop and the other one into a circular coil of N identical turns. If the same current is passed in both, the ratio of the magnetic field at the central of the loop (BL) to that at the centre of B the coil (BC), i.e. L will be BC 1 (A) N (B) N 1 (C) N2 (D) 2 N FIITJEE Ltd., FIITJEE House, 29-A, Kalu Sarai, Sarvapriya Vihar, New Delhi -110016, Ph 46106000, 26569493, Fax 26513942 website: www.fiitjee.com.
JEE-MAIN-2019 (9th Jan-Second Shift)-PCM-5 21.
A rod of mass ‘M’ and length ‘2L’ is suspended at its middle by a wire. It exhibits torsional oscillations; If two masses each of ‘m’ are attached at distance ‘L/2’ from its centre on both sides, it reduces the oscillation frequency by 20%. The value of ratio m/M is close to: (A) 0.77 (B) 0.57 (C) 0.37 (D) 0.17
22.
Charge is distributed within a sphere of radius R with a volume charge density A (r) 2 e 2r /a , where A and a are constants. If Q is the total charge of this charge r distribution, the radius R is: a 1 Q (A) a log 1 (B) log 2aA 2 1 Q 2aA a 1 a 1 (C) log (D) log 1 2 2 2aA 1 Q 2aA
23.
A parallel palate capacitor with square plates is filled with four dielectrics of dielectric constants K1, K2, K3, K4 arranged as shown in the figure. The effective dielectric constant K will be: (K 1 K 3 ) (K 2 K 4 ) K1 K 2 K 3 K 4 (K K 2 ) (K 3 K 4 ) (C) K 1 K1 K 2 K 3 K 4
(A) K
(K1 K 2 ) (K 3 K 4 ) 2(K 1 K 2 K 3 K 4 ) (K1 K 4 ) (K 2 K 3 ) (D) K 2(K 1 K 2 K 3 K 4 )
(B) K
24.
The pitch and the number of divisions, on the circular scale, for a given screw gauge are 0.5 mm and 100 respectively. When the screw gauge is fully tightened without any object, the zero of its circular scale lies 3 divisions below the mean line. The readings of the main scale and the circular scale for a thin sheet, are 5.5 mm and 48 respectively, the thickness of this sheet is (A) 5.755 mm (B) 5.950 mm (C) 5.725 mm (D) 5.740 mm
25.
A musician using an open flute of length 50 cm produces second harmonic sound waves. A person runs towards the musician from another end of a hall at a speed of 10 km/h. If the wave speed is 330 m/s, the frequency heard by the running person shall be close to: (A) 666 Hz (B) 753 Hz (C) 500 Hz (D) 333 Hz
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JEE-MAIN-2019 (9th Jan-Second Shift)-PCM-6 26.
In a car race on straight road, car A takes a time ‘t’ less than car B at the finish and passes finishing point with a speed ‘v’ more than that of car B. Both the cars start from rest and travel with constant acceleration a1 and a2 respectively. Then ‘v’ is equal to 2a1a1 (A) t (B) 2a1a 2 t a1 a 2 a a2 (C) a1a2 t (D) 1 t 2
27.
The magnetic field associated with a light wave is given, at the origin, by B = B0 [sin(3.14 × 107)ct + sin(6.28 × 107)ct] If this light falls on a silver plate having a work function of 4.7 eV, what will be the maximum kinetic energy of the photo electrons? (A) 6.82 eV (B) 12.5 eV (C) 8.52 eV (D) 7.72 eV
28.
In the given circuit the internal resistance of the 18 V cell is negligible. If R1 = 400 , R3 = 100 and R4 = 500 and the reading of an ideal voltmeter across R4 is 5 V, then the value of R2 will be (A) 300 (B) 450 (C) 550 (D) 230
29.
In a communication system operating at wavelength 800 nm, only one percent of source frequency is available as signal bandwidth. The number of channels accommodated for transmitting TV signals of band width 6 MHz are (Take velocity of light c = 3 × 108 m/s, h = 6.6 × 10–34 J-s) (A) 3.75 × 106 (B) 3.86 × 106 5 (C) 6.25 × 10 (D) 4.87 × 105
30.
The position co-ordinates of a particle moving in a 3-D coordinates system is given by x = a cost y = a sint and z = at The speed of the particle is: (A)
2 a
(B) a
(C)
3 a
(D) 2a
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JEE-MAIN-2019 (9th Jan-Second Shift)-PCM-7
PART –B (CHEMISTRY) 31.
The entropy change associated with the conversion of 1 kg of ice at 273 K to water vapours at 383 K is: (Specific heat of water liquid and water vapours are 4.2 kJ K–1 kg–1 and 2.0 kJ K–1 kg–1, heat of liquid fusion and vapourisation of water are 334 kJ kg–1 and 2491 kJ kg–1, respectively) (log 273 = 2.436, log 373 = 2.572, log 383 = 2.583) (A) 7.90 kJ K–1 kg–1 (B) 2.64 kJ K–1 kg–1 –1 –1 (C) 8.49 kJ K kg (D) 9.26 kJ K–1 kg–1
32.
For the following reaction the mass of water produced from 445 g of C57H110O6 is 2 C57H110O 6 s 163 O2 g 114 CO2 g 110H2O I (A) 490 g (B) 445 g (C) 495 g (D) 890 g
33.
The major product formed in the following reaction is:
(A)
(B)
(C)
(D)
34.
Which of the following conditions in drinking water causes methemoglobinemia? (A) > 50 ppm of lead (B) > 50 ppm of chloride (C) > 50 ppm of nitrate (D) > 100 ppm of sulphate
35.
The major product of the following reaction is:
(A)
(B)
(C)
(D)
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JEE-MAIN-2019 (9th Jan-Second Shift)-PCM-8 36.
37.
The major product obtained in the following reaction is:
(A)
(B)
(C)
(D)
The major product of the following reaction is:
(A)
(B)
(C)
(D)
38.
The correct match between item I and item II is Item I Item II (a) Benzaldehyde (p) Mobile phase (b) Alumina (q) Adsorbent (c) Acetonitrile (r) Adsorbate (A) a q, b p, c r (B) a r, b q, c p (C) a q, b r, c p (D) a p, b r, c q
39.
The metal that forms nitride by reacting directly with N2 of air is (A) K (B) Li (C) Rb (D) Cs
40.
For coagulation of arsenious sulphide sol, which one of the following salt solution will be most effective? (A) BaCl2 (B) AlCl3 (C) NaCl (D) Na3PO4 FIITJEE Ltd., FIITJEE House, 29-A, Kalu Sarai, Sarvapriya Vihar, New Delhi -110016, Ph 46106000, 26569493, Fax 26513942 website: www.fiitjee.com.
JEE-MAIN-2019 (9th Jan-Second Shift)-PCM-9 41.
The complex that has highest crystal field splitting energy() is (A) [Co(NH3)5(H2O)]Cl3 (B) K2[CoCl4] (C) [Co(NH3)5Cl]Cl2 (D) K3[Co(CN)6]
42.
The pH of rain water is approximately (A) 5.6 (C) 7.0
43.
(B) 7.5 (D) 6.5
Consider the following reversible chemical reactions: K1 A 2 g B 2 g ....... 1 2 AB g K2 3 A 2 g 3B2 g ....... 2 6 AB g
The relation between K1 and K2 is 1 (A) K1K2 = 3 3 (C) K 2 K1 44.
(B) K 2 K 13 (D) K1K2 = 3
The correct sequence of amino acids present in the tripeptide given below is
(A) Val – Ser – Thr (C) Leu – Ser – Thr
(B) Thr – Ser – Val (D) Thr – Ser – Leu
45.
For the reaction, 2A B products , when the concentrations of A and B both were doubled, the rate of the reaction increased from 0.3 mol L–1 s–1 to 2.4 mol L–1 s–1. When the concentration of A alone is doubled, the rate increased from 0.3 mol L–1 s–1 to 0.6 mol L–1 s–1. Which one of the following statements is correct? (A) Total order of the reaction is 4 (B) Order of the reaction with respect to B is 2 (C) Order of the reaction with respect to B is 1 (D) Order of the reaction with respect to A is 2
46.
The products formed in the reaction of cumene with O2 followed by treatment with dil. HCl are:
(A)
(B)
(C)
(D)
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JEE-MAIN-2019 (9th Jan-Second Shift)-PCM-10 47.
The tests performed on compound X and their inferences are: Test Interference (a) 2, 4-DNP test Colorued precipitate (b) Iodoform test Yellow precipitate (c) Azo-dye test No dye formation Compound ‘X’ is
(A)
(B)
(C)
(D)
48.
If the standard electrode potential for a cell is 2 V at 300 K, the equilibrium constant (K) for the reaction 2 Zn s Cu2 aq Zn aq Cu s At 300 K is approximately (R = 8 JK–1 mol–1, F = 96000 C mol–1) (A) e-80 (B) e–160 320 (C) e (D) e160
49.
The temporary hardness of water is due to (A) Na2SO4 (B) NaCl (C) Ca(HCO3)2 (D) CaCl2
50.
In which of the following processes, the bond order has increased and paramagnetic character has changed to diamagnetic? (A) NO NO (B) N2 N2 (C) O 2 O2
51.
(D) O 2 O 22
Which of the following combination of statements is true regarding the interpretation of the atomic orbitals? (1) An electron in an orbital of high angular momentum stays away from the nucleus than an electron in the orbital of lower angular momentum. (2) For a given value of the principal quantum number, the size of the orbit is inversely proportional to the azimuthal quantum number h (3) According to wave mechanics, the ground state angular momentum is equal to 2 (4) The plot of Vs r for various azimuthal quantum numbers, shows peak shifting towards higher r value (A) (1), (4) (B) (1), (2) (C) (1), (3) (D) (2), (3)
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JEE-MAIN-2019 (9th Jan-Second Shift)-PCM-11 52.
Which of the following compounds is not aromatic?
(A)
(B)
(C)
(D)
53.
Good reducing nature of H3PO2 is attributed to the presence of (A) Two P – OH bonds (B) One P – H bond (C) Two P – H bonds (D) One P – OH bond
54.
The correct statement regarding the given Ellingham diagram is
(A) At 1400oC, Al can be used for the extraction of Zn from ZnO (B) At 500oC, coke can be used for the extraction of Zn from ZnO (C) Coke cannot be used for the extraction of Cu from Cu2O (D) At 800oC Cu can be used for the extraction of Zn from ZnO 55.
The transition element that has lowest enthalpy of atomisation is (A) Fe (B) Cu (C) V (D) Zn
56.
The increasing basicity order of the following compounds is (1)
(2)
(3)
(4)
(A) (4) < (3) < (2) < (1) (C) (1) < (2) < (3) < (4)
(B) (4) < (3) < (1) < (2) (D) (1) < (2) < (4) < (3)
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JEE-MAIN-2019 (9th Jan-Second Shift)-PCM-12 57.
When the first electron gain enthalpy (egH) of oxygen is -141 kJ/mol, its second electron gain enthalpy is (A) a more negative value than the first (B) almost the same as that of the first (C) negative, but less negative than the first (D) a positive value
58.
At 100oC, copper(Cu) has FCC unit cell structure with cell edge length x A . What is the approximate density of Cu(in g cm–3) at this temperature? [Atomic mass of Cu = 63.55 u] 205 105 (A) 3 (B) 3 x x 211 422 (C) 3 (D) 3 x x
59.
A solution containing 62 g ethylene glycol in 250 g water is cooled to -10oC. If Kf for water is 1.86 K kg mol–1, the amount of water(in g) separated as ice is (A) 48 (B) 32 (C) 64 (D) 16
60.
Homoleptic octahedral complexes of a metal ion ‘M3+’ with three monodentate ligands L1, L2 and L3 absorb wavelengths in the region of green, blue and red respectively. The increasing order of the ligand strength is (A) L3 < L1 < L2 (B) L3 < L2 < L1 (C) L1 < L2 < L3 (D) L2 < L1 < L3
o
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JEE-MAIN-2019 (9th Jan-Second Shift)-PCM-13
PART–C (MATHEMATICS) 61.
The sum of the following series 9 12 22 3 2 12 12 22 32 42 15 12 22 .... 52 1 6 ... up to 15 terms, is: 7 9 11 (A) 7820 (B) 7830 (C) 7520 (D) 7510
62.
For each x R, let lim
x x x sin x x
x 0
x
be the greatest integer less than or equal to x. Then
is equal to
(A) sin1 (C) 1 63.
(B) 0 (D) sin 1
Let f : 0,1 R be such that f xy f x f y for all x, y 0,1 , and f 0 0 . If y y x satisfies the differential equation,
equal to (A) 4 (C) 5
dy 1 3 f x with y 0 1, then y y is dx 4 4 (B) 3 (D) 2
64.
If x sin1 sin10 and y cos1 cos10 , then y – x is equal to: (A) (B) 7 (C) 0 (D) 10
65.
If 0 x (A) 2 (C) 3
, then the number of values of x for which sin x – sin 2x + sin 3x = 0, is 2 (B) 1 (D) 4
66.
93 Let z0 be a root of the quadratic equation, x 2 x 1 0 . If z 3 6iz81 0 3iz 0 , then arg z is equal to (A) (B) 4 3 (C) 0 (D) 6
67.
The area of the region A x, y :0 y x x 1 and 1 x 1 in sq. units, is: (A)
2 3
(C) 2
1 3 4 (D) 3 (B)
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JEE-MAIN-2019 (9th Jan-Second Shift)-PCM-14 68.
x 4y 7z g, 3y 5z h, 2x 5y 9z k
If the system of linear equation consistent, then: (A) g h k 0 (C) g h 2k 0
is
(B) 2g h k 0 (D) g 2h k 0 3
69.
70.
1 t6 The coefficient of t 4 in the expansion of is 1 t (A) 12 (B) 15 (C) 10 (D) 14 If both the roots of the quadratic equation x 2 5x 4 0 are real and distinct and they lie in the interval 1, 5 , then m lies in the interval. (A) (4, 5) (C) (5, 6)
(B) (3, 4) (D) (–5, –4)
71.
Let S be the set of all triangle in the xy – plane, each having one vertex at the origin and the other two vertices lie on coordinate axes with integral coordinates. If each triangle in S has area 50 sq. units, then the number of elements in the set S is: (A) 9 (B) 18 (C) 32 (D) 36
72.
Let a, b and c be the 7th, 11th and 13th terms respectively of a non – constant A.P. If a these are also the three consecutive terms of a G.P. then is equal to: c 1 (A) (B) 4 2 7 (C) 2 (D) 13
73.
The logical statement ~ ~ p q p r ~ q r is equivalent to: (A) p r ~ q (B) ~ p ~ q r (D) p ~ q r
(C) ~ p r
x y z and perpendicular to the 2 3 4 x y z x y z plane containing the straight lines and is: 3 4 2 4 2 3 (A) x 2y 2z 0 (B) x 2y z 0 (C) 5x 2y 4z 0 (D) 3x 2y 3z 0
74.
The equation of the plane containing the straight line
75.
A data consists of n observations: n
x1, x 2 ,......, x n . If
x i 1
2
i
n
1 9n and
x
2
i
1 5n , then the standard deviation of this
i 1
data is: (A) 5
(B)
(C)
(D) 2
7
5
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JEE-MAIN-2019 (9th Jan-Second Shift)-PCM-15
76.
et If A e t et
77.
If f x
e t cos t
e t sin t
e cos t e sin t e sin t e cos t Then A is 2e t sin t 2e t cos t (A) Invertible only if t (B) not invertible for any t R 2 (C) invertible for all t R (D) invertible only if t t
t
5x 8 7x 6
x
2
1 2x7
2
t
t
dx, x 0 and f 0 0, then the value of f 1 is:
1 2 1 (C) 4
1 2 1 (D) 4
(A)
(B)
3
78.
Let f be a differentiable function R to R such that f x f y 2 x y 2 , for all x, y R . 1
If f 0 1 then
2
f x dx
is equal to
0
1 2 (D) 1
(A) 0
(B)
(C) 2 79.
d2 y at t , is: 2 dx 4 1 (B) 3 2 1 (D) 6 2
If x 3 tan t and y 3 sec t, then the value of (A)
3
2 2 1 (C) 6
80.
The number of natural numbers less than 7,000 which can be formed by using the digits 0, 1, 3, 7, 9 (repetition of digits allowed) is equal to: (A) 250 (B) 374 (C) 372 (D) 375
81.
If the circles x 2 y 2 16x 20y 164 r 2 and x 4 y 7 36 intersect at two distinct points, then: (A) 0 r 1 (B) 1 r 11 (C) r 11 (D) r 11
82.
A hyperbola has its centre at the origin, passes through the point (4, 2) and has transverse axis of length 4 along the x – axis. Then the eccentricity of the hyperbola is: 3 2 (A) (B) 2 3
2
(C)
3
2
(D) 2
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JEE-MAIN-2019 (9th Jan-Second Shift)-PCM-16
83.
Let A 4, 4 and B 9,6 be points on the parabola y 2 4x . Let C be chosen on the arc AOB of the parabola, where O is the origin, such that the area of ACB is maximum. Then, the area (in sq. units) of ACB , is: 3 (A) 31 (B) 32 4 1 1 (C) 30 (D) 31 2 4
84.
Let the equation of two sides of a triangle be 3x 2y 6 0 and 4x 5y 20 0 . If the orthocentre of this triangle is at (1, 1), then the equation of its third side is: (A) 122y 26x 1675 0 (B) 26x 61y 1675 0 (C) 122y 26x 1675 0 (D) 26x 122y 1675 0
85.
An urn contains 5 red and 2 green balls. A ball is drawn at random from the urn. If the drawn ball is green, then a red ball is added to the urn and if the drawn ball is red, then a green ball is added to the urn; the original ball is not returned to the urn. Now, a second ball is drawn at random from it. The probability that the second ball is red, is: 26 32 (A) (B) 49 49 27 21 (C) (D) 49 49
86.
If the lines x ay b, z cy d and x a ' z b ', y c ' z d' are perpendicular, then: (A) cc ' a a ' 0 (B) aa ' c c ' 0 (C) ab ' bc ' 1 0 (D) bb ' cc ' 1 0 ˆ b b ˆi b ˆj 2kˆ and c 5iˆ ˆj 2kˆ be three vectors such that the Let a ˆi ˆj 2k, 1 2 projection vector of b on a is a . If a b is perpendicular to c , then b is equal to:
87.
(A) (C) 88.
89.
The number of all possible positive integral values of for which the roots of the quadratic equation, 6x 2 11x 0 are rational numbers is: (A) 2 (B) 5 (C) 3 (D) 4 2x Let A x R : x is not a positive int eger Define a function f : A R as f x x 1 then f is (A) injective but nor surjective (B) not injective (C) surjective but not injective (D) neither injective nor surjective 3
90.
(B) 4 (D) 6
22 32
If
0
(A) 2 (C) 4
tan 2k sec
d 1
1 2
, k 0 , then the value of k is:
1 2 (D) 1 (B)
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JEE-MAIN-2019 (9th Jan-Second Shift)-PCM-17
JEE (Main) – 2019 ANSWERS PART A – PHYSICS 1.
C
2.
B
3.
D
4.
D
5.
D
6.
A
7.
C
8.
D
9.
B
10.
B
11.
D
12.
C
13.
C
14.
D
15.
C
16.
D
17.
B
18.
B
19.
A
20.
D
21.
C
22.
B
23.
A
24.
C
25.
A
26.
C
27.
D
28.
A
29.
C
30.
A
PART B – CHEMISTRY 31.
D
32.
C
33.
C
34.
C
35.
C
36.
D
37.
C
38.
B
39.
B
40.
B
41.
D
42.
A
43.
C
44.
A
45.
B
46.
C
47.
B
48.
D
49.
C
50.
A
51.
A
52.
A
53.
C
54.
A
55.
D
56.
B
57.
D
58.
D
59.
C
60.
A
PART C – MATHEMATICS 61.
A
62.
A
63.
B
64.
A
65.
A
66.
A
67.
C
68.
B
69.
B
70.
(Bonus)
71.
D
72.
B
73.
A
74.
B
75.
B
76.
C
77.
D
78.
D
79.
D
80.
B
81.
B
82.
A
83.
D
84.
D
85.
B
86.
B
87.
D
88.
C
89.
A
90.
A
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JEE-MAIN-2019 (9th Jan-Second Shift)-PCM-18
HINTS AND SOLUTIONS PART A – PHYSICS 1.
R = Ro e–t RB Ro e B t e ( B A )t e3t R A Ro e B t B A 3 n2 n2 3. TB n2 n2 TB 4
2.
Ps PP Es is = Eiip (0.9) (2300) (5) is = 45 A. (230)
3.
B
E C
1 oE 2 2 B2 E2 E2 UB ( o o ) UE 2o 2oC2 2 o
UE
4.
x = 3t2 + 5 v = 6t W = k 1 1 = (2) (30)2 2(0)2 2 2 = 900 J
5.
eE = evB
6.
eBr E B m eB 2r m E (1.6 1019 ) (0.5)2 (0.5 10 2 ) m 2 10 24 kg. 100
kq kq 10 ( 25) E 31 r1 32 r2 = k 10 6 ( ˆi 3ˆj) ( 4iˆ 3ˆj) r1 r2 125 10 10
1 1 = (9 103 ) ( ˆi 3 ˆj) ( 4iˆ 3 ˆj) 5 10
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JEE-MAIN-2019 (9th Jan-Second Shift)-PCM-19
1 4 ˆ 3 3 ˆ 7 ˆ 3 ˆ = (9 10 3 ) i i 9000 i j 10 10 10 5 10 5 = (63iˆ 27ˆj) (100)
7.
t = G a h b cc Mo Lo T = (M–1 L3 T–2)a (ML2T–1)b (LT–1)c –a + b = 0 a = b 3a + 2b + c = 0 c = –5a –2a – b – c = 1 1 1 5 a ; b ; c 2 2 2
8.
VOi 12 0.3 11.7 V VOf 12 0.7 11.3 V
9.
10.
Vo = –0.4 V
dV dV A A 2gh dt dt 2 0.74 2 (3.14) 2(9.8)h 60 100 h = 4.92 m E1
GMm GMm R h R 2
1 GM GMm E2 m 2 Rh 2(R h)
E1 = E2 ; h
R 2
11.
W1 = W2 600 – T2 = T2 – 400 T2 = 500 K
12.
E Pt
13.
PE = KE 1 1 m2 (A 2 x 2 ) = m2 x 2 2 2 A x 2
E2 (24)2 Rt (60) (60) 518 J. Z2 602 (8.33 2)2
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JEE-MAIN-2019 (9th Jan-Second Shift)-PCM-20 14.
T cos 45° = mg T sin 45° = F F = mg = 100 N.
15.
Q
16.
Xmax = d sin = 0.32 sin 30 = 0.16 mm Xmax 0.16 10 3 n 500 10 9 0.16 10 6 1600 = 320 500 5 Number of BFs = (2n + 1) = 641
f nRT 2 5 15 = (8.3) (1200 300) = 10000 J. 2 28
17.
= 60°
18.
mg
1 1 m 2 2 2 2 2 3
19. 20.
21.
3g 30 2
R = 530 k 5%
oi 2R oN i BC 2(R / N) BL 1 2 BC N BL
f
1 C 1 & 0.8 f 2 2 ML 2 3
C 2
ML mL2 2 3
ML2 mL2 25 2 3 2 16 ML 3
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JEE-MAIN-2019 (9th Jan-Second Shift)-PCM-21
22.
25 3m 1 16 2M 9 3m 16 2 M m 3 0.37 M 8
R A 2r Q 4 r 2 dr 2 e a (4r 2 ) dr 0r 2R a = 4A 1 e a 2 a Q R log 1 2 2Aa
L2 L2 0K 3 2 2 2 0L (K K ) C1 1 3 d d d 2 2 2 L L2 oK 2 oK 4 2 2 2 oL (K K ) C2 2 4 d d d 2 2 1 1 1 c c1 c 2 d d d 2 2 2 oKL oL (K1 K 3 ) oL (K 2 K 4 ) 0K 1
23.
0.5 mm 0.015 mm 100 0.5 MSR = 5.5 48 100 = 5.74 mm. Thickness = 5.74 – 0.015 = 5.725 mm
24.
Zero error = 0 3
25.
f
2 330 vs 660 Hz 2 0.5 50 330 18 s 50 f f 660 1 (660) 330 18 330 s = 666 Hz.
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JEE-MAIN-2019 (9th Jan-Second Shift)-PCM-22
2 2 t a2 a1
a1a 2 2 t a1 a2
2a1 2a 2
2 v
26.
27.
a1a2 t
1 a1 a2
( a1a 2 ) t
KEmax = hmax – =
(6.6 10 34 ) (6.28 107 ) (3 108 ) 4.7 1.6 10 19 2 3.14
= 12.37 – 4.7 = 7.67 eV 28.
5V
100
400
500
R2
12 6 6 400 600 R 2 1 1 1 200 600 R2 R2 = 300
18 V
29.
30.
c 3 108 3 1015 Hz 8 107 8 3 1013 (0.01) f 8 n 6 106 6 106 1 = 107 6.25 105 16 f
dx a sin t dt dy vy a cos t dt dz vz a dt vx
v v 2x v 2y v 2z a 2
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JEE-MAIN-2019 (9th Jan-Second Shift)-PCM-23
PART B – CHEMISTRY 31.
H2O s H2O H2O H2O g H2O g 1 kg 1 kg 1 kg at 273 K at 273 K at 373 K at 373 K at 383 K S = S1 + S2 + S3 + S4 334 373 2491 383 4.2n 2n = 9.267 kJ Kg–1 K-1 273 273 373 373
32.
2 C57H110O6 s 163 O 2 g 114 CO 2 g 110H2O I
Moles of C57H110O6 Moles of H2O 2 110 445 890
Mass of H2O 18
2 110 Mass of H2O = 495 g 33.
O
O
O
NaOH Ph C CH3 Ph C CH2 Ph C CH2
enolate ion
O
O
O
O
OH
O
H2O Ph C CH2 CH3 C H CH3 CH CH2 C Ph CH3 CH CH2 C Ph
34.
Fact based
35.
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JEE-MAIN-2019 (9th Jan-Second Shift)-PCM-24 36.
Nucleophilicity of NH2> OH OH
NH2
37.
OH
CH3CO 2 O/Py R.T
OH
O
NH2
NHCOCH3
NH2
Br2 /h
Br KOH
O
NH
CH3
38.
Acetonitrile is used as mobile phase for most of the reverse chromatography. Benzaldehyde is adsorbed on alumina.
39.
The only alkali metal which forms nitride by reacting directly with N2 is ‘Li’.
40.
As2S3 is a negatively charged sol. so AlCl3 will be most effective.
41.
As CN– is a strong field ligand. K3[Co(CN)6] will have maximum ‘’.
42.
Fact based.
43.
K1 A 2 g B2 g 2 AB g
....... 1
K2 3 A 2 g 3B 2 g ....... 2 6 AB g
Reaction(2) = -3 reaction(1) 3
1 K 2 K 2 K 13 K1
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JEE-MAIN-2019 (9th Jan-Second Shift)-PCM-25 44.
H 2O
Val 45.
46.
Ser
2A B products Rate = K[A]x[B]y r = K[A]x[B]y - - - - (i) 0.3 = K[A]x[B]y - - - - (1) 2.4 = K[2A]x[2B]y - - - - (2) 0.6 = K[2A]x[B]y - - - - (3) From (1), (2) & (3) x = 1, y = 2 Overall order = 2 + 1 = 3 Order w.r.t A = 1 Order w.r.t B = 2 H3C
CH3
O O
O2
47.
48.
Thr
OH
H
O dil.HCl
-COCH3 is present it will show both 2, 4-DNP & iodoform test. Due to steric inhibition of resonance. I.P of ‘N’ is not involved in delocalization so coupling reaction will not take place. 2 Zn s Cu2 aq Zn aq Cu s
-nFEcell = -RTnK
nK =
2 96500 2 = 160.83 8 300
K = e160 49.
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JEE-MAIN-2019 (9th Jan-Second Shift)-PCM-26
50.
NO NO B.O 0.5 3 Para Dia O2 O2 B.O 2 2.5 Para Para
51.
Refer Theory
N2 N2 B.O 3 2.5 Dia Para O2 O22 B.O 2 1 Para Dia
52. is anti aromatic
53.
Refer theory
54.
4 Al 6 ZnO 2 Al2O3 6 Zn
H for the above reaction is –ve. 55.
Due to weak metallic bonding.
56.
Correct order of basic strength is NH2(Et)2 > EtNH2 > NMC3 > Ph – NH – CH3
57.
2nd electron gain enthalpy of oxygen is positive.
58.
d
59.
ZM Na a3 4 63.55
6.023 1023 x 10
8 3
422 gm / cm3 x3
Let moles of H2O separated as ice = x gm Tf = iKfm 10 = 1 1.86
62 62 250 x 1000
x = 64 gm 60.
L1 L2 L3 Green Blue Red absorbed wave length Order of Red > Green > Blue L3 > L1 > L2 Strength of ligand 1/ Strength of ligand L2 > L1 > L3
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JEE-MAIN-2019 (9th Jan-Second Shift)-PCM-27
PART C – MATHEMATICS
3 n 1 3 1
2
61.
Tn
3.
n
2
2n 1 n 1 2n 1 6 2n 1
Tn S15
22 .... n2
1 15 3 n n2 2 n 1
n2 n 1
2 2 1 15 15 1 15 16 31 2 2 6
= 7820 62.
lim
x x x sin x x
x 0
x0 x 1 lim x x 1 sin 1 sin1 x 0 x x x 63.
f xy f x .f y f 0 1 as f 0 0 f x 1
dy f x 1 dx y xc At, x 0, y 1 c 1 y x 1 3 1 3 1 y y 1 1 3 4 4 4 4 64.
2
3
10
x sin 1 sin10 3 10
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JEE-MAIN-2019 (9th Jan-Second Shift)-PCM-28
2
3 10
y cos1 cos10 4 10 yx
65.
sin x sin 2x sin 3x 0 sin x sin 3x sin 2x 0 2 sin x.cos x sin 2x 0 sin 2x 2 cos x 1 0
sin 2x 0 or cos x 66.
z0 or 2 (where is a non – real cube root of unity) 81
z 3 6i 3i z 3 3i arg z 4 67.
1 x 0, 2 3
93
y
The graph is a follows 0
1
x
2
1
1 dx x 2 1 dx 2
y x2 1
0
y x2 1
68.
P1 x 4y 7z g 0 P2 3x 5y h 0 P3 2x 5y 9z k 0 Here 0 2P1 P2 P3 0 when 2g h k 0
69.
1 t 1 t 1 t 3t 3t 1 t
6 3
18
–1
1
3
6
12
3
coefficient of t 4 in 1 t
3
is
3 4 1
C4 6C2 15
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JEE-MAIN-2019 (9th Jan-Second Shift)-PCM-29 70.
x 2 mx 4 0 , 1,5 1
(1)
D 0 m2 16 0 m , 4 4,
(2)
f 1 0 5 m 0 m , 5
5
29 f 5 0 29 5m 0 m , 5 b m (4) 1 5 1 5 m 2,10 2a 2 m 4, 5 No option correct : Bonus * If we consider , 1, 5 then option (1) is correct. (3)
71.
Let A ,0 and B 0, be the vectors of the given triangle AOB 100 Number of triangles 4 (number of divisors of 100) 4 9 36
72.
a A 6d b A 10d c A 12d a, b, c are in G.P. 2
A 10d A 6d a 12d
A 14 d
A a A 6d d 6 14 4 A 12 14 c A 12d 12 d 6
73.
~ ~ p q p r ~ q r p ~ q p r ~ q r p ~ q r ~ q r p ~ q r p r ~ q
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JEE-MAIN-2019 (9th Jan-Second Shift)-PCM-30
74.
Vector along the normal to the plane containing the lines
x y z x y z and is 3 4 2 4 2 3
8iˆ ˆj 10kˆ . Vector perpendicular to the vectors 2iˆ 3 ˆj 4kˆ and 8iˆ ˆj 10kˆ is 26iˆ 52ˆj 26kˆ So, required plane is 26x 52y 26z 0 x 2y z 0 75.
x x
2
1 9n
i
……….(1)
2
1 5n
i
……….(2)
1 2 xi2 1 7n
x
2 i
6 n (1) . (2) 4 x i 4n
xi n
x
i
1 n variance = 6 – 1 = 5 standard diviation 5
1
76.
cos t
sin t
t
A e 1 cos t sin t sin t cos t 1
2 sin t
2cos t
e t 5 cos2 t 5 sin2 t t R
5e t 0 t R 77.
5x 8 7x 6
x
2
1 2x 7
2
dx
5x 6 7x 8 1 1 7 5 2 x x
2
As f 0 0, f x
f 1 78.
dx
1 C 1 1 2 5 7 x x
x7 2x7 x 2 1
1 4
f x f y 2 x y
3/2
divide both side by x y f x f y xy
2. x y
1/ 2
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JEE-MAIN-2019 (9th Jan-Second Shift)-PCM-31 Apply limit x y
f ' y 0 f ' y 0 f y c f x 1 1
1.dx 1 0
79.
dx 3 sec 2 t dt dy 3 sec t tan t dt dy tan t sin t dx sec t d2 y dt cos t 2 dx dx cos t cos3 t 1 1 2 3 sec t 3 3.2 2 6 2
80.
a1 a2 a3 Number of numbers 5 3 1 a 4 a1 a2 a3 2 ways for a 4 Numbers of numbers 2 53 Required number0020 53 2 53 1 = 374
81.
x 2 y 2 16x 20y 164 r 2 A 8,10 ,R1 r
x 4
2
2
y 7 36
B (4, 7), R2 6 R1 R 2 AB R1 R 2 1 r 11
82.
Given hyperbola is x2 y2 1 4 b2 Satisfying the point 4 b2 3 2 e 3
(4, 2)
(4,
2)
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JEE-MAIN-2019 (9th Jan-Second Shift)-PCM-32 83.
For maximum area, tangent at the point c must be parallel to chord BC. 1 t 2
B (9, 6) 2
c (t , 2t)
A (4, –4)
84.
A
Equation of AB is 3x 2y 6 0 Equation of AC is 4x 5y 20 0 . Equation of BE is 2x 3y 5 0 Equation of CF is 5x 4y 1 0 Equation of BC is 26x 122y 1675
(1,1)
B
85.
E
F
D
C
E1 : Event of drawing a Red ball and placing a green ball in the bag E 2 : Event of drawing a green ball and placing a red ball in the bag
E E E : Event of drawing a red ball in second draw P E P E1 P P E 2 P E1 E2 5 4 2 6 32 7 7 7 7 49 86.
87.
Line x ay b, z cy d x b y z d a 1 c Line x a ' z b ', y c ' z d' x b ' y d' z a' c' 1 Given both the lines are perpendicular aa ' c ' c 0 a.b Projection of b on a a a b1 b2 2 ………….(1) and a b c a b .c 0
5b1 b2 10 …………(2) from (1) and (2) b1 3 and b2 5 then b b12 b22 2 6
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JEE-MAIN-2019 (9th Jan-Second Shift)-PCM-33 88.
89.
D must be perfect square 121 24 2 maximum value of is 5 1 I 2 I 3 integral values 3 I 4 I 5 I
1 f x 2 1 x 1 2 f ' x 2 x 1
f is one – one but not onto 90.
1 2k
/3
0
1
2k k2
tan sec
d
2 cos
1 2k
/3 0
/3
0
sin cos
d
2 1 1 k 2
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FIITJEE Solutions to JEE(Main)-2019 Test Date: 10th January 2019 (First Shift)
PHYSICS, CHEMISTRY & MATHEMATICS Paper - 1 Time Allotted: 3 Hours
Maximum Marks: 360
Please read the instructions carefully. You are allotted 5 minutes specifically for this purpose.
Important Instructions: 1.
The test is of 3 hours duration.
2.
This Test Paper consists of 90 questions. The maximum marks are 360.
3.
There are three parts in the question paper A, B, C consisting of Physics, Chemistry and Mathematics having 30 questions in each part of equal weightage. Each question is allotted 4 (four) marks for correct response.
4.
Out of the four options given for each question, only one option is the correct answer.
5.
For each incorrect response 1 mark i.e. ¼ (one-fourth) marks of the total marks allotted to the question will be deducted from the total score. No deduction from the total score, however, will be made if no response is indicated for an item in the Answer Box.
6.
Candidates will be awarded marks as stated above in instruction No.3 for correct response of each question. One mark will be deducted for indicating incorrect response of each question. No deduction from the total score will be made if no response is indicated for an item in the answer box.
7.
There is only one correct response for each question. Marked up more than one response in any question will be treated as wrong response and marked up for wrong response will be deducted accordingly as per instruction 6 above.
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JEE-MAIN-2019 (10th Jan-First Shift)-PCM-2
PART –A (PHYSICS) 1.
In an electron microscope, the resolution that can be achieved is of the order of the wavelength of electrons used. To resolve a width of 7.5 × 1012 m, the minimum electron energy required is close to: (A) 500 keV (B) 100 keV (C) 1 keV (D) 25 keV
2.
To mop-clean a floor, a cleaning machine presses a circular mop of radius R vertically down with a total force F and rotates it with a constant angular speed about its axis. If the force F is distributed uniformly over the mop and the floor is , the torque, applied by the machine on the mop is: (A) FR/3 (B) FR/6 2 (C) FR/2 (D) FR 3
3.
A potentiometer wire AB having length L and resistance 12 r is joined to a cell D of emf and internal resistance r. A cell C having emf /2 and internal resistance 3r is connected. The length AJ at which the galvanometer as shown in figure shows no deflection is: 11 11 (A) L (B) L 12 24 13 5 (C) L (D) L 24 12
4.
A TV transmission tower has a height of 140 m and the height of the receiving antenna is 40 m. What is the maximum distance upto which signals can be broadcasted from this tower in LOS (Line of Sight) mode? (Given : radius of earth = 6.4 × 106 m). (A) 65 km (B) 48 km (C) 80 km (D) 40 km
5.
In the cube of side ‘a’ shown in the figure, the vector from the central point of the face ABOD to the central point of the face BEFO will be: 1 1 (A) a kˆ ˆi (B) a ˆi kˆ 2 2 1 1 (C) a ˆj ˆi (D) a ˆj kˆ 2 2
6.
A uniform metallic wire has a resistance of 18 and is bent into an equilateral triangle. Then, the resistance between any two vertices of the triangle is: (A) 4 (B) 8 (C) 12 (D) 2
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JEE-MAIN-2019 (10th Jan-First Shift)-PCM-3 7.
A block of mass m is kept on a platform which starts from rest with constant acceleration g/2 upward, as shown in figure. Work done by normal reaction on block in time t is: m g2 t 2 m g2 t 2 (A) (B) 8 8 3m g2 t 2 (C) 0 (D) 8
8.
In a Young’s double slit experiment with slit separation 0.1 mm, one observes a bright 1 fringe at angle rad by using light of wavelength 1. When the light of wavelength 2 is 40 used a bright fringe is seen at the same angle in the same set up. Given that 1 and 2 are in visible range (380 nm to 740 nm), their values are: (A) 625 nm, 500 nm (B) 380 nm, 525 nm (C) 380 nm, 500 nm (D) 400 nm, 500 nm
9.
Two guns A and B can fire bullets at speed 1 km/s and 2 km/s respectively. From a point on a horizontal ground, they are fired in all possible directions. The ratio of maximum areas covered by the bullets fired by the two guns, on the ground is: (A) 1 : 16 (B) 1 : 2 (C) 1 : 4 (D) 1 : 8
10.
The density of a material in SI units is 128 kg m3. In certain units in which the unit of length is 25 cm and the unit of mass 50 g, the numerical value of density of the material is: (A) 40 (B) 16 (C) 640 (D) 410
11.
A magnet of total magnetic moment 10 2 ˆi A-m2 is placed in a time varying magnetic field, B ˆi cos t where B = 1 Tesla and = 0.125 rad/s. The work done for reversing the direction of the magnetic moment at t = 1 second, is: (A) 0.01 J (B) 0.007 J (C) 0.028 J (D) 0.014 J
12.
A heat source at T = 103 K is connected to another heat reservoir at T = 102 K by a copper slab which is 1 m thick. Given that the thermal conductivity of copper is 0.1 WK1 m1, the energy flux through it in the steady state is: (A) 90 Wm2 (B) 120 Wm2 2 (C) 65 Wm (D) 200 Wm2
13.
A parallel plate capacitor is of area 6 cm2 and a separation 3 mm. The gap is filled with three dielectric materials of equal thickness (see figure) with dielectric constants K1 = 10, K2 = 12 and K3 = 14. The dielectric constant of a material which when fully inserted in above capacitor, gives same capacitance would be: (A) 4 (B) 14 (C) 12 (D) 36
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JEE-MAIN-2019 (10th Jan-First Shift)-PCM-4 14.
A charge Q is distributed over three concentric spherical shell of radii a, b, c (a < b < c) such that their surface charge densities are equal to one another. The total potential at a point at distance r from their common centre, where r < a, would be: Q a2 b2 c 2 Q ab bc ca (A) (B) 12 0 abc 4 0 a3 b3 c 3 (C)
15.
Q 4 0 a b c
(D)
Q a b C
4 0 a 2 b 2 c 2
Three Carnot engines operate in series between a heat source at a temperature T1 and a heat sink at temperature T4 (see figure). There are two other reservoirs at temperature T2 and T3, as shown, with T1 > T2 > T3 > T4. The three engines are equally efficient if: 1
(A) T2 T1T4 2 ;T3 T12T4
1 3
1
1 3
1
1 3
1
1 4
(B) T2 T12 T4 3 ;T3 T1T24 (C) T2 T1T42 3 ;T3 T12T4 (D) T2 T13 T4 4 ;T3 T1T43 16.
A satellite is moving with a constant speed in circular orbit around the earth. An object of mass ‘m’ is ejected from the satellite such that it just escapes from the gravitational pull of the earth. At the time of ejection, the kinetic energy of the object is: (A) 2 m2 (B) m2 1 3 (C) m2 (D) m2 2 2
17.
Water flows into a large tank with flat bottom at the rate of 104 m3s1. Water is also leaking out of a hole of area 1 cm2 at its bottom. If the height of the water in the tank remains steady, then this height is: (A) 5.1 cm (B) 1.7 cm (C) 4 cm (D) 2.9 cm
18.
A string of length 1 m and mass 5 g is fixed at both ends. The tension in the string is 8.0 N. The string is set into vibration using an external vibrator of frequency 100 Hz. The separation between successive nodes on the string is close to: (A) 10.0 cm (B) 33.3 cm (C) 16.6 cm (D) 20.0 cm
19.
A train moves towards a stationary observer with speed 34 m/s. The train sounds a whistle and its frequency registered by the observer is f 1. If the speed of the train is reduced to 17 m/s, the frequency registered is f 2. If speed of sound is 340 m/s, then the ratio f1/f2 is: (A) 18/17 (B) 19/18 (C) 20/19 (D) 21/20
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JEE-MAIN-2019 (10th Jan-First Shift)-PCM-5 20.
A plano convex lens of refractive index 1 and focal length f 1 is kept in contact with another plano concave lens of refractive index 2 and focal length f 2. If the radius of curvature of their spherical faces is R each and f1 = 2f2, then 1 and 2 are related as: (A) 1 + 2 = 3 (B) 21 + 2 = 1 (C) 32 + 1 = 1 (D) 22 + 1 = 1
21.
To get output ‘1’ at R, for the given logic gate circuit the input values must be: (A) X = 0, Y = 1 (B) X = 1, Y = 1 (C) X = 1, Y = 0 (D) X = 0, Y = 0
22.
If the magnetic field of a plane electromagnetic wave is given by (The speed of light = x 3 × 108 m/s) B 100 10 6 sin 2 2 1015 t then the maximum electric field c associated with it is: (A) 6 × 104 N/C (B) 3 × 104 N/C 4 (C) 4 × 10 N/C (D) 4.5 × 104 N/C
23.
Using a nuclear counter the count rate of emitted particles from a radioactive source is measured. At t = 0 it was 1600 counts per second and t = 8 seconds it was 100 counts per second. The count rate observed, as counts per second, at t = 6 seconds is close to: (A) 200 (B) 150 (C) 400 (D) 360
24.
A solid metal cube of edge length 2 cm is moving in a positive y-direction at a constant speed of 6 m/s. There is a uniform magnetic field of 0.1 T in the positive z-direction. The potential difference between the two faces of the cube perpendicular to the x-axis, is: (A) 12 mV (B) 6 mV (C) 1 mV (D) 2 mV
25.
Two electric dipoles A, B with respective dipole moments dA 4 qa ˆi and dB 2 qa ˆi are placed on the x-axis with a separation R, as shown in the figure. The distance from A at which both of them produce the same potential is: 2R R (A) (B) 2 1 2 1 (C)
R 2 1
(D)
2R 2 1
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JEE-MAIN-2019 (10th Jan-First Shift)-PCM-6 26.
In the given circuit the cells have zero internal resistance. The currents (in Amperes) passing through resistance R1 and R2 respectively, are: (A) 1, 2 (B) 2, 2 (C) 0.5, 0 (D) 0, 1
27.
A 2 W carbon resistor is color coded with green, black, red and brown respectively. The maximum current which can be passed through this resistor is: (A) 20 mA (B) 100 mA (C) 0.4 mA (D) 63 mA
28.
A piece of wood of mass 0.03 kg is dropped from the top of a 100 m height building. At the same time, a bullet of mass 0.02 kg is fired vertically upward, with a velocity 100 ms 1 , from the ground. The bullet gets embedded in the wood. Then the maximum height to which the combined system reaches above the top of the building before falling below is: (g = 10 ms2) (A) 20 m (B) 30 m (C) 40 m (D) 10 m
29.
An insulating thin rod of length has a linear charge density (x) 0
30.
A homogeneous solid cylindrical roller of radius R and mass M is pulled on a cricket pitch by a horizontal force. Assuming rolling without slipping, angular acceleration of the cylinder is: 3F F (A) (B) 2mR 3mR
x on it. The rod is rotated about an axis passing through the origin (x = 0) and perpendicular to the rod. If the rod makes n rotations per second, then the time averaged magnetic moment of the rod is: (A) n3 (B) n3 3 (C) n3 (D) n3 4
(C)
F 2mR
(D)
2F 3mR
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JEE-MAIN-2019 (10th Jan-First Shift)-PCM-7
PART –B (CHEMISTRY) 31.
The total number of isomers for a square planar complex [M(F)(Cl)(SCN)(NO2)] is (A) 16 (B) 8 (C) 4 (D) 12
32.
A process that has H = 200 J mol–1 and S = 40 JK–1 mol–1. Out of the values given below, choose the minimum temperature above which the process will be spontaneous: (A) 20 K (B) 12 K (C) 5 K (D) 4 K
33.
The values of Kp/Kc for the following reactions at 300 K are respectively: (At 300 K, Rt = 24.62 dm3 atm mol–1) N2 g O2 g 2NO g N2O 4 g 2NO 2 g N2 g 3H2 g 2NH3 g (A) 1, 24.62 dm3 atm mol–1, 606.0 dm6 atm2 mol–2 (B) 1, 24.62 dm3 atm mol–1, 1.65 10–3 dm–6 atm–2 mol2 (C) 1, 4.1 10–2 dm–3 atm–1 mol, 606.0 dm6 atm2 mol–2 (D) 24.62 dm3 atm mol–1, 606.0 dm6 atm2 atm2 mol–2, 1.65 10–3 dm–6 atm–2 mol2
34.
The total number of isotopes of hydrogen and number of radioactive isotopes among them, respectively are (A) 3 and 1 (B) 3 and 2 (C) 2 and 1 (D) 2 and 0
35.
Water filled in two glasses A and B have BOD values of 10 and 20 respectively. The correct statement regarding them is (A) B is more polluted than A (B) A is suitable for drinking, whereas B is not (C) both A and B are suitable for drinking (D) A is more polluted than B
36.
Which primitive unit cell has unequal edge length (a b c) and all axial angles different from 90o? (A) Triclinic (B) Hexagonal (C) Monoclinic (D) Tetragonal
37.
The major product of the following reaction is:
(A)
(B)
(C)
(D)
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JEE-MAIN-2019 (10th Jan-First Shift)-PCM-8 38.
Consider the given plots for a reaction obeying Arrhenius equation (0oC < T < 300oC) : (k and Ea are rate constant and activation energy respectively)
Choose the correct option: (A) I is right but II is wrong (C) I is wrong but II is right 39.
40.
(B) Both I and Ii are correct (D) Both I and II are wrong
The major product formed in the reaction given below will be:
(A)
(B)
(C)
(D)
Wilkinson catalyst is (A) [(Ph3P)3IrCl] (C) [(Ph 3P)3RhCl]
(B) [Et3P)3RhCl] (D) [(Et3P)3IrCl]
41.
If dichloromethane(DCM) and water(H2O) are used for differential extraction, which one of the following statements is correct? (A) DCM and H2O would stay as lower and upper layer respectively in the S.F (B) DCM and H2O will make turbid/colloidal mixture (C) DCM and H2O would stay as upper and lower layer respectively in the separating funnel(S.F) (D) DCM and H2O will be miscible clearly
42.
Which diccarboxylic acid in presence of a dehydrating agent is least reactive to give anhydride?
(A)
(B)
(C)
(D)
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JEE-MAIN-2019 (10th Jan-First Shift)-PCM-9 43.
The decreasing order of ease of alkaline hydrolysis for the following ester is
(A) III > II > IV > I (C) IV > II > III > I 44.
(B) III > II > I > IV (D) II > III > I > IV
Which of the graphs shown below does not represent the relationship between incident light and the electron ejected from metal surface?
(A)
(B)
(C)
(D)
45.
Which of the following is not an example of heterogeneous catalytic reaction? (A) Ostwald’s process (B) Combustion of coal (C) Hydrogenation of vegetable oils (D) Haber’s process
46.
The effect of lanthanoid contraction in the lanthanoid series of elements by the large means (A) increase in both atomic and ionic radii (B) decrease in atomic radii and increase in ionic radii (C) decrease in both atomic and ionic radii (D) increase in atomic radii and decrease in ionic radii
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JEE-MAIN-2019 (10th Jan-First Shift)-PCM-10 47.
Which hydrogen in compound(E) is easily replaceable during bromination reaction in presence of light?
(A) - hydrogen (C) - hydrogen 48.
49.
(B) - hydrogen (D) - hydrogen
The major product of the following reaction is
(A)
(B)
(C)
(D)
The correct structure of product ‘P’ in the following reaction is
(A)
(B)
(C)
(D)
50.
The type of hybridisation and number of lone pair(s) of electrons of Xe in XeOF4 respectively are (A) sp3d2 and 1 (B) sp3d and 2 3 2 (C) sp d and 2 (D) sp3d and 1
51.
The electronegativity of aluminium is similar to (A) carbon (B) beryllium (C) boron (D) lithium
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JEE-MAIN-2019 (10th Jan-First Shift)-PCM-11 52.
Consider the following Zn2 2e Zn s ;Eo 0.76 V Ca 2 2e Ca s ;Eo 2.87 V Mg2 2e Mg s ;Eo 2.36 V Ni2 2e Ni s ;Eo 0.25 V
The reducing power of the metals increases in the order (A) Ca < Zn < Mg < Ni (B) Ni < Zn < Mg < Ca (C) Zn < Mg < Ni < Ca (D) Ca < Mg < Zn < Ni 53.
The chemical nature of hydrogen peroxide is (A) oxidising agent in acidic medium, but not in basic medium (B) reducing agent in basic medium but not in acidic medium (C) oxidising and reducing agent in acidic medium but not in basic medium (D) oxidising and reducing agent in both acidic and basic medium
54.
A mixture of 100 m mol of Ca(OH)2 and 2 g of sodium sulphate was dissolved in water and the volume was made upto 100 mL. The mass of calcium sulphate formed and the concentration of OH– in resulting solution, respectively are (Molar mass of Ca(OH)2, Na2SO4 and CaSO4 are 74, 143 and 136 g mol–1 respectively; Ksp of Ca(OH)2 is 5.5 10–6) (A) 1.9 g, 0.28 mol L–1 (B) 13.6 g, 0.28 mol L–1 –1 (C) 1.9g, 0.14 mol L (D) 13.6 g, 0.14 mol L–1
55.
Liquids A and B form and ideal solution in the entire composition range. At 350 K, the vapour pressures of pure A and pure B are 7 103 Pa and 12 103 Pa, respectively. The composition of the vapour in equilibrium with a solution containing 40 mole percent of A at this temperature is (A) xA = 0.37; xB = 0.63 (B) xA = 0.28; xB = 0.72 (C) xA = 0.4; xB = 0.6 (D) xA = 0.76; xB = 0.24
56.
The major product ‘X’ formed in the following reaction is:
(A)
(B)
(C)
(D)
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JEE-MAIN-2019 (10th Jan-First Shift)-PCM-12
57.
The metal used for making X-ray tube window is (A) Mg (B) Na (C) Be (D) Ca
58.
The increasing order of pKa values of the following compounds is
(A) C < B < A < D (C) D < A < C < B 59.
(B) B < C < D < A (D) B < C < A< D
Hall-Herolut’s process is given by (A) Cu2 aq H2 g Cu s 2H aq (B) Cr2O3 2 Al Al2O3 2 Cr (C) 2Al2O3 3C 4 Al 3CO2 Coke, 1673 K (D) ZnO C Zn CO
60.
Two pi and half sigma bonds are present in (A) O 2 (B) N2 (C) O2
(D) N2
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JEE-MAIN-2019 (10th Jan-First Shift)-PCM-13
PART–C (MATHEMATICS) 61.
62.
An unbiased coin is tossed. If the outcome is a head then a pair of unbiased dice is rolled and the sum of the numbers obtained on the is noted. If the toss of the coin results in tail then a card from a well-shuffled pack of nine cards numbered 1, 2, 3,….., 9 is randomly picked and the number on the card is noted. The probability that the noted number is either 7 or 8 is: 13 15 (A) (B) 36 72 19 19 (C) (D) 72 36 3 The shortest distance between the point ,0 and the curve y x, x 0 , is: 2 5 3 (A) (B) 2 2 3 5 (C) (D) 2 4
63.
The plane passing through the point (4, 1, 2) and parallel to the lines x 2 y 2 z 1 x2 y3 z4 and also passes through the point: 3 1 2 1 2 3 (A) (1, 1< 1) (B) (1, 1, 1) (C) (1, 1, 1) (D) (1, 1, 1)
64.
The mean of five observations is 5 and their variance is 9.20. If three of the given five observations are 1, 3 and 8, then a ratio of other two observations is: (A) 10 : 3 (B) 4 : 9 (C) 5 : 8 (D) 6 : 7
65.
If 5, 5r, 5r2 are the lengths of the sides of a triangle, then r cannot be equal to: 3 5 (A) (B) 4 4 7 3 (C) (D) 4 2
66.
If
67.
3
20 Ci1 k 20 , then k equals: 20 Ci Ci1 21 i 1 (A) 400 (C) 200 20
(B) 50 (D) 100
3 The sum of all values of 0, satisfying sin2 2 cos4 2 is: 4 2 5 (A) (B) 4 3 (C) (D) 2 8 FIITJEE Ltd., FIITJEE House, 29-A, Kalu Sarai, Sarvapriya Vihar, New Delhi -110016, Ph 46106000, 26569493, Fax 26513942 website: www.fiitjee.com.
JEE-MAIN-2019 (10th Jan-First Shift)-PCM-14 68.
Consider the quadratic equation c 5 x 2 2cs c 4 0,c 5. Let S be the set of all integral values of c for which one root of the equation lies in the interval (0, 2) and its other root lies in the interval (2, 3). Then the number of elements in S is: (A) 18 (B) 12 (C) 10 (D) 11
69.
If
70.
In a class of 140 students numbered 1 to 140, all even numbered students opted Mathematics course, those whose number is divisible by 3 opted Physics course and those whose number is divisible by 5 opted Chemistry course. Then the number of students who did not opt for any of the three courses is: (A) 102 (B) 42 (C) 1 (D) 38
71.
If the third term in the binomial expansion of 1 xlog2 x
dy 3 1 4 y , x , and y , then y equals: 2 2 dx cos x cos x 3 3 4 3 4 1 1 (A) e 6 (B) 3 3 4 1 (C) (D) e3 3 3
value of x is: 1 (A) 4 1 (C) 8
5
equals 2560, then a possible
(B) 4 2 (D) 2 2
72.
If the parabolas y2 = 4b(x – c) and y2 = 8ax have a common normal, then which one of the following is a valid choice for the ordered triad (a, b, c)? 1 (A) ,2,3 (B) (1, 1, 3) 2 1 (C) ,2,0 (D) (1, 1, 0) 2
73.
If the system of equations xyz5 x 2y 3z 9 x 3y z has infinitely many solutions, then equals: (A) 21 (B) 8 (C) 18 (D) 5
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JEE-MAIN-2019 (10th Jan-First Shift)-PCM-15 74.
For each t R, let [t] be the greatest integer less than or equal to t. Then, 1 x sin 1 x sin 2 1 x lim x 1 1 x 1 x (A) equals 1 (C) equals 1
75.
2 4d sin 2 Let d R, and A 1 sin 2 d , [0, 2]. If the minimum 5 2 sin d sin 2 2d value of det(A) is 8, then a value of d is: (A) 5 (B) 7 (C) 2 2 1 (D) 2 2 2
76.
(B) equals 0 (D) does not exist
Let z1 and z2 be any two non-zero complex numbers such that 3 z1 4 z2 . If z
3z1 2z2 then: 2z2 3z1
(A) Re(z) = 0
b
a
(C)
79.
5 2
(D) Im(z) = 0
Let I x 4 2x 2 dx. If I is minimum then the ordered pair (a, b) is: (A) 0, 2
78.
(B) z
1 17 2 2
(C) z
77.
2, 2
(D)
(B) 2,0
2, 2
A point P moves on the line 2x – 3y + 4 = 0. If Q(1, 4) and R(3, 2) are fixed points, then the locus of the centroid of PQR is a line: 3 (A) with slope (B) parallel to x-axis 2 2 (C) with slope (D) parallel to y-axis 3
max | x |, x 2 , x 2 Let f(x) . Let S be the set of points in the interval (4, 4) at 2 x 4 8 2 x , which f is not differentiable. Then S: (A) is an empty set (B) equals {2, 1, 0, 1, 2} (C) equals {2, 1, 1, 2} (D) equals {2, 2}
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JEE-MAIN-2019 (10th Jan-First Shift)-PCM-16 80.
If a circle C passing through the point (4, 0) touches the circle x2 + y2 + 4x – 6y = 12 externally at the point (1, 1), then the radius of C is: (A) 2 5 (B) 4 (C) 5
(D)
57
81.
Let f : R R be a function such that f(x) = x3 + x2f(1) + xf(2) + f(3), x R. Then f(2) equals: (A) 4 (B) 30 (C) 2 (D) 8
82.
Let
83.
ˆ 4iˆ 3 ˆj 6kˆ and c 3iˆ 6 ˆj 1 kˆ be three vectors a 2iˆ 1ˆj 3k,b 2 3 such that b 2a and a is perpendicular to c. Then a possible value of (1, 2, 3) is:
(A) (1, 3, 1)
1 (B) ,4,0 2
1 (C) ,4, 2 2
(D) (1, 5, 1)
Let A be a point on the line r 1 3 ˆi 1 ˆj 2 5 kˆ and B(3, 2, 6) be a point in the space. Then the value of for which the vector AB is parallel to the plane x – 4y + 3z = 1 is: 1 1 (A) (B) 4 8 1 1 (C) (D) 2 4
sin
n
84.
Let n 2 be a natural number and 0 < < /2. Then
sin n 1
sin
1 n
cos
d is equal to:
(where C is a constant of integration) n 1 (A) 2 1 n 1 sinn1
n 1 n
n 1 (C) 2 1 n 1 sinn1
n 1 n
C
n 1 (B) 2 1 n 1 sinn1
C
n 1 (D) 2 1 n 1 sinn1
n 1 n
n 1 n
C
C
85.
Consider a triangular plot ABC with sides AB = 7 m, BC = 5 m and CA = 6 m. A vertical lamp-post at the mid point D of AC subtends an angle 30° at B. The height (in m) of the lamp-post is: 3 2 (A) 21 (B) 21 2 3 (C) 2 21 (D) 7 21
86.
The equation of a tangent to the hyperbola 4x2 – 5y2 = 20 parallel to the line x – y = 2 is: (A) x – y + 1 = 0 (B) x – y + 7 = 0 (C) x – y + 9 = 0 (D) x – y – 3 = 0 FIITJEE Ltd., FIITJEE House, 29-A, Kalu Sarai, Sarvapriya Vihar, New Delhi -110016, Ph 46106000, 26569493, Fax 26513942 website: www.fiitjee.com.
JEE-MAIN-2019 (10th Jan-First Shift)-PCM-17 87.
If the line 3x + 3y – 24 = 0 intersects the x-axis at the point A and the y-axis at the point B, then the incentre of the triangle OAB, where O is the origin, is: (A) (3, 4) (B) (2, 2) (C) (4, 3) (d) (4, 4)
88.
The sum of all two digit positive numbers which when divided by 7 yield 2 or 5 as remainder is: (A) 1256 (B) 1465 (C) 1365 (D) 1356
89.
If the area enclosed between the curves y = kx2 and x = ky2, (k > 0), is 1 square unit. Then k is: 3 1 (A) (B) 2 3 2 (C) 3 (D) 3
90.
Consider the statement: “ P n : n2 n 41 is prime.” Then which one of the following is true? (A) Both P(3) and P(5) are true (B) P(3) is false but P(5) is true (C) Both P(3) and P(5) are false (D) P(5) is false but P(3) is true
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JEE-MAIN-2019 (10th Jan-First Shift)-PCM-18
JEE (Main) – 2019 ANSWERS PART A – PHYSICS 1.
D
2.
D
3.
C
4.
A
5.
C
6.
A
7.
D
8.
A
9.
A
10.
A
11.
BONUS
12.
A
13.
C
14.
D
15.
B
16.
B
17.
A
18.
D
19.
B
20.
A
21.
C
22.
B
23.
A
24.
A
25.
A
26.
C
27.
A
28.
C
29.
C
30.
D
PART B – CHEMISTRY 31.
D
32.
C
33.
B
34.
A
35.
A
36.
A
37.
A
38.
A
39.
BONUS
40.
C
41.
A
42.
A
43.
B
44.
C
45.
B
46.
C
47.
B
48.
D
49.
B
50.
A
51.
C
52.
B
53.
D
54.
A
55.
B
56.
D
57.
C
58.
D
59.
C
60.
D
PART C – MATHEMATICS 61.
C
62.
A
63.
B
64.
B
65.
C
66.
D
67.
C
68.
D
69.
A
70.
D
71.
A
72.
B
73.
B
74.
B
75.
A
76.
No option is correct
77.
D
78.
C
79.
B
80.
C
81.
C
82.
B
83.
A
84.
A
85.
B
86.
A
87.
B
88.
D
89.
B
90.
A
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JEE-MAIN-2019 (10th Jan-First Shift)-PCM-19
HINTS AND SOLUTIONS PART A – PHYSICS 2
1.
2
P h 2m 2m 2 (6.62 1034 )2 1 = eV 31 12 2 2 9.1 10 (7.5 10 ) 1.6 1019
k
Alternate method 150 V
7.5 × 10–12 =
150 V
80 kV 3 80 Energy = keV 25 keV 3 V
2.
x
dx
d = ( dN)x F = 2 2 dx x R R
d 0
3.
VAJ I R AJ E E x 12r 3 12r 3r L
4.
5.
D 2 hTR 2hRR
a a 1 0, , 2 2 a a 2 , ,0 2 2 aˆ a ˆ r2 r1 i k 2 2 ˆi kˆ Unit vector = 2
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JEE-MAIN-2019 (10th Jan-First Shift)-PCM-20 6. 6
6
Rnet = 4
6
7.
W = F s cos g 1 g = m g t2 2 2 2
8.
For maxima, dsin = n d ( Sin when is small) n 2500 = nm n Take n = 4, 5 for = 625 nm and 500 nm.
9.
Area (Range)2 v4 4 A1 1 A2 2
10.
11.
12.
13.
128 kg 64 (2kg) m3 10 cm3 64 (2 kg) 503 = 6 3 10 (50 cm) 64 50 50 50 2 kg (2 kg) = = 8 3 100 100 100 (50 cm) (50 cm3 ) Work done by external agent = Uf – Ui = 2 MB
Q kA (T2 T1 ) t 1 Q k (T2 T1 ) A t Cnet = C1 + C2 + C3
kAEo AE AE AE k1 o k 2 o k 3 o d 3 d 3 d 3 d
k
k1 k 2 k 3 12 3
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JEE-MAIN-2019 (10th Jan-First Shift)-PCM-21 14.
Q1 + q 2 + Q3 = Q Q3 Q1 Q2 =k 2 2 4a 4 b 4 c 2
…(1) …(2)
Subs. Q1, Q2, Q3 in (1)
15.
k
Q 4 (a b 2 c 2 )
v
kQ1 kQ2 kQ3 a b c
2
n1 = n2 = n3 T T T 1 2 1 3 1 4 T1 T2 T3
T2 T3 T4 T1 T2 T3
T2T3 = T1T4 and
T32 T4 T2
Solve for T2 and T3. 16.
Initially, kinetic energy =
1 1 GMe mv 2 m 2 2 r
By conservation of M.E., GMem KE 0 r GMem KE = mv 2 r 17.
18.
dV a 2gh 0 (for maximum height) dt 2 10 8 h = 5.1 cm 2ga2 2 9.8 10 8 n T 2 m On solving, n = 5, 5 loops are formed in 1 m. f
19.
Separation between successive nodes =
340 f1 fo ; f2 fo 340 17 f1 306 18 17 18 or or f2 323 19 17 19
1 m = 20 cm 5
340 340 34
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JEE-MAIN-2019 (10th Jan-First Shift)-PCM-22
20.
21.
1 1 1 1 1 1 (1 1) ; (2 1) f1 R f 2 R 1 ( 1) 1 1 ; 2 f1 R f2 R 1 2 f2 2f1 ; f1 f2 1 1 2( 2 1) R R 1 + 22 = 3 R = P Q (x y) (xy)
= x y xy = x y xy = xy 22.
E = CB
23.
In 8 seconds count rate becomes 4 half lives = 8s 1 half lie = 2s
In 6s or 3 half lives, count rate = 24.
25.
1600 = 200 23
E = vB v = Ed = dvB P1
P2 r
OV
v1 = v 2 k(2qa) k(4qa) x2 (R x)2 R–x= 2x R x= 2 1
x
26.
1 times. 16
OV
OV
20
20
O
0.5 A
OV
10 V
10 V
10 V
OV
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JEE-MAIN-2019 (10th Jan-First Shift)-PCM-23
27.
By colour coding, R = 50 102 I2R = P P I 20 mA R
28.
u=0
Ycm from ground =
40 m
0.02 100 = 40 m/s 0.05 V2 40 40 H = cm = 80 m 2g 20 Height above building = 80 – 40 = 40 m
COM
Vcm = 60 m
100 m/s
29.
0.03 100 = 60 m 0.05
dM = di A dq 2 = x 2 = dx x 2 2 L
M dM 0
30.
F – f = ma 1 fR = mR2 2 a = R
PART B – CHEMISTRY 31.
Cl
F
Cl
M
= 3 GI Total = 12 GI
NO 2
F
Cl
M
M SCN
NO 2
Cl
F NCS
ONO
= 3 GI
F M
SCN = 3 GI
ONO
NCS = 3 GI
H T
32.
S
33.
KP = KC(RT)ng
34.
Fact based
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JEE-MAIN-2019 (10th Jan-First Shift)-PCM-24 35.
More polluted water has high biological oxygen demand.
36.
Fact based.
37. Major product will be
Due to E2 elimination & extended
conjugation
38.
K Ae Ea /RT (i) is true but (ii) is false (ii) should be
K
T
39.
40.
Fact based
41.
Density of DCM is higher than H 2O.
42.
Adipic acid will give unstable 7 membered anhydride.
43.
Electron withdrawing group enhances the rate of hydrolysis.
44.
K.E = h - h0, So (B) is not correct.
45.
Fact based.
46.
Atomic & ionic radii decreases
47.
Allylic radical will form in presence of sunlight.
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JEE-MAIN-2019 (10th Jan-First Shift)-PCM-25 48.
49.
O
50. F
F Xe
F
F
51.
Diagonal relationship.
52.
Lesser the SRP value higher the reducing power.
53.
Fact based.
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JEE-MAIN-2019 (10th Jan-First Shift)-PCM-26 54.
55.
PA APA0 YAPT PT APA0 BPB0 0.4 7 103 + 0.6 12 103 = 104 0.4 7 103 = YA 104 YA = 0.28 YB = 1 -0.28 = 0.72
56.
NaBH4 don’t reduce ester.
57.
Fact based
58.
Higher the Ka value lower the pKa value.
59.
Hall-Herolut’s process is given by 2Al2O3 3C 4 Al 3CO2 3 2 2Al2O3 4Al 6O
At cathode: 4 Al3 12e 4Al At anode: 6O2 3 O 2 g 12e
2C 3 O2 3 CO2 60
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JEE-MAIN-2019 (10th Jan-First Shift)-PCM-27
PART C – MATHEMATICS 61.
P (7 or 8)
P H P 7 or 8 P T P 7 or 8
62.
1 11 1 2 11 1 19 2 36 2 9 72 9 72
3 Let P be the point nearest to , 0 , then normal 2 3 at P will pass through , 0 . 2
t2 t Let Co – ordinates of P be s , 4 2 Hence equation of normal is y tx
t t2 2 4
3 The line passes through , 0 2 3 3t t t t2 (–2, 0 are rejected) 2 2 4 hence nearest point is (1, 1) 2
5 2 3 2 1 1 0 2
distance
63.
ˆi ˆj kˆ n n1 n2 3 1 2 7iˆ 7ˆj 7kˆ 1
2
3
Equation of plane is 7 x 4 7 y 1 7 z 2 0 7x 7y 7z 7 0 x y z 1 64.
1 3 8 x y 5 25 12 x y x y 13
2
x
i
………..(1)
2
N 1 9 64 x 2 y 2 9.2 25 5 34.2 5 74 x 2 y 2
171 74 x 2 y 2 97 x 2 y 2
……………(2)
2
x 2 y 2 2xy 169 97 2xy xy 36
x y
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JEE-MAIN-2019 (10th Jan-First Shift)-PCM-28 T = 4, 9 So ratio is 65.
4 9 or 9 4
5,5r,5r 2 sides of triangle, 5 5r 5r 2 ………………(1) 2 5 5r 5r ………………(2) 2 5r 5r 5 ………………(3) 2 From r r 1 0 1 5 1 5 , ………..(4) r 2 2 From (2) r2 r 1 0 r R ……….(5) From (3) r2 r 1 0 1 5 1 5 r So, r 0 2 2 1 5 1 5 …………(6) r , , 2 2 1 5 1 5 From (4), (5), (6) r , 2 2 now check options 3
20 CI1 20 20 CI CI1 i 1 20 20 CI1 CI1 I Now 20 20 21 CI CI1 CI 21 Let given sum be S, so 3 2 20 i 1 20.21 100 S 3 3 I1 21 21 2 21 20
66.
Given S
67.
k k 100 21
sin2 2 cos4 2
3 4
Let cos2 2 t
1 cos2 2 cos4 2 t
3 4
1 1 cos2 2 2 2
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JEE-MAIN-2019 (10th Jan-First Shift)-PCM-29
2cos2 2 1 0 cos 4 0 4 2n 1 2 3 2n 1 , 0, 8 8 8 2 Sum of values of is 2 68.
Case – 1 c 50 f 0 0
………….(i) ……………(ii)
c40 f 2 0
4 c 5 4c c 4 0
c 24 f 2 0
………..(iii)
9 c 5 6c c 4 0
49 ………(iv) 4 Here i ii iii iv 4c 49 0 c
49 c , 24 4 Case – II …………….(i) c 50 f 0 0
……………..(ii) c4 f 2 0 c 24 ………..(iii) f 3 0 c 49 ………..(iv) 4 c
49 c , 24 4
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JEE-MAIN-2019 (10th Jan-First Shift)-PCM-30
69.
dy 3 sec 2 x y sec 2 x dx This is linear differential equation 3 sec 2 x dx Integrating factor e e3 tan x
Hence y.e3 tan x e
3 tan x
.sec 2 x dx
e3 tan x c 3 1 y Ce3 tan x 3 4 4 1 Given y Ce 3 3 3 4 3 3 Ce 1 1 Hence y e3 .e3 e6 3 3 4 y e3 tan x
70.
140 n p 46 3 140 n C 28 5 140 n M 70 2 n P C M n P n C n M n P C n C M n M P n P M C 140 140 140 140 46 28 70 15 10 6 30 144 9 14 23 4 102 So required number of student 140 102 38
71.
In the expansion of 1 xlog2 x
third term say T3 5C2 xlog2 x
xlo2 x
2
5
2
2560
256
taking lograthium to the base 2 on both sides 2
2 log2 x 8 log2 x 2
1 4 1 Here x 4 x 4,
72.
Normal to there 2 curves are y m x c 2bm bm3
y mx 4am 2am3 FIITJEE Ltd., FIITJEE House, 29-A, Kalu Sarai, Sarvapriya Vihar, New Delhi -110016, Ph 46106000, 26569493, Fax 26513942 website: www.fiitjee.com.
JEE-MAIN-2019 (10th Jan-First Shift)-PCM-31 If they how a common normal (c + 2b) – 4a = (2a – b) m2 (m = 0 corresponds to axis) c m2 20 2a b c 2 2a b According to the question answer is (1, 2, 3, 4) but my be in question they want common normal other than x – axis, hence answer is (B) 73.
xyz5 x 2y 3z 9, x 3y z 1 1 1 D 1 2 3 0 2 9 3 3 2 0 5 1 3 0 1 1 5
Now, D3 1 2 9 0 2 27 9 5 3 2 0 13 1 3
at 5, b 13 above 3 planes from common line
74.
Lim
1 x sin 1 x sin 1 x 2 1 x 1 x
x 1
Lim
1 x sin x 1 sin 2
x 1 1 x 1 sin x 1 Lim 1 1 0 x 1 x 1 x 1
2 75.
sin 2
4d
d sin 2 2 sin d sin 2 2d 4 d sin 2 d sin
A 1 5 2 1 1
0
(New R3 R3 2R2 R1 )
0 2
4 d d sin2 4 d 2 sin2 2
Because minimum value of A 8 d 2 9 d 1 or –5
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JEE-MAIN-2019 (10th Jan-First Shift)-PCM-32
76.
3z1 2 2z2 3z1 2cos 2 sin i 2z2 2z 1 1 2 cos sin 3z1 2 2 2z 3z 5 3 Given, z 1 2 cos sin i 3z2 2z1 2 2 Let
Which is neither purely real nor purely imaginary and z depends on .
b
77.
x
4
2x 2 dx
a
From figure min area is 2, 2
78.
Let point P is , and centroid of PQR is (h, k), then 3h 1 3 and 3k 4 2 3h 4 and 3k 2 Because , lies on 2x 3y 4 0 2 3h 4 3 k 2 4 0
locus is 6x 9y 2 0 whose slope is
2 3
79.
From the graph we can easily conclude that f x is non – derivable at x 2, 1,0,1,2
80.
Tangent at (1, –1) is x 1 y 1 2 x 1 3 y 1 12 0 3x 4y 7 2
2
Required circle is x 1 y 1 3x 4y 7 0 It pass through (4, 0) 9 1 12 7 0 2
required circle is x 2 y 2 8x 10y 16 0 Radius 16 25 16 5 81.
f x x 3 ax 2 bx c f ' x 3x 2 2ax b f " x 6x 2a
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JEE-MAIN-2019 (10th Jan-First Shift)-PCM-33 f "' x 6 a f ' 1 3 2a b
a b 3
b f " 12 2a 2a b 12 c f "' 3 c 6 and a 5, b 2 f x x 2 5x 2 2x 6 f 2 8 20 4 6 2
82.
Because b 2a , so 3 2 21
……….(i)
Because a is perpendicular to c so 6 61 3 3 1 0
………..(ii)
1, 2 , 3 1,3 21, 1 21 where 1 R
1 , 4, 0 satisfied above triplet. 2 83.
Let A is 1 3, 1, 2 5 AB 3 2 i 3 j 4 5 kˆ which is parallel to plane x 4y 3z 1 1 3 2 4 3 3 4 5 0
8 2 0
84.
sinn sin
1 n
cos
sinn1
t
n
1 4
d
n
t dt
(Put sin t )
t n 1 1
1
1 n 1 n t 1 n1 1 n1 t t n1 dt dt t tn n 1 1 Put 1 n 1 z dt dz, t tn 1
1
n 1 2 1 n 1 sinn1
h 1 n
n 1
1 n 1 t1n 1 zn n I z dz c n 1 n2 1 1 1 n 1 n
c
c
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JEE-MAIN-2019 (10th Jan-First Shift)-PCM-34
85.
Length of median BM A
1 1 2 2 BC2 BA 2 AC 2 25 49 36 2 2 T (Top of tower)
3 7
M
h
3 30o
B
5
C
1 112 112 28 2 7 2 4
Let h be height of tower, given tan30o
86.
M
2 7
Hyperbola is
h 2 h
h2
7 28 3 3
x2 y2 1 5 4
Equation of its tangent in slope from is y mx 5m2 4 Hence tangent with slope 1 is y x 1 87.
ax bx 2 cx3 ay1 by 2 cy 2 I 1 , abc abc 8 60 6 0 10 0 10 0 6 8 8 0 , 2, 2 24 24
B (0, 6)
10 6 8 C (0, 0)
88.
A (8, 0)
Two digit numbers of the form 7 2 are 16, 23 …………….,93 Two digit numbers of the form 7 5 are 12, 19,…………….,96 12 13 Sum of all the above numbers equals to 16 93 12 96 654 702 1356 2 2
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JEE-MAIN-2019 (10th Jan-First Shift)-PCM-35
89.
y kx 2 , x ky 2 3
1 1 x k k 2 x 4 x 0 or x 3 x , 0 k k 1 1 Point of intersection are , and 0,0 k k
1 1 , k k
1/k
1/k
x 1 x 3/2 kx 3 Area kx1 dx 1 k 3 k 3/2 0 2 1 1 2 1 k2 3k 3 1 k 3
90.
P n n2 41 P 3 9 3 41 47 P 5 25 5 41 61
Hence P 3 and P 5 are both prime
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FIITJEE Solutions to JEE(Main)-2019 Test Date: 10th January 2019 (Second Shift)
PHYSICS, CHEMISTRY & MATHEMATICS Paper - 1 Time Allotted: 3 Hours
Maximum Marks: 360
Please read the instructions carefully. You are allotted 5 minutes specifically for this purpose.
Important Instructions: 1.
The test is of 3 hours duration.
2.
This Test Paper consists of 90 questions. The maximum marks are 360.
3.
There are three parts in the question paper A, B, C consisting of Physics, Chemistry and Mathematics having 30 questions in each part of equal weightage. Each question is allotted 4 (four) marks for correct response.
4.
Out of the four options given for each question, only one option is the correct answer.
5.
For each incorrect response 1 mark i.e. ¼ (one-fourth) marks of the total marks allotted to the question will be deducted from the total score. No deduction from the total score, however, will be made if no response is indicated for an item in the Answer Box.
6.
Candidates will be awarded marks as stated above in instruction No.3 for correct response of each question. One mark will be deducted for indicating incorrect response of each question. No deduction from the total score will be made if no response is indicated for an item in the answer box.
7.
There is only one correct response for each question. Marked up more than one response in any question will be treated as wrong response and marked up for wrong response will be deducted accordingly as per instruction 6 above.
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JEE-MAIN-2019 (10th Jan-Second Shift)-PCM-2
PART –A (PHYSICS) 1.
The Wheatstone bridge shown in Fig. here, gets balanced when the carbon resistor used as R1 has the colour code (Orange, Red, Brown). The resistors R2 and R4 are 80 and 40 , respectively. Assuming that the colour code for the carbon resistors gives their accurate values, the colour code for the carbon resistor, used as R3 would be: (A) Brown, Blue, Brown (B) Brown, Blue, Black (C) Red, Green, Brown (D) Grey, Black, Brown
2.
Consider the nuclear fission Ne20 2He4 C12 .Given that the binding energy / nucleon of Ne20, He4 and C12 are, respectively, 8.03 MeV, 7.07 MeV, and 7.86 MeV, identify the correct statement: (A) energy of 12.4 MeV will be supplied (B) 8.3 MeV energy will be released (C) energy of 3.6 MeV will be released (D) energy of 11.9 MeV has to be supplied
3.
A hoop and a solid cylinder of same mass and radius are made of a permanent magnetic material with their magnetic moment parallel to their respective axes. But the magnetic moment of hoop is twice of solid cylinder. They are placed in a uniform magnetic field in such a manner that their magnetic moments make a small angle with the field. If the oscillation periods of hoop and cylinder are Th and TC respectively, then : (A) Th TC (B) Th 2TC (C) Th 1.5TC
(D) Th 0.5C
4.
An unknown metal of mass 192 g heated to a temperature of 100°C was immersed into a brass calorimeter of mass 128 g containing 240 g of water at a temperature of 8.4°C. Calculate the specific heat of the unknown metal if water temperature stabilizes at 21.5°C. (Specific heat of brass is 394 J kg–1 K–1) (A) 458 J kg–1 K–1 (B) 1232 J kg–1 K–1 –1 –1 (C) 916 J kg K (D) 654 J kg–1 K–1
5.
Two identical spherical balls of mass M and radius R each are stuck on two ends of a rod of length 2R and mass M (see figure). The moment of inertia of the system about the axis passing perpendicularly through the centre of the rod is : 137 17 (A) MR2 (B) MR 2 15 15 209 152 (C) MR2 (D) MR 2 15 15
6.
The self produced emf of a coil is 25 volts. When the current in it is changed at uniform rate from 10 A to 25 A in 1 s, the change in the energy of the inductance is : (A) 740 J (B) 437.5 J (C) 540 J (D) 637.5 J
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JEE-MAIN-2019 (10th Jan-Second Shift) PCM-3 7.
The actual value of resistance R, shown in the figure is 30 . This is measured in an experiment as shown using the standard V formula R , where V and I are the readings of the voltmeter I and ammeter, respectively. If the measured value of R is 5% less, then the internal resistance of the voltmeter is : (A) 600 (B) 570 (C) 35 (D) 350
8.
At some location on earth the horizontal component of earth’s magnetic field is 18 10–6 T. At this location, magnetic needle of length 0.12 m and pole strength 1.8 A m is suspended from its mid-point using a thread, it makes 45° angle with horizontal in equilibrium. To keep this needle horizontal, the vertical force that should be applied at one of its ends is : (A) 3.6 105 N (B) 1.8 105 N (C) 1.3 105 N
9.
Two vectors A and B have equal magnitudes. The magnitude of A B is ‘n’ times the magnitude of A B . The angle between A and B is:
2
n 1 (A) cos1 2 n 1 n2 1 (C) sin1 2 n 1 10.
11.
12.
(D) 6.5 10 5 N
n 1 (B) cos1 n 1 n 1 (D) sin1 n 1
A metal plate of area 1 10–4 m2 is illuminated by a radiation of intensity 16 mW/m2. The work function of the metal is 5eV. The energy of the incident photons is 10 eV and only 10% of it produces photo electrons. The number of emitted photo electrons per second and their maximum energy, respectively, will be : [1 eV = 1.6 10–19 J] (A) 1014 and 10 eV (B) 1012 and 5 eV 11 (C) 10 and 5 eV (D) 1010 and 5 eV A particle which is experiencing a force, given by F 3 i 12 j, undergoes a displacement of d 4 i . If the particle had a kinetic energy of 3 J at the beginning of the displacement, what is its kinetic energy at the end of the displacement? (A) 9 J (B) 12 K (C) 10 J (D) 15 J
Consider a Young’s double slit experiment as shown in figure. What should be the slit separation d in terms of wavelength such that the first minima occurs directly in front of the slit (S1)? (A) (B) 2( 5 2) ( 5 2) (C) (D) 2(5 2) (5 2) FIITJEE Ltd., FIITJEE House, 29-A, Kalu Sarai, Sarvapriya Vihar, New Delhi -110016, Ph 46106000, 26569493, Fax 26513942 website: www.fiitjee.com.
JEE-MAIN-2019 (10th Jan-Second Shift)-PCM-4 13.
The eye can be regarded as a single refracting surface. The radius of curvature of this surface is equal to that of cornea (7.8 mm). This surface separates two media of refractive indices 1 and 1.34. Calculate the distance from the refracting surface at which a parallel beam of light will come to focus. (A) 1 cm (B) 2 cm (C) 4.0 cm (D) 3.1 cm
14.
A current of 2 mA was passed through an unknown resistor which dissipated a power of 4.4 W. Dissipated power when an ideal power supply of 11 V is connected across it is : (A) 11 105 W (B) 11 103 W (C) 11 104 W
(D) 11 105 W
15.
The diameter and height of a cylinder are measured by a meter scale to be 12.60.1 cm and 34.2 0.1 cm, respectively. What will be the value of its volume in appropriate significant figures? (A) 426481 cm3 (B) 426481.0 cm3 (C) 426080 cm3 (D) 430080 cm3
16.
For equal point charges Q each are placed in the xy plane at (0, 2), (4, 2), (4, –2) and (0, –2). The work required to put a fifth change Q at the origin of the coordinate system will be : Q2 1 Q2 1 (A) 1 (B) 1 4 0 4 0 3 3 (C)
Q2 2 2 0
(D)
Q2 4 0
17.
The modulation frequency of an AM radio station is 250 kHz, which is 10% of the carrier wave. If another AM station approaches you for license what broadcast frequency will you allot? (A) 2750 kHz (B) 2900 kHz (C) 2250 kHz (D) 2000 kHz
18.
A closed organ pipe has a fundamental frequency of 1.5 kHz. The number of overtones that can be distinctly heard by a person with this organ pipe will be : (Assume that the highest frequency a person can hear is 20,000 Hz) (A) 6 (B) 4 (C) 7 (D) 5
19.
For the circuit shown below, the current through the Zener diode is :
(A) 9 mA (C) Zero
(B) 5 mA (D) 14 mA
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JEE-MAIN-2019 (10th Jan-Second Shift) PCM-5 20.
The electric field of a plane polarized electromagnetic wave in free space at time t = 0 is given by an expression E(x, y) 10 ˆj cos[(6x 8z)] The magnetic field B (x, z, t) is given by : (c is the velocity of light) 1 ˆ 1 ˆ (A) 6k 8iˆ cos[(6x 8z 10 ct)] (B) 6k 8iˆ cos[(6x 8z 10ct)] c c 1 ˆ 1 ˆ (C) 6k 8iˆ cos[(6x 8z 10 ct)] (D) 6k 8iˆ cos[(6x 8z 10 ct)] c c
21.
Charges – q and + q located at A and B, respectively, constitute an electric dipole. Distance AB = 2a, O is the mid point of the dipole and OP is perpendicular to AB. A charge Q is placed at P where OP = y and y >> 2a. The charge Q experiences an electrostatic force F. If Q is now moved along the equatorial line to P such that y y OP ' , the force on Q will be close to: 2a 3 3
(A) 3F (C) 9F
F 3 (D) 27F (B)
22.
Two stars of masses 3 1031 kg each, and at distance 2 1011 m rotate in a plane about their common centre of mass O. A meteorite passes through O moving perpendicular to the star’s rotation plane. In order to escape from the gravitational field of this double star, the minimum speed that meteorite should have at O is : (Take Gravitational constant G = 6.67 10–11 Nm2 kg–2) (A) 2.4 104 m/s (B) 1.4 105 m/s (C) 3.8 104 m/s (D) 2.8 105 m/s
23.
Half mole of an ideal monoatomic gas is heated at constant pressure of 1 atm from 20°C to 90°C. Work done by has is close to: (Gas constant R = 8.31 J/molK) (A) 581 J (B) 291 J (C) 146 J (D) 73 J
24.
A parallel plate capacitor having capacitance 12 pF is charged by a battery to a potential difference of 10 V between its plates. The charging battery is now disconnected and a porcelain slab of dielectric constant 6.5 is slipped between the plates. The work done by the capacitor on the slab is : (A) 692 pJ (B) 508 pJ (C) 560 pJ (D) 600 pJ FIITJEE Ltd., FIITJEE House, 29-A, Kalu Sarai, Sarvapriya Vihar, New Delhi -110016, Ph 46106000, 26569493, Fax 26513942 website: www.fiitjee.com.
JEE-MAIN-2019 (10th Jan-Second Shift)-PCM-6 25.
A particle starts from the origin at time t = 0 and moves along the positive x-axis. The graph of velocity with respect to time is shown in figure. What is the position of the particle at time t = 5s? (A) 10 m (B) 6 m (C) 3 m (D) 9 m
26.
A rigid massless road of length 3l has tow masses attached at each end as shown in the figure. The rod is pivoted at point P on the horizontal axis (see figure). When released from initial horizontal position, its instantaneous angular acceleration will be :
g 13l g (C) 2l (A)
g 3l 7g (D) 3l (B)
27.
Two forces P and Q, of magnitude 2F and 3F, respectively, are at an angle with each other. If the force Q is doubled, then their resultant also gets doubled. Then, the angle is : (A) 120° (B) 60° (C) 90° (D) 30°
28.
A cylindrical plastic bottle of negligible mass of filled with 310 ml of water and left floating in a pond with still water. If pressed downward slightly and released, it starts performing simple harmonic motion at angular frequency . If the radius of the bottle is 2.5 cm then is close to: (density of water = 103 kg/m3) (A) 3.75 rad s–1 (B) 1.25 rad s–1 –1 (C) 2.50 rad s (D) 5.00 rad s–1
29.
A particle executes simple harmonic motion with an amplitude of 5 cm. When the particle is at 4 cm from the mean position, the magnitude of its velocity is SI units is equal to that of its acceleration. Then, its periodic time in seconds is : 4 3 (A) (B) 3 8 8 7 (C) (D) 3 3
30.
Two kg of a monoatomic gas is at a pressure of 4 104 N/m2. The density of the gas is 8 kg/m3. What is the order of energy of the gas due to its thermal motion? (A) 103 J (B) 105 J 4 (C) 10 J (D) 106 J
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JEE-MAIN-2019 (10th Jan-Second Shift) PCM-7
PART –B (CHEMISTRY) 31.
The ground state energy of hydrogen atom is –13.6 eV. The energy of second excited state of He+ ion in eV is: (A) –54.4 (B) –3.4 (C) –6.04 (D) –27.2
32.
Haemoglobin and gold sold are examples of: (A) positively and negatively charged sols, respectively (B) positively charged sols (C) negatively charged sols (D) negatively and positively charged sols, respectively
33.
The major product of the following reaction is:
(A)
(B)
(C)
(D)
34.
The amount of sugar (C12H22O11) required to prepare 2 L of its 0.1 M aqueous solution is: (A) 136.8 g (B) 17.1 g (C) 68.4 g (D) 34.2 g
35.
Among the following reactions of hydrogen with halogens, the one that requires a catalyst is: (A) H2 I2 2HI (B) H2 Cl2 2HCl (C) H2 Br2 2HBr (D) H2 F2 2HF
36.
5.1 g NH4SH is introduced in 3.0 L evacuated flask at 327°C. 30% of the solid NH4SH decomposed to NH3 and H2S as gases. The Kp of the reaction at 327°C is (R = 0.082 L atm mol–1 K–1, Molar mass of S = 32 g mol–1, molar mass of N = 14 g mol–1) (A) 0.242 10–4 atm2 (B) 1 10–4 atm2 –3 2 (C) 4.9 10 atm (D) 0.242 atm2
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JEE-MAIN-2019 (10th Jan-Second Shift)-PCM-8 37.
The reaction that is NOT involved in the ozone layer depletion mechanism in the stratosphere is :
uv (A) CF2Cl2 (g) Cl(g) CF2Cl(g)
(B) Cl O(g) O(g) Cl(g) O2 (g)
(C) CH4 2 O3 3CH2 O 3H2O
hv (D) HOCl(g) OH(g) C l(g)
38.
In the cell Pt(s) H2 (g,1bar) HCl(aq) AgCl(s) Ag(s) Pt(s) the cell potential is 0.92 V when a 10–6 molal HCl solution is used. The standard electrode potential of (AgCl / Ag,Cl ) electrode is : 2.303RT 0.06V at 298K Given, F (A) 0.94 V (B) 0.76 V (C) 0.40 V (D) 0.20 V
39.
The 71st electron of an element X with an atomic number of 71 enters into the orbital: (A) 6p (B) 4f (C) 5d (D) 6s
40.
The correct match between item ‘I’ and item ‘II’ is: Item ‘I’ Item ‘II’ (compound) (reagent) (a) Lysine (p) 1-naphtol (b) Furfural (q) Ninhydrin (c) Benzyl alcohol (r) KMnO4 (d) Styrene (s) Ceric ammonium nitrate (A) (a)(q); (b) (p); (c) (s); (d) (r) (B) (a)(q); (b) (p); (c) (r); (d) (s) (C) (a)(r); (b) (p); (c) (q); (d) (s) (D) (a)(q); (b) (r); (c) (s); (d) (p)
41.
An aromatic compound ‘A‘ having molecular formula C7H6O2 on treating with aqueous ammonia and heating forms compound ‘B’. The compound ‘B’ on reaction with molecular bromine and potassium hydroxide provides compound ‘C’ having molecular formula C6H7N. The structure of ‘A’ is : (A) (B)
(C)
42.
(D)
The process with negative entropy change is: (A) Dissociation of CaSO4(s) to CaO(s) and SO3(g) (B) Sublimation of dry ice (C) Dissolution of iodine in water (D) Synthesis of ammonia from N2 and H2
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JEE-MAIN-2019 (10th Jan-Second Shift) PCM-9 43.
An ideal gas undergoes isothermal compression from 5 m3 to 1 m3 against a constant external pressure of 4 NM–2. Heat released in this process is used to increase the temperature of 1 mole of Al. If molar heat capacity of Al is 24 J mol–1K–1, the temperature of Al is increases by: 3 (A) K (B) 2 K 2 2 (C) K (D) 1 K 3
44.
Elevation in the boiling point for 1 molal solution of glucose is 2 K. The depression in the freezing point for 2 molal solution of glucose in the same solvent is 2 K. The relation between Kb and Kf is : (A) K b 1.5K f (B) K b K f (C) K b 0.5K f
45.
46.
47.
(D) K b 2K f
The major product of the following reaction is :
(A)
(B)
(C)
(D)
Sodium metal on dissolution in liquid ammonia gives a deep blue solution due to the formation of: (A) sodium-ammonia complex (B) sodamide (C) sodium ion-ammonia complex (D) ammoniated electrons k1 For an elementary chemical reaction, A 2 2A, the expression for k 1
(A) k1[A 2 ] k 1[A]2
(B) 2k1[A 2 ] k 1[A]2
(C) k1[A 2 ] k 1[A]2
(D) 2k1[A 2 ] 2k 1[A]2
d[A] is dt
48.
Which of the following tests cannot be used for identifying amino acids? (A) Biuret test (B) Barfoed test (C) Ninhydrin test (D) Xanthoproteic test
49.
The difference in the number of unpaired electrons of a metal ion in its high-spin and low-spin octahedral complexes is two. The metal ion is : (A) Ni2 (B) Fe2 (C) Co2 (D) Mn2
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JEE-MAIN-2019 (10th Jan-Second Shift)-PCM-10 50.
51.
The major product obtained in the following reaction is:
(A)
(B)
(C)
(D)
The pair that contains two P – H bonds in each of the oxoacids is : (A) H4P2O5 and H4P2O6 (B) H3PO2 and H4P2O5 (C) H3PO3 and H3PO2
52.
Which is the most suitable reagent for the following transformation?
(A) Tollen’s reagent (C) CrO 2 Cl2 / Cs2 53.
(D) H4P2O5 and H3PO3
(B) I2 / NaOH (D) alkaline KMnO4
What is the IUPAC name of the following compound?
(A) 3-Bromo-1, 2-dimethylbut-1-ene (B) 3-Bromo-3-methyl-1, 2-dimethylprop-1-ene (C) 2-Bromo-3-methylpent-3-ene (D) 4-Bromo-3-methylpent-2-ene 54.
The number of 2-centre-2-electron and 3-centre-2-electron bonds in B2H6, respectively, are : (A) 2 and 1 (B) 4 and 2 (C) 2 and 2 (D) 2 and 4
55.
In the reaction of oxalate with permanganate in acidic medium, the number of electrons involved in producing one molecule of CO2 is : (A) 1 (B) 10 (C) 2 (D) 5
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JEE-MAIN-2019 (10th Jan-Second Shift) PCM-11 56.
A reaction of cobalt(III) chloride and ethylenediamine in a 1 : 2 mole ratio generates two isomeric products A (violet coloured) and B (green coloured). A can show optical activity, but, B is optically inactive. What type of isomers does A and B represent? (A) Geometrical isomers (B) Coordination isomers (C) Linkage isomers (D) Ionisation isomers
57.
The electrolytes usually used in the electroplating of gold and silver, respectively, are : (A) [Au(CN2 ] and [Ag(CN)2 ] (B) [Au(Cn)2 ] and [AgCl2 ] (C) [Au(OH)4 ] and [Ag(OH)2 ]
(D) [Au(NH3 )2 ] and [Ag(CN)2 ]
58.
A compound of formula A2B3 has the hcp lattice. Which atom forms the hcp lattice and what fraction of tetrahedral voids is occupied by the other atoms : 2 1 (A) hcp lattice - A, Tetrahedral voids - B (B) hcp lattice - A, Tetrahedral voids - B 3 3 2 1 (C) hcp lattice - B, Tetrahedral voids - A (D) hcp lattice - B, Tetrahedral voids – A 3 3
59.
The major product of the following reaction is :
60.
(A)
(B)
(C)
(D)
What will be the major product in the following mononitration reaction?
(A)
(B)
(C)
(D)
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JEE-MAIN-2019 (10th Jan-Second Shift)-PCM-12
PART–C (MATHEMATICS) 61.
The value of such that sum of the squares of the roots of quadratic equation, x 2 (3 )x 2 has the lest value is : 15 (A) (B) 1 8 4 (C) (D) 2 9
62.
The value of cos
cos 3 cos 10 sin 10 is : 2 2 2 2 2
1 512 1 (C) 256
1 1024 1 (D) 2
(A)
(B)
63.
The curve amongst the family of curves represented by the differential equation, (x 2 y 2 )dx 2xy dy 0 which passes through (1, 1), is : (A) a circle with centre on the x-axis (B) an ellipse with major axis along the y-axis (C) a circle with centre on the y-axis (D) a hyperbola with transverse axis along the x-axis
64.
Let f : (–1, 1)R be a function defined by f(x) max x , 1 x 2 . if K be the set of all
points at which f is not differentiable, then K has exactly : (A) five elements (B) one element (C) three elements (D) two elements 10
65.
66.
The positive value of for which the co-efficient of x2 in the expression x 2 x 2 is x 720, is : (A) 4 (B) 2 2 (C) 5 (D) 3 2
The tangent to the curve, y xe x passing, through the point (1, e) also passes through the point : 4 (A) (2, 3e) (B) ,2e 3 5 (C) ,2e (D) (3, 6e) 3
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JEE-MAIN-2019 (10th Jan-Second Shift) PCM-13 67.
Let N be the set of natural numbers and two functions f and g be defined as f, g : NN such that n 1 if n is odd f(n) 2 n if n is even 2 and g(n) n ( 1)n . Then fog is : (A) onto but not one-one. (B) one-one but not onto. (C) both one-one and onto. (D) neither one-one nor onto.
68.
The number of values of (0, ) for which the system of linear equations x 3y 7z 0 x 4y 7z 0 (sin 3)x (cos 2)y 2z 0 has a non-trivial solution, is : (A) three (B) two (C) four (D) one Let ( 2) a b and (4 2) a 3 b be two given vectors where a and b are non collinear. The value of for which vectors and are collinear, is : (A) –4 (B) –3 (C) 4 (D) 3
69.
70.
Two sides of a parallelogram are along the lines, x + y = 3 and x – y + 3 = 0. If its diagonals intersect at (2, 4), then one of its vertex is : (A) (3, 5) (B) (2, 1) (C) (2, 6) (D) (3, 6)
71.
If
x
1
2 2 ' f(t) dt x t f(t) dt, then f 12 is : 0
x
24 25 4 (C) 5
18 25 6 (D) 25
(A)
(B)
5
72.
73.
5
3 i 3 i Let z . If R(z) and I(z) respectively denote the real and 2 2 2 2 imaginary parts of z, then : (A) I(z) = 0 (B) R(z) > 0 and I(z) > 0 (C) R(z) < 0 an I(z) > 0 (D) R(z) = –3 1 , then the minimum 3 number of independent shots at the target required by him so that the probability of 5 hitting the target at least once is greater than , is : 6 (A) 3 (B) 6 (C) 5 (D) 4
If the probability of hitting a target by a shooter, in any shot, is
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JEE-MAIN-2019 (10th Jan-Second Shift)-PCM-14
74.
If
5
x e
4x3
dx
1 4 x3 e f(x) C, where C is a constant of integration, then f(x) is equal 48
to : (A) 2x 3 1 (C) 2x 3 1
(B) 4x 3 1 (D) 4x 3 1
75.
If the area of an equilateral triangle inscribed 2 2 x y 10x 12y c 0 is 27 3 sq. units then c is equal to : (A) 13 (B) 20 (C) –25 (D) 25
76.
Consider the following three statements : P : 5 is a prime number. Q : 7 is a factor of 192. R : L.C.M. of 5 and 7 is 35. Then the truth value of which one of the following statements is true? (A) (~ P) (Q R) (B) (P Q) (~ R) (C) (~ P) (~ Q R) (D) P (~ Q R)
77.
The length of the chord of the parabola x2 = 4y having equation x 2y 4 2 0 is : (A) 3 2 (C) 8 2
78.
circle,
b 1 2 det(A) 2 Let A b b 1 b where b > 0. Then the minimum value of is : b 1 b 2 (A) 2 3 (B) 2 3
(D)
3
y2 x2 Let S (x, y) R2 : 1 , where r 1. Then S represents : 1 r 1 r 2 (A) a hyperbola whose eccentricity is , when 0 r 1. 1 r 2 (B) an ellipse whose eccentricity is , when r 1. r 1 2 (C) a hyperbola whose eccentricity is , when 0 r 1. r 1 1 (D) an ellipse whose eccentricity is , when r 1. r 1 25
80.
the
(B) 2 11 (D) 6 3
(C) 3 79.
in
if
50
Cr 50 r C25 r K
50
C25 , then K is equal to :
r 0
(A) (25)2 (C) 224
(B) 225 – 1 (D) 225
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JEE-MAIN-2019 (10th Jan-Second Shift) PCM-15 81.
82.
83.
The plane which bisects the line segment joining the points (–3, –3, 4) and (3, 7, 6) at right angles, passes through which one of the following points? (A) (–2, 3, 5) (B) (4, –1, 7) (C) (2, 1, 3) (D) (4, 1, –2) n 19 The value of cot cot 1 1 2p is : n1 p 1 21 (A) 19 22 (C) 23
19 21 23 (D) 22
(B)
If mean and standard deviation of 5 observations x1, x 2 , x3 , x 4 , x 5 are 10 and 3, respectively, then the variance of 6 observations x1, x 2 ,, x 3 and –50 is equal to : (A) 509.5 (B) 586.5 (C) 582.5 (D) 507.5
84.
Let f be a differential function such that 3 f(x) 1 f '(x) 7 ,(x 0) and f(1) 4. Then lim x f : x 0 4 x x 4 (A) exists and equals . (B) exists and equals 4. 7 (C) does not exist. (D) exists and equals 0.
85.
Two vertices of a triangle are (0, 2) and (4, 3). If its orthocenter is at the origin, then its third vertex lies in which quadrant? (A) third (B) second (C) first (D) fourth
86.
The value of
2
to t, is : 1 (A) (7 5) 12 3 (C) (4 3) 20 87.
dx , where [t] denotes the greatest integer less than or equal [x] [sin x] 4
2
1 (7 5) 12 3 (D) (4 3) 10
(B)
On which of the following lines lies the point of intersection of the line, x4 y5 z3 and the plane, x + y + z = 2? 2 2 1 x 3 4 y z 1 x4 y5 z5 (A) (B) 3 3 2 1 1 1 x 1 y 3 z 4 x2 y3 z3 (C) (D) 1 2 5 2 2 3
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JEE-MAIN-2019 (10th Jan-Second Shift)-PCM-16 88.
Let a1,a 2 ,a3 , ,a10 be in G.P. with ai 0 for i 1, 2, ,10 and S be the set of pairs (r,k),r,k N (the set of natural numbers) for which loge a1r a2k
loge a2r a3k
loge a3r a4k
loge a4r a5k loge a7r a8k
loge a5r a 6r loge a8r a9k
loge a 6r a7k 0 loge a9r a10k
Then the number of elements in S, is : (A) 4 (C) 2
(B) infinitely many (D) 10
89.
With the usual notation, in ABC, if A + B = 120°, a 3 1, then the ratio A : B, is : (A) 7 : 1 (B) 5 : 3 (C) 9 : 7 (D) 3 : 1
90.
A helicopter is flying along the curve given by y x
3
2
7, (x 0). A solider positioned at
1 the point ,7 wants to shoot down the helicopter when it is nearest to him. Then this 2 nearest distance is : 5 1 7 (A) (B) 6 3 3 1 7 1 (C) (D) 6 3 2
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JEE-MAIN-2019 (10th Jan-Second Shift) PCM-17
JEE (Main) – 2019 ANSWERS PART A – PHYSICS 1.
A
2.
No option is correct
3.
A
4.
B
5.
A
6.
A
7.
B
8.
D
9.
A
10.
C
11.
D
12.
A
13.
D
14.
A
15.
B
16.
B
17.
D
18.
A
19.
A
20.
B
21.
D
22.
D
23.
B
24.
B
25.
D
26.
A
27.
A
28.
B
29.
C
30.
C
PART B – CHEMISTRY 31.
C
32.
A
33.
C
34.
C
35.
A
36.
A
37.
C
38.
D
39.
C
40.
A
41.
A
42.
D
43.
C
44.
D
45.
D
46.
D
47.
D
48.
B
49.
C
50.
C
51.
B
52.
B
53.
D
54.
B
55.
B
56.
A
57.
A
58.
A
59.
B
60.
D
PART C – MATHEMATICS 61.
D
62.
A
63.
A
64.
C
65.
A
66.
B
67.
A
68.
B
69.
A
70.
D
71.
A
72.
D
73.
C
74.
B
75.
D
76.
D
77.
D
78.
A
79.
B
80.
B
81.
D
82.
A
83.
D
84.
B
85.
B
86.
C
87.
C
88.
B
89.
A
90.
C
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JEE-MAIN-2019 (10th Jan-Second Shift)-PCM-18
HINTS AND SOLUTIONS PART A – PHYSICS 1.
R1 = 32 × 10 = 320 Ry 40 320 R1 R3 = = 160 R2 80 Colour code of R3 be Brown, Blue, Brown.
2.
Q = (B.E.)R – (B.E.)P = 20 × 8.03 – (8 × 7.07 + 12 × 7.86) = 160.6 – (56.56 + 94.32) Q = +9.72 meV 9.72 MeV released.
3.
T 2
I B
Th I R c Tc Ic h
2
=
1 1 2
Th = T c 4.
5.
6.
Heat loss = Heat gain 192 × S(100 – 21.5) = (128 × 0.394 + 240 × 4.2) (21.5 – 8.4) 192 × 78.5 × S = 1058.432 × 13.1 S = 0.91995 J/g K–1 S = 919.95 J/kg K–1
4R2 2 I 2 MR 2 M4R2 M 12 5 137 1 4 = MR2 8 = MR2 3 5 15
LT 25 t 25 1 5 15 3 1 2 2 1 5 U L If I1 (252 10 2 ) 2 2 3 5 = 525 4.7.5 J 6
L
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JEE-MAIN-2019 (10th Jan-Second Shift) PCM-19
7.
8.
9.
30 R 30 00.95 R 30 R = 570 = F × 0.06 = 1.8 × 0.012 × 18 × 10–6 F = 6.48 × 10–5 A B n A B
A2 + B2 + 2AB cos = n2(A2 + B2 – 2AB cos ) 2a2 (n2 1) cos (1 n2 ) 2a 2 n2 1 cos 2 n 1 10.
11.
12.
13.
14.
A B a
16 10 3 10 4 1.6 1011 10 10 10 19 (K.E.)max = (10 – 5) eV = 5eV n
K.E. – 3 = F d K.E. = 3 + (3iˆ 12ˆj) (4iˆ ) K.E. = 3 + 12 = 15 J
(2d)2 (d)2 2d
( 5 2)d
d
2
2
2( 5 2)
2 1 2 1 v u R 1.34 1 0.34 v 7.8 1.34 7.8 v nm = 3.074 cm 0.34 P 4.4 2 I 4 10 6 V 2 11 11 4 106 P W R 4.4 = 11 × 10–5 W R
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JEE-MAIN-2019 (10th Jan-Second Shift)-PCM-20
15.
2 Dh 4 = 4260 cm2 v D h 2 v D h v R2h
0.1 0.1 = 2 v 12.6 34.2
2x426 426 12.6 34.2 = 67.61 + 12.459 = 80.075 v = 4260 80 cm3 =
16.
Y
(0, 2)
Q
Q (4, 2)
0 Q Q (0, –2)
17.
18.
19.
20.
1 1 1 2 Q2 4 0 2 2 2 5 Q2 1 1 4 0 5
W VQ
x
0 Q (4, –2)
The interval between two carrier frequencies should be at least two times of AM frequency.
250 kHz 13.33 1.5 kHz Possible hormones 1, 3, 5, 7, 9, 11, 13 i.e 6.
50 5 mA 10k 120 50 i5 k = 14 mA 5k i2 (14 5) mA = 9 mA i10 k
E 10ˆj cos(6x 8z 10ct) E 10 Bo o C C W = 10 C ˆ Eˆ Bˆ C
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JEE-MAIN-2019 (10th Jan-Second Shift) PCM-21 ˆi
ˆj
kˆ
0
1
0
Bx
By
3 4 Bz ˆi 0 ˆj Bxkˆ ˆi ˆj 5 5 3 4 Bz , By 0, Bz 5 5 1 ˆ cos (6x 8z 10 ct) B ( 8iˆ 6k) C
21.
22.
Bz
6iˆ 8 ˆj 10
1 r3 Required force = 27 F.
F
1 2( GMm) mv 2 0 2 r 4GM 4 6.67 10 11 3 1031 V2 r 2 1011 V 20 2 10 4 m / s = 2.828 × 105 m/s
23.
W = nRT 1 = 8.31 70 2 = 290.85 J
24.
W
25.
rt = 5 = area 1 = 2 2 2 2 3 1 m 2 = (2 + 4 + 3) m = 9 m.
Q2 Q2 2c 2ck Q2 1 = 1 2c k 1 1 = 12 100 pJ 1 2 6.5 12 100 11 = pJ = 507.69 pJ 2 13
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JEE-MAIN-2019 (10th Jan-Second Shift)-PCM-22
I 5Mog 4Mo g = 5Mo 2 2Mo 42 Mo g = 13 Mo 2 g = 13
26.
27.
2 P Q P 2Q 13 + 12 cos = 10 + 6 cos 1 cos = 2 = 120°.
28.
Agx = Frestoring r2gx = n2x r 2g v
g 3.14 10 2.5 10 2 v 310 10 6 = 2.5 × 10–2 × 102 10 =r
= 2.5 10 29.
f
2.5 10 = 1.25 2
v 4 a4
w
A 2 x2
4
(w 2 x)4
w 25 16 w 2 4 3 w 4 2 4 8 T 2 w 3 3
30.
1 M Vm2 2 1 3P = 2 2
E
=
3 4 10 4 = 1.5 × 104 J 8
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JEE-MAIN-2019 (10th Jan-Second Shift) PCM-23
PART B – CHEMISTRY 2
31.
EHe
32.
n eV z2 4 13.6 9
E 13.6
Fact based O O
33.
HO - C - C - OH HO
OH
OH O O
O
C
O
n
HO
34.
0.1
nC12H22O11
2 nC12H22O11 0.2
Wt C12H22O11 0.2 342 68.4
35.
First reaction will be requiring a catalyst among halogens oxidizing power decrease down the group.
36.
NH4HS s NH3 g H2S g 5.1 g 0.1 g mol – 0.03
0.03 mol
0.03 mol
0.98 0.98 V = 3L, T = 327oC 2 2 K P PNH3 PH2S PV = nRT KP
0.98 0.98 2 2
P 3 = 0.06 0.0821 600
0.06 0.0821 200 3 P = 0.98
P KP = 0.243 37.
CH4 is not present in stratosphere.
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JEE-MAIN-2019 (10th Jan-Second Shift)-PCM-24
38.
H Cl E Eo 0.06 log 1/2 H2
106.10 6 11/2 o –12 0.92 = E - 0.06 log10 Eo + 0.06 12 Eo = 0.92 - 0.06 12 Eo = 0.92 – 0.72 EoAg / AgCl 0.20 0.92 Eo 0.06 log
39.
The electron configuration is [Xe]4f 145d16s2
40.
Lysine is an amino acid nindydrin test is used for amino acids. Furfural reacts with 1-napthol to give violet colouration. Benzyl alcohol undergoes reaction with ceric ammonium nitrate to give red colouration. CH = CH 2
Styrene
41.
, discharges color of KMnO4
CH3
CONH2
NH3 ,
[A]
42.
NH2
Br2 , KOH
[B]
CaSO 4 s CaO s SO3 g CO 2 s CO2 g I2 I2 aq N2 g 3 H2 g 2NH3 g N2 g 3 H2 2NH3 g S 2SNH3 SN2 3 SH2 There is decrease in number of moles of NH3 entropy is decreasing.
43.
q = PV q = 16 CP = 24 qP CP T 16 2 T K K 24 3 FIITJEE Ltd., FIITJEE House, 29-A, Kalu Sarai, Sarvapriya Vihar, New Delhi -110016, Ph 46106000, 26569493, Fax 26513942 website: www.fiitjee.com.
JEE-MAIN-2019 (10th Jan-Second Shift) PCM-25 44.
Tb = Kbm Tb = Kb 1 2 = Kb Tf = Kfm 2 = 2Kf = Kf Kf 1 Kb 2
45.
NaBH4 reduces both carbonyl group and imine.
46.
Na s x y NH3 Na NH3 x
e NH3 y Blue colour ammoniated electrons
47.
1 dA 2 K 1 A 2 K 1 A 2 dt d A 2 2K1 A 2 2K 1 A dt
48.
Barfoed test is used to detect monosaccirides.
49.
Co2+ high spin t 52g eg2 ‘3’ unpaired electrons Co2+ low spin t 62g e1g ‘1’ unpaired electron Difference is 3 – 1 = 2
50.
O
O
O
O 1. NaOEt 2.
O
O
OC2H5
O
H2O
HO COOEt
HO
O
O
P
P
P
H H3PO 2
52.
O COOEt
O
H
O
O NaOEt HO
51.
OC2H5
HO
OH
O
OH
COOEt
H
H4P 2O 5
Iodoform reaction can be used for this transformation. FIITJEE Ltd., FIITJEE House, 29-A, Kalu Sarai, Sarvapriya Vihar, New Delhi -110016, Ph 46106000, 26569493, Fax 26513942 website: www.fiitjee.com.
JEE-MAIN-2019 (10th Jan-Second Shift)-PCM-26 53.
4-Bromo-3-methyl pentane 2c - 2e
54.
H
H B H
H B
H 3c - 2e
55.
bond
H bond
5 C2O24 2KMnO 4 H 10 CO2 2Mn2 H2O After balancing 10 gain or loss of electron.
56.
57.
[Au(CN2 ] and [Ag(CN)2 ] both are soluble complexes.
58.
Formula of the compounds No. of octahedral voids are equal to the number of atoms forming lattice A occupy octahedral void i.e. 2/3 of them B forms crystal lattice A 2/3B A 2B3
59.
60.
‘EAS’ first ring is activated and second is deactivated NO2 attack at para position of activated ring.
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JEE-MAIN-2019 (10th Jan-Second Shift) PCM-27
PART C – MATHEMATICS 61.
2
2
2
2 2
3 2 2 2 9 6 4 2 2 4 5 For least value 2 sin 2n A cos A cos 2 A cos 22 A...........cos 2n1 A n 2 sin A
62.
Using formula
63.
x 2 dx 2xydy y 2 dx 0 x 2 dx y 2 dx 2xy dy
x
2
y 2 dx 2xy dy 0
x.2y dy y 2 dx dx x2 Integrals y2 x c x x 2 y 2 cx y
64.
A, B, C are sharp edges 1
–1
A
x
C
10
65.
x2 x 2 x Consider constant term r 10 r 10 Cr x 2 x 10 r 2r 0 2 10 5r 0 r2 10C2 2 720 4
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JEE-MAIN-2019 (10th Jan-Second Shift)-PCM-28
66.
2
y xe x (1, e) lies on this 2 2 dy Now xe x .2x e x .1 dx Put x = 1 m 2e e 3e Equation of tangent at (1, e) y e 3e x 1 y e 3ex 3e y 3ex 2e
4 3 , 2e satisfies it Answer is B
67.
f g 1 1 Many one f g 2 1
f g 2k k f g 2k 1 k 1 Onto
68.
sin3
1 1
cos 2
4
3 0
2 7 7 7 sin3 14 cos 2 14 0
sin3 2cos 2 2 0, sin 69.
1 2
2 a b 4 2 a 3b and are collinear
2
1
0 4 2 3 3 6 4 2 0 4 0 4
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JEE-MAIN-2019 (10th Jan-Second Shift) PCM-29 D (3, 6)
70.
C
Intersection point is A (0, 3) M 4, 6 B 1, 2 , D 3, 6
A (0, 3)
71.
x+y=3
B (1, 2)
Differentiability we get f x 2x x 2 f x
1 x2 2x f x f " x 2 2 1 x2 1 x2
1 24 f ' 2 25 5
72.
5
3 i i b e ei5 /6 2 2 5 3 z 2 cos 2 3 2 6 n
73.
5 2 1 6 3 n
1 2 6 3 n5
74.
4 x 3
5
x .e dx x .x e 2
3
4x 3
dx
3
4x t 12x 2dx dt 1 t et dt 12 4 1 t e t dt 48 1 t te 1.et c 48 3 1 4x3 e . 4x 3 e 4 x c 48
75.
r 25 36 c 36 c 25 FIITJEE Ltd., FIITJEE House, 29-A, Kalu Sarai, Sarvapriya Vihar, New Delhi -110016, Ph 46106000, 26569493, Fax 26513942 website: www.fiitjee.com.
JEE-MAIN-2019 (10th Jan-Second Shift)-PCM-30 76.
P is True Q is False R is True ~ Q is True ~ Q R is True P ~ Q R is True.
77.
x 2y 4 2
x 2 4y Solving we get point of intersection A 2 2, 2 , B 4 2,8
6 2 6 6 2
AB 78.
79.
2
3
Det A b2 3 det A 3 b b b Least value 2 3 y2 x2 1 1 r 1 r r 1 ellipse
2 r 1 e 1 r 1 r 1 25
80.
50
r 50 r r 1
25
r 1
81.
50 r 25 r 25
50 r 25 r 25
50 25 1 25 r 1 r 25 r
50 25 25 25 r 1
25
Cr 50 C25 225 1
A (–3, 3, 4), B (3, 7, 6) Mid point 0,2,5 n AB 6iˆ 10ˆj 2kˆ Equation of plane r .n a.n r . 6iˆ 10ˆj 2kˆ 0iˆ 2ˆj 5kˆ . 6iˆ 10ˆj 2kˆ
3x 5y z 15 (4, 1, –2) satisfies it Answer is D FIITJEE Ltd., FIITJEE House, 29-A, Kalu Sarai, Sarvapriya Vihar, New Delhi -110016, Ph 46106000, 26569493, Fax 26513942 website: www.fiitjee.com.
JEE-MAIN-2019 (10th Jan-Second Shift) PCM-31
82.
n 19 cot cot 1 1 2p p 1 n 1 19 cot cot 1 1 n2 n n1 19 1 cot tan1 2 1 n n n1
19 cot tan 1 n 1 tan1 1 n1 1 1 cot tan 20 tan 1
19 cot tan1 21 21 19 83.
x 50 2
ex 1 3 ex 2 5 5 1 2500 9 x2 5 5 x 2 545 2
New variable
84.
1 0 3045 507.5 6 6
3 f x . , x0 4 x 3 f 'x f x 7 4x f ' x 7
3
f x .e
4 x dx
(Linear)
3
7.e
4x dx
c
f x . x3/4 7.x 3/4 c x 7/4 c 7 4 f x 4x cx 3/4 7
1 4 f cx3/4 x x 1 Lt xf Lt 4 c x 7/4 4 x 0 x x 0
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JEE-MAIN-2019 (10th Jan-Second Shift)-PCM-32 (h, k)
85.
k 3 0 h4 k 4 0 h 32
k3
4h k
h
(0, 0)
3 4 (0, 2)
/2
86.
(4, 3)
dx
x sin x 4
/2
0
2 dx dx / 2 x 1 4 0 x 4
1
0
1
dx dx dx / 2 2 1 4 1 1 1 4 0 4
/2
dx
1 4 1
1 1 1 2 1 2 4 52 9 3 5 20 87.
Put 2 4, 2 5, 3 in x y z 2 2 4 2 5 3 2 5 10 2 P 0, 1, 1 Now put in options Answer is C
88.
For any value of r determinant is zero.
89.
a 3 1 b 3 1 sin A 3 1 3 1 2 3 2 3 sinB 2 3 1 sin A 32 sin 120 A
sin A 32 sin12cos A cos12 sin A 1 3 2 3 1 cot A 2 2 FIITJEE Ltd., FIITJEE House, 29-A, Kalu Sarai, Sarvapriya Vihar, New Delhi -110016, Ph 46106000, 26569493, Fax 26513942 website: www.fiitjee.com.
JEE-MAIN-2019 (10th Jan-Second Shift) PCM-33
3 cot A 1 1 2 32 3 2 1 3 cot A 1 3 2 2 3 cot A 4 2 3 1 3 cot A 3 2 3 cot A 3 2 cot A 2 3 tan15
A 105o B 15o 90.
dy 3 x dx 2 2 Slope of normal 3 x 3/2 Let point is x1, x1 2
y x 3/2 2
Normal y x13/2
2 2 x x 3 x 1
1
Now put (1, 7) and solve it. 1 x1 3 1 1 P , 7 , A 1, 7 3 3 3
AD
1 7 6 3
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FIITJEE Solutions to JEE(Main)-2019 Test Date: 11th January 2019 (First Shift)
PHYSICS, CHEMISTRY & MATHEMATICS Paper - 1 Time Allotted: 3 Hours
Maximum Marks: 360
Please read the instructions carefully. You are allotted 5 minutes specifically for this purpose.
Important Instructions: 1.
The test is of 3 hours duration.
2.
This Test Paper consists of 90 questions. The maximum marks are 360.
3.
There are three parts in the question paper A, B, C consisting of Physics, Chemistry and Mathematics having 30 questions in each part of equal weightage. Each question is allotted 4 (four) marks for correct response.
4.
Out of the four options given for each question, only one option is the correct answer.
5.
For each incorrect response 1 mark i.e. ¼ (one-fourth) marks of the total marks allotted to the question will be deducted from the total score. No deduction from the total score, however, will be made if no response is indicated for an item in the Answer Box.
6.
Candidates will be awarded marks as stated above in instruction No.3 for correct response of each question. One mark will be deducted for indicating incorrect response of each question. No deduction from the total score will be made if no response is indicated for an item in the answer box.
7.
There is only one correct response for each question. Marked up more than one response in any question will be treated as wrong response and marked up for wrong response will be deducted accordingly as per instruction 6 above.
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JEE-MAIN-2019 (11th Jan-First Shift)-PCM-2
PART –A (PHYSICS) 1.
A body is projected at t = 0 with a velocity 10 ms–1 at an angle of 600 with the horizontal. The radius of curvature of its trajectory at t = 1s is R. Neglecting air resistance and taking acceleration due to gravity g = 10 ms–2, the radius of R is: (A) 10.3 m (B) 2.8 m (C) 2.5 m (D) 5.1 m
2.
A hydrogen atom, initially in the ground state is excited by absorbing a photon of wavelength 980 Å. The radius of the atom in the excited state, in terms of Bohr radius a0, will be: (hc = 12500 eV- Å) (A) 25a0 (B) 9a0 (C) 16a0 (D) 4a0
3.
In the given circuit the current through Zener Diode is close to: (A) 0.0 mA (B) 6.7 mA (C) 4.0 mA (D) 6.0 mA
4.
There are two long co – axial solenoids of same length I. The inner and outer coils have radii r1 and r2 and number of turns per unit length n1 and n2 respectively. The ratio of mutual inductance to the self – inductance of the inner – coil is:
5.
(A)
n1 n2
(B)
n2 r1 . n1 r2
(C)
n2 r2 2 . n1 r12
(D)
n2 n1
A particle undergoing simple harmonic motion has time dependent displacement given by x t A sin be (A) 1/9 (C) 2
6.
t . The ratio of kinetic to potential energy o the particle at t = 210s will 90 (B) 1 (D) 3
A satellite is revolving in a circular orbit at a height h from the earth surface, such that h < < R where R is the radius of the earth. Assuming that the effect of earth’s atmosphere can be neglected the minimum increase in the speed required so that the satellite could escape from the gravitational field of earth is: (A)
2gR
(B)
gR
(C)
gR 2
(D)
gR
2 1
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JEE-MAIN-2019 (11th Jan-First Shift)-PCM-3
7.
x2 ; where x is kt
The force of interaction between two atoms is given by F = exp
the distance, k is the Boltzmann constant and T is temperature and and are two constants. The dimension of is: 0 2 4 2 4 (A) M L T (B) M LT 2 2 2 2 (C) MLT (D) M L T 8.
In a Young’s double slit experiment, the path difference, at a certain point on the screen, between two interfering waves is 1/8th of wavelength. The ratio of the intensity at this point to that at the centre of a bright fringe is close to: (A) 0.74 (B) 0.85 (C) 0.94 (D) 0.80
9.
In the circuit shown, The switch S1 is close at time t = 0 and the switch S2 is kept open. At some later time (t0), the switch S1 is opened and S2 is closed. The behaviour of the current I as a function of time ‘t’ is given by:
10.
(A)
(B)
(C)
(D)
Three charges Q, +q and +q are placed at the vertices of a right – angle isosceles triangle as shown below. The net electrostatic energy of the configuration is zero, if the value of Q is:
(A) +q (C)
q 1 2
(B)
2q 2 1
(D) –2q
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JEE-MAIN-2019 (11th Jan-First Shift)-PCM-4
11.
12.
The variation of refractive index of a crown glass thin prism with wavelength of the incident light is shown. Which of the following, graph is the correct one, if Dm is the angle of minimum deviation? (A)
(B)
(C)
(D)
The equilateral triangle ABC is cut from a thin solid sheet of wood. (See figure) D, E and F are the mid points of its sides as shown and G is the centre of the triangle. The moment of inertia of the triangle about an axis passing through G and perpendicular to the plane of the triangle is I0. If the smaller triangle DEF is removed from ABC, the moment of inertia of the remaining figure about the same axis is I. Then:
15 I0 16 9 (C) I I0 16
3 I0 4 I (D) I 0 4
(A) I
(B) I
13.
A slab is subjected to the two forces F1 and F2 of same magnitude F as shown in the figure. Force
F2 is in XY – plane while force F1 acts along z – axis at the point 2 i 3 j . The moment of these
forces about point O will be:
(C) 3ˆi 2 ˆj 3kˆ F (A) 3ˆi 2 ˆj 3kˆ F
(D) 3ˆi 2 ˆj 3kˆ F (B) 3ˆi 2 ˆj 3kˆ F
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JEE-MAIN-2019 (11th Jan-First Shift)-PCM-5 14.
Ice at –200 C is added to 50 g of water at 400 C. When the temperature of the mixture reaches 0o C, it is found that 20 g of ice is still unmelted. The amount of ice added to the water was close to (Specific heat of water = 4.2 J/g/0C) Heat of fusion of water at 0oC = 334 J/g) (A) 50 g (B) 100 g (C) 60 g (D) 40 g
15.
A particle is moving along a circular path with a constant speed of 10 ms–1. What is the magnitude of the change in velocity of the particle, when it moves through an angle of 60o+ around the centre of the circle? (A) 10 3 m / s (B) 0 (C) 10 2 m / s
16.
(D) 10 m/s
An amplitude modulated signal is given by
V t 10 1 0.6 cos 2.2 104 t sin 5.5 105 t . Here t is in seconds. The seconds. The sideband frequencies (in kHz) are. [Given = 22/7] (A) 1785 and 1715 (B) 1785 and 1715 (C) 89.25 and 85.75 (D) 892.5 and 857.5 17.
An object is at a distance of 20 m from a convex lens of focal length 0.3 m. The lens forms an image of the object. If the object moves away from the lens at a speed of 5 m/s, the speed and direction of the image will be: (A) 2.26 x 10–3 m/s away from the lens (B) 0.92 x 10–3 m/s away from the lens -3 (C) 3.22 x 10 m/s towards the lens (D) 1.16 x 10-3 m/s towards the lens
18.
A gas mixture consists of 3 moles of oxygen and 5 moles or argon at temperature T. Considering only translational and rotational modes, the total internal energy of the system is: (A) 15 RT (B) 12 RT (C) 4 RT (D) 20 RT
19.
Two equal resistances when connected in series to a battery, consume electric power of 60 W. If these resistances are now connected in parallel combination to the same battery, the electric power consumed will be. (A) 60 W (B) 240 W (C) 120 W (D) 30 W
20.
The resistance of the meter bridge AB in given figure is 4 . With a cell of emf = 005 V and rheostat resistance Rh = 2 the null point is obtained at some point J. When the cell is replaced by another one of emf = 2 the same null point J is found for Rh = 6 . The emf 2 is:
(A) 0.4V (C) 0.6V
(B) 0.3V (D) 0.5V
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JEE-MAIN-2019 (11th Jan-First Shift)-PCM-6 21.
An electromagnetic wave of intensity 50 Wm–2 enters in a medium of refractive index’ n’ without any loss . The ratio of the magnitudes of electric fields, and the ratio of the magnitudes of magnetic fields of the wave before and after entering into the medium are respectively. Given by:
1 1 , n n 1 (C) n, n (A)
22.
(B)
n, n
1 , n n
(D)
In the figure shown below, the charge on the left plate of the F capacitor is –30C. The charge on the right place of the 6 F capacitor is:
(A) –12C (C) –18C
(B) +12C (D) +18C
23.
A rigid diatomic ideal gas undergoes an adiabatic process at room temperature. The rational between temperature and volume for the process is TVx = constant, then x is: (A) 3/5 (B) 2/5 (C) 2/3 (D) 5/3
24.
A liquid of density is coming out of a hose pipe of radius a with horizontal speed and hits a mesh. 50% of the liquid passes through the mesh unaffected. 25% looses all of its momentum and 25% comes back with the same speed. The resultant pressure on the mesh will be:
1 2 4 1 (C) 2 2 (A)
25.
3 2 4
(D) 2
If the deBroglie wavelength of an electron is equal to 10-3 times the wavelength of a photon of frequency 6 x1014 Hz, then the speed of electron is equal to: (Speed of light = 3 x 10 = –34 -31 8 = m/s ; Planck’s constant = 6.63 x 10 J.s ; Mass of electron = 9.1 x 10 kg) (A) 1.1 x 106 m/s (C) 1.8 x 106 m/s
26.
(B)
(B) 1.7 x 106 m/s (D) 1.45 x 106 m/s
The give graph shown variation (with distance r from centre) of: (A) Electric field of a uniformly charged sphere (B) Potential of a uniformly charged spherical shell (C) Potential of a uniformly charged sphere (D) Electric field of a uniformly charged spherical shell
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JEE-MAIN-2019 (11th Jan-First Shift)-PCM-7 27.
Equation of travelling wave on a stretched string of linear density 5 g/m is y = 0.03 sin (450 t – 9x) where distance and time are measured in SI united. The tension in the string is: (A) 10 N (B) 7.5 N (C) 12.5 N (D) 5 N
28.
In an experiment, electrons are accelerated, from rest, by applying, a voltage of 500 V. Calculate the radius of the path if a magnetic field 100 mT is then applied. [Charge of the electron = 1.6 x 10-19 C Mass of the electron = 9.1 x 10-31 kg] (A) 7.5 x 10-3 m (B) 7.5 x 10-2 m (C) 7.5 m (D) 7.5 x 10-4 m
29.
A body of mass 1 kg falls freely from a height of 100 m, on a platform of mass 3 kg which is mounted on a spring having spring constant k = 1.25 x 106 N/m. The body sticks to the platform and the spring’s maximum compression is found to be x. Given that g = 10 ms-2, the value of x will be close to: (A) 40 cm (B) 4 cm (C) 80 cm (D) 8 cm
30.
In a Wheatstone bridge (see figure) Resistance P and Q are approximately equal. When R = 400 , the bridge is balanced. On interchanging P and Q, the value of R, for balance is 405 . The value of X is close to: (A) 401.5 ohm (B) 404.5 ohm (C) 403.5 ohm (D) 402.5 ohm
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JEE-MAIN-2019 (11th Jan-First Shift)-PCM-8
PART –B (CHEMISTRY) 31.
Among the following compounds, which one is found in RNA?
(A)
(B)
(C)
(D)
32.
The correct match between items I and II is: Item – I Item – II (Mixture) (Separation method) (a) H2O : Sugar p. Sublimation (b) H2O : Aniline q. Recrystallization (c) H2O : Toluene r. Stem distillation s. Differential extraction (A) a - d, b – r, c – p (B) a – q, b – r, c – s (C) a – r, b – p, c – s (D) a – q, b – r, c – p
33.
Which compounds out of the following is/are not aromatic? (a)
(b)
(c)
(d)
(A) b, c and d (C) b 34.
(B) c and d (D) a and c
A solid having density of 9 x 103 kg m
-3
forms face centred cubic crystale of edge length
200 2 pm. What is the molar mass of the solid? [Avogadro constant 6 x 1023 and mol–1, 3] (A) 0.0432 kg mol–1 (C) 0.0305 kg mol–1
(B) 0.0216 kg mol–1 (D) 0.4320 kg mol-1
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JEE-MAIN-2019 (11th Jan-First Shift)-PCM-9
35.
For the cell Zn s Zn 2 aq Mx aq M s , different half cells and their standard electrode potentials are given below:
If E
0 zn2 / zn
0.76 V, which cathode will given a maximum value of E0cell per electron
transferred? (A) Ag / Ag
(B) Fe 3 /Fe 2
(C) Au3 /Au
(D) Fe 2 / Fe
36.
The correct match between item I and item II is: Item – I Item – II a. Norethindrone p. Anti – biotic b. Ofloxacin q. Anti – fertility c. Equanil r. Hypertention s. Analgesics (A) a – q, b – r, c – s (B) a – q, b – p, c – r (C) a – r, b – p, c – s (D) a – r, b – p, c – r
37.
The polymer obtained from the following reactions is:
38.
(A)
(B)
(C)
(D)
NaH is an example of: (A) electron rich hydride (C) saline hydride
(B) metallic hydride (D) molecular hydride
39.
The element the usually does NOT show variable oxidation states is: (A) Cu (B) Ti (C) Sc (D) V
40.
An example of solid sol is: (A) Paint (C) Butter
41.
(B) Gem stones (D) Hair cream
The concentration of dissolved oxygen (DO) is cold water can go upto: (A) 14 ppm (B) 8 ppm (C) 10 ppm (D) 16 ppm
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JEE-MAIN-2019 (11th Jan-First Shift)-PCM-10
42.
The correct statements among (a) to (d) regarding H2 as a fuel are: (a) it produces less pollutants than petrol (b) A cylinder of compressed dihydrogen weighs 30 times more than a petrol tank producing the same amount of energy. (c) Dihydrogen is stored in tanks of metal alloys like NaNi5 (d) On combustion, values of energy released per gram of liquid dihydrogen and LPG are 50 and 142 kJ, respectively. (A) (b) and (d) only (B) (a) and (d) only (C) (b), (c) and (d) only (D) (a), (b) and (c) only
43.
The major product of the following reaction is:
(A)
(B)
(C)
(D)
44.
The freezing point of diluted milk sample is found to be – 0.20C, while it should have been 0.50C for pure milk. How much water been added to pure milk to make the diluted sample? (A) 1 cup of water to 2 cups of pure milk (B) 3 cups of water to 2 cups of pure milk (C) 1 cup of water to 3 cups of pure milk (D) 2 cups of water to 3 cups of pure milk
45.
If a reaction follows the Arrhenius equation, the plot Ink v 1/(RT) gives straight line with a gradient (–y) unit. The energy required to activate the reactant is: (A) y/R unit (B) y unit (C) yR unit (D) –y unit
46.
A 10 g effervescent tablet containing sodium bicarbonate and oxalic acid releases 0.25 ml of CO2 at T = 298.15 K and p = 1 bar. If molar volume of CO2 is 25.0 L under such condition, what is the percentage of sodium bicarbonate in each tablet? [Molar mass of NaHCO3 = 84 g mol–1] (A) 0.84 (B) 33.6 (C) 16.8 (D) 8.4
47.
The amphoteric hydroxide is: (A) be (OH)2 (C) Mg(OH)2
48.
(B) Ca(OH)2 (D) Sr(OH)2
Peroxyacetyl nitrate (PAN), an eye irritant is produced by: (A) classical smog (B) acid rain (C) organic waste (D) photochemical smog FIITJEE Ltd., FIITJEE House, 29-A, Kalu Sarai, Sarvapriya Vihar, New Delhi -110016, Ph 46106000, 26569493, Fax 26513942 website: www.fiitjee.com.
JEE-MAIN-2019 (11th Jan-First Shift)-PCM-11 49.
The major product of the following reactions is:
(A)
(B)
(C)
(D)
50.
An organic compound is estimated through Dumus method and was found to evolve 6 mole of CO2 4 moles of H2O and 1 mole of nitrogen gas. The formula of the compound is: (A) C12H8N (B) C12H8N2 (C) C6H8N2 (D) C6H8N
51.
Consider the reaction
N2 g 3H2 g 2NH3 g The equilibrium constant of the above reaction is K3. If pure ammonia is left to dissociate, the partial pressure of ammonia at equilibrium is given by (Assume that
PNH3 Ptotal at equilibrium) (A) (C)
33 / 2 K1p/ 2 P2 16 K P2 1/ 2 p
4
(B)
K1p/ 2 P 2 16 3 K1p/ 2 P2 3/2
(D)
4
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JEE-MAIN-2019 (11th Jan-First Shift)-PCM-12 52.
53.
The major product of the following reaction is:
(A)
(B)
(C)
(D)
Two blocks of the same metal having same mass and at temperature T1 and T2 respectively, are brought in contact with each other and allowed to attain thermal equilibrium at constant pressure. The change in entropy, S, for this process is:
T1 T2 2 (A) Cp In 4T1T2 T1 T2 4T1T2
(C) 2Cp In
1 2 T T 1 2 (B) 2Cp In T1T2 T1 T2 (D) 2Cp In 2T1T2
54.
Match the metal column I with the coordination compounds/enzymes column II Column I Column II Metals Coordination compounds/enzymes a. Co i. Wilkinson catalyst b. Zn ii. Chlorophyll c. Rh iii. Vitamin B12 d. Mg iv. Carbonic anhydrase (A) a – iii, b – iv, c – i, d - ii (B) a – i, b – ii, c – iii, d – iv (C) a – ii, b – i, c – iv, d – iii (D) a – iv, b – iii, c – i, d – ii
55.
For the chemical reaction X Y , the standard reaction Gibbs energy depends on temperature T (in K) as
3 r G0 in kJ mol1 120 T 8
The major component of the reaction mixture at T is: (A) Y if T = 300 K (B) Y if T = 280 K (C) X if T = 350 K (D) X if T = 315 K
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JEE-MAIN-2019 (11th Jan-First Shift)-PCM-13
56.
Match the ores (column A) with the metals (column B) Column A Column B Ores Metals I. Siderite a. Zinc II. Kaolinite b. Copper III. Malachite c. Iron IV. Calamine d. Aluminium (A) I – a, II – b, III – c, IV – d (B) I – c, II – d, III – b, IV – a (C) I – c, II – d, III – a, IV – b (D) I – b, II – c, III – d, IV – a
57.
Heat treatment of muscular pain involves radiation of wavelength of about 900 nm. Which spectral line of H – atom is suitable for this purpose?
RH 1 105 cm 1,h 6.6 10 34 Js c 3 108 ms 1 (A) Paschen, 3 (C) Balmer, 2
(B) Paschen, 5 3 (D) Lyman, 1
58.
The chloride that CANNOT get hydrolysed is (A) PbCI4 (B) CCI4 (C) SnCI4 (D) SiCI4
59.
The correct order of the atomic radii of C, Cs, AI and S is: (A) C < S < AI < Cs (B) S < C < Cs < AI (C) S < C < AI, Cs (D) C < S < Cs < AI
60.
The major product of the following reaction is COCH3 i KMnO 4 /KOH, ii H2SO4 dil H3C COCOOH (A)
COOH (B)
HOOC
OHC COOH
(C)
COCH3 (D)
HOOC
HOOC
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JEE-MAIN-2019 (11th Jan-First Shift)-PCM-14
PART–C (MATHEMATICS) 61.
If the system of linear equations
2x 2 y 3z a 3x y 5z b x 3y 2z c Where a, b, c are non zero real numbers, has more than one solution, then: (A) b – c + a = 0 (B) b – c – a = 0 (C) a + b + c = 0 (D) b + c –a = 0 62.
Let fk x
1 sink x cosk x for k = 1, 2, 3, … Then for all x R, the value of k
f4 x f6 (x) is equal to: 1 12 1 (C) 12
1 4 5 (D) 12
(A)
63.
A square is inscribed in the circle x 2 y 2 6 x 8y 103 0 with its sides parallel to the coordinate axes. Then the distance of the vertex of the square which is nearest to the origin is: (A) 6 (B) 137 (C)
64.
(B)
41
(D) 13
If q is false and p q r is true, then which one of the following statements is a tautology? (A) p r p r (B) p r p r (C) p r
(D) p r
65.
The area (in sq. units) of the region bounded by the curve x 2 4 y and the straight line x 4 y 2 is: (A) 5/4 (B) 9/8 (C) 7/8 (D) 3/4
66.
The value of the integral
2
or equal to x) is: (A) 0 (C) 4
sin2 x x 1 dx (where [x] denotes the greatest integer less than 2 2 (B) sin 4 (D) 4 – sin 4
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JEE-MAIN-2019 (11th Jan-First Shift)-PCM-15
67.
68.
1 d each, 2 1 1 10 items gave outcome each and the remaining 10 items gave outcome d each. If 2 2 4 the variance of this outcome data is then d equals: 3 2 (A) (B) 2 3 5 (C) (D) 2 2 x 3 y 2 z 1 The plane containing the line and also containing its projection on 2 1 3 The outcome of each of 30 items was observed; 10 items gave an outcome
the plane 2x + 3y – z = 5, contains which one of the following points? (A) (2, 2, 0) (B) (–2, 2, 2) (C) (0, –2, 2) (D) (2, 0, –2) 69.
Two integers are selected at random from the set {1, 2, …, 11}. Given that the sum of selected numbers is even, the conditional probability that both the numbers are even is: 7 1 (A) (B) 10 2 2 3 (C) (D) 5 5
70.
Two circles with equal radii intersecting at the points (0, 1) and (0, –1). The tangent at the point (0, 1) to one of the circles passes through the centre of the other circle. Then the distance between the centres of these circles is: (A) 1 (B) 2 (C) 2 2 (D) 2
71.
The value of r for which (A) 15 (C) 11
72.
Cr 20 C0 20 Cr 120C1 20 Cr 2 20C2 ... 20 C0 20Cr is maximum is: (B) 20 (D) 10
If tangents are drawn to the ellipse x 2 2 y 2 2 at all points on the ellipse other than its four vertices than the mid points of the tangents intercepted between the coordinate axes lie on the curve: 1 1 x2 y2 (A) 1 (B) 1 4x2 2y 2 4 2 (C)
73.
20
1 1 2 1 2 2x 4y
(D)
x2 y2 1 2 4
1, 2 x 0 Let f x 2 and g x f x f x , Then, in the interval 2, 2 ,g is: x 1, 0, x 2 (A) differentiable at all points (B) not continuous (C) not differentiable at two points (D) not differentiable at one point
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JEE-MAIN-2019 (11th Jan-First Shift)-PCM-16
74.
If x loge loge x x 2 y 2 4 y 0 , then (A) (C)
1 2e
dy at x = e is equal to: dx (B)
2 4 e2 1 2e
2e 1 2 4 e2
(D)
4 e2
e 4 e2
m 1 x2 2 dx A x 1 x C , for a suitable chosen integer m and a function A(x), x4 where C is a constant of integration, then (A(x))m equals: 1 1 (A) (B) 9 27x 3x 3 1 1 (C) (D) 6 27x 9x 4
75.
If
76.
Let [x] denote the greatest integer less than or equal to x. Then: lim
tan sin2 x x sin x x
2
x2
x 0
(A) does not exist (C) equal + 1 77.
(B) equals (D) equals 0
The maximum value of the function f x 3x 3 18x 2 27x 40 on the set S =
x R : x
2
30 11x is
(A) – 122 (C) 122 78.
Let f : R R be defined by f x 1 1 (A) , 2 2 1 1 (C) R , 2 2
79.
(B) –222 (D) 222 x , x R. Then the range of f is: 1 x2
(B) R 1,1 (D) 1,1 0
The sum of an infinite geometric series with positive terms is 3 and the sum of the cubes 27 of its terms is . Then the common ratio of this series is: 19 1 2 (A) (B) 3 3 2 4 (C) (D) 9 9
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JEE-MAIN-2019 (11th Jan-First Shift)-PCM-17 80.
The straight line x + 2y = 1 meets the coordinate axes at A and B. A circle is drawn through A, B and the origin. Then the sum of perpendicular distances from A and B on the tangent to the circle at the origin is: 5 (A) (B) 2 5 2 5 (C) (D) 4 5 4
81.
In a triangle, the sum of lengths of two sides is x and the product of the lengths of the same two sides is y. If x 2 c 2 y , where c is the length of the third side of the triangle, then the circumradius of the triangle is: 3 c (A) y (B) 2 3 c y (C) (D) 3 3
82.
Equation of a common tangent to the parabola y 2 4x and the hyperbola xy = 2 is: (A) x y 1 0 (B) x 2y 4 0 (C) x 2y 4 0 (D) 4x 2y 1 0
83.
The sum of the real values of x for which the middle term in the binomial expansion of 8
x3 3 equals 5670 is: 3 x (A) 0 (C) 4
84.
(B) 6 (D) 8 a3 a Let a1, a2, …, a10 be a G.P. If 25 , then 9 equals: a1 a5 (B) 4 52
3
(D) 2(52)
(C) 5
85.
(A) 54
0 2q r Let A p q r . If AAT = I3 , P then p is: p q r 1 1 (A) (B) 5 3 1 1 (C) (D) 2 6
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JEE-MAIN-2019 (11th Jan-First Shift)-PCM-18
86.
If y (x) is the solution of the differential equation
dy 2x 1 2x y e , x 0 where y(1)= dx x
1 2 e , then: 2
87.
(B) y loge 2
1 (C) y(x) is decreasing in ,1 2
(D) y(x) is decreasing in (0, 1)
The direction ratios of normal to the plane through the points (0, –1, 0) and (0, 0, 1) and making an angle with the plane y – z + 5 = 0 4 (A) 2, –1, 1 (B) 2, 2, 2 (C)
88.
89.
loge 2 4
(A) y loge 2 loge 4
(D) 2 3,1, 1
2,1, 1
ˆ ˆi ˆj 4kˆ and c 2iˆ 4 ˆj 2 1 kˆ be coplanar vectors. Then the Let a ˆi 2ˆj 4k,b non – zero vector a c is: (A) 10iˆ 5ˆj (B) 14iˆ 5ˆj (C) 14iˆ 5 ˆj (D) 10iˆ 5 ˆj
If one real root of the quadratic equation 81x 2 kx 256 0 is cube of the other root, then a value of k is: (A) – 81 (B) 100 (C) 144 (D) – 300 3
90.
1 x iy Let 2 i 3 27 (A) 91 (C) 85
i
1 , where x and y are real numbers, then y – x equals:
(B) – 85 (D) – 91
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JEE-MAIN-2019 (11th Jan-First Shift)-PCM-19
JEE (Main) – 2019 ANSWERS PART A – PHYSICS 1.
B
2.
C
3.
A
4.
D
5.
No match (Correct answer is
6.
D
7.
B
8.
B
9.
No match
10.
B
11.
A
12.
A
13.
B
14.
D
15.
D
16.
C
17.
D
18.
A
19.
B
20.
B
21.
D
22.
D
23.
B
24.
B
25.
D
26.
B
27.
C
28.
D
29.
No match (Correct answer is 2 cm)
30.
D
1 ) 3
PART B – CHEMISTRY 31.
B
32.
B
33.
B
34.
C
35.
C
36.
B
37.
C
38.
C
39.
C
40.
B
41.
C
42.
D
43.
D
44.
B
45.
B
46.
A
47.
A
48.
D
49.
A
50.
C
51.
A
52.
B
53.
A
54.
A
55.
D
56.
B
57.
A
58.
B
59.
A
60.
C
PART C – MATHEMATICS 61.
B
62.
A
63.
C
64.
B
65.
B
66.
A
67.
D
68.
D
69.
C
70.
B
71.
B
72.
C
73.
D
74.
B
75.
A
76.
A
77.
C
78.
A
79.
B
80.
A
81.
B
82.
C
83.
A
84.
A
85.
C
86.
C
87.
BC
88.
D
89.
D
90.
A
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JEE-MAIN-2019 (11th Jan-First Shift)-PCM-20
HINTS AND SOLUTIONS PART A – PHYSICS 1.
5 3
at t = 1 ux = 5, uy = 5 3
10 o
60
vy = 5 3 10 ;
30o
5
tan (2 3 ) = –30°
a
a g 10
2.
v2 102 a (10 cos 30o ) 10 20 = 2 = m 3 3 52 (10 5 3 )2 200 100 3 = 2.8 m 10 cos 10 0.965
R
12 – 500(2i1 + i2) – 10 = 0 2 1 2i1 i2 500 250 < i1 when zenor has break down. So, i2 = 0.
2i1 – i2
R1 = 500
i1 i1
12V R2
i2
R2 = 500
M 0nn1n2 n12 L 0 n12r12
=
5.
t
Energy supplied 12400 E = 12.65 eV 900 En – E1 = 12.65 1 (13.6) 1 2 = 12.65 n 2 n 14.3 n4 r n2
3.
4.
vx 5
K
n2 n1
1 1 1 mv 2 ; U kx 2 m2 x2 2 2 2 cos(wt) k v2 2 2 U x sin (wt) = cot2 210 90
2
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JEE-MAIN-2019 (11th Jan-First Shift)-PCM-21
= cot 2 2 3 2
1 1 = . 3 3
6.
v = vf – vi 2gMe gMe = Re Re = ( 2 1) gR e
7.
Power of e should be dimensionless. So, [] = (Tk) L2 = [] (ML2 T–2) () = (M–1 T2) 1 E KT 2 2 2 (ML T ) ; (E) = [KT) () = (F) (M–1 T2) () = (MLT–2)
8.
x =
8
2 8 4 R Ires = I + I + 2I cos 4 1 = 2I 1 2I 1.7 2 I 2I 1.7 0.85 res Iman 4I Phase P
I
9.
t
Above is the correct graph for growth and decay of current. 10.
kQq kq2 kQq =0 a a a 2 q Q 1 1 2 UTotal
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JEE-MAIN-2019 (11th Jan-First Shift)-PCM-22 11.
Prism formula Dm = Sm = (n – 1) A (for thin prism) So, answer is 1.
12.
I m2
(let = mass present area)
I1 4
and I2 2 I So, I2 16
…(1) 4
…(2)
Moment of inertia of remaining sheet = I = 13.
0 1 2
I 16
15 I 16
F F 3 ˆ ˆ = 6iˆ ˆi j + (2iˆ 3ˆj) (Fk) 2 2 ˆ ˆ ˆ = 3F k( 2Fj 3Fi )
ˆ = F(3iˆ 2ˆj 3k)
14.
15.
16.
m × 0.5 × 20 + (m – 20) × 80 = 50 × 1 × 40 90 m – 1600 = 2000 90 m = 3600 m = 40 gm v = 2v sin 2 = 2 × 10 × sin(30o) = 10 m/s
fso = fc fm
c m 2 (5.5 0.22) 105 = 22 2 7 = 89.25, 85.75 =
17.
1 1 1 V 20 30 1 1 1 200 3 197 V 0.30 20 60 60
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JEE-MAIN-2019 (11th Jan-First Shift)-PCM-23 dV
0
dV V 2 du 3 2 = ( 5) 0.00116 m / s dt u dt 197
dt V
2
du
dt u2
2
18.
Utotal = UO2 UAr
3 5 RT 5 3 RT 2 2 = 15 RT =
19.
Assuming constant voltage supply
P 60
V2 R1 R 2
And P
20.
(1)
V 2 V 2 2V 2 = 4P = 4 × 60 R1 R2 R1 = 240 W
0.5 =
6 x (2 L)
…(1)
E2
6 x (6 L)
…(2)
So dividing equation (1) and (2) E2 2 4 3 0.5 6 4 5 21.
E2 = 0.3 volt.
Ei C …(1) Bi Ef c …(2) Bf n EiB f 1 Ef Bi n Ei 1 Bi Ef n B f 1 n e 0 r 1 : n n
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JEE-MAIN-2019 (11th Jan-First Shift)-PCM-24 22.
–18
4V + 6V = 30 V=3
+18
6F –30 +30
–12
+12
–30 +30
4F
23.
24.
Equation of adiabatic process TV 2/f = constant 2 2 =x f 5 Area = A
1 1 Av 2 2av 2 4F 3 4 Av 2 A 4 F
100% 50%
25% 25%
25.
–3
e = photon × 10 h c 103 s mv h v mC 10 3 =
6.62 10 34 6 1014
9.1 10 31 3 108 10 3 = 1.45 × 106
26.
27.
The potential inside a uniformly charged shell is constant, while it decrease hyperbolically outside. 9x y = 0.03 sin 450 t 450 450 So, v = 50 m/s 9 T Also, v
50
T
5 103 T = 2500 × 5 × 10–3 = 12.5 N 28.
ke
p2 500 e 2Me
& R
…(i)
p 1000 me e 10 10 9.1 10 31 eB eB 1.6 10 19
= 100 × 7.541 × 10–6
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JEE-MAIN-2019 (11th Jan-First Shift)-PCM-25 29.
v 2 10 100 = 20 5
1 kg
1 20 5 4 =5 5
100 m
COLM v p
3 kg
COTME 2 1 1 2 6 30 4 (5 5 ) 1.25 10 2 2 k 30 1 2 = 4 10 n kn k 2 900 1200 1 2 250 40x kx 6 2 1.25 10 k 2 x 2cm 30.
P 400 …(1) Q S Q 405 and …(2) P 5 Solving S2 = 400 × 405 S = 402.5
PART B – CHEMISTRY 31.
Fact based
32.
Fact based
33.
b antiaromatic d Non aromatic (no cyclic conjugation)
34.
Density
ZM NAV a3
Here Z = 4, a = 200 2
3 Kg NAV = 6.02 1023 , d 9 10 m3
35.
o EoCell E Cathode EoZn2 / Zn
Since Au3+ Au has maximum value. 36. 37
38.
Fact based. NaNO HCl
2 NH2
H O
2 N2 (unstable)
OH
ester polymer
NaH is an example of saline hydride. FIITJEE Ltd., FIITJEE House, 29-A, Kalu Sarai, Sarvapriya Vihar, New Delhi -110016, Ph 46106000, 26569493, Fax 26513942 website: www.fiitjee.com.
JEE-MAIN-2019 (11th Jan-First Shift)-PCM-26 39.
Sc3+ has noble gas configuration hence only +3 exists.
40.
An example of solid sol is gem stones.
41.
In cold water, dissolved oxygen can reach a concentration upto 10 ppm
42.
Fact based.
43.
O
O
O OEt
CH2NH2
CN
DIBAL - H
44.
45.
46.
NH
OEt
H2 /Ni
N
1 1 0.5 ,0.2 2 x 0.5 x Here ; x = 5, hence 3 cup 0.2 2 Ea nA RT Slope = - Ea = -y nK
Let NaHCO3 = x gm Then, H2C2O4 = (10 – x) gm x 84 2 NaHCO3 Na2CO3 H2O CO2
nNaHCO3
nCO2
x 168
Total CO2 =
10 x nH2C2 O4 90 H2C2O4 H2O CO2 CO 10 x nCO2 90
x 10 x 0.25 168 90 25
On solving ‘x’ %
x 100 = 10 x 10
47.
Be – O – H Both bond has same dissociation energy.
48.
Fact based. FIITJEE Ltd., FIITJEE House, 29-A, Kalu Sarai, Sarvapriya Vihar, New Delhi -110016, Ph 46106000, 26569493, Fax 26513942 website: www.fiitjee.com.
JEE-MAIN-2019 (11th Jan-First Shift)-PCM-27 49.
–SO3H will be replaced by Br this is called ipso effect.
50.
CO2 = 6 mole, N1 = 1 mole Catom= 6, Natom = 2 Hence C6H8N2
51.
N2 3H2 2NH3 m equ x 3x P1 2 P1 PT = 4x K p x 27 3 P x 4 P1 27x 4K p 2
1/ 2
27 K p
52.
33/ 2 K1/p 2p 2 PT 4 16
Cl
Cl
Br Alc KOH
HBr
O
O
O
T1 T2 C
53.
STotal CP n
54.
Co Vitamin B12 Zn Carbonic anhydrase Rh Wilkinson catalyst Mg Chlorophyll
55.
2 T1
Tautomerise
P
n
HO
T1 T2 2 T2
3 Go 120 T 0 8 Then T = 320 K Hence T > 320 K Y formed T < 320 K X formed
56.
Siderite Iron Kaolinite Aluminium Malachite Copper Calamine Zinc
57.
[i] 900 nm = 9000 A It is in far infra red region hence paschen.
58.
Central atom has no vacant orbital.
59.
On moving down size increases.
60.
It is the case of side chain oxidation.
o
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JEE-MAIN-2019 (11th Jan-First Shift)-PCM-28
PART C – MATHEMATICS 61.
2x 2y 3z a 3x y 5z b x 3y 2z c
(1) (2) (3)
2x 2y 3z x 3y 2z 3x y 5z 0 ac b 0
62.
sin4 x cos4 x 1 2 sin2 x .cos 2 x 1 1 2 sin x.cos2 x 4 4 4 2 2 2 2 2 6 6 sin x cos x 1 3 sin x.cos x sin x cos x F6 x 6 6 1 1 sin2 x.cos2 x 6 2 1 1 64 2 1 F4 x f6 x 4 6 24 24 12 F4 x
63.
Centre (3, –4) Radius 9 16 103 128 8 2 5, 4 will be nearer to the (0, 0) Ans.
(3, 4)
(–5, 4) 8
5 2 42 25 16 41
8 (–5, –12)
64.
As p q r is true If p q is true and r is true. As q is false p q can not be true This case is not possible
Or
(11, 4)
8 (3, –4) 8
(3, –12)
(11, –12)
p q is F and r is F.
Their will be two cases for p, q, r. T, F, F, or F, F, F respectively Now check the options.
65.
x 2 4y ……..(1) x 2 4y ……..(2) Solve (1) and (2) x 2 x2 x2 x 2 0 x 2 x 1 0 x 2, 1
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JEE-MAIN-2019 (11th Jan-First Shift)-PCM-29 2
x 2 x2 Area dx 4 4 1 2
1 x2 x3 2x 4 2 3 1 1 8 1 1 2 4 2 4 3 2 3 1 10 3 12 2 1 10 7 4 3 6 4 3 6 1 27 9 4 6 8 sin2 x 2 sin x 0 1 x 1 x dx 2 2 2
66.
2 sin2 x 2 sin x dx dx 0 x 1 x 0 1 0 1 2 2
67.
Variance remains some if same number is subtracted from each observation. (subtract 10 from each observation) 2
2
10 d 10 0 10 d 30
2
2
10 d 10 0 10 d 4 30 3
20d2 4 30 3 d2 2 d 2
68.
Equation of required plane is a x 3 b y 2 c z 1 0 [as it contains the point (3, –2, 1)] 2a b 3c 0 2a 3b c 0 a b c 8 8 8 a b c 1 1 1 equation of plane is x 3 y 2 z 1 0 xyz4
(3, –2, 1)
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JEE-MAIN-2019 (11th Jan-First Shift)-PCM-30
69.
P
both number is even number of ways selecting two numbers such that their sum is even. 5
70.
5
C2 10 10 2 6 C2 C2 10 15 25 5
y
The two circle will be orthogonal OD = 1 OA OB OD 1 AB 2
D (0, 1) r
r 45o
A
O
x
B
(0, –1)
71.
20
Cr .
20
C0 20Cr 1 . 20C1 ...... 20 C0 . 20Cr
Selecting r student from 20 boys and 20 girls 40Cr 40
72.
Cr will be maximum if r = 20.
Equation of tangent is x y cos sin 1 a b a A is , 0 cos
B
b B is 0, sin Let P (h, k) is mid point a 2h cos
2k
a cos ,b sin P A
b sin
cos2 sin2 1 a2 b2 2 2 1 4h 4k 2 1 2 2 1 4x 4y 1 1 2 2 1 2x 4y
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JEE-MAIN-2019 (11th Jan-First Shift)-PCM-31
73.
2 x 0 1 f x 2 x 1 0 x 2 1 2 x 0 fx 2 x 1 0 x 2
f x x 2 1 2 x 2
1 x 2 1 2 x 0 g x f x f x : 2 2 x 1 x 1 0 x 2 x2 g x 0 2 2 x 1
2 x 0 0 x 1 1 x 2
y
O
1
x 2
g x is not differentiable at x 1
74.
1 1 dy . loge loge x 2x 2y. 0 n x x dx Put x e dy 1 2e 2y 0 dx dy 2e 1 (1) dx 2y Put x = e in original equation O e2 y 2 4 x.
y 2 4 e2 Now put in (i) dy 2e 1 dx 2 4 e 2 2
75.
1 x dx x4 1 Let 2 1 t x
x
1 1 1 x2 dx 3 4 x x
1 1dx x2
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JEE-MAIN-2019 (11th Jan-First Shift)-PCM-32
2 dx dt x3 3/2
Integration becomes
1 1 t t dt 2 2 3 2
3/ 2
1 1 2 1 3x 3/2 1 3 1 x2 3x 3 1 3 1 x2 3x 3 3 1 1 A x 3 27x 9 3x
76.
Lt
tan sin2 x sin2 x
x 0
. sin
2
x2
x
x sin x x x
2
2
sin x x x x 1 . 1 Lt 1 . …………(i) x 0 x x x x x x Lt 1 1 x 0 x x 2
Lt
x 0
x x
x
x 0 x
0
Put in equation (i) Limit does not exist.
77.
x 2 11x 30 0 x 6 x 5 0 3
6
5 5x6
2
f x 3x 18x 27x 40 f ' x 9x 2 36x 27
9 x 2 4x 3 9 x 3 x 1 Nature of f(x)
f x will be maximum when x 6
+ 1 3
+
– 3
2
f 6 3 6 18 6 27 6 40 36 18 18 162 40 122
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JEE-MAIN-2019 (11th Jan-First Shift)-PCM-33
78.
79.
x x 1 2 yx x y 0 D0 1 4y 2 0 1 y2 4 1 1 y . 2 2 y
2
a 3 1 r Cube both sides a3 27 ……….(1) 3 1 r and
a3 27 3 1 r 19
(1) / (2) gives
r 80.
81.
……….(2) 1 r3 3
1 r
19
2 3
1 Equation of circle x x 1 y y 0 2 y x2 y2 x 0 2 Equation of tangent at (0, 0) x0 y0 x .0 y .0 0 2 22 2x y 0 ……..(1) Sum of distance of A and B from Line (i) is 1 2 5 5 2 2 5 5 2 5
B 1 0, 2
(1, 0) (0, 0)
A x + 2y = 1
x2 c 2 y
a b
2
c 2 ab
a 2 b 2 c 2 ab a2 b 2 c 2 1 2ab 2 1 cos c 2
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JEE-MAIN-2019 (11th Jan-First Shift)-PCM-34
c
2 3
sin C
3 2
c c c 2R R sinc 2 sinc 3 82.
Equation of tangent to parabola y 2 4x is y mx
1 m
………(1)
Now solve it with xy 2
1 mx m x 2 1 mx 2 x 2 0 m For common tangent D = 0 1 2 8m 0 m 1 m 2 1 Put it in equation (i) y x 2 2 2x y 4 0 83.
5th term will be the middle term. 4
4
x3 3 t 41 8C4 5670 3 x 8C 4 . x 8 5670 8765 8 x 5670 43 2 567 x8 81 7 x 8 81 0 Real value of x 3 84.
a3 a1r 2 r2 a1 a1
r 2 25 a a r8 2 Now 9 1 4 r 4 25 5 4 a5 a1r 85.
A is orthogonal matrix 4q2 r 2 p2 q2 r 2 1
……..(1)
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JEE-MAIN-2019 (11th Jan-First Shift)-PCM-35
p2 q2 r 2 0 2
……..(2)
2
and 2q r 0 Solving (1), (2) and (3) 1 p2 2 1 p 2 1
86.
………(3)
2 dx I.F. e x e 2x .x Solution will be y xe 2x e 2x . xe 2x .dx c
xy e 2x
x2 c 2
1 1 1 1 e 2 .e 2 c (Put x = 1 and y e 2 ) 2 2 2 c 0 x y e 2x 2 dy 2xe 2x e 2x e2x 1 2x dx 2 2 dy 1 0 x dx 2 dy 1 0 if x , 1 dx 2 dr = (a, b, c)
87.
d. r. of AB = (0, 1, 1) a.0 b.1 c.1 0 bc 0
B (0, –1, 0) (0, 0, 1) A
……….(i) bc 0 a.0 b 1 c 1 cos 4 a 2 b2 c 2 . 2 1
bc
a b2 c 2 b2 c 2 2bc a2 2bc a2 2b b 2
a 2 2b2
2
. 2
……….(ii) a 2b
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JEE-MAIN-2019 (11th Jan-First Shift)-PCM-36 a b 2 1
From (i) a
OR
b 1
2 a b c 2 1 1 a b c 2 1 1
dr. will be 88.
a 1 1 2
a b c 2 1 1
from (i)
2, 1, 1 or
b c 0 2 4 4 2 4 1R
2, 1, 1
0
3 R3 2R1
1 2
4
1
4
0
2
0 0 9
2
9 2 0
2 OR 2 5 ˆi ˆj kˆ a c 1 2 4
2 4 2 1
ˆi 2 2 2 16 ˆj 2 1 8
2 9 2iˆ ˆj 5 2iˆ ˆj
89.
3
K 81
256 81 4 3 From (1) and (2) 4 64 K 3 27 81 K 300
(put 2 )
…………(1)
4
………..(2)
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JEE-MAIN-2019 (11th Jan-First Shift)-PCM-37 3
90.
i x iy 2 3 27 i3 i2 x iy 3 2 i 1 23 3 2 3 2 . 27 9 3 27 i 2 x iy 8 4i 27 3 27 x 2 8 27 3 y 1 and 4 27 27 yx 1 2 48 27 27 3 1 27 4 18 27 109 18 27 91 27 y x 91
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FIITJEE Solutions to JEE(Main)-2019 Test Date: 11th January 2019 (Second Shift)
PHYSICS, CHEMISTRY & MATHEMATICS Paper - 1 Time Allotted: 3 Hours
Maximum Marks: 360
Please read the instructions carefully. You are allotted 5 minutes specifically for this purpose.
Important Instructions: 1.
The test is of 3 hours duration.
2.
This Test Paper consists of 90 questions. The maximum marks are 360.
3.
There are three parts in the question paper A, B, C consisting of Physics, Chemistry and Mathematics having 30 questions in each part of equal weightage. Each question is allotted 4 (four) marks for correct response.
4.
Out of the four options given for each question, only one option is the correct answer.
5.
For each incorrect response 1 mark i.e. ¼ (one-fourth) marks of the total marks allotted to the question will be deducted from the total score. No deduction from the total score, however, will be made if no response is indicated for an item in the Answer Box.
6.
Candidates will be awarded marks as stated above in instruction No.3 for correct response of each question. One mark will be deducted for indicating incorrect response of each question. No deduction from the total score will be made if no response is indicated for an item in the answer box.
7.
There is only one correct response for each question. Marked up more than one response in any question will be treated as wrong response and marked up for wrong response will be deducted accordingly as per instruction 6 above.
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JEE-MAIN-2019 (11th Jan-Second Shift)-PCM-2
PART –A (PHYSICS) 1.
A particle moves from the point
5.0iˆ 4.0ˆj
2.0iˆ 4.0ˆj m,
at t = 0 with an initial velocity
ms–1+. It is acted upon by a constant force which produces a constant
acceleration 4.0iˆ 4.0 ˆj ms–2. What is the distance of the particle from the origin at time 2s? (A) 15m (C) 5m
(B) 20 2 m (D) 10 2 m
2.
A thermometer graduated according to a linear scale reads a value x0 when in contact with boiling water, and x0/3 when in contact with ice. What is the temperature of an object in 0C, if this thermometer in the contact with the object reads x0/2? (A) 25 (B) 60 (C) 40 (D) 45
3.
A galvanometer having a resistance of 20 and 30 divisions on both sides has figure of merit 0.005 ampere /division. The resistance that should be connected in series such that it can be used as a voltmeter upto 15 volt is: (A) 100 (B) 120 (C) 80 (D) 125
4.
In the experimental setup of metre bridge shown in the figure, the null point is obtained at a distance of 40cm from A. If a 10 resistor is connected in series with R1, the null point shifts by 10cm. The resistance that should be connected in parallel with (R1 + 10) such that the null point shifts back to its initial position is:
(A) 20 (C) 60 5.
(B) 40 (D) 30
A circular disc D1 of mass M and radius R has two identical discs D2 and D3 of the same mass M and radius R attached rigidly as its opposite ends (see figure). The moment of inertia of the system about the axis OO’, passing through the centre of D1 as shown in the figure, will :
(A) MR2 4 (C) MR2 5
(B) 3MR2 2 (D) MR 2 3
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JEE-MAIN-2019 (11th Jan-Second Shift)-PCM-3
6.
The magnitude of torque on a particle of mass 1kg is 2.5 Nm about the origin. If the force acting on it is IN, and the distance of the particle from the origin is 5m, the angle between the force and the position vector is (in radians): (A) (B) 6 3 (C) (D) 8 4
7.
A copper wire is wound on a wooden frame, whose shape is that of an equilateral triangle. If the linear dimension of each side of the frame is increased by a factor of 3, keeping the number of turns of the coil per unit length of the frame the same, then self inductance of the coil: (A) decreases by a factor of 9 (B) increases by a factor of 27 (C) increases by a factor of 3 (D) decreases by a factor of 9 3
8.
A particle of mass m is moving in a straight line with momentum p. Starting at time t = 0, a force F = kt acts in the same direction on the moving particle during time interval T so that its momentum changes from p to 3p. Here k is a constant. The value of T is: k p (A) 2 (B) 2 p k (C)
2k p
(D)
2p k
9.
A paramagnetic substance in the form of a cube with sides 11 cm has a magnetic dipole moment of 20 x 10–6 J/T when a magnetic intensity of 60 x 103 A/m is applied. Its magnetic susceptibility is: (A) 3.3 x 10–2 (B) 40.3 x 10-2 (C) 2.3 x 10–2 (D) 3.3 x 10-4
10.
A simple pendulum of length 1 m is oscillating with an angular frequency 10 rad/s. The support of the pendulum starts oscillating up and down with a small angular frequency of 1 rad/s and an amplitude of 10–2 m. The relative change in the angular frequency of the pendulum is best given by: (A) 10-3 rad/s (B) 1 rad/s -1 (C) 10 rad/s (D) 10-5 rad/s
11.
The circuit shown below contains two ideal diodes, each with a forward resistance of 50. If the battery voltage is 6V, the current through the 100 resistance (in Amperes) is:
(A) 0.036 (C) 0.027
(B) 0.020 (D) 0.030
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JEE-MAIN-2019 (11th Jan-Second Shift)-PCM-4 12.
An electric field of 1000V/m is applied to an electric dipole at angle of 45°. The value of electric dipole moment is 10-29 C.m. What is the potential energy of the electric dipole? (A) 20 1018 J (B) 7 10 27 J 29 (C) 10 10 J (D) 9 10 20 J
13.
A metal bal of mass 0.1 kg is heated upto 5000C and dropped into a vessel of heat capacity 800 JK–1 and containing 0.5 kg water. The initial temperature of water and vessel is 300C. What is the approximate percentage increment in the temperature of the water? [Specific heat Capacities of water and metal are, respectively 4200 Jkg1K 1 and 400 jkg1K 1 ] (A) 15% (B) 30% (C) 25% (D) 20%
14.
The region between y = 0 and y = d contains a magnetic field B Bzˆ . A particle of mass mv m and charge q enters the region with a velocity v v ˆi . If d = , the acceleration of 2qB the charged particle at the point of its emergence at the other side is: qvB 1 ˆ 3 ˆ qvB 3 ˆ 1 ˆ (A) j (B) i j i m 2 2 m 2 2 qvB ˆj ˆi qvB ˆj ˆi (C) (D) m 2 m 2
15.
A pendulum is executing simple harmonic motion and its maximum kinetic energy is K1. If the length of the pendulum is doubled and it performs simple harmonic motion with the same amplitude as in the first case, its maximum kinetic energy is K2 then: K (A) K 2 2K1 (B) K 2 1 2 K (C) K 2 1 (D) K 2 K1 4
16.
Two rods A and B of identical dimensions are at temperature 30oC. If a heated upto 180oC and B upto T0C, then the new lengths are the same. If the ratio of the coefficients of linear expansion of A and B is 4:3, then the value of T is: (A) 2300C (B) 270oC (C) 200oC (D) 250oC
17.
If speed (V), acceleration (A) and force (F) are considered as fundamental units, the dimension of Young’s modulus will be: (A) V 2 A 2F2 (B) V 2 A 2F2 (C) V 4 A 2F (D) V 4 A 2F
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JEE-MAIN-2019 (11th Jan-Second Shift)-PCM-5
18.
A string is wound around a hollow cylinder of mass 5 kg and radius 0.5m. If the string is now pulled with a horizontal force of 40 N, and the cylinder is rolling without slipping on a horizontal surface (see figure), then the angular acceleration of the cylinder will be (Neglect the mass and thickness of the string)
(A) 20 rad/s2 (C) 12 rad/s2
(B) 16 rad/s2 (D) 10 rad/s2
19.
A 27 mW lager beam has a cross – sectional area of 10 mm2. The magnitude of the maximum electric field in this electromagnetic wave is given by: [Given permittivity of space 0 9 10 12 SI units, speed of light c = 3 x 108 m/s] (A) 2 kV/m (B) 0.7 kV/m (C) 1 kV/m (D) 1.4 kV/m
20.
In the circuit shown, the potential difference between A and B is:
(A) 1V (C) 3V
(B) 2V (D) 6V
21.
The mass and the diameter of a planet are three times the respective values for the Earth. The period of oscillation of a simple pendulum on the Earth is 2s. The period of oscillation of the same pendulum on the planet would be: 3 2 (A) s (B) s 2 3 3 (C) s (D) 2 3 s 2
22.
An amplitude modulated signal is plotted below:
Which one of the following best described the above signal? (A) 9 sin 2.5 105 t sin 2 10 4 t V (B) 1 9 sin 2 10 4 t sin 2.5 105 t V
(C) 9 sin 2 10 t sin 2.5 10 t V 4
5
(D) 9 sin 4 10 t sin 5 10 t V 4
5
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JEE-MAIN-2019 (11th Jan-Second Shift)-PCM-6 23.
In a process, temperature and volume of one mole of an ideal monoatomic gas are varied according to the relation VT = K, where I is a constant. In this process the temperature of the gas is increased by T. The amount of heat absorbed by gas is (R is gas constant) 1 1 (A) RT (B) KRT 2 2 3 2K (C) RT (D) T 2 3
24.
When 100g of a liquid A at 1000C is added to 50g of a liquid B at temperature 75oC, the temperature of the mixture becomes 90oC. The temperature of the mixture, if 100g of liquid A at 100oC is added to 50g of liquid B at 50oC, will be: (A) 85oC (B) 60oC o (C) 80 C (D) 70oC
25.
In a hydrogen like atom, when an electron jumps from the M – shell to the L-shell the wavelength of emitted radiation is . If an electron jumps from N-shell to the L-shell the wavelength of emitted radiation will be: 27 16 (A) (B) 20 25 25 20 (C) (D) 16 27
26.
A monochromatic light is incident at a certain angle on an equilateral triangular prism and suffers minimum deviation. If the refractive index of the material of the prism is 3 , then the angle of incidence (A) 90o (B) 30o o (C) 60 (D) 45o
27.
In a double – slit experiment, green light (5303 A ) falls on a double slit having a separation of 19.44 m and a width of 4.05m. The number of bright fringes between the first and the second diffraction minima is: (A) 10 (B) 05 (C) 04 (D) 09
28.
Seven capacitors, each of the capacitance 2F, are to be connected in a configuration to 6 obtain an effective capacitance of F. Which of the combinations, shown in figures 13 below, will achieve the desired value?
0
(A)
(B)
(C)
(D)
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JEE-MAIN-2019 (11th Jan-Second Shift)-PCM-7
29.
A particle of mass m and charge q is in an electric and magnetic field given by: E 2iˆ 3 ˆj;B 4 ˆj 6kˆ The charged particle is shifted from the origin to the point P(x = 1; y = 1) along a straight path. The magnitude of the total work done is: (A) (0.35)q (B) 5q (C) (2.5)q (D) (0.15)q
30.
In a photoelectric experiment, the wavelength of the light incident on a metal is changed from 300 nm to 400 nm. The decrease in the stopping potential is close to hc 1240 nm V e (A) 0.5 V (B) 1.5 V (C) 1.0 V (D) 2.0 V
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JEE-MAIN-2019 (11th Jan-Second Shift)-PCM-8
PART –B (CHEMISTRY) 31.
The reaction:
MgO s C s Mg s CO g , for which r H0 491.1 kJ mol1 and r S0 198.0 JK 1 mol1 is not feasible at 298 K. Temperature above which reaction will be feasible is: (A) 2040.5 K (C) 2480. K 32.
(B) 1890.0K (D) 2380.K
The correct match between Item I and Item II is Item I Item II A. Allosteric effect P. Molecule binding to the active site of enzyme B. Competitive inhibitor Q. Molecule crucial for communication in the body C. Receptor R. Molecule binding to a site other than the active site of enzyme D. Poison S. Molecule binding to the enzyme covalently (A) A R, B P, C Q, D S (C) A R, B P, C S, D Q
(B) A P, B R, C Q, D S (D) A P, B R, C S, D Q
33.
The coordination number of Th in K 4 Th C2O 4 4 OH2 2 is: C2 O 42 Oxalato (A) 14 (B) 6 (C) 8 (D) 10
34.
The major product obtained in the following reaction is:
(A)
(B)
(C)
(D)
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JEE-MAIN-2019 (11th Jan-Second Shift)-PCM-9
35.
The standard reaction Gibbs energy for a chemical reaction at an absolute temperature T is given by
r G0 A BT Where A and B are non – zero constant. Which of the following is TRUE about this reaction? (A) Endothermic if A > 0 (B) Exothermic if A > 0 and B < 0 (C) Endothermic if A < 0 and B > 0 (D) Exothermic if B < 0 36.
The radius of the largest sphere which fits properly at the centre of the edge of a body centered cubic unit cell is: (Edge length is represented by ‘a’) (A) 0.0027a (B) 0.047 a (C) 00137a (D) 0.07a
37.
The hydride that is NOT electron deficient is: (A) SiH4 (B) B 2H6 (C) GaH3
38.
(D) AIH3
Given the equilibrium constant: KC of the reaction:
0
Cu s 2 Ag aq Cu2 aq 2 Ag s is 10 x 1015=, calculate the Ecell of the reaction
of 298 K 2.303
RT at 298K 0.059 V F
(A) 0.04736 mV (C) 0.4736 V
(B) 0.4736 mV (D) 0.04736 V
39.
The correct option with respect to the Pauling electronegativity values of the elements is: (A) Te > Xe (B) Ga > Ge (C) Si > AI (D) P > S
40.
Which of the following compounds will produce a precipitate with AgNO3? (A) (B)
(C)
41.
(D)
The de Broglie wavelength () associated with a photoelectron varies with the frequency (v) of the incident radiation as, [v0 is threshold frequency]: (A)
(C)
1 v v0
(B)
1
v v0 4
1
v v0
1
3 2
(D)
1 1
v v0 2
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JEE-MAIN-2019 (11th Jan-Second Shift)-PCM-10 42.
Which of the following compounds reacts with ethylmagnesium bromide and also decolourizes bromine water solution: (A) (B)
(C)
43.
(D)
In the following compound
The favourable site/s for protonation is/are (A) a and e (B) b, c and d (C) a and d (D) a 44.
Taj Mahal is being slowly disfigured and discoloured. This is primarily due to: (A) global warming (B) acid rain (C) water pollution (D) soil pollution
45.
The relative stability of +1 oxidation state of group 13 elements follows the order: (A) AI Ga TI In (B) TI In Ga AI (C) Ga AI In TI (D) AI Ga In TI
46.
For the equilibrium
2H2O H3O OH , the value of G0 at 298 K is approximately: (A) 100 kJ mol–1 (C) 80 kJ mol–1 47.
(B) – 80 kJ mol-1 (D) –100 kJ mol-1
The reaction that does NOT define calcinations is: (A) Fe 2 O3 .XH2O Fe 2 O3 XH2O (B) 2Cu2 S 3O 2 2Cu2 O 2SO 2 (C) ZnCO3 ZnO CO 2 (D) CaCO3 .MgCO3 CaO MgO 2CO 2
48.
A compound ‘X’ on treatment with Br2/NaOH, provided C3H9N, which gives positive carbylamine test. Compound ‘X’ is: (A) CH3COCH2NHCH3 (B) CH3CH2COCH2NH2 (C) CH3CH2CH2CONH 2
(D) CH3CON CH3 2
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JEE-MAIN-2019 (11th Jan-Second Shift)-PCM-11
49.
Among the colloids cheese (C), milk (M) and smoke (S), the correct combination of the dispersed phase and dispersion medium, respectively is: (A) C: liquid in solid; M: liquid in solid; S: solid in gas (B) C : liquid in solid; M: liquid in liquid; S: solid in gas (C) C : solid in liquid ; M : liquid in liquid ; S : gas in solid (D) C : solid in liquid ; M : solid in liquid; S : solid in gas
50.
The homopolymer formed from 4 – hydroxybutanoic acid is: (A)
(B)
(C)
(D)
51.
K2HgI4 is 40% ionized in aqueous solution. The value of its van’t Hoff factor (i) is: (A) 1.6 (B) 1.8 (C) 2.0 (D) 2.2
52.
25 mL of the given HCI solution requires 20 mL of 0.1 M sodium carbonate solution. What is the volume of this HCI solution required to titrate 30 mL of 0.0 M aqueous NaOH solution? (A) 25 mL (B) 75 mL (C) 50 mL (D) 12.5 mL
53.
The reaction 2X B is a zeroth order reaction. If the initial concentration of X is 0.2M, the half life is 6 h. When the initial concentration of X is 0.5 M, the time required to reach its final concentration of 0.2 M will be: (A) 9.0 h (B) 12.0 h (C) 18.0 h (D) 7.2 h
54.
Match the following items in column I with the corresponding items in column II. Column I Column II I. Na2CO3 . 10 H2O A. Portland cement ingredient II. Mg (HCO3)2 B. Castner – Kellner process III. NaOH C. Solvay process IV. Ca3AI2O6 D. Temporary hardness (A) I – B, II – C, III – A, IV – D (B) I – C, II – B, III – D, IV – A (C) I – D, II – A, III – B, IV – C (D) I – C, II – D, III – B, IV – A
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JEE-MAIN-2019 (11th Jan-Second Shift)-PCM-12 55.
56.
The major product obtained in the following conversion is:
(A)
(B)
(C)
(D)
The major product of the following reaction is:
(A)
(B)
(C)
(D)
57.
The higher concentration of which gas in air can cause stiffness of flower buds? (A) NO2 (B) CO2 (C) SO2 (D) CO
58.
The correct match between Item I and Item II is Item I Item II A. Ester test P. Tyr B. Carbylamine test Q. Asp C. Phthalein dye test R. Ser S. Lys (A) A – Q, B – S, C – P (B) A – R, B – Q, C – P (C) A – R, B – S, C – Q (D) A – Q, B – S, C – R
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JEE-MAIN-2019 (11th Jan-Second Shift)-PCM-13
59.
The number of bridging CO ligand(s) and Co-Co bond(s) in Co2 (CO)8 respectively are: (A) 2 and 1 (B) 2 and 0 (C) 0 and 2 (D) 4 and 0
60.
4KOH,O2 A 2B 2 H2O
Green
4HCI 3B 2C MnO 2 2H2O
Purple
H2O,KI 2C 2 A 2KOH D
In the above sequence of reactions, A and D, respectively are: (A) Ki and KMnO4 (B) MnO2 and KIO3 (C) KIO3 and MnO2 (D) KI and K2MnO4
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JEE-MAIN-2019 (11th Jan-Second Shift)-PCM-14
PART–C (MATHEMATICS) 61.
lim x 0
x cot 4 x sin2 x cot 2 2x
is equal to:
(A) 0 (C) 4 62.
63.
(B) 2 (D) 1
2
7 cot 1 x 10 0 , lie in the interval:
(A) ,cot 5 cot 4,cot 2
(B) cot 2,
(C) ,cot 5 cot 2,
(D) cot 5,cot 4
If a hyperbola has length of its conjugate axis equal to 5 and the distance between its foci is 13, then the eccentricity of the hyperbola is:
13 12 13 (C) 6 (A)
64.
1
All x satisfying the inequality cot x
(B) 2 (D)
13 8
If the area of the triangle whose one vertex is at the vertex of the parabola, y 2 4 x a2 0 and the other two vertices are the points of intersection of the
parabola and y – axis, is 250 sq. units, then a value of ‘a’ is:
1/ 3
(B) 5 2
(A) 5 5 (C) 10 65.
2/3
Two lines
(D) 5
x 3 y 1 z 6 x5 y2 z3 and intersect at the point R. The 1 3 1 7 6 4
reflection f R in the xy – plane has coordinates: (A) (2, –4, –7) (B) (2, 4, 7) (C) (2, –4, 7) (D) (–2, 4, 7) 66.
Contrapositive of the statement “If two numbers are not equal, then their squares are not equals” is: (A) If the squares of two numbers are not equal, then the numbers are equal (B) If the squares of two numbers are equal, then the numbers are not equal (C) If the squares of two numbers are equal, then the numbers are equal (D) If the squares of two numbers are not equal, then the numbers are not equal
67.
If in a parallelogram ABDC, the coordinates of A, B and C are respectively (1, 2), (3, 4) and (2, 5), then the equation of the diagonal AD is: (A) 5x 3y 1 0 (B) 5x 3y 11 0 (C) 3x 5y 7 0 (D) 3x 5y 13 0
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JEE-MAIN-2019 (11th Jan-Second Shift)-PCM-15
68.
69.
The integral
/4
/ 6
dx equals: sin 2x tan5 x cot 5 x
(A)
1 1 tan1 20 9 3
(C)
40
1 1 tan1 9 l 3 10 4 1 1 (D) tan1 5 4 3 3 (B)
Let x, y be positive real numbers and m, n positive integers. The maximum value of the expression
xm yn is: 1 x 2m 1 y 2n
1 2 mn (D) 6mn
(A) 1 (C)
(B)
1 4
2
70.
n
q 1 q 1 q 1 Let Sn 1 q q ... q and Tn 1 ... where q is a 2 2 2 real number and q 1. If 101 C1 101 C2 .S1 ... 101 C101.S100 T100 then is equal to: 2
n
99
(A) 2 (C) 200 71.
(B) 202 (D) 2100
Let and be the roots of the quadratic equation n 1 n x sin x sin cos 1 cos 0 0 45 , and < . Then n is n0
2
0
equal to:
1 1 1 cos 1 sin 1 1 (C) 1 cos 1 sin
1 1 1 cos 1 sin 1 1 (D) 1 cos 1 sin
(A)
72.
(B)
A bag contains 30 white balls and 10 red balls. 16 balls are drawn one by one randomly from the bag with replacement. If X be the number of white balls drawn, then
mean of X is equal to: s tandard deviation of X (A) 4
(B) 4 3
(C) 3 2
(D)
4 3 3
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JEE-MAIN-2019 (11th Jan-Second Shift)-PCM-16
73.
Let z be a complex number such that z z 3 i where i
1 . Then z is equal
to: (A) (C)
74.
34 3 41 4
5 3 5 (D) 4 (B)
a bc
2a
2a
2b
bca
2b
2c
2c
c ab
If
2
= a b c x a b c , x = and a + b+ +c 0, then x is equal to: (A) abc (C) 2(a + b + c) 75.
Let
(B) – (a + b + c) (D) –2 (a + b+ c)
3ˆi j, ˆi 3ˆj and ˆi 1 ˆj respectively be the position vectors of the points A,
B and C with respect to the origin O. If the distance of C from the bisector of the acute angle between OA and OB is
3 , then the sum of all possible values of is 2
(A) 4 (C) 2
(B) 3 (D) 1
76.
If 19th terms of non – zero A.P. is zero, then its (49th term) : (29th term) is: (A) 4:1 (B) 1:3 (C) 3:1 (D) 2:1
77.
If
x 1 dx f x 2x 1 C , where C is a constant of integration of integration, 2x 1
then f(x) is equal to:
1 x 1 3 2 (C) x 4 3
2 x 2 3 1 (D) x 4 3
(A)
78.
(B)
Let a function f : 0, 0, be defined by f x 1 (A) not injective but it is surjective (C) neither injective nor surjective
79.
1 . Then f is: x
(B) injective only (D) both injective as well as surjective
Let K be the set of all real values of x where the function f x sin x x 2 x cos x is not differentiable. Then the set K is equal to: (A) (en empty set) (C) {0}
(B) {} (D) {0, }
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JEE-MAIN-2019 (11th Jan-Second Shift)-PCM-17
80.
The area (in sq. units) in the first quadrant bounded by the parabola, y x 2 1 , the tangent to it at the point (2, 5) and the coordinate axes is:
8 3 187 (C) 24
37 24 14 (D) 3
(A)
81.
Given
(B)
cos A cos cosC bc c a ab for a ABC with usual nation. If , 11 12 13
then the ordered tried (, , ) has a value: (A) (7, 19, 25) (B) (3, 4, 5) (C) (5, 12, 13) (D) (19, 7, 25) 82.
The solution of the differential equation (A) loge
2x xy 2y
(C) loge 83.
1 x y xy2 1 x y
(D) loge
2y 2 y 1 2x
Let the length of the latus rectum of an ellipse with its major axis long x – axis and center at the origin, be 8. If the distance between the foci of this ellipse is equal to the length of the length of its minor axis, then which one of the following points lies on it?
(C) 4,
3
(D) 4
(A) 4, 2, 2 2
84.
dy 2 x y when y(1) = 1, is: dx 1 x y (B) loge 1 x y 2 x 1
3, 2
3
(B) 4 3, 2 2
2, 2
let S = {1, 2, … 20}. A subset B of S is said to be “nice”, if the sum of the elements of B is 203. Then the probability that a randomly chosen subset of S is ‘nice’ is:
7 2 20 4 (C) 20 2
5 2 20 6 (D) 20 2
(A)
(B)
85.
If the point (2, , ) lies on the plane which passes through the points (3, 4, 2) and (7, 0, 6) and is perpendicular to the plane 2x 5y 15 , then 2 3 is equal to: (A) 12 (B) 7 (C) 5 (D) 17
86.
Let x 10 x 10
50
(A) 12.50 (C) 12.25
50
a0 a1x a 2 x 2 ... a50 x50 , for xR; then
a2 is equal to: a0
(B) 12.00 (D) 12.75
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JEE-MAIN-2019 (11th Jan-Second Shift)-PCM-18 87.
The number of functions f from {1, 2, 3, …, 20} only {1, 2, 3, …, 20} such that f(k) is a multiple of 3, whenever k is a multiple of 4, is: (A) 65 15 ! (B) 5! 6 ! (C) 15 ! 6 !
(D) 56 15
88.
A circle cuts a chord of length 4a on the x – axis and passes through a point on the y – axis, distant 2b from the origin. Then the locus of the center of this circle, is: (A) a hyperbola (B) an ellipse (C) a straight line (D) a parabola
89.
Let f x
x 2
a x
2
dx b2 d x
2
,r R f, where a, b and d are non – zero real
constant. Then: (A) f is an increasing function of x (B) f is a decreasing function of x (C) f is not a continuous function of x (D) f is neither increasing nor decreasing function of x 90.
Let A and B be two invertible matrices of order 3 x 3. If det (ABAT) = 8 and det(AB–1) = 8, then det (BA–1 BT) is equal to:
1 4 1 (C) 16 (A)
(B) 1 (D) 16
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JEE-MAIN-2019 (11th Jan-Second Shift)-PCM-19
JEE (Main) – 2019 ANSWERS PART A – PHYSICS 1.
B
2.
A
3.
C
4.
C
5.
B
6.
A
7.
C
8.
B
9.
D
10.
A
11.
B
12.
B
13.
D
14.
BONUS
15.
A
16.
A
17.
D
18.
B
19.
D
20.
B
21.
D
22.
B
23.
A
24.
C
25.
D
26.
C
27.
B
28.
B
29.
B
30.
C
PART B – CHEMISTRY 31.
C
32.
A
33.
D
34.
C
35.
A
36.
D
37.
A
38.
C
39.
B
40.
B
41.
D
42.
B&C
43.
B
44.
B
45.
D
46.
C
47.
B
48.
C
49.
B
50.
D
51.
B
52.
A
53.
C
54.
D
55.
A
56.
D
57.
C
58.
A
59.
A
60.
B
PART C – MATHEMATICS 61.
D
62.
B
63.
A
64.
D
65.
A
66.
C
67.
A
68.
B
69.
C
70.
D
71.
C
72.
B
73.
B
74.
B
75.
D
76.
C
77.
D
78.
C
79.
A
80.
B
81.
A
82.
B
83.
B
84.
B
85.
B
86.
C
87.
C
88.
D
89.
A
90.
C
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JEE-MAIN-2019 (11th Jan-Second Shift)-PCM-20
HINTS AND SOLUTIONS PART A – PHYSICS 1.
2.
3.
1 S (5iˆ 4 ˆj)2 (4ˆj 4ˆj)4 2 = 10iˆ 8ˆj 8iˆ 8ˆj r2 r1 18iˆ 16ˆj r2 20iˆ 20ˆj rf 20 2 x t x0 100o C x100 x 0 x0 x0 2 3 100o C x0 x 3 = 25°C
t
Rg = 20 NL = Ng = N = 30 1 FOM = = 0.005 A/Div.
1 Current sentivity = CS = 0.005 1 Igmax = 0.005 × 30 = 15 × 10–2 = 0.15 15 = 0.15[20 + R] 100 = 20 + R R = 80. 4.
R1 2 …(i) R2 3 R1 10 1 R2 R1 + 10 = R2 …(ii) 2R 2 R 10 = R2 ; 10 2 3 3 R2 = 30 & R1 = 20 30 R 30 R 2 30 3 R = 60
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JEE-MAIN-2019 (11th Jan-Second Shift)-PCM-21
MR 2 MR2 2 MR2 2 4 2 2 MR MR = 2MR2 = 3 MR2 2 2
5.
I
6.
2.5 = 1 × 5 sin 1 Sin = 0.5 = 2 6
7.
Total length L will remain constant L = (3a) N (N = total turns) And length of winding = (d) N = (d = diameter of wire) Self inductance = 0n2A
a
a a
3 a2 = 0n 2 dN 4
a2 N a So self inductance will become 3 times.
8.
dp F kt dt
3P
P
2p
9.
10.
T
dP kt dt 0
KT 2 p ; T2 2 k
I H Magnetic moment I Volume 20 10 6 I = 20 N/m2 6 10 20 1 103 3 60 10 3 = 0.33 × 10–3 = 3.3 × 10–4
Angular frequency of pendulum geff 1 geff 2 geff
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JEE-MAIN-2019 (11th Jan-Second Shift)-PCM-22
1 g 2 g [s = angular frequency of support] 1 g 2 g
1 2(As5 ) 2 10 1 10 2 = 10–3 10
11. 12.
I
6 0.002 (D2 is in reverse bias) 300
U = P E = –PE cos = –(10–29) (103) cos 45° = –0.707 × 10–26 J = –7 × 10–27 J
13.
0.1 × 400 × (500 – T) = 0.5 × 4200 × (T – 30) + 800 (T – 30) 40(500 – T) = (T – 30) (2100 + 800) 20000 – 40T = 2900 T – 30 × 2900 20000 + 30 × 2900 = T(2940) T = 30.4°C T 6.4 100 100 20% T 30
14.
BONUS
15.
Maximum kinetic energy at lowest point B is given by K = mg (1 – cos )
m
where = angular amp. K1 = mg (1 – cos )
A
K2 = mg(2) (1 – cos ) K2 = 2K1 16.
B
1 = 2
1T1 = 2T2
1 T1 4 T 30 ; 2 T2 3 180 30 T = 230°C
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JEE-MAIN-2019 (11th Jan-Second Shift)-PCM-23
17.
F F y ; [Y] A A Now from dimension ML F F 2 ; L T2 T M 4 2 V F V L2 2 T M A A 2 4 F v L2 2 2 2 F = MA MA A V4 L2 2 A [F] [Y] F1V 4 A 2 [A]
18.
40 + f = m(R) …(i) 40 × R – f × R = mR=2 40 – f = mR …(ii) From (i) and (ii) 40 = 16 mR
19.
Intensity of EM wave is given by Power 1 I 0E02C Area 2 27 10 3 1 = 9 10 2 E 2 3 108 10 106 2 E = 2 103 kV / m = 1.4 kV/m
20.
Potential difference across AB will be equal to battery equivalent across CD. E1 E 2 E3 1 2 3 r1 r2 r3 VAB VCD 1 1 1 1 1 1 1 1 1 r1 r2 r3 1 1 1 6 = = 2V 3
21.
g
GM R2 2
2 Mp R e 1 1 3 ge Me Rp 3 3 1 Also, T g
gp
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JEE-MAIN-2019 (11th Jan-Second Shift)-PCM-24
22.
Tp Te
ge 3 gp
Tp 2 3s
Analysis of graph says (1) Amplitude varies as 8 – 10 V or 9 1 (2) Two time period 100 s (signal wave) and 8 s (carrier wave) 2t 2t Hence signal is 9 1 sin sin T1 T2 = 9 1 sin (2 × 104t) sin 2.5 × 105 t
23.
VT = K
PV V k nR PV2 = K R C CV (For polytropic process) 1 x R 3R R C 1 2 2 2 Q = nC T
24.
100 × SA × [100 – 90] = 50 × SB × (90 – 75) 2SA = 1.5 SB 3 S A SB 4 Now, 100 × SA × [100 –T] = 50 × SB (T – 50) 3 2 (100 T) (T 50) 4 300 – 3T = 2T – 100 400 = 5T T = 80
25.
For M L steel 1 1 K5 1 K 2 2 3 36 2 for N L 1 1 K 3 1 K 2 2 4 16 2 20 27
26.
i=e r1 = r2 =
A 300 2
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JEE-MAIN-2019 (11th Jan-Second Shift)-PCM-25
by Snell’s law 1 sini 3
1 3 2 2
i = 60 27.
28.
For diffraction Location of 1st minima D y1 0.2469D a Location of 2nd minima 2D y2 0.4938D a Now for interference Path for interference Path difference at P. dy 4.8 D Path difference at P. dy 9.6 D So orders of maxima in between P and Q is 5, 6, 7, 8, 9 So 5 bright fringes all present between P & Q.
y2
P y1
6 F 13 Therefore three capacitors most be in parallel to get 6 in 1 1 1 1 1 1 Ceq 3C C C C C Ceq
Ceq 29.
Q
3C 6 F 13 13
Fnet dE q(v B) = (2q ˆi 3q ˆi) q(v B) W = Fnet S
= 2q + 3q = 5q 30.
hc eV1 1 hc eV2 2 (i) – (ii)
…(i) …(ii)
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JEE-MAIN-2019 (11th Jan-Second Shift)-PCM-26
1 1 hc e(V1 V2 ) 1 2 hc 2 1 V1 V2 e 1 2 = (1240 nm V) =
100 nm 300 nm 400 nm
12.4 1 V. 12
PART B – CHEMISTRY 31.
In order to be spontaneous GO should be –ve GO = HO –TSO 0 = 491.1 103 – T 198 491100 T 2480 198 If temp is above 2480 K, the reaction will be spontaneous.
32.
Fact based, go through definition.
33.
Th is a metal having large size and oxalate is a bidentate ligand hence its co-ordination number in given complex is 10. O
34.
Since LiAlH4 is a strong reducing agent, it will reduce –NO2, C O H and ketonic group but cannot reduce normal alkene >C=C<
35.
GO = HO – TSO GO = A – BT In endothermic reaction H = +ve. Hence, A = +ve
ll
36.
3a 4R 3a 4 2(R + r) = a 3a 2 r a 4 R
3a 2r a 2 3a 2a 3a 2r a 2 2 2a 1.7329 .268a r = .067a 4 4
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JEE-MAIN-2019 (11th Jan-Second Shift)-PCM-27
37.
SiH4 has complete octet hence it is not an electron deficient hydride.
38.
Go nFEocell 2.303RT log K eq 2F Eocell 2.303RT log K eq
2.303RT log 10 1015 F .059 16 8 .059 .472 2
2 E ocell Eocell
39.
Electronegativity increases from left to right in a period and decreases down the group. Br
40. On ionization
41.
, this compound will produce Aromatic cation, which is stable.
h mv According to Einstein’s theory of photoelectric effect: h ho + KE
1 h = h o + mv 2 2 2h o mv 2 2h o m
v2 1
v o 2
h 1
m o 2
1 1
o 2 42.
Option B and option D both will react with Grignard reagent and decolourizes Br2/H2O. (IIT has given option D only)
43.
After protonation at b or c or d the conjugate acid is stabilized by resonance.
44.
Acid rain reacts with marble. Hence, The Taj Mahal which made up of marble is discoloured.
45.
Inert pair effect gradually increases down the group. Hence, stability of lower oxidation state increases down the group.
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JEE-MAIN-2019 (11th Jan-Second Shift)-PCM-28 46.
Go 2.303RT log K eq
= –2.303 8.314 298 log 10–14 = –2.303 8.314 298 –14 = 79,881.87 80 KJ mol–1 47.
Calcination takes place in absence of air. Hence step 2 is not defining it.
48.
Br2/NaOH converts amide into primary amine having one carbon atom less, which gives carbylamine test.
49.
Go through different types of colloid and their examples.
50.
4-hydroxy butanoic acid undergoes intermolecular esterification to give polymer.
51.
K 2 Hg 4 2K Hg 4
2
1
2
Total number of particle = 1 + 2 1 2 Hence, Van’t Hoff factor = 1 1 2 0.4 = 1 0.8 1.8 1 52.
Apply law of equivalence: 25 N = 30 0.1 2 30 0.2 6 1.2 NHCl 0.2 25 5 5 nd For the 2 titration 1.2 VHCl 30 0.2 5 6 5 30 VHCl = 25 ml 1.2 1.2
53.
For zero order reaction Co – Ct = Kt 0.5M – 0.2M = Kt 0.3 = Kt ‘K’ can be calculated by C t1/2 o 2K 0.2 6 2K
----- (1)
0.2 2 10 1 1 12 12 60 Putting the value of K in eq (1) 0.3 0.3 t 60 0.3 18 Hr 1 K 60 K
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JEE-MAIN-2019 (11th Jan-Second Shift)-PCM-29
54.
Fact based
55.
Attack on C
C
is more preferred than benzene ring.
56.
First Markonikov’s addition one alkene followed by intramolecular Friedel-Craft alkylation takes place.
57.
SO2 gas causes stiffness of flower buds?
58.
Go through structure of amino acids.
59.
Go through structure of [Co2(CO)8] O OC OC
C
CO
Co Co CO
OC
C
CO
O
60.
4KOH, O2 MnO2 2K 2MnO4 2H2O
Green
4 HCl
3 K 2MnO 4 2KMnO 4 MnO 2 2H2O Purple
H2O, K
2KMnO 4 2MnO 2 2KOH KO3
PART C – MATHEMATICS 61.
62.
x cos 4x sin2 2x sin2 x.cos2 2x.sin 4x 4x cos 4x . cos2 x sin 4x cos2 2x 1 as x 0
cot x 5 cot x 2 0 1
1
cot 1 x ,2 5,
Put 0 cot 1 x cot 1 x 0,2 x cot 2,
63.
2b 5 and 2ae 13 2
ae
a 2 b2 2
a2 ae b2
169 25 4 4
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JEE-MAIN-2019 (11th Jan-Second Shift)-PCM-30 a6 ae 13 e a 12
64.
y 2 4 x a2
Vertices of triangle are a2 ,0 and (0, 2a) and (0, –2a)
1 2 a 2 a3 125
4a 250
Area
65.
Points on the given lines are 3, 3 1, 6 and 7 5, 6 2, 4 3 3 7 5 3 1 6 2 1, 1 Point R is (2, –4, 7) Image of R under xy – plane is (2, –4, –7)
66.
Contra positive of p q is ~ q ~ p Answer is C C (2, 5)
67.
5 9 E is , 2 2
E
5 Slope of AD 3 Equation of AD is y 2
5 x 1 3
5x 3y 1 0 4
68.
I 6
D
A (1, 2)
B (3, 4)
sec 2 x dx
2 tan x tan5 x cot 5 x
Put tan x t 1
1
3
1
dt 1 2t t 5 5 t 4 t dt
2t 1
10
1
3
Put t5 y 1 1 I tan1 y 52 3 10 1 1 1 tan 5/ 2 10 4 3 FIITJEE Ltd., FIITJEE House, 29-A, Kalu Sarai, Sarvapriya Vihar, New Delhi -110016, Ph 46106000, 26569493, Fax 26513942 website: www.fiitjee.com.
JEE-MAIN-2019 (11th Jan-Second Shift)-PCM-31
69.
xm y n
1 x 1 y 2m
2n
1 1 n 1 m x m y n x y 1 Put x m m 2 x 1 1 1 2 m x xm
Maximum Value
1 4
101
70.
101
Cr sr 1
r 1
101
101Cr r 1
qr 1 q 1
101 1 101 101 r 101 C q Cr r q 1 r 1 r 1 1 101 1 q 1 2101 1 q 1
101 101 1 q 2 100 2 q1 100 2
71.
Using quadratic formula, 2
x
cos sin 1 cos sin 1
4 sin cos
2 sin
cos sin 1
2
cos sin 1
2 sin cos , cos ec cos , cos ec
n0
n
1
n
n
n
n
cos ec sin n0
n 0
1 1 1 cos 1 sin (C) is the correct answer.
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JEE-MAIN-2019 (11th Jan-Second Shift)-PCM-32 72.
There are 30 white balls and 10 red balls 30 3 P (white ball) p 40 4 1 q 4 mean x np s tan darddeviation x npq np q
73.
3 16 4 4 3 1 4
z x iy x 2 y 2 x iy 3 i y 1 x 2 1 x 3 x 2 1 9 6x x 2 4 x 3 5 z x2 y2 3
74.
R1 R1 R2 R3 1
a b c 2b 2c
1
1
bac
2b
2c
c ab
c 3 c 3 c1, c 2 c 2 c1 1
0
0
a b c 2b a b c 0 2c 0 a b c
a b c
3 2
a b c x a b c
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JEE-MAIN-2019 (11th Jan-Second Shift)-PCM-33
75.
y
Equation of angle bisector of DA and OB is y=x 1 3 Given that, 2 2 2 1 3 2, 1 Sum of values of 1
A
B 1, 3
a 18d 0 a 18d t 49 a 48d 18d 48d t 29 a 28d 18d 28d
77.
x
O
76.
3, 1
30d 3 10d
Put 2x 1 t 2 x 1 dx 2x 1 t2 3 t 3 3t dt C 6 2 2
t2 3 t C 6 2 x4 2x 1 C 3 78.
1 x Neither one - one nor Onto y 1
y=1
(1, 0)
79.
At x or x f x sin x x 2 x cos x f x is differentiable For x 0 f x sin x x 2 x cos x
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JEE-MAIN-2019 (11th Jan-Second Shift)-PCM-34 f ' x cos x 1 2 cos x 2 x sin x f ' 0 2 For x 0 f x sin x x 2 x cos x f ' x cos x 1 2 cos x 2 x sin x f ' 0 2 LHD = RHD f is differentiable at x = 0.
80.
Equation of tangent at (2, 5) is
y5 x 2 1 2
(2, 5)
or y 4x 3 2
Required Area x 2 1 dx 0
1 3 2 . 5 2 4
x3 25 2 x 0 3 8 8 9 37 3 8 24
(3/4,0) (0, –3)
81.
bc c a ab abc 11 12 13 18 a 7k, b 6k, c 5k b2 c 2 a2 1 2bc 5 19 5 cosB , cos C 25 7 1 19 5 5 35 7 7 19 25 35 35 35 : : 7 : 19 :25 cos A
82.
uxy du dy 1 dx dx du 1 u2 dx du 1 u2 dx
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JEE-MAIN-2019 (11th Jan-Second Shift)-PCM-35
du dx 1 u2 1 1 u log xc 2 1 u 1 1 x y log xc 2 1 x y Put x 1 c 1 1 x y log 2 x 1 1 x y
83.
x2 y2 1 a2 b2 Given that 2b 2ae 2b2 8 b ae and a a 1 e 2 4, a2 e2 a2 1 e2
Consider
1 2 a 8, b 4 2 e2
Hence equation of ellipse is
4 84.
x2 y2 1 64 32
3, 2 2 lies on this
Sum of all elements of S = 210 So X be a nice set if x S 7 , S 1, 6 , S 2, 5 ,S 3, 4 , S 1,2,4
5 220 (2) is the answer. P x
85.
A 7, 0, 6 and B (3, 4, 2) AB 4iˆ 4 ˆj 4kˆ Also 2iˆ 5ˆj is parallel to the plane
ˆi
ˆj
kˆ
Normal perpendicular to the required plane is 1 1 1 5iˆ 2jˆ 3kˆ 2 5 0 Equation of the plane is 5 x 7 2 y 0 3 z 6 0 5x 2y 3z 17
2, ,
lies on this 10 2 3 17 2 3 7
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JEE-MAIN-2019 (11th Jan-Second Shift)-PCM-36
86.
50
10 x
a0 1050
50
10 x
2
a 2 50C2 10
a2 a0 87.
48
2 48 50 C2 10 2 1052 (2)
12.25
k 4,8,12,16,20 f k can takes the values 3,6,9,12,15,18
Number of ways 6C5 .5! Total number of onto functions 6C5 .5! 15! 6!15!
88.
k 2 4a2 r 2 and 2
h 0
2
k 2b r 2 2
h2 k 2b k 2 4a2 h2 4bk 4b 2 4a2 Locus is x 2 4 by b2 a 2
r
(0, 2b)
h (k)
r
Parabola.
k 2a
89.
f x x x3 a
f ' x
1 2 2
a
x
2
a2
d x b2 d x
1 2 2
2
x 2 a2
b2
b
2
d x
2
2
b2 d x
ve 90.
2
ABA T A .B . A T A B
AB 1 8 A 8 B 1
BA B
B
2
A
B
2
8B
B 8
1 16
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FIITJEE Solutions to JEE(Main)-2019 Test Date: 12th January 2019 (First Shift)
PHYSICS, CHEMISTRY & MATHEMATICS Paper - 1 Time Allotted: 3 Hours
Maximum Marks: 360
Please read the instructions carefully. You are allotted 5 minutes specifically for this purpose.
Important Instructions: 1.
The test is of 3 hours duration.
2.
This Test Paper consists of 90 questions. The maximum marks are 360.
3.
There are three parts in the question paper A, B, C consisting of Physics, Chemistry and Mathematics having 30 questions in each part of equal weightage. Each question is allotted 4 (four) marks for correct response.
4.
Out of the four options given for each question, only one option is the correct answer.
5.
For each incorrect response 1 mark i.e. ¼ (one-fourth) marks of the total marks allotted to the question will be deducted from the total score. No deduction from the total score, however, will be made if no response is indicated for an item in the Answer Box.
6.
Candidates will be awarded marks as stated above in instruction No.3 for correct response of each question. One mark will be deducted for indicating incorrect response of each question. No deduction from the total score will be made if no response is indicated for an item in the answer box.
7.
There is only one correct response for each question. Marked up more than one response in any question will be treated as wrong response and marked up for wrong response will be deducted accordingly as per instruction 6 above.
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JEE-MAIN-2019 (12th Jan-First Shift)-PCM-2
PART –A (PHYSICS) 1.
An ideal gas occupies a volume of 2 m3 at a pressure of 3 × 106 Pa. The energy of the gas is: (A) 9 × 106J (B) 6 × 104J 8 (C) 10 J (D) 3 × 102J
2.
A travelling harmonic wave is represented by the equation y(x, t) = 103sin(50t + 2x), where x and y are in meter and t is in seconds. Which of the following is a correct statement about the wave? (A) The wave is propagating along the negative x-axis with speed 25 ms1. (B) The wave is propagating along the positive x-axis with speed 100 ms1. (C) The wave is propagating along the positive x-axis with speed 25 ms1. (D) The wave is propagating along the negative x-axis with speed 100 ms1.
3.
An ideal battery of 4 V and resistance R are connected in series in the primary circuit of a potentiometer of length 1 m and resistance 5 . The value of R, to give a difference of 5 mV across 10 cm of potentiometer wire, is: (A) 490 (B) 480 (C) 395 (D) 495
4.
In a meter bridge, the wire of length 1 m has a nondR uniform cross-section such that, the variation of its d dR 1 resistance R with length is . Two equal d resistances are connected as shown in the figure. The galvanometer has zero deflection when the jockey is at point P. What is the length AP? (A) 0.2 m (B) 0.3 m (C) 0.25 m (D) 0.35 m
5.
A passenger train of length 60 m travels at a speed of 80 km/hr. Another freight train of length 120 m travels at a speed of 30 km/hr. The ratio of times taken by the passenger train to completely cross the freight train when: (i) they are moving in the same direction, and (ii) in the opposite directions is: 11 5 (A) (B) 5 2 3 25 (C) (D) 2 11
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JEE-MAIN-2019 (12th Jan-First Shift)-PCM-3 6.
Two electric bulbs, rated at (25 W, 220 V) and (100 W, 220 V), are connected in series across a 220 V voltage source. If the 25 W and 100 W bulbs draw powers P1 and P2 respectively, then: (A) P1 = 16 W, P2 = 4 W (B) P1 = 16 W, P2 = 9 W (C) P1 = 9 W, P2 = 16 W (D) P1 = 4 W, P2 = 16 W
7.
A straight rod of length L extends from x = a to x = L + a. The gravitational force it exerts on a point mass ‘m’ at x = 0, if the mass per unit length of the rod is A + Bx2, is given by: 1 1 1 1 (A) Gm A BL (B) Gm A BL aL a a a L 1 1 (C) Gm A BL aL a
1 1 (D) Gm A BL a a L
8.
What is the position and nature of image formed by lens combination shown in figure? (f1, f2 are focal lengths) (A) 70 cm from point B at left; virtual (B) 40 cm from point B at right; real 20 (C) cm from point B at right; real 3 (D) 70 cm from point B at right ; real
9.
A point source of light, S is placed at a distance L in front of the centre of plane mirror of width d which is hanging vertically on a wall. A man walks in front of the mirror along a line parallel to the mirror, at a distance 2 L as shown below. The distance over which the man can see the image of the light source in the mirror: (A) d (B) 2d d (C) 3d (D) 2
10.
A light wave is incident normally on a glass slab of refractive index 1.5. If 4% of light gets reflected and the amplitude of the electric field of the incident light is 30 V/m, then the amplitude of the electric field for the wave propagating in the glass medium will be: (A) 30 V/m (B) 10 V/m (C) 24 V/m (D) 6 V/m
11.
For the given cyclic process CAB as shown for a gas, the work done is: (A) 30 J (B) 10 J (C) 1 J (D) 5 J
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JEE-MAIN-2019 (12th Jan-First Shift)-PCM-4 12.
13.
14.
15.
Determine the electric dipole moment of the system of three charges, placed on the vertices of an equilateral triangle, as shown in the figure: ˆj ˆi ˆi ˆj (A) 3 q (B) q 2 2 (C) 2 q ˆj (D) 3 q ˆj The position vector of the centre of mass r cm of an asymmetric uniform bar of negligible area of cross-section as shown in figure is: 13 5 5 13 (A) r cm L xˆ L yˆ (B) r cm L xˆ L yˆ 8 8 8 8 3 11 11 3 (C) r cm L xˆ L yˆ (D) r cm L xˆ L yˆ 8 8 8 8 A person standing on an open ground hears the sound of a jet aeroplane, coming from north at an angle 60° with ground level. But he finds the aeroplane right vertically above his position. If is the speed of sound, speed of the plane is: 3 2 (A) (B) 2 3 (C) (D) 2 Two light identical springs of spring constant k are attached horizontally at the two ends of a uniform horizontal rod AB of length l and mass m. the rod is pivoted at its centre ‘O’ and can rotate freely in horizontal plane. The other ends of the two springs are fixed to rigid supports as shown in figure. The rod is gently pushed through a small angle and released. The frequency of resulting oscillation is: 1 3k 1 2k (A) (B) 2 m 2 m (C)
16.
1 6k 2 m
1 k 2 m
A simple pendulum, made of a string of length l and a bob of mass m, is released from a small angle 0. It strikes a block of mass M, kept on a horizontal surface at its lowest point of oscillations, elastically. It bounces back and goes up to an angle 1. Then M is given by: 1 m 1 (A) 0 (B) m 0 2 0 1 0 1 1 (C) m 0 0 1
17.
(D)
(D)
m 0 1 2 0 1
A 100 V carrier wave is made to vary between 160 V and 40 V by a modulating signal. What is the modulation index? (A) 0.3 (B) 0.5 (C) 0.6 (D) 0.4
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JEE-MAIN-2019 (12th Jan-First Shift)-PCM-5 18.
A cylinder of radius R is surrounded by a cylindrical shell of inner radius R and outer radius 2R. The thermal conductivity of the material of the inner cylinder is K1 and that of the outer cylinder is K2. Assuming no loss of heat, the effective thermal conductivity of the system for heat flowing along the length of the cylinder is: K K2 (A) 1 (B) K1 K 2 2 2K1 3K 2 K 3K 2 (C) (D) 1 5 4
19.
The least count of the main scale of a screw gauge is 1 mm. The minimum number of divisions on its circular scale required to measure 5 m diameter of a wire is: (A) 50 (B) 200 (C) 100 (D) 500
20.
The output of the given logic circuit is: (A) AB AB (B) AB AB (C) AB (D) AB
21.
As shown in the figure, two infinitely long, identical wires are bent by 90° and placed in such a way that the segments LP and QM are along the x-axis, while segments PS and QN are parallel to the y-axis. If OP = OQ = 4 cm, and the magnitude of the magnetic field at O is 104 T, and the two wires carry equal current (see figure), the magnitude of the current in each wire and the direction of the magnetic field at O will be (0 = 4 × 107 NA2): (A) 20 A, perpendicular out of the page (B) 40 A, perpendicular out of the page (C) 20 A, perpendicular into the page (D) 40 A, perpendicular into the page
22.
A particle of mass m moves in a circular orbit in a central potential field U(r)
23.
A proton and an -particle (with their masses in the ratio of 1 : 4 and charges in the ratio of 1:2 are accelerated from rest through a potential difference V. If a uniform magnetic field (B) is set up perpendicular to their velocities, the ratio of the radii rp : r of the circular paths described by them will be: (A) 1: 2 (B) 1 : 2 (C) 1 : 3 (D) 1: 3
1 2 kr . If 2 Bohr’s quantization conditions are applied, radii of possible orbits and energy levels vary with quantum number n as: 1 (A) rn n, En n (B) rn n, En n 1 (C) rn n, En n (D) rn n2 , En 2 n
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JEE-MAIN-2019 (12th Jan-First Shift)-PCM-6 24.
Let the moment of inertia of a hollow cylinder of length 30 cm (inner radius 10 cm and outer radius 20 cm), about its axis be I. The radius of a thin cylinder of the same mass such that its moment of inertia about its axis is also I, is: (A) 12 cm (B) 16 cm (C) 14 cm (D) 18 cm
25.
In the figure shown, a circuit contains two identical resistors with resistance R = 5 and an inductance with L = 2 mH. An ideal battery of 15 V is connected in the circuit. What will be the current through the battery long after the switch is closed? (A) 5.5 A (B) 7.5 A (C) 3 A (D) 6 A
26.
In the figure shown, after the switch ‘S’ is turned from position ‘A’ to position ‘B’, the energy dissipated in the circuit in terms of capacitance ‘C’ and total charge ‘Q’ is: 1 Q2 3 Q2 (A) (B) 8 C 8 C 2 5Q 3 Q2 (C) (D) 8 C 4 C
27.
A satellite of mass M is in a circular orbit of radius R about the centre of the earth. A meteorite of the same mass, falling towards the earth collides with the satellite completely inelastically. The speeds of the satellite and the meteorite are the same, just before the collision. The subsequent motion of the combined body will be: (A) such that it escapes to infinity (B) in an elliptical orbit (C) in the same circular orbit of radius R (D) in a circular orbit of a different radius
28.
The galvanometer deflection, when key K1 is closed but K2 is open, equals 0 (see figure). On closing K2 also and adjusting R2 to 5 , the deflection in galvanometer becomes 0 . The resistance of the galvanometer is, then, given by 5 [Neglect the internal resistance of battery]: (A) 5 (B) 22 (C) 25 (D) 12
29.
A particle A of mass ‘m’ and charge ‘q’ is accelerated by a potential difference of 50 V. Another particle B of mass ‘4 m’ and charge ‘q’ is accelerated by a potential difference of 2500 V. The ratio of de-Broglie wavelength A is close to: B (A) 10.00 (B) 0.07 (C) 14.14 (D) 4.47
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JEE-MAIN-2019 (12th Jan-First Shift)-PCM-7 30.
There is a uniform spherically symmetric surface charge density at a distance Ro from the origin. The charge distribution is initially at rest and starts expanding because of mutual repulsion. The figure that represents best the speed V(R(t)) of the distribution as a function of its instantaneous radius R(t) is: (A) (B)
(C)
(D)
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JEE-MAIN-2019 (12th Jan-First Shift)-PCM-8
PART –B (CHEMISTRY) 31.
In the Hall-heroult process, aluminium is formed at the cathode. The cathode is made out of (A) Pure Aluminium (B) Carbon (C) Copper (D) Platinum
32.
The correct order for acid strength of compounds: CH CH, CH3 – C CH, CH2 = CH2 is as follows: (A) CH CH > CH2 = CH2 > CH3 – C CH (B) CH3 – C CH > CH CH > CH2 = CH2 (C) CH3 – C CH > CH2 = CH2 > HC CH (D) HC CH > CH3 – C CH > CH2 = CH2
33.
K In a chemical reaction A 2B 2C D , the initial concentration of B was 1.5 times of A but the equilibrium concentrations of A and B were found to be equal. The equilibrium, constant(K) for the aforesaid chemical reaction is (A) 4 (B) 16 (C) 1/4 (D) 1
34.
Given Gas: H2 CH4 CO2 SO2 Critical: 33 190 304 630 Temp/K On the basis of data given, predict which of the following gases shows least adsorption on a definite amount of charcoal (A) SO2 (B) CH4 (C) CO2 (D) H2
35.
Mn2(CO)10 is an organometalic compound due to the presence of (A) Mn – C bond (B) Mn – Mn bond (C) Mn – O bond (D) C – O bond
36.
The major product of the following reaction is:
(A)
(B)
(C)
(D)
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JEE-MAIN-2019 (12th Jan-First Shift)-PCM-9 37.
A metal on combustion in excess air forms X, X upon hydrolysis with water yields H2O2 and O2 along with another product. The metal is: (A) Na (B) Rb (C) Mg (D) Li
38.
The molecules that has minimum/no role in the formation of photochemical smog is: (A) N2 (B) CH2 = O (C) O3 (D) NO
39.
The pair of metal ions that can give a spin-only magnetic moment of 3.9 BM for the complex [M(H2O)6]Cl2 is (A) V2+ and Co2+ (B) V2+ and Fe2+ 2+ 2+ (C) Co and Fe (D) Cr2+ and Mn2+
40.
For a disatomic ideal gas in a closed system, which of the following plots does not correctly describe the relation between various thermodynamic quantities?
(A)
(B)
(C)
(D)
41.
Among the following compounds most basic amino acid is (A) Asparagine (B) Lysine (C) Serine (D) Histidine
42.
The increasing order of reactivity of the following compounds towards reaction with alkyl halide directly is:
(a) (A) (b) < (a) < (c) < (d) (C) (b) < (a) < (d) < (c)
(c)
(b)
(d)
(B) (a) < (b) < (c) < (d) (D) (a) < (c) < (d) < (b)
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JEE-MAIN-2019 (12th Jan-First Shift)-PCM-10 43.
Which of the following has lowest freezing point?
(A)
(B)
(C)
(D)
44.
50 mL of 0.5 M oxalic acid is needed to neutralize 25 ml of sodium hydroxide solution. The amount of NaOH in 50 mL of the given sodium hydroxide solution is (A) 40 g (B) 10 g (C) 20 g (D) 80 g
45.
The volume of gas A has is twice than that of gas B. The compressibility factor of gas A is thrice that that of gas B at same temperature. The pressure of gases for equal per moles are (B) 2PA = 3PB (A) 3PA = 2PB (C) PA = 3 PB (D) PA = 2PB
46.
The hardness of water sample (in terms of equivalents of CaCO3) containing 10–3 M CaSO4 i s (molar mass of CaSO4 = 136 g mol–1) (A) 10 ppm (B) 50 ppm (C) 90 ppm (D) 100 ppm
47.
(A) CH3CH2COCH3 + PhMgX (C) PhCOCH3 + CH3CH2MgX
(B) PhCOCH2CH3 + CH3MgX (D) HCHO + PhCH(CH3)CH2MgX
48.
Freezing point of a 4% aqueous solution of X is equal to freezing point of 12% aqueous solution of Y. If molecular weight of X is A then molecular weight of Y is (A) 3A (B) 2A (C) A (D) 4A
49.
The metal d-orbital that can directly facing the ligands in K3[Co(CN)6] are (A) d xy and dx2 y2 (B) dx2 y2 and dz2 (C) dxz ,dyz and dz2
(D) dxy, dxz and dyz
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JEE-MAIN-2019 (12th Jan-First Shift)-PCM-11
50.
51.
Decomposition of X exhibits a rate constant for 0.05 g/year. How many years are required for the decomposition of 5 g of X into 2.5 g? (A) 50
(B) 25
(C) 20
(D) 40
dE dT The standard electrode potential E– and its temperature coefficient for a cell are 2 V and -5 10–4 VK–1 at 300 K respectively. The cell reaction is : 2 Zn s Cu2 aq Zn aq Cu s Standard reaction enthalpy (rH–)…….. (A) -412.8 (C) 1920
52.
(B) -384.0 (D) 206.4 kJ
The major product of the following reaction is
(A)
(B)
(C)
(D)
53.
The element with Z = 120(not yet discovered) will be an/a (A) Inner transition metal (B) Alkaline earth metal (C) Alkali metal (D) Transition metal
54.
In the following reaction HCl Aldehyde Alcohol Acetal Aldehyde Alcohol t HCHO BuOH CH3CHO MeOH The best combination is (A) CH3CHO and tBuOH (C) CH3CHO and MeOH
(B) HCHO and MeOH (D) HCHO and tBuOH
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JEE-MAIN-2019 (12th Jan-First Shift)-PCM-12 55.
Two solids dissociate as follows 2 A s B g C g ;Kp1 x atm 2 D s C g E g ;K p2 y atm
The total pressure when both the solids dissociate simultaneously is (A) x y atm (B) 2 x y atm
(C) (x + y) atm 56.
(D) x2 + y2 atm
In the following reactions, products A and B are:
(A)
(B)
(C)
(D)
o
57.
What is the work function of the metal if the light of wavelength 4000 A generates photoelectrons of velocity 6 105 ms–1 from it? (Mass of electron = 9 10–31 kg Velocity of light = 3 108 ms–1 Planck’s constant = 6.626 10–34 Js Charge of electron = 1.6 10–19 JeV–1) (A) 0.9 eV (B) 3.1 eV (C) 2.1 eV (D) 4.0 eV
58.
Iodine reacts with concentrated HNO3 to yield along with other products. The oxidation state of iodine in Y is (A) 5 (B) 7 (C) 3 (D) 1
59.
Poly - -hydroxybutyrate-co- -hydroxyvalerate(PHBV) is a copolymer of ________ (A) 3-hydroxybutanoic acid and 4-hydroxypentanoic acid (B) 2-hydroxybutanoic acid and 3-hydroxypentanoic acid (C) 3-hydroxybutanoic acid and 2-hydroxypentanoic acid (D) 3-hydroxy butanoic acid and 3-hydroxypentanoic acid
60.
Water sample with BOD vales of 4 ppm and 18 ppm respectively are (A) Clean and clean (B) Highly polluted and clean (C) Clean and highly polluted (D) Highly polluted and highly polluted. FIITJEE Ltd., FIITJEE House, 29-A, Kalu Sarai, Sarvapriya Vihar, New Delhi -110016, Ph 46106000, 26569493, Fax 26513942 website: www.fiitjee.com.
JEE-MAIN-2019 (12th Jan-First Shift)-PCM-13
PART–C (MATHEMATICS) 61.
An ordered pair (, ) for which the system of linear equations (1 + )x + y + z = 2 x + (1 + )y + z = 3 x + y + 2z = 2 has a unique solution, is: (A) (2, 4) (B) (3, 1) (C) (4, 2) (D) (1, 3)
62.
The product of three consecutive terms of a G.P. is 512. If 4 is added to each of the first and the second of these terms, the three terms now form an A.P. Then the sum of the original three terms of the given G.P. is: (A) 36 (B) 32 (C) 24 (D) 28
63.
The Boolean expression ((p q) (p q)) ( p q) is equivalent to: (A) p q (B) p (q) (C) (p) (q) (D) p (q)
64.
Consider three boxes, each containing 10 balls labelled 1, 2, ….., 10. Suppose one ball is randomly drawn from each of the boxes. Denote by ni, the label of the ball drawn from the ith box, (i = 1, 2, 3). Then, the number of ways in which the balls can be chosen such that n1 < n2 < n3 is: (A) 120 (B) 82 (C) 240 (D) 164
65.
A tetrahedron has vertices P(1, 2, 1), Q(2, 1, 3), R(1, 1, 2) and O(0, 0, 0). The angle between the faces OPQ and PQR is: 17 19 (A) cos1 (B) cos1 31 35 9 7 (C) cos1 (D) cos1 35 31
66.
The maximum area (in sq. units) of a rectangle having its base on the x-axis and its other two vertices on the parabola, y = 12 – x2 such that the rectangle lies inside the parabola, is: (A) 36 (B) 20 2 (C) 32 (D) 18 3
67.
If the straight line, 2x – 3y + 17 = 0 is perpendicular to the line passing through the points (7, 17) and (15, ), then equals: 35 (A) (B) 5 3 35 (C) (D) 5 3
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JEE-MAIN-2019 (12th Jan-First Shift)-PCM-14 68.
The sum of the distinct real values of ˆ ˆi ˆj k, ˆ ˆi ˆj kˆ are co-planar, is: ˆi ˆj k, (A) 1 (B) 0 (C) 1 (D) 2
69.
Let P(4, 4) and Q(9, 6) be two points on the parabola, y2 = 4x and let X be any point on the arc POQ of this parabola, where O is the vertex of this parabola, such that the area of PXQ is maximum. Then this maximum Area (in sq. units) is: 75 125 (A) (B) 2 4 625 125 (C) (D) 4 2 1 0 0 Let P 3 1 0 and Q = [qij] be two 3 × 3 matrices such that Q – P5 = I3. Then 9 3 1
70.
q21 q31 is equal to: q32 (A) 10 (C) 15
71.
,
for
which
the
vectors,
(B) 135 (D) 9
Let y = y(x) be the solution of the differential equation, x = loge 4 – 1, then y(e) is equal to: e (A) 2 e (C) 4
(B) (D)
dy y x loge x,(x 1). If 2y(2) dx
e2 2
e2 4
72.
The area (in sq. units) of the region bounded by the parabola, y = x2 + 2 and the lines, y = x + 1, x = 0 and x = 3, is: 15 21 (A) (B) 4 2 17 15 (C) (D) 4 2
73.
In a random experiment, a fair die is rolled until two fours are obtained in succession. The probability that the experiment will end in the fifth throw of the die is equal to: 200 150 (A) 5 (B) 5 6 6 225 175 (C) 5 (D) 5 6 6
74.
Let C1 and C2 be the centres of the circles x2 + y2 – 2x – 2y – 2 = 0 and x2 + y2–6x– 6y+14 =0 respectively. If P and Q are the points of intersection of these circles, then the area (in sq. units) of the quadrilateral PC1QC2 is: (A) 8 (B) 6 (C) 9 (D) 4 FIITJEE Ltd., FIITJEE House, 29-A, Kalu Sarai, Sarvapriya Vihar, New Delhi -110016, Ph 46106000, 26569493, Fax 26513942 website: www.fiitjee.com.
JEE-MAIN-2019 (12th Jan-First Shift)-PCM-15
75.
The maximum value of 3 cos 5 sin for any real value of is: 6 79 (A) 19 (B) 2 (C) 34 (D) 31
76.
Considering only the principal values of inverse functions, the set A x 0 : tan 1 2x tan1 3x 4 (A) contains two elements (B) contains more than two elements (C) is a singleton (D) is an empty set
77.
If be the ratio of the roots of the quadratic equation in x, 3m2x2 + m(m – 4)x + 2 = 0, 1 then the least value of m for which 1, is: (A) 2 3 (B) 4 3 2 (C) 2 2
(D) 4 2 3
78.
If a variable line, 3x + 4y – = 0 is such that the two circles x2 + y2 – 2x – 2y + 1 = 0 and x2 + y2 – 18x – 2y + 78 = 0 are on its opposite sides, then the set of all values of is the interval: (A) (2, 17) (B) [13, 23] (C) [12, 21] (D) (23, 31)
79.
For x > 1, if (2x)2y = 4e2x – 2y, then 1 loge 2x x loge 2x loge 2 x x loge 2x loge 2 (C) x
(A)
80.
dy is equal to: dx
(B) loge 2x (D) x loge 2x
The integral cos loge x dx is equal to: (where C is a constant of integration) x sin loge x cos loge x C 2 x (C) cos loge x sin loge x C 2
(A)
81.
2
(B) x cos loge x sin loge x C (D) x cos loge x sin loge x C
A ratio of the 5th term from the beginning to the 5th term from the end in the binomial 1 1 expansion of 2 3 is: 1 3 2 3 1
1
(A) 1: 2 6 3
(B) 1: 4 16 3
1
(C) 4 36 3 : 1
1
(D) 2 36 3 : 1
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JEE-MAIN-2019 (12th Jan-First Shift)-PCM-16 1 2 3 ..... k 5 2 . If S12 S 22 ..... S10 A, then A equal to: k 12 (A) 283 (B) 301 (C) 303 (D) 156
82.
Let Sk
83.
The perpendicular distance from the origin to the plane containing the two lines, x2 y2 z5 x 1 y 4 z 4 and , is: 3 5 7 1 4 7 (A) 11 6 (B) 11 6 (D) 6 11
(C) 11 84.
If the sum of the deviations of 50 observations from 30 is 50, then the mean of these observations is: (A) 30 (B) 51 (C) 50 (D) 31
85.
Let S be the set of all points in (, ) at which the function, f(x) = min {sinx, cosx} is non-differentiable. Then S is a subset of which of the following? 3 3 (A) ,0, (B) , , , 4 4 4 4 4 4 3 3 (C) , , , (D) , , , 2 2 4 2 4 4 2 4
86.
Let f and g be continuous functions on [0, a] such that f(x) = f(a – x) and g(x) + g(a – x) = a
4, then
f(x)g(x)dx is equal to: 0
a
a
(A) 4 f(x)dx
(B)
(C) 2 f(x)dx
(D) 3 f(x)dx
0 a
0
a
0
87.
0
cot3 x tan x is: x 4 cos x 4 lim
(A) 4 (C) 8 2 88.
f(x)dx
(B) 4 2 (D) 8
z ( R) is a purely imaginary number and |z| = 2, then a value of is: z (A) 2 (B) 1 1 (C) (D) 2 2
If
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JEE-MAIN-2019 (12th Jan-First Shift)-PCM-17 89.
Let S = {1, 2, 3, ….., 100}. The number of non-empty subsets A of S such that the product of elements in A is even is: (A) 2100 – 1 (B) 250 (250 – 1) (C) 250 – 1 (D) 250 + 1
90.
If the vertices of a hyperbola be at (2, 0) and (2, 0) and one of its foci be at (3, 0), then which one of the following points does not lie on this hyperbola? (A) 6,2 10 (B) 2 6,5
(C) 4,
15
(D) 6,5 2
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JEE-MAIN-2019 (12th Jan-First Shift)-PCM-18
JEE (Main) – 2019 ANSWERS PART A – PHYSICS 1.
Insufficient information
2.
C
3.
C
4.
C
5.
A
6.
A
7.
D
8.
D
9.
C
10.
C
11.
B
12.
D
13.
A
14.
D
15.
C
16.
C
17.
C
18.
D
19.
B
20.
C
21.
C
22.
A
23.
A
24.
B
25.
D
26.
B
27.
B
28.
B
29.
C
30.
C
PART B – CHEMISTRY 31.
B
32.
D
33.
A
34.
A
35.
A
36.
C
37.
B
38.
D
39.
A
40.
A
41.
B
42.
A
43.
D
44.
Bonus
45.
B
46.
D
47.
A
48.
A
49.
B
50.
A
51.
A
52.
A
53.
B
54.
B
55.
C
56.
A
57.
C
58.
A
59.
D
60.
C
PART C – MATHEMATICS 61.
A
62.
D
63.
C
64.
A
65.
B
66.
C
67.
D
68.
A
69.
B
70.
A
71.
C
72.
D
73.
D
74.
D
75.
A
76.
C
77.
B
78.
C
79.
A
80.
C
81.
C
82.
C
83.
B
84.
D
85.
B
86.
C
87.
D
88.
A
89.
B
90.
D
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JEE-MAIN-2019 (12th Jan-First Shift)-PCM-19
HINTS AND SOLUTIONS PART A – PHYSICS 1.
Cannot determine, degree of freedom must be given.
2.
y = 10–3 sin (50t + 2x) Wave is travelling along negative x-axis 50 Wave speed = = 25 m/s. k 2
3. 4V
i
R
4 5R
VAB = i(5) = i
VAB 20 0.1 2 (0.1) L 5 R 1 5R 2 Now, = 5 × 10–3 5R R = 395 VAP
P
0.1 V
A
4.
20 5R
5 1m
dR k d
B
k = constant
d R = 2k resistance of wire AB. R/ 2 L d Again, dR k L Length AP 0 0 R k2 L1/2 ; k = k2 L1/2 2 1 L m = 0.25 m 4
R
0
5.
dR k
1
0
0.06 km
PASSENGER TRAIN 30 km/hr
80 km/hr 0.12 km
FREIGHT TRAIN
Time taken if both moving in same direction t1
distance 0.12 0.06 0.18 speed 80 30 50
Time taken if moving in opposite direction. distance 0.12 0.06 0.18 t2 = speed 80 30 110 t1 11 t2 5 FIITJEE Ltd., FIITJEE House, 29-A, Kalu Sarai, Sarvapriya Vihar, New Delhi -110016, Ph 46106000, 26569493, Fax 26513942 website: www.fiitjee.com.
JEE-MAIN-2019 (12th Jan-First Shift)-PCM-20
6.
V2 P (220)2 R1 1936 25 (220)2 R2 484 100 220 1 i A R1 R 2 11 Power dissipated through R1 = P1 = i2R1 = 16 W Power dissipated through R2 = P2 = i2R2 = 4 W
Resistance, R
7.
x=a x=0
x
R1
R2
i 220 V
L
dx
Mass of element = dm = (A + Bx2)dx Field due to element at x = 0 G(dm) GA dE 2 GB dx 2 x x Total field a L 1 aL E GA dx GB dx 2 a a x 1 1 = G A BL a aL So, force = mE 1 1 = Gm A BL a aL 8.
Image by convex lens:
1 1 1 v u f
;
1 1 1 v 20 5
20 cm 3 Image by concave lens: v
20 14 u 2 cm 3 3
1 1 1 ; v u f v = 70 cm
1 3 1 v 14 5
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JEE-MAIN-2019 (12th Jan-First Shift)-PCM-21 9.
h = d
d
d
L
S
d/2 L
d d d = 3d. 2 2
h
d/2 d
10.
1 0E02C 2 1 0.96 I = E02 V 2 I
…(i) …(ii) 2
E v 0.96 0 E0 0 C 2
E 1 0.96 0 E0 0 1.5 2
E 1 0.96 0 r …(iii) 1.5 E0 1 C and v ; v o o r r r r
C ; r 1.5 v From equation (iii) r r
2
E 0.96 o (1.5)2 Eo Eo 24 V /m 11.
; r 1 for transparent medium.
1 1.5
Work done = Area of loop 1 = (4) (5) = 10 J 2
12.
P qL Pnet 2P cos 30o ( ˆj) = P 3 ( ˆj)
Y
X
P
30° 30°
=
3 qL ( ˆj)
P
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JEE-MAIN-2019 (12th Jan-First Shift)-PCM-22 13.
Three parts of rod can be considered as point masses. Y
2m (L. L)
m
L 2L, 2
m 5L
2m r1 m r2 m r3 rcm 4m rcm
X
2 , 0
ˆ m 2L ˆi L ˆj m 5L ˆi 2m(L ˆi Lj) 2 2 4m
13 ˆ 5 ˆ L i Lj 2 = 2 4 13 ˆ 5 ˆ = Li Lj 8 8
14.
B
A 60°
60°
15.
kx
kx
VP Speed of plane V Speed of sound V cos 60° = VP V VP 2
C
Torque on rod at displacement from mean position is very small. L x 2 L L2 kL2 2kx 2k 2 4 2 Now, I
6k kL2 mL2 ; 2 12 m 1 6k 2 2 m
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JEE-MAIN-2019 (12th Jan-First Shift)-PCM-23 16.
Just before collision speed of m v 2gL(1 cos o )
Just after collision speed of M v1 2gL(1 cos 1 )
Mm And v1 v ; M m
v1 M m v Mm
1 cos 1 M m 1 cos o M m
Sin ( 1 / 2) M m Sin (0 / 2) M m 1 M m o M m M1 + m1 = Mo – mo o M m 1 o 1 17.
AC = 100 AC + Am = 160 AC – Am = 40 AC = 100, Am = 60 A m 0.6 AC
18.
Equivalent thermal resistance,
1 cos 2 2 sin2
1 1 1 R R1 R2 2 2 k(2R)2 k1R2 k 2 (2R) R L L L
19.
4k = k1 + 3k2 k 3k 2 k 1 4
Least count = 5 10 6
Pitch No. of divisions on circular scale
103 N
N = 200
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JEE-MAIN-2019 (12th Jan-First Shift)-PCM-24
20.
Y A.AB AB.B
A. A B AB .B AB 0
21.
i B 2 o (cos 1 cos 2 ) 4nd B = 10–4
1 = 90°
o = 4 × 10 d = 4 × 10–2
22.
U
–7
2 = 180°
i = 20 A (into the page)
1 2 kr 2
Force, F
dU kr dr
For circular motion
mv 2 kr r
…(i)
nh 2 nh r2 2 km mvr
And
…(ii)
r n Total energy, E = k + U 1 1 = mv 2 kr 2 2 2
1 2 1 2 kr kr 2 2 2 E = kr En =
23.
mP = m qp = q m = 4 m q = 2q Radius of circular path, r
[From equation (i)]
kp = qV = k k = 2qV = 2k
mv 2km qB qB
rP 1 r 2
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JEE-MAIN-2019 (12th Jan-First Shift)-PCM-25 24.
r = 10 cm R = 20 cm
R
mass per unit area m (R 2 r 2 )
Mass = m r
Consider an element of radius x and thickness dx
x
r R
Mass of element dm = 2 x(dx) Moment of inertia of element, dI = (dm)x2
R
I = 2 x3 dx r
2 4 4 (R r ) 4 m = (R4 – r4) 2 2 (R r ) 2 m 2 2 I = (R r ) …(i) 2 Moment of inertia of thin cylinder of same mass, I = mro2 …(ii) =
25.
m 2 2 (R r ) 2 ro2 = 250 ; ro 16 cm. mro2
After long time, inductor will behave like resistance less wire, 15 15 i Req (5 / 2)
i 5 15 V
=6A
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5
JEE-MAIN-2019 (12th Jan-First Shift)-PCM-26 26.
Before switch is shifted: Energy stored, 1 Ui CE2 2 and charge stored, Q = CE
E
C
After switch is shifted: N C
4C
3C
M
Q CE E 4C 4C 4 2 1 1 E Energy stored, Uf (4C) CE 2 2 8 4 3 Energy dissipated = Ui – Uf = CE2 8 VM VN
27.
v
vy v
m
m
After Collision
vx
2m
2mvx = mv
v 2 v vy 2 vx
v 2 So, path will be elliptical. v net v x 2 v y 2
28.
When K1 closed and K2 is open
i
V 220 R
220
R
i
G
…(i)
When k1 and k2 are closed. KVL i iR i 25 5 220 5 R V i iR i 25 5 220 5 R i(220 R) R = 22
V
5 220
iR 25
iR i 25 5
G i/5
From eq. (i)
R
V
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JEE-MAIN-2019 (12th Jan-First Shift)-PCM-27
29.
P
30.
h 2mqV 4 1 2500 10 2 14.14 1 1 50
Ui + Ki = Uf + Kf kQ 2 kQ 2 1 O mv 2 2R0 2R 2 V
kQ2 m
1 1 R0 R
PART B – CHEMISTRY 31.
Cathode is made up of carbon.
32.
Acidic strength Stability of conjugate base E.N. sp carbon > sp2 carbon > sp3 carbon
33.
A 2 B 2C D Initially conc. a 1.5a at eq. a–x 1.5(a–2x) 2x x at equilibrium a–x = 1.5a – 2x 0.5a = x a = 2x 2 2x x 4x 2 .x kC 4 2 2 a x 1.5a 2x x x
34.
Amount of gas adsorbed TC
35.
Organometalic compound contains at least one chemical bond between carbon and a metal.
36.
CN
CHO O i DBAL H E Cb
OH
1 O ii H3 O
CHO
Nitriles are selectively reduced by DIBAL–H to imines followed by hydrolysis to aldehydes similarly, esters are also reduced to aldehyde with DIBAL–H. 37.
Rb form super oxides on reaction with excess air Rb + O2 RbO2 2RbO2 + H2O 2RbOH + H2O2 + O2
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JEE-MAIN-2019 (12th Jan-First Shift)-PCM-28 38.
Photochemical smog is produced when ultraviolet light from sun reacts with oxides of nitrogen in atmosphere.
39.
H2O is weak field ligand and magnetic moment = Moment n n 2 BM Solving for ‘n’ we get it as 3; i.e., no. of unpaired electron = 3 Fe2+ = t2g4eg2 Co2+ = t2g5eg2 V2+ = t2g3eg0 M is V, Co
40.
CP does not changes with change in pressure
41.
Compound Histidine Serine Lysine Asparagine
42.
Lone pair over nitrogen will act as the reacting centre. So, more nucleophilic nitrogen will be more reactive towards alkyl halide.
43.
Stronger intermolecular forces result in higher melting point.
44.
Eq. of (COOH)2 = Eq. of NaOH 50 0.5 2 = 25 M 1 50 2 Mass of NaOH in 50 mL = 40 4 g 1000
45.
Z = PV/nRT ZnRT P V
Pl value 7.6 5.7 9.8 5.4
Z V PA Z A VB 3 1 3 PB ZB VA 1 2 2 2PA = 3PB at constant T and mol P
46.
As 1L solution have 10–3 mol CaSO4 Eq. of CaSO4 = eq. of CaCO3 In 1L solution nCaSO4 v.f. nCaCO3 v.f. 10 3 2 nCaCO3 2 nCaCO3 10 3 mol in 1 L
w CaCO3 100 10 3 g in 1 L solution hardness in terms of CaCO3 w CaCO3 100 10 3 g = 106 106 = 100 ppm w Total 1000g FIITJEE Ltd., FIITJEE House, 29-A, Kalu Sarai, Sarvapriya Vihar, New Delhi -110016, Ph 46106000, 26569493, Fax 26513942 website: www.fiitjee.com.
JEE-MAIN-2019 (12th Jan-First Shift)-PCM-29 47.
The reaction in option (A) will yield 1o-alcohol.
48.
Tf X Tf Y k f m X k tm Y 4 1000 12 1000 A 96 M 88 M = 3.27 A = 3A
49.
Given K3[Co(CN)6] is inner orbital complex with hybridization d2sp3 and octahedral geometry. Ligands are approaching metal along the axes. Hence, dx2 y2 , dz2 orbitals are directly in front of the ligands.
50.
According to unit of rate constant it is a zero order reaction then half life of reaction will be C 5g t1/2 0 = 50 years 2k 2 0.05g / year
51.
G nFEcell 2 96500 2 386 kJ dE 4 o S nF 2 96500 5 10 J / C 96.5 kJ dT at 298 K TS = 298 (–96.5 J) = – 28.8 kJ at constant T (=248 K) and pressure G = H – TS H = G + TS = –386 – 28.8 = -412.8 kJ
Cl
52. H3 CO
H3 CO Cl / CCl
2 4
Cl Cl Anhyd. AlCl
3
H3 CO
H3 CO
Cl
53.
[Og118] 8s2 is configuration for Z = 120, where Og is oganesson As per the configuration it is in IInd group.
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JEE-MAIN-2019 (12th Jan-First Shift)-PCM-30 1 Steric Crowding
54.
Rate
55.
A s B g C g P1
P1 P2
D s C g E g P1 P2
P2
k P1 x atm2 k P2 y atm2
k P1 P1 P1 P2 k P2 P2 P1 P2
kP1 kP2 P1 P2
2
x + y (P1 + P2)2 P1 P2 x y 2 P1 P2 x y
PTotal PB PC PE = 2 P1 P2 2 x y 56.
57.
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JEE-MAIN-2019 (12th Jan-First Shift)-PCM-31 58.
2 + 10HNO3 2 HO3 + 10NO2 + 4H2O Iodine in HIO3 has +5 oxidation state.
59.
PHBV is obtained by copolymerization of 3-Hydroxybutanioic acid & 3-Hydroxypentanoic acid.
60.
Clean water has BOD value of less than 5 ppm and highly polluted water has a BOD value of 17 ppm or more.
PART C – MATHEMATICS 61.
1 x y z 0 x 1 y z 0 x y 2z 0 1 1 D 1 1 2
C1 C1 C2 C3 1
1
D 2 1 1 1 1
2
R2 R 2 R1, R 3 R3 R1 1 1 D 2 0 1 0 2 0 0 1
62.
63.
For unique solution 2 0 2 a Let the numbers be ,a,ar r 3 Given a 512 a 8 8 Now given 4, 12,8r are in A.P. r 2 2r 5r 2 0 1 r or 2 2 Numbers are 4, 8, 16 or 16, 8, 4 Sum of numbers = 4 + 8 + 16 = 28
p q p p q ~ q ~ p q p p ~ q q ~ q ~ p q p p ~ q ~ p q p ~ q ~ p q ~ p ~ q
64.
10
C3 120
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JEE-MAIN-2019 (12th Jan-First Shift)-PCM-32 65.
Vector perpendicular to face ˆi ˆj kˆ
O
OAB 1 2 1 5iˆ ˆj 3kˆ 2 1 3
Vector perpendicular to face ˆi ˆj kˆ ABC 2
1
1 1
1 ˆi 5ˆj 3kˆ 2
Angle between two faces 5 5 9 19 cos 35 35 35
A (1, 2, 1)
B (2, 1, 3)
19 cos1 35 66.
A 2 12 2
C (–1, 1, 2)
(0, 12)
dA 0 2 12 3 2 0 2 d A 4 12 4 32
,12
,12
2
2
,0 67.
,0
Line perpendicular to 2x 3y 5 0 is 3x 2y c 0 Which is satisfied by point (7, 17) 3 7 2 17 c 0 c 55 equation of line is 3x 2y 55 0 3 15 2 55 0 2 55 45 5
1 1
68.
1 1 1
D 1 1 , R1 R2 R3 2 1 1 , 1 1
1 1 1
0
0 2
C3 C3 C1 and C2 C2 C1, 2 1 1 0 2 1 0 1 0 1 Hence sum of distinct value of 2 1 1
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JEE-MAIN-2019 (12th Jan-First Shift)-PCM-33
69.
m
PQ
64 2 94
X
1 1 , X , 1 4 4 (Coordinates of points X for maximum area) 125 area sq. units. 4
Q (9, 6)
2yy ' 4 2y 2 4 y 1, x
70.
71.
1 0 0 1 0 0 3 P 6 1 0 P 9 1 0 . P5 27 6 1 54 9 1 0 0 1 0 0 2 1 Q 15 1 0 0 1 0 15 135 15 1 0 0 1 135 q q31 15 135 10 So 21 q32 15 2
P (4, –4)
0 0 1 15 1 0 135 15 1 0 0 2 0 15 2
dy 1 y log x dx x dx
I.F. e x x yx xn x dx x2 x dx 2 2 2 2 x x xy n C 2 4 Putting x 2 n 4 1 1 c n2 2 2 2 c 0 x x y n x 2 4 e e e y e 2 4 4 xy nx
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JEE-MAIN-2019 (12th Jan-First Shift)-PCM-34
72.
Required area 3
x
2
0
3
2 x 1 dx x 2 x 1 dx 0
15 2
(0, 11) y x2 2
(0, 2) (0, 1) (0, 0)
73.
(3, 0)
P _ _ _ 44 P 4 _ _ 44 P not4 _ _ 44
74.
1 5 5 1 1 5 5 1 1 25 25 175 1 5 4 5 6 6 6 6 6 6 6 6 6 6 6 6
Since circles are orthogonal and have equal radii therefore the quadrilateral PC1QC 2 is a square. Hence area 2 2 4 sq.units
P
2
2 C2
C1
2
2 Q
75.
5 sin 3cos 6 5 sin cos cos sin 3 cos 6 6
5 3 1 sin cos 2 2 2
76.
5 3 1 2 76 Maximum value is 19 2 2 4 tan1 2x tan1 3x 4 2x 3x Taking tangent on both side, we get 1 1 6x 2 6x 2 5x 1 0 x 1 6x 1 0
1 {–1 is rejected as it does not satisfies the given equation} 6 Hence number of element in S is one. x
77.
Let roots are and now 1 1 1 2 2
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JEE-MAIN-2019 (12th Jan-First Shift)-PCM-35 2
2 2 m m 4 3.
2 2 2 3m 3m m2 8m 2 0 m 43 2 So least value of m 4 3 2
78.
3x 4y 0
7 31 0 (since centres lie opposite side) 7,31 ……………(1) 7 31 1 and 5 5 7 5 and 31 2 or 12 and 21 or 41 1 2 3
2 10 …………….(2) …………….(3)
12, 21
79.
2yn 2x n4 2x 2y 2y 1 n2x n 4 2x n 2 x . 1 n 2x 2 y 2 1 n 2x n 2x
n 2x
80.
n2 xn 2x n 2 x x
By parts
x I x cos log x sin log x dx x I x cos log x sin log x dx
I x cos log x x sin log x cos log x dx c I
x cos log x sin log x c 2
1 C4 2 31/3 5 th term frombegining 5 th term from end 4/3 10 C4 2 2
4
10
81.
26/3 6 1 31/3
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JEE-MAIN-2019 (12th Jan-First Shift)-PCM-36
82.
22 223 4/ 3 4
4/3 6
2 2
3
1/3
32/3 .28/3 4. 36
2
k 1 2 2 10 5 k 1 A 12 k 1 2
Sk
5A 3 11 12 23 5A 1 6 3 3 505 A 5 A = 303 22 32 .......112
x2 y2 z5
83.
Equation of plane is
3
5
7
1
4
7
0
x 2 7 y 2 14 z 5 7 0 x 2y z 11 0
The perpendicular distance from the origin to the plane is
0 0 0 11 1 4 1
11 6
50
84.
Given
x
i
30 50
i 1
x i 30 50 50
85.
x
i
50
31
Hence number of points where f(x) is non – differentiable are 2 3 which are and 4 4
3 4 4
a
86.
a
I f x g x dx f a x g a x dx 0
b
0
b
f x dx f a b x dx a a a
a
a
f x 4 g x dx I 4 f x dx I I 2 f x dx 0
0
0
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JEE-MAIN-2019 (12th Jan-First Shift)-PCM-37 87.
Using LH rule 3cot 2 x cos ec 2 x sec 2 x im 8 x 4 sin x 4 z z z z
88.
zz z z 2 zz z z 2
2 z 2
2
89.
Product is even when atleast one element of subset is even Hence required number of subsets = total subsets – number of subsets all whose elements are odd 2100 250
90.
Equation of hyperbola is
x2 y2 1 and ae = 3 4 b2 We know that a 2e 2 a2 b 2 9 4 b2 b 2 5 x2 y2 Hence equation of hyperbola is 1 4 5 x2 y2 Hence 6, 5 2 does not lie on 1 4 5
(-3, 0)
(-2, 0)
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FIITJEE Solutions to JEE(Main)-2019 Test Date: 12th January 2019 (Second Shift)
PHYSICS, CHEMISTRY & MATHEMATICS Paper - 1 Time Allotted: 3 Hours
Maximum Marks: 360
Please read the instructions carefully. You are allotted 5 minutes specifically for this purpose.
Important Instructions: 1.
The test is of 3 hours duration.
2.
This Test Paper consists of 90 questions. The maximum marks are 360.
3.
There are three parts in the question paper A, B, C consisting of Physics, Chemistry and Mathematics having 30 questions in each part of equal weightage. Each question is allotted 4 (four) marks for correct response.
4.
Out of the four options given for each question, only one option is the correct answer.
5.
For each incorrect response 1 mark i.e. ¼ (one-fourth) marks of the total marks allotted to the question will be deducted from the total score. No deduction from the total score, however, will be made if no response is indicated for an item in the Answer Box.
6.
Candidates will be awarded marks as stated above in instruction No.3 for correct response of each question. One mark will be deducted for indicating incorrect response of each question. No deduction from the total score will be made if no response is indicated for an item in the answer box.
7.
There is only one correct response for each question. Marked up more than one response in any question will be treated as wrong response and marked up for wrong response will be deducted accordingly as per instruction 6 above.
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JEE-MAIN-2019 (12th Jan-Second Shift)-PCM-2
PART –A (PHYSICS) 1.
2.
The moment of inertia of a solid sphere, about an axis parallel to its diameter and at a distance of x from it, is ‘I(x)’. Which one of the graphs represents the variation of I(x) with x correctly?
(A)
(B)
(C)
(D)
A load of mass M kg is suspended from a steel wire of length 2 m and radius 1.0 mm in Searle’s apparatus experiment. The increase in length produced in the wire is 4.0 mm. Now the load is fully immersed in a liquid of relative density 2. The relative density of the material of load is 8. The new value of increase in length of the steel wire is: (A) 3.0 mm (B) 4.0 mm (C) 5.0 mm (D) zero
3.
3 3 F,R 20 ,L H and R1 10 . Current in L-R1 path is 2 10 1 and C-R2 path is 2. The voltage of A.C source is given by, V 200 2 sin(100 t) volts. The phase difference between 1 and 2 is : (A) 60° (B) 30° (C) 90° (D) 0°
In the above circuit, C
4.
An ideal gas is enclosed in a cylinder at pressure of 2 atm and temperature, 300 K. The mean time between two successive collisions is 6 10–8 s. If the pressure is doubled and temperature is increased to 500 K, the mean time between two successive collisions will be close to : (A) 2 107 s (B) 4 10 8 s (C) 0.5 10 8 s (D) 3 106 s
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JEE-MAIN-2019 (12th Jan-Second Shift)-PCM-3 5.
In the figure, given that VBB supply can vary from 0 to 5.0 V, VCC = 5 V, dc = 200, RB = 100, k, RC = 1 k an VBE = 1.0 V, The minimum base current and the input voltage at which the transistor will go to saturation, will be respectively : (A) 25 A and 3.5 V (B) 20 A and 3.5 V (C) 25 A and 2.8 V (D) 20 A and 2.8 V
6.
In the circuit shown, find C if the effective capacitance of the whole circuit is to be 0.5 F. All values in the circuit are in F. 7 6 (A) F (B) F 11 5 7 (C) 4 F (D) F 10
7.
An alpha-particle of mass m suffers 1-deminsional elastic collision with a nucleus at rest of unknown mass. It is scattered directly backwards losing, 64% of its initial kinetic energy. The mass of the nucleus is : (A) 2 m (B) 3.5 m (C) 1.5 m (D) 4 m
8.
A 10 m long horizontal wire extends from North East to South West. It is falling with a speed of 5.0 ms–1, at right angles to the horizontal component of the earth’s magnetic field, of 0.3 10–4 Wb/m2. The value of the induced emf in wire is : (A) 1.5 10 3 V (B) 1.1 10 3 V (C) 2.5 10 3 V (D) 0.3 10 3 V
9.
To double the covering range of a TV transition tower, its height should be multiplied by : 1 (A) (B) 2 2 (C) 4 (D) 2
10.
A plano-convex lens (focal length f 2, refractive index 2, radius of curvature R) fits exactly into a plano-concave lens (focal length f 1, refractive index 1, radius of curvature R). Their plane surfaces are parallel to each other. Then, the focal length of the combination will be : R (A) f1 f2 (B) 2 1 2f1f2 (C) (D) f1 f2 f1 f2
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JEE-MAIN-2019 (12th Jan-Second Shift)-PCM-4 11.
A vertical closed cylinder is separated into two parts by a frictionless piston of mass m and of negligible thickness. The piston is free to move along the length of the cylinder .The length of the cylinder above the piston is l1, and that below the piston is l2, such that l1 > l2. Each part of the cylinder contains n moles of an ideal gas at equal temperature T. If the piston is stationary, its mass, m, will be given by: (R is universal gas constant and g is the acceleration due to gravity) RT l1 3l2 RT 2l1 l2 (A) (B) ng l1l2 g l1l2 nRT 1 1 nRT l1 l2 (C) (D) ng l2 l1 g l1l2
12.
Two satellites, A and B, have masses m and 2m respectively. A is in a circular orbit of radius R, and B is in a circular orbit of radius 2R around the earth. The ratio of their kinetic energies, TA/TB, is : 1 (A) (B) 1 2 1 (C) 2 (D) 2
13.
A long cylindrical vessel is half filled with a liquid. When the vessel is rotated about its own vertical axis, the liquid rises up near the wall. If the radius of vessel is 5 cm and its rotational speed is 2 rotations per second, then the difference in the heights between the centre and the sides, in cm, will be : (A) 2.0 (B) 0.1 (C) 0.4 (D) 1.2
14.
A block kept on a rough inclined plane, as shown in the figure, remains at rest upto a maximum force 2 N down the inclined plane. The maximum external force up the inclined plane that does not move the block is 10 N. The coefficient of static friction between the block and the plane is : [Take g = 10 m/s2] 3 3 (A) (B) 2 4 1 2 (C) (D) 2 3
15.
In a Frank-hertz experiment, an electron of energy 5.6 eV passes through mercury vapour and emerges with an energy 0.7 eV. The minimum wavelength of photons emitted by mercury atoms is close to : (A) 1700 nm (B) 2020 nm (C) 220 nm (D) 250 nm
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JEE-MAIN-2019 (12th Jan-Second Shift)-PCM-5 16.
A particle of mass 20 g is released with an initial velocity 5 m/s along the curve from the point A, as shown in the figure. The point A is at height h from point B. The particle slides along the frictionless surface. When the particle reaches point B, its angular momentum about O will be : [Take g = 10 m/s2] (A) 2 kg-m2/s (B) 8 kg-m2/s 2 (C) 6 kg-m /s (D) 3 kg-m2/s
17.
A galvanometer, whose resistance is 50 ohm, has 25 divisions in it. When a current of 4 10–4 A passes through it, its needle (pointer) deflects by one division. To use this galvanometer as a voltmeter of range 2.5 V, it should be connected to a resistance of : (A) 250 ohm (B) 200 ohm (C) 6200 ohm (D) 6250 ohm
18.
A soap bubble, blown by a mechanical pump at the mouth of a tube, increases in volume, with time, at a constant rate. The graph that correctly depicts the time dependence of pressure inside the bubble is given by :
19.
(A)
(B)
(C)
(D)
In the given circuit diagram, the currents, 1 = –0.3 A, 4 = 0.8 A and 5 = 0.4 A, are flowing as shown. The currents 2, 3 and 6, respectively are :
(A) 1.1 A, –0.4 A, 0.4 A (C) 0.4 A, 1.1 A, 0.4 A
(B) 1.1 A, 0.4 A, 0.4 A (D) –0.4 A, 0.4 A, 1.1 A
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JEE-MAIN-2019 (12th Jan-Second Shift)-PCM-6 20.
A resonance tube is old and has jagged end. It is still used in the laboratory to determine velocity of sound in air. A tuning fork of frequency 512 Hz produces first resonance when the tube is filled with water to a mark 11 cm below a reference mark, near the open end of the tube. The experiment is repeated with another fork of frequency 256 Hz which produces first resonance when water reaches a mark 27 cm below the reference mark. The velocity of sound in air, obtained in the experiment, is close to : (A) 322 ms–1 (B) 341 ms–1 –1 (C) 335 ms (D) 328 ms–1
21.
A paramagnetic material has 1028 atoms/m3. Its magnetic susceptibility at temperature 350 K is 2.8 10–4. Its susceptibility at 300 K is : (A) 3.267 104 (B) 3.672 10 4 (C) 3.726 104 (D) 2.672 104
22.
In a radioactive decay chain, the initial nucleus is
232 90
Th . At the end there are 6 -
particles and 4 -particles with are emitted. If the end nucleus is by : (A) A = 208; Z = 80 (B) A = 202; Z = 80 (C) A = 208; Z = 82 (D) A = 200; Z = 81 23.
X, A and Z are given
Let l, r, c and v represent inductance, resistance, capacitance and voltage, respectively. 1 The dimension of in S units will be : rcv (A) [LA 2 ] (B) [A 1 ] (C) [LTA]
24.
A Z
(D) [LT 2 ]
Formation of real image using a biconvex lens is shown below :
If the whole set up is immersed in water without disturbing the object and the screen positions, what will one observe on the screen? (A) Image disappears (B) Magnified image (C) Erect real image (D) No change 25.
When a certain photosensitive surface is illuminated with monochromatic light of frequency v, the stopping potential for the photo current is –V0/2. When the surface is illuminated by monochromatic light of frequency v/2, the stopping potential is –V0. The threshold frequency for photoelectric emission is : 5v 4 (A) (B) v 3 3 3v (C) 2 v (D) 2
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JEE-MAIN-2019 (12th Jan-Second Shift)-PCM-7 26.
A simple harmonic motion is represented by : y 5(sin 3 t 3 cos 3 t) cm The amplitude and time period of the motion are : 2 3 (A) 10 cm, s (B) 10 cm, s 3 2 3 2 (C) 5 cm, s (D) 5 cm, s 2 3
27.
Two particles A, B are moving on two concentric circles of radii R1 and R2 with equal angular speed . At t = 0, their positions and direction of motion are shown in the figure :
The relative velocity v A vB at t is given by : 2 (A) (R1 R2 ) ˆi (B) (R1 R2 ) ˆi (C) (R R ) ˆi (D) (R R ) ˆi 2
1
1
2
28.
The mean intensity of radiation on the surface of the Sun is about 108 W/m2. The rms value of the corresponding magnetic field is closet to : (A) 1 T (B) 102 T –2 (C) 10 T (D) 10–4 T
29.
A parallel plate capacitor with plates of area 1 m2 each, are at a separation of 0.1 m. If the electric field between the plates is 100 N/C, the magnitude of charge on each plate is: (A) 7.85 10 10 C (B) 6.85 10 10 C (C) 8.85 10 10 C (D) 9.85 10 10 C
30.
The charge on a capacitor plate in a circuit, as a function of time, is shown in the figure :
What is the value of current at t = 4 s? (A) zero (C) 2 A
(B) 3 A (D) 1.5 A
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JEE-MAIN-2019 (12th Jan-Second Shift)-PCM-8
PART –B (CHEMISTRY) 31.
8 g of NaOH is dissolved in 18 g of H2O. Mole fraction of NaOH in solution and molality (in mol kg–1) of the solution respectively are : (A) 0.2, 22.20 (B) 0.2, 11.11 (C) 0.167, 11.11 (D) 0.167, 22.20
32.
The magnetic moment of an octahedral homoleptic Mn() complex is 5.9 BM. The suitable ligand for this complex is : (A) ethylenediamine (B) CN (C) NCS (D) CO
33.
The element that does NOT show catenation is : (A) Ge (B) Si (C) Sn (D) Pb
34.
Among the following, the false statement is : (A) It is possible to cause artificial rain by throwing electrified sand carrying charge opposite to the one on clouds from an aeroplane. (B) Tyndall effect can be used to distinguish between a colloidal solution and a true solution. (C) Lyophilic sol can be coagulated by adding an electrolyte. (D) Latex is a colloidal solution of rubber particles which are positively charged.
35.
The correct structure of histidine in a strongly acidic solution (pH = 2) is :
36.
(A)
(B)
(C)
(D)
The major product of the following reaction is:
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JEE-MAIN-2019 (12th Jan-Second Shift)-PCM-9
37.
(A)
(B)
(C)
(D)
m for NaCl, HCl and NaA are 126.4, 425.9 and 100.5 S cm2mol–1, respectively. If the conductivity of 0.001 M HA is 5 105 Scm1, degree of dissociation of HA is : (A) 0.50 (B) 0.25 (C) 0.125 (D) 0.75
38.
Molecules of benzoic acid (C6H5COOH) dimerise in benzene. ‘w’ g of the acid dissolved in 30 g of benzene shows a depression in freezing point equal to 2 K. If the percentage association of the acid to form dimer in the solution is 80, then w is : (Given that Kf = 5 K kg mol–1, Molar mass of benzoic acid = 122 g mol–1) (A) 2.4 g (B) 1.0 g (C) 1.5 g (D) 1.8 g
39.
Chlorine on reaction with hot and concentrated sodium hydroxide gives : (A) Cl and ClO3 (B) Cl and ClO (C) ClO 3 and ClO 2
40.
(D) Cl and ClO2
The major product of the following reaction is :
(A)
(B)
(C)
(D)
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JEE-MAIN-2019 (12th Jan-Second Shift)-PCM-10
41.
The combination of plots which do not represent isothermal expansion of an ideal gas is:
(a)
(c) (A) (b) and (d) (C) (b) and (c) 42.
43.
(b)
(d) (B) (a) and (c) (D) (a) and (d)
The major product in the following conversion is :
(A)
(B)
(C)
(D)
The major product of the following reaction is :
(A)
(B)
(C)
(D)
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JEE-MAIN-2019 (12th Jan-Second Shift)-PCM-11 44.
The correct order of atomic radii is : (A) Nd > Ce > Eu > Ho (C) Ce > Eu > Ho > Nd
(B) Ho > Nd > Eu > Ce (D) Eu > Ce > Nd > Ho
45.
The element that shows greater ability to form p p multiple bonds, is: (A) Sn (B) C (C) Ge (D) Si
46.
The two monomers for the synthesis of Nylon 6, 6 are : (A) HOOC(CH2)4COOH, H2N(CH2)6NH2 (B) HOOC(CH2)6COOH, H2N(CH2)6NH2 (C) HOOC(CH2)4COOH, H2N(CH2)4NH2 (D) HOOC(CH2)6COOH, H2N(CH2)4NH2
47.
The aldehydes which will not form Grignard product with one equivalent Grignard reagents are: (a)
(b)
(c)
(d)
(A) (b), (d) (C) (b), (c), (d)
(B) (b), (c) (D) (c), (d)
48.
Given : (i) C(graphite) O 2 (g) CO2 (g); rH x kJ mol1 1 (ii) C(graphite) O 2 (g) CO(g); rH y kJ mol1 2 1 (iii) CO(g) O 2 (g) CO2 (g); rH z kJ mol1 2 Based on the above thermochemical equations, find out which one of the following algebraic relationships is correct? (A) x = y + z (B) z = x + y (C) y = 2z – x (D) x = y – z
49.
The volume strength of 1M H2O2 is : (Molar mass of H2O2 = 34 g mol–1) (A) 5.6 (B) 16.8 (C) 11.35 (D) 22.4
50.
The correct statement(s) among I to III with respect to potassium ions that are abundant within the cell fluids is/are : I. They activate many enzymes II. They participate in the oxidation of glucose to produce ATP III. Along with sodium ions, they are responsible for the transmission of nerve signals (A) I and II only (B) I and III only (C) I, II and III (D) III only
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JEE-MAIN-2019 (12th Jan-Second Shift)-PCM-12 51.
The compound that is NOT a common component of photochemical smog is : (A) O3 (B) H3C C OONO 2 O
(C) 52.
CH2 CHCHO
(D)
CF2 Cl2
For a certain reaction consider the plot of nk versus 1/T given in the figure. If the rate constant of this reaction at 400 K is 10–5 s–1, then the rate constant at 500 K is:
(A) 10–6 s–1 (C) 10 4 s1
(B) 2 10 4 s1 (D) 4 10 4 s1
53.
An open vessel at 27°C is heated until two fifth of the air (assumed as an ideal gas) in it has escaped from the vessel. Assuming that the volume of the vessel remains constant, the temperature at which the vessel has been heated is : (A) 500°C (B) 500 K (C) 750°C (D) 750 K
54.
The major product of the following reaction is :
55.
(A)
(B)
(C)
(D)
The increasing order of the reactivity of the following with LiAIH4 is : (a)
(b)
(c)
(d)
(A) (b) < (a) < (c) < (d) (C) (a) < (b) < (d) < (c)
(B) (b) < (a) < (d) < (c) (D) (a) < (b) < (c) < (d)
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JEE-MAIN-2019 (12th Jan-Second Shift)-PCM-13 56.
The pair that does NOT require calcination is : (A) ZnO and MgO (B) ZnO and Fe2O3xH2O (C) ZnCO3 and CaO (D) Fe2O3 and CaCO3MgCO3
57.
The major product of the following reaction is :
58.
(A)
CH3CH C CH2
(B)
(C)
CH3CH CHCH2NH2
(D)
CH3CH2 CH
CH2
NH2 NH2
CH3 CH2C CH
If K sp of Ag2 CO3 is 8 1012 , the molar solubility of Ag2CO3 in 0.1 M AgNO3 is : (A) 8 1012 M (C) 8 1010 M
(B) 8 10 11M (D) 8 1013 M
59.
The upper stratosphere consisting of the ozone layer protects us from the sun’s radiation that falls in the wavelength region of : (A) 200 – 315 nm (B) 400 – 550 nm (C) 0.8 – 1.5 nm (D) 600 – 750 nm
60.
If the de Brogile wavelength of the electron in nth Bohr orbit in a hydrogenic atom is equal to 1.5a0 (a0 is Bohr radius), then the value of n/z is : (A) 0.40 (B) 1.50 (C) 1.0 (D) 0.75
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JEE-MAIN-2019 (12th Jan-Second Shift)-PCM-14
PART–C (MATHEMATICS) 61.
2 sin1 x
lim
1
(A)
2 2
(C) 62.
63.
64.
2
(D)
Let a, b, and c be three unit vectors, out of which vectors b and c are non-parallel. If and are the angles which vector a makes with vectors b and c respectively and 1 a b c b, then is equal to : 2 (A) 30° (B) 90° (C) 60° (D) 45°
The integral
(A)
(C)
66.
(B)
Let f be a differentiable function such that f(1) 2 and f '(x) f(x) for all xR. If h(x) f(f(x)), then h '(1) is equal to : (A) 2e2 (B) 4e (C) 2e (D) 4e2 2x x e x e The integral loge x dx is equal to : x 1 e 1 1 1 1 1 (A) e 2 (B) 2 2 e 2 e 2e 3 1 1 3 1 (C) 2 (D) e 2 2 e 2e 2 2e
65.
is equal to :
1 x
x 1
3x13 2x11
2x
4
3x 2 1
x4
6 2x 4 3x 2 1 x4
3
2x 4 3x 2 1
3
C
C
4
dx is equal to (where C is a constant of integration)
(B)
(D)
x12
6 2x 4 3x 2 1 x12
3
2x 4 3x 2 1
3
C
C
If sin4 4 cos4 2 4 2 sin cos ; , [0, ], then cos( ) is equal to : (A) 0 (B) –1 (C) 2 (D) 2
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JEE-MAIN-2019 (12th Jan-Second Shift)-PCM-15 67.
If a curve passes through the point (1, –2) and has slope of the tangent at any point (x, x 2 2y y) on it as , then the curve also passes through the point : x (A) (3, 0) (B) 3, 0
(D) 2, 1
(C) (–1, 2)
68.
If an angle between the line,
x 1 y 2 z 3 and the plane, x 2y kz 3 is 2 1 2
2 2 cos1 , then a value of k is : 3 5 3 3 (C) 5
(A)
69.
3 5 5 (D) 3
(B)
n n 1 n lim 2 2 2 2 is equal to 2 2 n 3 5n n 1 n 2 (A) (B) tan 1(3) 4 (C) (D) tan 1(2) 2 x
70.
In a game, a man wins Rs. 100 if he gets 5 or 6 on a throw of a fair die and loses Rs. 50 for getting any other number on the die. If he decides to throw the die either till he gets a five or a six or to a maximum of three throws, then his expected gain/loss (in rupees) is : 400 (A) loss (B) 0 9 400 400 (C) gain (D) loss 3 3
71.
If a straight line passing through the point P(–3, 4) is such that its intercepted portion between the coordinate axes is bisected at P, then its equation is : (A) 3x 4y 25 0 (B) 4x 3y 24 0 (C) x y 7 0 (D) 4x 3y 0
72.
There are m men and two women participating in a chess tournament. Each participant plays two games with every other participant. If the number of games played by the men between themselves exceeds the number of games played between the men and the women by 84, then the value of m is : (A) 12 (B) 11 (C) 9 (D) 7
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JEE-MAIN-2019 (12th Jan-Second Shift)-PCM-16 73.
The tangent to the curve y x 2 5x 5, parallel to the line 2y 4x 1, also passes through the point : 7 1 1 (A) , (B) , 7 2 4 8 1 1 7 (C) ,7 (D) , 8 4 2
74.
Let S be the set of all real values of such that a plane passing through the points ( 2 ,1,1), (1, 2 ,1) and (1,1, 2 ) also passes through the point (–1, –1, 1). Then S is equal to: (A) 3 (B) 3, 3
(C) 1, 1 75.
(D) 3, 3
Let z1 and z2 be two complex numbers satisfying z1 9 and z2 3 4i 4 . Then the minimum value of z1 z 2 is : (A) 0 (C) 1
76.
(B) 2 (D) 2
The total number or irrational terms in the binomial expansion of 7 (A) 55 (C) 48
1 5
3
60 1 10
is :
(B) 49 (D) 54
77.
The equation of a tangent to the parabola, x 2 8y, which makes an angle with the positive direction of x-axis, is : (A) y x tan 2 cot (B) y x tan 2cot (C) x y cot 2 tan (D) x y cot 2 tan
78.
In a class of 60 students, 40 opted for NCC, 30 opted for NSS and 20 opted for both NCC and NSS. If one of these students is selected at random, then the probability that the student selected has opted neither for NCC nor for NSS is : 1 1 (A) (B) 6 3 2 5 (C) (D) 3 6
79.
The mean and the variance of five observations are 4 and 5.20, respectively. If three of the observations are 3, 4 and 4; then the absolute value of the difference of the other two observations, is : (A) 7 (B) 5 (C) 1 (D) 3
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JEE-MAIN-2019 (12th Jan-Second Shift)-PCM-17
80.
81.
sin 1 1 3 5 If A sin 1 sin ; then for all , , det (A) lies in the interval : 4 4 1 sin 1 5 5 (A) 1, (B) , 4 2 2 3 3 (C) 0, (D) , 3 2 2
If a circle of radius R passes through the origin O and intersects the coordinate axes at A and B, then the locus of the foot of perpendicular from O on AB is : (A) (x 2 y 2 )2 4R 2 x 2 y 2 (B) (x 2 y 2 )3 4R 2 x 2 y 2 (C) (x 2 y 2 )2 4Rx 2 y 2
(D) (x 2 y 2 )(x y) R2 xy
82.
The set of all values of for which the system of linear equations x 2y 2z x x 2y z y x y z (A) is a singleton (B) contains exactly two elements (C) is an empty set (D) contains more than two elements
83.
If the function f given by f(x) x3 3(a 2)x 2 3ax 7, for some aR is increasing in (0, f(x) 14 1] and decreasing in [1, 5), then a root of the equation, 0 (x 1) is : (x 1)2 (A) –7 (B) 5 (C) 7 (D) 6
84.
Let Z be the set of integers. If 2 A {x Z : 2( x 2)( x 5 x 6) 1} and B {x Z : 3 2x 1 9}, then the number of subsets of the set A B, is : (A) 215 (B) 218 12 (C) 2 (D) 210
85.
If n C4 , nC5 and nC6 are in A.P., then n can be : (A) 9 (B) 14 (C) 11 (D) 12
86.
Let S and S be the foci of an ellipse and B be any one of the extremities of its minor axis. If SBS is a right angled triangle with right angle at B and area (SBS) = 8 sq. units, then the length of a latus rectum of the ellipse is : (A) 4 (B) 2 2 (C) 4 2 (D) 2
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JEE-MAIN-2019 (12th Jan-Second Shift)-PCM-18 87.
If the angle of elevation of a cloud from a point P which is 25 m above a lake be 30° and the angle of depression of reflection of the cloud in the lake from P be 60°, then the height of the cloud (in meters) from the surface of the lake is : (A) 60 (B) 50 (C) 45 (D) 42
88.
The expression ~(~ p q) is logically equivalent to : (A) ~ p ~ q (B) p ~ q (C) ~ p q (D) p q 3
89.
90.
3
3
3
3 1 1 3 If the sum of the first 15 terms of the series 1 2 33 3 ..... is 4 2 4 4 equal to 225 k, then k is equal to : (A) 108 (B) 27 (C) 54 (D) 9
The number of integral values of m for which the quadratic expression, (1 + 2m)x2 – 2(1 + 3m)x + 4(1 + m), xR, is always positive, is : (A) 3 (B) 8 (C) 7 (D) 6
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JEE-MAIN-2019 (12th Jan-Second Shift)-PCM-19
JEE (Main) – 2019 ANSWERS PART A – PHYSICS 1.
D
2.
A
3.
No option is correct
4.
B
5.
A
6.
A
7.
D
8.
B
9.
C
10.
B
11.
D
12.
B
13.
A
14.
A
15.
D
16.
C
17.
B
18.
No option is correct
19.
B
20.
D
21.
A
22.
C
23.
B
24.
A
25.
D
26.
A
27.
C
28.
D
29.
C
30.
A
PART B – CHEMISTRY 31.
C
32.
C
33.
D
34.
C
35.
C
36.
A
37.
C
38.
A
39.
A
40.
B
41.
A
42.
B
43.
B
44.
D
45.
B
46.
A
47.
A
48.
A
49.
C
50.
D
51.
A
52.
C
53.
C
54.
D
55.
C
56.
A
57.
D
58.
C
59.
A
60.
D
PART C – MATHEMATICS 61.
B
62.
B
63.
D
64.
A
65.
B
66.
D
67.
B
68.
A
69.
D
70.
B
71.
B
72.
A
73.
B
74.
B
75.
A
76.
D
77.
C
78.
A
79.
A
80.
D
81.
B
82.
A
83.
C
84.
A
85.
B
86.
A
87.
B
88.
A
89.
B
90.
C
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JEE-MAIN-2019 (12th Jan-Second Shift)-PCM-20
HINTS AND SOLUTIONS PART A – PHYSICS 1.
I = ICM + Mx2 2 ICM = MR 2 5 2 I MR2 Mx 2 5
2.
In first case: mg A Y 1 In second case:
1
mg B A Y 1
3 mg mg B 4 2 YA YA
3.
2
mg YA
3 1 = 3 mm 4
For the first branch: 1 3 10 6 100 2 XC 103 tan 1 R2 20
1 –90°. For the second branch:
tan 2
4.
XL 3 R
2 60°. Phase difference between current in branch 1 and 2 = 150°. No option is correct. P The mean time between two collision T
T t1 P1 2 t 2 P2 T1
6 10 8 3 2 5 t 2
t 2 6 10 8
1 5 4 10 8 sec . 2 3
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JEE-MAIN-2019 (12th Jan-Second Shift)-PCM-21 5.
6.
When switched on: VCE = 0 VCC – RCIC = 0 V iC CC 5 10 3 A RC
IC = BiB iB = 25 A VBB = iBRB – VBE = 0
VBB = VBE + iBRB = 3.5 V
4 1 3 C 7C 1 Ceq = 0.5 F 7 7 2 C C 3 3 14C 7 C 3 3 7 C 11
7.
vo
(I) (II)
M
m
mvo = –mv1 + mv2 vo = v 1 + v2
2mv o v2 mM
m v2
M
2
1 1 M m 2 36 1 KEf mv12 m mv 02 v0 2 2 Mm 100 2
v1
M = 4 m.
8.
Emf = Bv sin 45° = (0.3 10 4 ) (10) 5
1 2
= 1.1 × 10–3 V 9.
Covering range of transition power = 2hR To double the range make height 4 times.
10.
1 (1 1) (1 2 ) f R R 1
R
1 (1 2 ) f R
2
R
f
R 1 2
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JEE-MAIN-2019 (12th Jan-Second Shift)-PCM-22 11. 1
nT P1A
m P2A
2
12.
nT
P2A – P1A = mg 1 P A P A m 1 1 2 2 g 1 2 1 nRT nRT m g 1 2 nRT 1 1 m g 1 2
1 GM m 2 R 1 GM KEB = (2m) 2 2R KE A 1 KEB KE A
w 2 x 2 (2 2)2 (0.05)2 2g 2g –4 = 25 × 8 × 10 = 2 cm
13.
y
14.
mg sin + 2 = mg cos 10 – mg sin = mg cos On adding: (1) + (2) 3 12 12 = 2mg ; mg 2 3 On (1) – (2): 1 8 2 mg ; mg = 8 ; 2
…(1) …(2)
3 2
12410 eV A o 12410 eV A o 4.9 eV 250 nm
15.
5.6 eV 0.7 eV 4.9 eV
16.
v 52 2gh 52 2 10 10 225 = 15 m/s h rmv 20 (20 103 kg) (15) = 6 kg m2/sec
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JEE-MAIN-2019 (12th Jan-Second Shift)-PCM-23 17.
Vo = igo (RG R) igo = 4 × 10–4 × 25 = 10–2 A
Vo = 2.5 V V 2.5 Rg R o 2 = 250 igo 10 18.
R = 200 .
V = kt =
4 3 R 3 1/3
3k R t 4
4T R Bonus
Pin = Patm + 19.
I3 + I5 = I4 I1 + I2 = I4 I3 + I6 = I2 + I1
20.
1 4(11 e)
I3 = I4 – I5 = 0.4 A I2 = I4 – I1 = 1.1 A I6 = I2 + I1 – I3 = 0.4 A
v 512
2 4(27 e)
v 256
11 e 1 27 e 2 22 + 2e = 27 + e e=5 21.
For paramagnetic materials
1 R
1 T2 2 T1
T1 350 1 2.8 10 4 T2 300 = 3.267 × 10–4
2
22.
232 90
6 Th
208 78
4 Y
208 82
23.
L [A 1 ] RCV
24.
If the water is filled focal length will decrease and image will disappear.
X
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JEE-MAIN-2019 (12th Jan-Second Shift)-PCM-24 25.
eVs h eVo h 2 2 h eVo 2 h 0 = 2h 2 2 3h 3 th 2 2
26.
…(1) …(2)
y 5(sin 3 t 3 cos 3 t) cm y = 10 sin(3t + ) A = 10 cm 2 T = sec 3
27.
R1
A
A at t = 0
at t =
2
B
R2 B
v A v B R1ˆi R2 ˆi = (R R ) ˆi 2
28.
I
1
B2o C 2o
B2o I 2o C 103 2 4 10 7 3 108 Bo 10 4 T B2o
29.
30.
q A o q = EAo q = 100 × 1 × 8.85 × 10–12 q = 8.85 × 10–10 C E
Current = slope of q – t graph = 0. [at t = 4 sec]
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JEE-MAIN-2019 (12th Jan-Second Shift)-PCM-25
PART B – CHEMISTRY 8 0.2 40 18 Moles of H2O = 1 18 0.2 Mole fraction of NaOH = 0.167 1.2 8 1000 Molality = 11.11 40 18
31.
Moles of NaOH =
32.
Homoleptic complexes contain identical ligands, e.g., [Mn(NCS)6]4–.
33.
Due to the lowest bond energy of Pb – Pb bond.
34.
Lyophobic sols are coagulated by adding electrolysis.
35.
The COO– group absorbs H+ in acidic medium (pH = 2)
36.
37.
0m (HA) = 100.5 + 425.9 – 126.4 = 400 K 1000 5 10 5 103 50 M 103 50 0.125 400 m0
38.
2A A2 1 2 0.8 1 0.8 2 i = 1 – 0.8 +
0.8 0.6 2
Tf = Kf i m = 5 0.6
x 1000 2 Since Tf 2 122 30
x = 2.44 g
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JEE-MAIN-2019 (12th Jan-Second Shift)-PCM-26 39.
6NaOH Cl2 5 NaCl NaClO3 3H2O
40.
41.
The plot (b) and (d) are incorrect. The correct ones are given below:
P
U
0
Vm
(b)
0
Vm
(c)
42
43
44.
The atomic radii are Eu = 199 pm Ce = 183 pm Nd = 181 pm Ho = 176 pm
45.
It is due to the smallest atomic size of carbon in the given options.
46.
The monomers are hexamethylene diamine and adipic acid.
47.
Each of (b) and (d) will react two moles of Grignard reagent.
48.
(ii) + (iii) = (i) Y+z=x
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JEE-MAIN-2019 (12th Jan-Second Shift)-PCM-27 49.
Volume strength = 11.35 M = 11.35 (STP)
50.
Active transport proteins exchanges Na+ ions for K+ ions across the plasma membrane of animal cells.
51.
O3 is not common component of London and Los Angeles smog. It is present only in Los Angeles smog
52.
n nA
Ea 4606 nA RT T k Ea 500 400 n 5 10 R 500 400 1 k n 5 4606 2.303 n10 2000 10 k n 5 n10 10 k 10 4
53.
n1T1 = n2T2 2n n 300 = n T2 5 3 300 T2 T2 = 500 K 5
54.
55.
The order is identical to nucleophilic substitution order: Acid chloride > Acid anhydride > Acid > Amide
56.
In calcination the ore is converted to metal oxide. ZnO and MgO are already in oxide form.
57.
Br EtOH Br
NaNH
2 Br
H CH3 CH2 C CH Br
EtO
58.
8 10–12 = (2S’ + 0.1)2S’ or S’ = 8 10–10 M
59.
In ozone layer the wavelength of U.V radiation is 200 – 340 nm.
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JEE-MAIN-2019 (12th Jan-Second Shift)-PCM-28 60.
2r = n 2r 2n2 a0 n 2 a0 n n Z Z 1.5a0 n a0 1.5a0 Z n 1.5 0.75 Z 2 2
PART C – MATHEMATICS 61.
2 sin1 x
lim
1 x
x 1
2 sin1 x
2 sin1 x 2
lim
x 1
1 x
2sin1 x
1
lim
2 sin1 x
2cos x
x 1
.
1
1 x 2 Assuming x cos 2
lim
62.
.
1
2 sin 2 2
0
f ' x f x
2
1 x R
Integrate and use f 1 2 f x 2e x 1 f ' x 2e x 1
h x f f x h' x f ' f x f ' x h' 1 f ' f 1 f 1 f ' 2 f ' 1 2e.2 4e e
63.
2x
e
x e 1 e loge x.dx 1 x loge x.dx 2x
x
x e Let t, v e x 1 1 1 1 3 1 1 dt dv 1 2 1 e 2 e 2 1 2 2 e 2 2e e e
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JEE-MAIN-2019 (12th Jan-Second Shift)-PCM-29
64.
a.c b a.b .c 21 b
b and c are linearly independent 1 a.c and a.b 0 2 (All given vectors are unit vectors) a c 60o and a b 90o
65.
30o
3x13 2x11
2x
4
3x 2 1
4
dx
2 3 x 3 x5 dx 3 1 4 2 x2 x 4
3 1 Let 2 2 4 t x x 1 dt 1 x12 4 3 C C 3 2 t 6t 6 2x 4 3x 2 1
66.
A.M. G.M. 1 sin4 4 cos 4 1 1 sin4 . 4 cos 4 .1.1 4 4 4 sin 4 cos2 2 4 2 sin cos given that sin4 4 cos4 2 4 2 sin cos
A.M. G.M. sin4 1 4 cos4 1 sin 1,cos , As , 0, 2 1 sin 1, cos 2 1 sin as 0, 2 cos cos 2 sin sin 2
67.
dy x 2 2y (Given) dx x dy y 2 x dx x 2
dx I.F. e x x 2 y . x 2 x . x 2 dx C
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JEE-MAIN-2019 (12th Jan-Second Shift)-PCM-30
x4 C y
Hence b passes through 1, 2 C
9 4
x4 9 4 4 Checking options, only option (B) satisfies. yx 2
68.
Direction Ratio of line are 2, 1, –2 Normal vector of plane is ˆi 2ˆj kkˆ 2iˆ ˆj 2kˆ . ˆi 2ˆj kkˆ sin 3 1 4 k2 2k sin …………(1) 3 k2 5
cos
2 2 3
(Given)
…………(2)
Using sin2 cos2 1 k 2
5 3
2n
69.
n 2 r 1 n r 2n 1 lim Using D.I. as limit of sum, we get x r2 r 1 n 1 2 n lim
x
2
2
dx tan1 2 2 1 x 0
70.
2 1 Let w denotes probability that outcome 5 or 6 w 6 3 4 2 Let, L denotes probability that outcome 1, 2, 3, 4 L 6 3 Expected Gain/Loss w 100 Lw 50 100 L2 w 50 50 100 L3 150 2
3
1 2 1 2 1 2 100 . 50 0 150 0 3 3 3 3 3 3
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JEE-MAIN-2019 (12th Jan-Second Shift)-PCM-31
71.
72.
x y 1 a b a b 3, 4 , 2 2 a 6, b 8 equation of line is 4x 3y 24 0 Let the line be
B (0, b) (–2, 4)
A (a, 0)
Let m – men, 2 – women m C2 2 mC1 2C1 .2 84 m2 5m 84 0 m 12 m 7 0 m = 12
73.
y x 2 5x 5 dy 7 2x 5 2 x dx 2 7 1 at x , y 2 4 29 7 1 Equation of tangent at , is 2x y 0 4 2 4 Now check options 1 x , y7 8
74.
All four points are coplanar so 1 2 2 0 2
2 1
0
2
2
2
2
2
0
1
3 0
1
2
3 75.
z1 9, z2 3 4i 4 C1 0,0 radius r1 9 C2 3, 4 , radius r2 4 C1C2 r1 r2 5 Circle touches internally z1 z 2 min 0 60 r
76.
r
General term Tr 1 60 Cr ,7 5 3 10 for rational term, r = 0, 10, 20, 30, 40, 50, 60 number of rational terms = 7 number of irrational terms = 54 FIITJEE Ltd., FIITJEE House, 29-A, Kalu Sarai, Sarvapriya Vihar, New Delhi -110016, Ph 46106000, 26569493, Fax 26513942 website: www.fiitjee.com.
JEE-MAIN-2019 (12th Jan-Second Shift)-PCM-32
x 2 8y dy x tan dx 4 x1 4 tan
77.
y1 2 tan2 Equation of tangent :y 2 tan2 tan x 4 tan x y cot 2 tan 78.
A opted NCC B opted NSS
P (neither A nor B)
79.
B
A
10 1 z 60 6
20
20
20
Mean x 4, 2 5.2,n 5, x1 3 x 2 4 x 3
x
i
20
x 4 x5 9 2 i
x x x
……….(i) 2
2 xi2 106
x 24 x 52 65
……….(ii) 2
Using (i) and (ii) x 4 x5 49 x4 x5 7
1
80.
sin
1
1
sin
sin
1
A sin 1
2 1 sin2
1 1 3 5 , sin 2 2 4 4 1 0 sin2 2 A 2, 3
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JEE-MAIN-2019 (12th Jan-Second Shift)-PCM-33
h k Equation of AB is hx ky h2 k 2 Slope of AB
81.
P (h, k)
h2 k 2 h2 k 2 A , 0 ,B 0, k h As, AB 2R
x
2
1
82.
1
3
y
h2 k 2
2
A
4R2h2k 2
2 3
B
4R 2 x 2 y 2
2
2 1 0
1
1
1
3
1 0 1 83.
f ' x 3x 2 6 a 2 x 3a f ' x 0 x 0, 1 f ' x 0 x 1, 5
f ' x 0 at x 1 a 5 2
f x 14 x 1 x 7 f x 14
x 1
2
x7
Hence root of equation
84.
x 2 x2 5x 6
A x z:2 x 2 x 2 5x 6
2
f x 14
x 1
2
0 is 7.
1
20 x 2,2,3
A 2,2,3 B x Z : 3 2x 1 9 B 0,1,2,3,4 Hence, A B has is 15 elements. So number of subsets of A B is 215 .
85.
2. nC5 nC4 nC6
n n n 5n5 4 n4 6n6 2 1 1 1 . 5 n 5 n 4 n 5 30
2.
n = 14 satisfying equation.
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JEE-MAIN-2019 (12th Jan-Second Shift)-PCM-34 86.
mSB .mS'B 1 b 2 a2 e 2 …………(i) 1 S 'B.SB 8 2 a 2e 2 b 2 16 …………..(ii) 2 2 b a 1 e2
B
S’
…………..(iii) using (i), (ii), (iii) a = 4 b2 2 1 e 2 2b 2 L.R 4 a 87.
x y 3x y 25 x 25 tan 60o y tan 30o
S (a, e 0)
(–a, e 0)
Cloud
………..(i)
3y 50 x 3x 50 x x 25m Height of cloud from surface 25 25 50m
30o
x
60o 25 m
25 m
Surface
88. p
q
~p
~p q
~ ~ p q
~ p ~ q
T F T F
T T F F
F T F T
T T T F
F F F T
F F F T
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JEE-MAIN-2019 (12th Jan-Second Shift)-PCM-35 3
89.
3
3
3
3 6 9 12 S ..........15 term 4 4 4 4 27 15 3 r 64 r 1
27 15 15 1 . 64 2
2
= 225 K (Given in question) K 27 90.
Expression is always positive it 2m 1 0 m
1 2
and D 0 m2 6m 3 0 …….(iii) 3 12 m 3 12 Common interval is 3 12 m 3 12 Integral value of m 0,1,2,3, 4,5,6
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FIITJEE Solutions to JEE(Main)-2019 Test Date: 8th April 2019 (First Shift)
PHYSICS, CHEMISTRY & MATHEMATICS Paper - 1 Time Allotted: 3 Hours
Maximum Marks: 360
Please read the instructions carefully. You are allotted 5 minutes specifically for this purpose.
Important Instructions: 1.
The test is of 3 hours duration.
2.
This Test Paper consists of 90 questions. The maximum marks are 360.
3.
There are three parts in the question paper A, B, C consisting of Physics, Chemistry and Mathematics having 30 questions in each part of equal weightage. Each question is allotted 4 (four) marks for correct response.
4.
Out of the four options given for each question, only one option is the correct answer.
5.
For each incorrect response 1 mark i.e. ¼ (one-fourth) marks of the total marks allotted to the question will be deducted from the total score. No deduction from the total score, however, will be made if no response is indicated for an item in the Answer Box.
6.
Candidates will be awarded marks as stated above in instruction No.3 for correct response of each question. One mark will be deducted for indicating incorrect response of each question. No deduction from the total score will be made if no response is indicated for an item in the answer box.
7.
There is only one correct response for each question. Marked up more than one response in any question will be treated as wrong response and marked up for wrong response will be deducted accordingly as per instruction 6 above.
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JEE-MAIN-2019 (8th Apr-First Shift)-PCM-2
PART –A (PHYSICS) 1.
If 10 22 gas molecules each of mass 10 26 kg collide with a surface (perpendicular to it) elastically per second over an area 1 m2 with a speed 10 4 m/s, the pressure exerted by the gas molecules will be of the order of: N N (A) 108 2 (B) 103 2 m m 4 N 16 N (C) 10 2 (D) 10 m m2
2.
A particle moves in one dimension from rest under the influence of a force that varies with the distance traveled by that varies with the distance traveled by the particle as shown in the figure. The kinetic energy of the particle after it has traveled 3 m is: (A) 2.5 J (B) 4 J (C) 5 J (D) 6.5 J
3.
An upright object is placed at a distance of 40 cm in front of a convergent lens of focal length 20 cm. A convergent mirror of focal length 10 cm is placed at a distance of 60 cm on the other side of the lens. The position and size of the final image will be: (A) 40 cm from the convergent mirror, same size as the object (B) 20 cm from the convergent mirror, same size as the object (C) 40 cm from the convergent lens, twice the size of the object (D) 20 cm from the convergent mirror, twice the size of the object
4.
Four particles A, B, C and D with masses mA m, mB 2m, mc 3m and mD 4m are at the corners of a square. They have accelerations of equal magnitude with directions as shown. The acceleration of the centre of mass of the particles is: a ˆ ˆ (A) ij (B) Zero 5 a ˆ ˆ (C) ij (D) a ˆi ˆj 5
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JEE-MAIN-2019 (8th Apr-First Shift)-PCM-3 5.
Two identical beakers A and B contain equal volumes of two different liquids at 60o C each and left to cool down. Liquid in A has density of 8 10 2 kg / m3 and specific heat of 2000 Jkg1 K 1 while liquid in B has density of 103 kgm3 and specific heat of 4000JKg1K 1 . Which of the following best describes their temperature versus time graph schematically? (assume the emissivity of both the beakers to be the same)
(A)
(B)
(C)
(D)
6.
A thin strip 10 cm long is on a U shaped wire of negligible resistance and it is connected to a spring of spring constant 0.5Nm1 (see figure). The assembly is kept in a uniform magnetic field of 0.1 T. If the strip is pulled from its equilibrium position and released, the number of oscillations it performs before its amplitude decreases by a factor of e is N. If the mass of the strip is 50 grams, its resistance 10 and air drag negligible, N will be close to: (A) 50000 (B) 10000 (C) 1000 (D) 5000
7.
A 20 Henry inductor coil is connected to a 10 ohm resistance in series as shown in figure. The time at which rate of dissipation of energy (Joule’s heat) across resistance is equal to the rate at which magnetic energy is stored in the inductor, is 2 ln 2 1 (C) ln 2 2
(A)
(B) ln 2 (D) 2ln 2
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JEE-MAIN-2019 (8th Apr-First Shift)-PCM-4 8.
A wire of length 2L is made by joining two wires A and b of same lengths but different radii r and 2r and made of the same material. It is vibrating at a frequency such that the joint of the two wires forms a node. If the number of antinodes in wire A is p and that in B is q then the ratio p : q is:
(A) 1 : 4 (C) 3 : 5
(B) 1 : 2 (D) 4 : 9
9.
A steel wire having a radius of 2.0 mm, carrying a load of 4 kg, is hanging from a ceiling. Given that g 3.1 ms2 , what will be the tensile stress that would be developed in the wire? (A) 6.2 10 6 Nm2 (B) 4.8 106 Nm2 6 2 (C) 5.2 10 Nm (D) 3.1 106 Nm2
10.
Voltage rating of a parallel plate capacitor is 500V. Its dielectric can withstand a V maximum electric field of 106 . The plate area is 10 4 m2 . What is the dielectric m constant if the capacitance is 15 pF? (given 0 8.86 10 12 C2 / Nm2 ) (A) 3.8 (B) 6.2 (C) 4.5 (D) 8.5
11.
An alternating voltage v t 220 sin100l volt is applied to a purely resistive load of 50 . The time taken for the current to rise from half of the peak value of the peak value is: (A) 2.2 ms (B) 3.3 ms (C) 5 ms (D) 7.2 ms
12.
The wavelength of the carrier waves in a modern optical fiber communication network is close to: (A) 1500 nm (B) 600 nm (C) 2400 nm (D) 900 nm
13.
Water from a pipe is coming at a rate of 100 litres per minute. If the radius of the pipe is 5 cm, the Reynolds number for the flow is of the order of : (density of water = 1000 kg/m3, coefficient of viscosity of water = 1 mPa s) (A) 103 (B) 10 6 2 (C) 10 (D) 10 4
14.
A boy’s catapult is made of rubber cord which is 42 cm long, with 6 mm diameter of cross – section and of negligible mass. The boy keeps a stone weighing 0.02 kg on it and stretches the cord by 20 cm by applying a constant force. When released, the stone flies off with a velocity of 20 ms–1. Neglect the change in the area of cross section of the cord while stretched. The Young’s modulus of rubber is closest to: (A) 103 Nm2 (B) 10 6 Nm2 (C) 108 Nm2
(D) 10 4 Nm2
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JEE-MAIN-2019 (8th Apr-First Shift)-PCM-5 15.
Two particles move at right angle to each other. Their de Broglie wavelengths are 1 and 2 respectively. The particles suffere perfectly inelastic collision. The de Broglie wavelength , of the final particle, is given by: 2 (A) 1 2 (B) 1 2 2 1 1 1 1 1 (C) (D) 2 2 2 1 2 1 2
16.
Four identical particles of mass M are located at the corners of a square of side ‘a’. What should be their speed if each of them revolves under the influence of other’s gravitational field in a circular orbit circumscribing the square? GM GM (A) 1.35 (B) 1.16 a a GM GM (C) 1.41 (D) 1.21 a a
17.
In figure, the optical fiber is l 2m long and has a diameter of d = 20 m. If a ray of light is incident on one end of the fiber at angle 1 40o , the number of reflection it makes before emerging from the other end is close to:
(A) 57000 (C) 66000 18.
19.
(B) 45000 (D) 55000
A circular coil having N turns and radius r carries a current. It is held in the XZ plane in a magnetic field Biˆ . The torque on the coil due to the magnetic field is: Br 2 1 (A) (B) zero N Br 2 1 (C) (D) Br 2IN N Ship A is sailing towards north – east with velocity 30 ˆi 50ˆj km/hr where ˆi points east and ˆj , north. Ship B is at a distance of 80 km east and 150 km north of Ship A and is sailing towards west at 10 km/hr. A will be at minimum distance from B ins: (A) 2.2 hrs. (B) 4.2 hrs. (C) 2.6 hrs. (D) 3.2 hrs.
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JEE-MAIN-2019 (8th Apr-First Shift)-PCM-6 20.
A plane electromagnetic wave travels in free space along the x – direction. The electric field component of the wave at a particular point of space and time is E Vm1 along y – direction. Its corresponding magnetic filed component, B would be: (A) 2 10 8 T along z – direction (B) 6 10 8 T along x – direction (C) 6 10 8 T along z- direction (D) 2 10 8 T along y – direction
21.
A thermally insulted vessel contains 150 g of water at 0o C . Then the air from the vessel is pumped out a adiabatically. A fraction of water turns into ice and the rest evaporates at 0o C itself. The mass of evaporated water will be closes to: (Latent heat of vaporization of water = 2.10 10 6 Jkg1 and Laten heat of Fusion of water 3.36 105 Jkg1 ) (A) 35 g (C) 130 g
(B) 150 g (D) 20 g
22.
Radiation coming from transition n = 2 to n = 1 of hydrogen atoms fall of He ions in n = 1 and n = 2 states. The possible. Transition of helium ions as they absorb energy from the radiation is: (A) n 2 n 4 (B) n 2 n 5 (C) n 2 n 3 (D) n 1 n 4
23.
A 200 resistor has a certain color code. If one replaces the red color by green in the code, the new resistance will be: (A) 500 (B) 400 (C) 300 (D) 100
24.
The reverse breakdown voltage of a Zener diode is 5.6 V in the given circuit .
The current Iz through the Zener is: (A) 10 mA (C) 7 mA 25.
(B) 15 mA (D) 17 mA
A thin circular plate of mass M and radius R has its density varying as p r p0 r with P0 as constant and r is the distance from its center. The moment of Inertia of the circular plate about an axis perpendicular to the plate and passing through its edge is I aMR2 . The value of the coefficient a is: 8 1 (A) (B) 5 2 3 3 (C) (D) 5 2 FIITJEE Ltd., FIITJEE House, 29-A, Kalu Sarai, Sarvapriya Vihar, New Delhi -110016, Ph 46106000, 26569493, Fax 26513942 website: www.fiitjee.com.
JEE-MAIN-2019 (8th Apr-First Shift)-PCM-7
26.
In SI units, the dimensions of (A) AT 3ML3/2 (C) A 2 T 3M1L2
27.
0 is: 0
(B) A 1TML3 (D) AT 2M1L1
For the circuit shown, with R1 1.0 , R 2 2.0 ,E1 2 V and E2 E 3 4 V, the potential difference between the points ‘a’ and ‘b’ is approximately (in V):
(A) 3.3 (C) 3.7
(B) 2.3 (D) 2.7
28.
A solid conducting sphere, having a charge Q, is surrounded by an uncharged conducting hollow spherical shell. Let the potential difference between the surface of the solid sphere and that of the outer surface of the hollow shell be V. If the shell is now given a charge of –4 Q, the new potential difference between the same two surface is: (A) 2 V (B) –2V (C) 4 V (D) V
29.
In an interference experiment the ratio of amplitudes of coherent waves is
a1 1 . The a2 3
ratio of maximum and minimum intensities of fringes will be: (A) 9 (B) 4 (C) 18 (D) 2 30.
The bob of a simple pendulum has mass 2g and a charge of 5.0 C . It is at rest in a V uniform horizontal electric field of intensity 2000 . At equilibrium, the angle that the m m pendulum makes with the vertical is: (take g 10 2 ) s (A) tan 1 2.0 (B) tan 1 0.2 (C) tan 1 5.0
(D) tan 1 0.5
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JEE-MAIN-2019 (8th Apr-First Shift)-PCM-8
PART –B (CHEMISTRY) 31.
Element ‘B’ forms ccp structure and ‘A’ occupies half of the octahedral voids, while oxygen atoms occupy all the tetrahedral voids. The structure of bimetallic oxide is (A) A 2B 2 O (B) AB 2 O 4 (C) A 4B 2 O (D) A 2BO 4
32.
Coupling of benzene diazonium chloride with 1 – naphthol in alkaline medium will give:
(A)
(B)
(C)
(D)
33.
The size of the iso-electronic species CI , Ar and Ca2+ is affected by (A) Principal quantum number of valence shell (B) Azimuthal quantum number of valence shell (C) electron – electron interaction in the outer orbitals (D) nuclear charge
34.
In the following compounds, the decreasing order of basic strength will be: (A) C 2H5NH2 NH3 C2H5 2 NH (B) NH3 C 2H5NH2 C2H5 2 NH (C) C2H5 2 NH C 2H5NH2 NH3
35.
(D) C2H5 2 NH NH3 C2H5NH2
The correct order of the spin only magnetic moment of metal ions in the following low 4
4
3
2
spin complexes, V CN6 , Fe CN6 , Ru NH3 6 , and Cr NH3 6 , is: (A) Cr 2 Ru3 Fe2 V 2 (B) V 2 Cr 2 Ru3 Fe2 2 2 3 2 (C) Cr V Ru Fe (D) V 2 Ru3 Cr 2 Fe2
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JEE-MAIN-2019 (8th Apr-First Shift)-PCM-9 36.
An organic compound ‘X’ showing the following solubility profile is:
(A) Oleic acid (C) Benzamide 37.
(B) o – Toluidine (D) m – Cresol
Diborane B 2H6 reacts independently with O2 and H2 O to produce, respectively: (A) H3BO3 and B2O3
(B) B2O3 and H3BO3
(C) HBO2 and H3BO3
(D) B2O3 and BH4
38.
The lanthanoide that would show colour is: (A) Gd3 (B) La3 3 (C) Lu (D) Sm3
39.
The major product of the following reaction
40.
(A)
(B)
(C)
(D)
100 mL of a water sample contains 0.81 g of calcium bicarbonate and 0.73 g of magnesium bicarbonate. The hardness of this water sample expressed in terms of equivalents of CaCO3 is: (molar mass of calcium bicarbonate is 162 gmol1 and magnesium bicarbonate is 146 g
mol1 ) (A) 10, 000 ppm (C) 5,000 ppm
(B) 1, 000 ppm (D) 100 ppm
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JEE-MAIN-2019 (8th Apr-First Shift)-PCM-10 41.
42.
An organic compound neither reacts with neutral ferric chloride solution nor with Fehling solution. It however, reacts with Grignard reagent and gives positive iodoform test. The compound is:
(A)
(B)
(C)
(D)
For the reaction 2A B C, the values of initial rate at different reactant concentrations are given in the table below: The rate law for the reaction is: Initial Rate (mol L1s1 ) A molL1 B molL1
0.05
0.05
0.045
0.10
0.05
0.090
0.20
0.10
0.72
(A) Rate = k[A]2[B]2 (C) Rate = k[A][B] 43.
(B) Rate = k[A][B]2 (D) Rate = k[A]2[B]
The major product of the following reaction is:
(A)
(B)
(C)
(D)
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JEE-MAIN-2019 (8th Apr-First Shift)-PCM-11 44.
Which is wrong with respect to our responsibility as a human being to protect our environment? (A) Restricting the use of vehicles (B) Using plastic bags (C) Setting up compost tin in gardens (D) Avoiding the use of floodlighted facilities.
45.
The major product of the following reaction
46.
(A)
(B)
(C)
(D)
For silver CP JK 1 mol1 23 0.01T . If the temperature (T) of 3 moles of silver is raised from 300 K to 1000 K at 1 atm pressure, the value of H will be close to: (A) 13 kJ (B) 62 kJ (C) 16 kJ (D) 21 kJ
47.
The correct order of hydration enthalpies of alkali metal ions is: (A) Li Na K Cs Rb (B) Na Li K Rb Cs (C) Na Li K Cs Rb (D) Li Na K Rb Cs
48.
Adsorption of a gas follows Freundlich adsorption isotherm. x is the mass of the gas x adsorbed on mass m of the adsorbent. The plot of log versus log p is shown in the m x given graph, is proportional to: m
(A) p2/3
(B) p2
(C) p3
(D) p3/2
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JEE-MAIN-2019 (8th Apr-First Shift)-PCM-12 49.
The vapour pressures of pure liquids A and B are 400 and 600 mm Hg, respectively at 298 K. On mixing the two liquids, the sum of their initial volumes is equal to the volume of the final mixture. The mole fraction of liquid B is 0.5 in the mixture. The vapour pressure of the final solution, the mole fractions of components A and B in vapour phase, respectively are: (A) 500 mmHg, 0.4, 0.6 (B) 500 mmHg, 0.5, 0.5 (C) 450 mmHg, 0.5, 0.5 (D) 450 mmHg, 0.4, 0.6
50.
0 Given that E0O2 /H2O 1.23V; E0S O2 /SO2 2.05 V;EBr 2
51.
8
2 /Br
4
strongest oxidizing agent is: (A) O2
(B) S2O82
(C) Au3
(D) Br2
1.09 V;E0Au3 /Au 1.4V . The
The quantum number of four electrons are given below: I. n 4,l 2,ml 2, ms 1/ 2 II. n 3,l 2, ml 1,m8 1/ 2 III. n 4,l 1, ml 0, ms 1/ 2 IV. n 3,l 1, mI 1; ms 1/ 2 The correct order of their increasing energies will be: (A) I < III < II < IV (B) I < II < III < IV (C) IV < II < III < I (D) IV < III < II < I
52.
If solubility product of Zr3 PO 4 4 is denoted by K SP and its molar solubility is denoted by S, then which of the following relation between S and K SP is correct? 1/7
K (A) S SP 6912
1/9
K Sp (C) S 929
1/6
K (B) S SP 144
1/7
K (D) S SP 216
53.
Which of the following amines can be prepared by Gabriel phthalimide reaction? (A) t – butylamine (B) n – butylamine (C) neo – pentylamine (D) triethylamine
54.
Which one of the following equations does not correctly represent the first law of thermodynamics for the given processes involving an ideal gas? (Assume non – expansion work is zero) (A) Adiabatic process: U w (B) Isochoric process: U q (C) Cyclic process: q w (D) Isothermal process: q w
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JEE-MAIN-2019 (8th Apr-First Shift)-PCM-13 55.
The following ligand is:
(A) tridentate (C) tetradentate 56.
(B) bidentate (D) hexadentate
The IUPAC name of the following compound is
(A) 3-Hydroxy – 4 – methylpentaonic acid (B) 4 – Methyl – 3 – hydroxypentanoic acid (C) 2 – Methyl – 3 – hydroxylpentan-5-oic acid (D) 4, 4 – Dimethyl – 3 – hydroxybutanoic acid 57.
In order to oxidize a mixture of one mole of each of FeC 2 O4 , Fe2 C2 O 4 3 ,FeSO 4 and Fe2 SO 4 3 in acidic medium, the number of moles of KMnO4 required is:
(A) 1 (C) 2
(B) 1.5 (D) 3
58.
Assertion : Ozone is destroyed by CFCs in the upper stratosphere. Reason : Ozone holes increase the amount of UV radiation reaching the earth. (A) Assertion and reason are incorrect (B) Assertion is false, but the reason is correct (C) Assertion and reasons are both correct, and the reason is the correct explanation for the assertion. (D) Assertion and reason are correct, but the reason is not the explanation for the assertion.
59.
Maltose on treatment with dilute HCI gives: (A) D – Fructose (B) D – Galactose (C) D – Glucose and D – Fructose (D) D – Glucose
60.
With respect to an ore, Ellingham diagram helps to predict the feasibility of its (A) Electrolysis (B) Thermal reduction (C) Vapour phase refining (D) Zone refining
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JEE-MAIN-2019 (8th Apr-First Shift)-PCM-14
PART–C (MATHEMATICS) 61.
The magnitude of the projection of the vector the plane containing the vectors ˆi ˆj kˆ and
3 2 3 (D) 2
(A) 3 6 (C)
2iˆ 3 ˆj kˆ on the vector perpendicular to ˆi 2ˆj 3kˆ , is:
(B)
6
62.
The shortest distance between the line y x and the curve y 2 x 2 is: 11 (A) (B) 2 4 2 7 7 (C) (D) 8 4 2
63.
If and be the roots of the equation x 2 2x 2 0 , then the least value of n for n
which 1 is: (A) 4 (C) 5
64.
(B) 2 (D) 3
All possible numbers are formed using the digits 1, 1, 2, 2, 2, 2, 3, 4, 4 taken all at a time. The number of such numbers in which the odd digits occupy even places is: (A) 180 (B) 175 (C) 162 (D) 160 5x 2x dx is equal to: sin 2 (where c is a constant of integration). (A) x 2 sin x 2 sin 2x c (C) x 2 sin x sin 2x c sin
65.
66.
(B) 2x sin x 2 sin 2x c (D) 2x sin x sin 2x c
Let O (0, 0) and A (0, 1) be two fixed points. Then the locus of a point P such that the perimeter of AOP , is 4, is: (A) 9x 2 8y 2 8y 16 (B) 8x 2 9y 2 9y 18 (C) 9x 2 8y 2 8y 16
(D) 8x 2 9y 2 9y 18
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JEE-MAIN-2019 (8th Apr-First Shift)-PCM-15
67.
68.
69.
3 5 If cos ,sin and 0 , , then tan 2 is equal to: 5 13 4 63 33 (A) (B) 52 52 63 21 (C) (D) 16 16
cos sin 0 1 32 Let A , R such that A . Then a value of is: sin cos 1 0 (A) 0 (B) 16 (C) (D) 32 64 1 x 2x If f x loge is equal to: , x 1, then f 2 1 x 1 x
(B) f x
(A) 2f x
(C) 2f x 2
2
(D) 2f x 2
70.
71.
3 cos x sin x dy If 2y cot 1 , x 0, then is equal to cos x 3 sin x 2 dx (A) x (B) x 6 3 (C) x (D) 2x 6 3
The sum of the solutions of the equation (A) 9 (C) 10
72.
lim x 0
73.
x 4 2 0, x 0 is equal to:
(B) 4 (D) 12 sin2 x
2 1 cos x
(A) 2 (C) 4
x 2 x
equals (B) 4 2 (D) 2 2
2. 20C0 5. 20 C1 8. 20C2 11. 20C3 .... 62. 20 C20 is equal to
(A) 223 (C) 224
(B) 226 (D) 225
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JEE-MAIN-2019 (8th Apr-First Shift)-PCM-16
74.
such that y 0 0 . If
ay 1
1 2 1 (C) 16
If f x
2
dy 2x x 2 1 y 1 dx
, then the value of ‘a’ is 32
(A)
75.
Let y y x be the solutions of the differential equation, x 2 1
(B) 1 (D)
1 4
2 x cos x and g x loge x, x 0 then the value of the integral 2 x cos x
/4
g f x dx is:
/ 4
76.
(A) loge 1
(B) loge 2
(C) loge e
(D) loge 3
The area (in sq. units) of the region A 26 3 53 (C) 6
(A)
x, y R R 0 x 3, 0 y 4, y x (B)
2
3x is:
59 6
(D) 8
77.
The mean and variance of seven observations are 8 and 16, respectively. If 5 of the observations are 2, 4, 10, 12, 14, then the product of the remaining two observations is: (A) 40 (B) 45 (C) 49 (D) 48
78.
The length of the perpendicular from the point (2, –1, 4) on the straight line, x3 y2 z is: 10 7 1 (A) greater than 2 but less than 3 (B) less than 2 (C) greater than 4 (D) greater than 3 but less than 4
79.
The contrapositive of the statement “If you are born in India, then you are a citizen of India”, is: (A) If you are a citizen of India, then you are born in India (B) If your are not a citizen of India, then you are not born in India (C) If you are no born in India, then you are not a citizen of India (D) If you are born in India, then you are not a citizen of India
80.
The sum of all natural numbers ‘n’ such that 100 n 200 and H.C. F (91, n) > 1 is: (A) 3221 (B) 3303 (C) 3203 (D) 3121
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JEE-MAIN-2019 (8th Apr-First Shift)-PCM-17 81.
If S1 and S 2 are respectively the sets of local minimum and local maximum points of the function. f x 9x 4 12x 3 36x 2 25, x R, then (A) S1 2, 1 ;S 2 0 (B) S1 2,0 ;S2 1 (C) S1 2; S2 0,1 (D) S1 1 ;S2 0,2
82.
The sum of the co – efficient of all even degree terms in x in the expansion of
6
6
x x3 1 x x3 1 , x 1 is equal to:
(A) 26 (C) 32
(B) 24 (D) 29
83.
A point on the straight line, 3x 5y 15 which is equidistant from the coordinate, axes will lie only in: (A) 4th quadrant (B) 1st , 2nd and 4th quadrants (C) 1st quadrant (D) 1st and 2nd quadrants
84.
If the tangents on the ellipse 4x 2 y 2 8 at the points (1, 2) and (a, b) are perpendicular to each other, then a 2 is equal to: 2 4 (A) (B) 17 17 64 128 (C) (D) 17 17
85.
3 1 If cos1 , tan1 , where 0 , , then is equal to: 2 5 3 9 9 (A) sin1 (B) cos1 5 10 5 10 9 9 (C) tan 1 (D) tan 1 14 5 10
86.
The equation of a plane containing the line of intersection of the planes 2x y 4 0 and y 2z 4 0 and passing through the point (1, 1, 0) is: (A) x 3y z 4 (B) 2x z 2 (C) x 3y 2z 2 (D) x y z 0
87.
The sum of the squares of the lengths of the chords intercepted on the circle, x 2 y 2 16, by the lines, x y n,n N , where N is the set of all natural numbers is: (A) 320 (B) 160 (C) 105 (D) 210
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JEE-MAIN-2019 (8th Apr-First Shift)-PCM-18 88.
The greatest value of c R for which the system of linear equations x cy cz 0 cx y cz 0 cx cy z 0 has a non – trivial solution, is: 1 (A) –1 (C) 2 (B) 2 (D) 0
89.
Let f : 0,2 R be a twice differentiable function such that f " x 0, for all x 0,2 . If x f x f 2 x , then is: (A) increasing on (0, 2) (B) decreasing on (0, 2) (C) decreasing on (0, 1) and increasing on (1, 2) (D) increasing on (0, 1) and decreasing on (1, 2)
90.
Let A and b be two non – null events such that A B . Then, which of the following statements is always correct? (A) P A B 1 (B) P A B P A (C) P A B P B P A
(D) P A B P A
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JEE-MAIN-2019 (8th Apr-First Shift)-PCM-19
JEE (Main) – 2019 ANSWERS PART A – PHYSICS 1. 5. 9. 13. 17. 21. 25. 29.
Bonus A D D Bonus D A B
2. 6. 10. 14. 18. 22. 26. 30.
D D D B D B C D
3. 7. 11. 15. 19. 23. 27.
Bonus D B D C A A
4. 8. 12. 16. 20. 24. 28.
A B A B A A D
34. 38. 42. 46. 50. 54. 58.
C D B B B B D
64. 68. 72. 76. 80. 84. 88.
A D B B D A B
PART B – CHEMISTRY 31. 35. 39. 43. 47. 51. 55. 59.
B B D A D C C D
32. 36. 40. 44. 48. 52. 56. 60.
C D A B A A A B
33. 37. 41. 45. 49. 53. 57.
D B C D A B C
PART C – MATHEMATICS 61. 65. 69. 73. 77. 81. 85. 89.
D C A D D A A C
62. 66. 70. 74. 78. 82. 86. 90.
C C D C D B D D
63. 67. 71. 75. 79. 83. 87.
A C C A B D D
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JEE-MAIN-2019 (8th Apr-First Shift)-PCM-20
HINTS AND SOLUTIONS PART A – PHYSICS 1.
Pressure is defined as normal force per unit area. Force is calculated as change in momentum/ time. By this answer is 2N/m2 None of the option matches so this question must be Bonus. Detailed solution is as following: Magnitude of change in momentum per collision = 2 mv Force N(2mv) Pressure = Area 1 22 26 10 2 10 10 4 = 1 = 2 N/m2
2.
According to work energy theorem. Work done by force on the particle = Change in KE Work done = Area under F-x graph = F dx =
(2 3) 1 2 W = KEfinal – KEinitial = 6.5 KEinitial = 0 KEfinal = 6.5 J 22
3.
There will be 3 phenomenon (i) Refraction from lens (ii) Reflection from mirror (iii) Refraction from lens After these phenomena. Image will be on object and will have same size. None of the option depicts so this question is Bonus.
1st refraction u = –40 cm ; f = +20 cm v = +40 cm (image I1) and m1 = –1 for reflection u = –20 cm ; f = –10 cm
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JEE-MAIN-2019 (8th Apr-First Shift)-PCM-21 v = –20 cm (image I2) and m2 = –1 2nd refraction u = –40 cm ; f = +20 cm v = +40 cm (image I3) and m3 = –1 Total magnification = m1 × m2 × m3 = –1 and final image is formed at distance 40 cm from convergent lens and is of same size as the object. 4.
a A a ˆi ; aB ajˆ aC a ˆi ; aD ajˆ ma aa mb ab mc ac mda d acm ma mb mc md ma ˆi 2mjˆ 3ma ˆi 4majˆ acm 10m a a a ˆ ˆ 2ma ˆi 2majˆ = = ˆi ˆj = ij 10m 5 5 5
5.
6.
dT eA(T 4 T04 ) dt dT eA 4 dT 4eAT03 (T T04 ) ; ( T) dt ms dt ms T = T0 + (Ti – T0) e–kt 4eAT03 where k ms 4eAT03 dT k ; k vs dt dT 1 dt s ASA = 2000 × 8 × 102 = 16 × 105 BSB = 4000 × 103 = 4 × 106 ASA < BSB dT dT dt A dt B ms
T0 2
m 2 k 10
A = A0e–1/
A0 , t= e 2m 2m 2 2 = 104 s t== b B R t 10 4 No of oscillation 5000. T0 2 / 10 for A
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JEE-MAIN-2019 (8th Apr-First Shift)-PCM-22 7.
LIDI = I2R E 1 E L ( e t / 2 ) (1 e t / 2 ) 10 10 2 10 e 1/ 2 1 e1/ 2 ; t = 2n2
8.
Let mass per unit length of wires are 1 and 2 respectively. Materials are same, so density is same. r 2L 4r 2L and 2 4 L L Tension in both are same = T, let speed of wave in wires are V1 and V2 V V V V V1 1 & f02 2 2L 2L 2L 4L Frequency at which both resonate is L.C.M. of V both frequencies i.e. . 2L
1
Hence number of loops in wires are 1 and 2 respectively
So, ratio of number of antinodes is 1 : 2. 9.
Tensile stress in wire will be Tensile force = Cross section Area mg 4 3.1 = Nm2 = 3.1 × 106 Nm–2 2 R 4 10 6
10.
A = 10–4 m2 Emax = 106 V/m C = 15 F k A Cd k C 0 ; d 0 A 15 1012 500 10 6 15 5 = = 8.465 8.86 1012 10 4 8.86 k 8.5 k
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JEE-MAIN-2019 (8th Apr-First Shift)-PCM-23 11.
V(t) = 220 sin(100 t) volt time taken, 1 t 3 sec 100 300 = 3.3 ms
12.
To minimize attenuation, wavelength of carrier waves is close to 1500 nm.
13.
Reynolds number =
14.
vd Volume flow rate = v × r2 100 103 1 v 60 25 104 2 v m /s 3 103 2 10 10 2 Reynolds number = = 2 × 104 103 3 Order 104 2 1 Energy of catapult = Y A 2 1 = Kinetic energy of the ball = mv 2 2 2 1 20 1 Therefore, Y 32 10 6 42 10 2 2 10 2 (20)2 2 42 2 6 2 Y = 3 × 10 Nm
15.
h h ˆ P1 ˆi and P2 j 1 2 Using momentum conservation h ˆ h ˆ i j P P1 P2 = 1 2
P
2
h h 1 2 2
h h 1 2 1 1 1 2 2 2 1 2 h
2
2
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JEE-MAIN-2019 (8th Apr-First Shift)-PCM-24 16.
Net force on particle towards centre of circle is GM2 GM2 FC 2 2 2a 2 a GM2 1 = 2 2 a 2 This force will act as centripetal force. Distance of particle from centre of circle is a . 2 a mv 2 r , FC r 2 2 2 mv GM 1 2 2 a a 2 2 GM 1 v2 1 a 2 2
v2 17.
GM GM 1.35 ; v 1.16 a a
If we approximate the angle 2 as 30° initially then answer will be closer to 57000. but if we solve thoroughly, answer will be close to 55000. So both the answers must be awarded. Detailed solution as following. Exact solution By Snell’s law 1.sin 40° = (1.31) sin 2 .64 64 sin 2 .49 1.31 131 64 64 64 d Now tan 2 13065 114.3 x (131)2 (64)2 Now number of reflections 2 64 64 105 = 114.3 20 10 6 114.3 55991 55000
Approximate solution By Snell’s law 1.sin 40° = (1.31)sin 2
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JEE-MAIN-2019 (8th Apr-First Shift)-PCM-25
0.64 64 0.49 1.31 131 If assume 2 30° d Tan 300 = x 3d x Now number of reflections 2 105 = 3d 3 20 10 6 3 57735 57000 Sin 2 =
18.
Magnetic moment of coil = NIA ˆj = NI ( r 2 ) ˆj Torque on loop (coil) = M B ˆ = NI( r 2 ) B sin 90o ( k) ˆ = NIr 2B ( k)
19.
If we take the position of ship ‘A’ as origin then positions and velocities of both ships can be given as: v A (30iˆ 50ˆj)km / hr vB 10iˆ km / hr ; rA 0iˆ 0ˆj rB (80iˆ 150 ˆj) km Time after which distance between them will be minimum r v t BA 2BA ; v BA Where rBA (80iˆ 150 ˆj) km VBA 10iˆ (30iˆ 50ˆj) ( 40iˆ 50 ˆj) km / hr ˆ (80iˆ 150ˆj) ( 40i 50h) t= 2 40i 50hˆ
= 20.
3200 7500 10700 hr hr 2.6 hrs 4100 4100
The direction of propagation of an EM wave is direction of E B. ˆi ˆj Bˆ Bˆ kˆ
E E 6 B B C 3 108 B = 2 × 10–8 T along z direction. C
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JEE-MAIN-2019 (8th Apr-First Shift)-PCM-26 21.
Suppose ‘m’ gram of water evaporates then, heat required Qreq = mLv Mass that converts into ice = (150 – m) So, heat released in this process Qrel = (150 – m) Lf Now, Qrel = Qreq (150 – m) Lf = mLv M(Lf + Lv) = 150 Lf
m
150L f ; m = 20 g Lf Lv
22.
Energy released for tension n = 2 to n = 1 of hydrogen atom 1 1 E 13.6Z2 2 2 n1 n2 Z = 1, n1 = 1, n2 = 2 1 1 E 13.6 1 2 2 1 2 3 E 13.6 eV 4 + For He ion z = 2 (A) n = 1 to n = 4 15 1 1 E 13.6 22 2 2 13.6 eV 4 1 4 (B) n = 2 to n = 4 3 1 1 E 13.6 22 2 2 13.6 eV 4 1 4 (C) n = 2 to n = 5 1 21 1 E 13.6 22 2 2 13.6 eV 5 25 2 (D) n = 2 to n = 5 1 5 1 E 13.6 22 2 2 13.6 eV 3 9 2
23.
When red is replace with green 1st digit changes to 5 so new resistance will be 500 .
24.
9 = Vz + VR1 VZ = 5.6 V VR1 9 5.6 VR1 3.4 IR1
VR1 R
3.4 ; IR1 = 17 mA 200
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JEE-MAIN-2019 (8th Apr-First Shift)-PCM-27 Vz VR2 IR2 (R2 )
5.6 IR2 ; IR2 7 mA 800 Iz = (17 – 7) mA = 10 mA R
25.
M 0r(2rdr) 0
0 2 R 3 3
R
I0 (MOI about COM)
0r(2rdr) r 2 0
0 2 R 5 5
By parallel axis theorem I = I0 + MR2 0 2R5 0 2R3 8 R2 0 2R5 5 3 15 8 2 = MR 5
=
26.
Dimension of
0 0
[0] = [M–1L–3T4A2] [0] = [MLT–2A–2] 1
Dimension of
0 M1L3 T 4 A 2 2 0 MLT 2 A 2
= [M–2L–4T6A4]1/2 = [M–1L–2T3A2]
27.
E1 E2 E3 2R1 R 2 2R1 Eeq 1 1 1 2R1 R 2 2R1
2 2 1 2
28.
4 2 1 2
4 2 5 10 = 3.3 1 3 3 2 2
As given in the first condition: Both conducting spheres are shown. kQ kQ Vin Vout r1 r2
1 1 = kQ V r1 r2
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JEE-MAIN-2019 (8th Apr-First Shift)-PCM-28 In the second condition: Shell is now given charge –4Q. kQ 4kQ kQ 4kQ Vin Vout r2 r2 r2 r1 =
kQ kQ r1 r2
1 1 = kQ V r1 r2 Hence, we also obtain that potential difference does not depend on charge of outer sphere. P. d. remains same. 29.
Given
a1 1 a2 3 2
I a 1 Ratio of intensities, 1 1 I2 a2 9 2
2 Imax I1 I2 1 3 Now, 4 Imin I1 I2 1 3
30.
qE 5 10 6 2000 mg 2 10 3 10 1 tan = tan1 (0.5) 2 tan
PART B – CHEMISTRY 31.
For cubic unit cell, only FCC has octahedral and tetrahedral voids. 1 ZB = 4, ZA = 4 2,Zo 8 4 Formula = A2B2O8 = AB 2 O 4
32.
Electrophilic substitution reaction takes place.
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JEE-MAIN-2019 (8th Apr-First Shift)-PCM-29 33.
For isoelectronic species the size is compared by nuclear charge. 1 Size Z Nuclear Charge
34.
Basic strength order (CH3CH2)2NH > CH3CH2NH2 > NH3 More the number of +I groups, higher is the basic strength.
35.
Since CN– and NH3 are strong field ligands, low spin complexes are formed. Complex Configuration No. of unpaired electrons 3 0 [V(CN)6]4– 3 t e 2g g
2+
t 42g eg0
2
[Ru(NH3)6]3+
t52g eg0
1
[Fe(CN)6]4–
t 62g eg0
0
[Cr(NH3)6]
Magnetic moment is directly proportional to number of unpaired electrons. 36.
m-cresol is the right answer. 37.
B2H6 3H2O 2H3BO3 3 H2 B2H6 3 O2 B 2O3 3H2O
38.
Sm3+(4f5) = yellow colour, other ions have stable electron configurations with half filled or full-filled electron configuration.
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JEE-MAIN-2019 (8th Apr-First Shift)-PCM-30 H
39.
O - CH3
OCH3
Br SN 2
Conc.HBr excess
CH = CH2
CH = CH2 OH
OH HBr
CH - CH3
CH3Br
CH = CH2
Br OH
Br
40.
neq. CaCO3 = neqCa(HCO3)2 + neqMg(HCO3)2 W 0.81 0.73 or, 2 2 2 100 162 146 w = 1.0 Volume of water = 100 mL Mass of water = 100 g 1.0 Hardness = 106 = 10000 ppm 100 OH
41.
(neq = Number of equivalent)
Neutral ve phenolic group is absent FeCl 3
CH3 C - C2H5 O
42.
x
RMgX vereaction Grignard Re agent
Acidic H C
NaOH I
2 vereaction Iodoform
are present
O
Test Fehling' s solution
ve no CHO group ispresent
y
r = K[A] [B] 0.045 = K(0.05)x (0.05)y 0.090 = K(0.10)x (0.05)y 0.72 = K(0.20)x (0.10)y Dividing (1) by (2) we get
. . . . (1) . . . . (2) . . . . (3)
x
0.045 0.05 x 1 0.090 0.10 Dividing (2) by (3)
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JEE-MAIN-2019 (8th Apr-First Shift)-PCM-31 x
y
0.090 0.10 0.05 y2 0.720 0.20 0.10 Hence, r = K[A][B]2 43.
Fridel-craft acylation. –Cl group is an ortho & para directing 44.
Plastics are non-biodegradable.
45.
Reduction followed by substitution reaction T2
46.
H n Cp.m dT 3
T1
1000
23 0.01T dT
300
3 23 1000 300
0.01 1000 2 300 2 2
= 61950 J 62 kJ 47.
48.
Hydration enthalpy depends upon ionic potential (charge/size). As ionic potential increases hydration enthalpy increases. q hydH0 r
x Kp1/n m Taking log from both sides
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JEE-MAIN-2019 (8th Apr-First Shift)-PCM-32 x 1 logK logP m n 1 2 slope n 3 x Kp2/3 m log
49.
Ptotal= X A PA0 XB PB0 0.5 400 0.5 600 500 mmHg Now, mole fraction of A in vapour P 0.5 400 YA A 0.4 and mole fraction of B in vapour Ptotal 500 YB = 1 – 0.4 = 0.6
50.
For strongest oxidising agent, standard reduction potential should be highest. Peroxy oxygen (–O – O–) is reduced to oxide (O 2–) in the change
51.
According to Aufbau principle, the energy sequence is 3p < 3d < 4p < 4d
52.
53.
Gabriel phthalimide synthesis:
For branched chain RX, elimination reaction takes place.
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JEE-MAIN-2019 (8th Apr-First Shift)-PCM-33 54.
According to the first law of thermodynamics q = U – w
55.
Both nitrogen & oxygen are donating atoms.
56.
The priority of COOH is higher that OH. COOH is the functional group.
3-Hydorxy-4-methylpentanoic acid 57.
n-factors of KMnO4 = 5, n-factor of FeSO4 = 1 n-factors of FeC2O4 = 3, Fe2(SO4)3 does not react n-factors of Fe2(C2O4)3 = 6, neq KMnO4 = neq[FeC2O4 + Fe2(C2O4)3 + FeSO4] or, x 5 = 1 3 + 1 6 + 1 1 x=2
58.
The upper stratosphere consists of ozone (O3), which protect us from harmful ultraviolet (UV) radiations coming from sun. The layer get depleted by CFC’s
59.
60.
Ellingham diagram which are the curves of the graph between G and T helps in predicting the feasibility of thermal reduction of ores.
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JEE-MAIN-2019 (8th Apr-First Shift)-PCM-34
PART C – MATHEMATICS 61.
Vector perpendicular to plane containing the vectors ˆi ˆj kˆ d ˆi 2ˆj 3kˆ is parallel to vector ˆi ˆj kˆ 1 1 1 ˆi 2ˆj kˆ 1 2 3 Required magnitude of projection 2iˆ 3 ˆj kˆ . ˆi 2ˆj kˆ ˆi 2ˆj kˆ
2 6 1 6
3
6
3 2
62.
We have dy dy 1 2y 1 1 dx dx P 2 t 2 , t 2t 1 t 2 9 1 P , 4 2 So, shortest distance 9 2 7 4 4 2 4 2
63.
x 1
2
1 0 x 1 i, 1 i
n
n 1 i 1 n (least natural number) = 4
64. 4th place
2nd place
8th place
6th place
(even places) 3! 6! Number of such numbers 4C3 180 2! 2! 4!
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JEE-MAIN-2019 (8th Apr-First Shift)-PCM-35 5x 5x x 2 sin cos 2 2 dx 2x dx x x sin 2 sin cos 2 2 2 sin3x sin 2x dx sin x sin
65.
3 sin x 4 sin3 x 2 sin x cos x dx sin x
3 4 sin2 x 2cos x dx 3 2 1 cos 2x 2cos x dx 1 2cos 2x 2cos x dx
x sin 2x 2 sin x c 66.
AP OP AO 4 2
h2 k 1 h2 k 2 1 4 2
h2 k 1 h2 k 2 3 2
h2 k 1 9 h2 k 2 6 h2 k 2 2k 8 6 h2 k 2 k 4 3 h2 k 2 k 2 16 8k 9 h2 k 2
2
2
9h 8k 8k 16 0 Locus of P is 9x 2 8y 2 8y 16 0 67.
and 2 4 4 3 4 5 5 If cos then tan and if sin then tan 5 3 13 12 (since here lies in the first quadrant) 0
Now tan 2 tan 4 5 63 3 12 4 5 16 1 tan .tan 1 . 3 12 tan tan
68.
cos sin A sin cos cos sin cos sin A2 sin cos sin cos
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JEE-MAIN-2019 (8th Apr-First Shift)-PCM-36 cos 2 sin 2 sin 2 cos 2 cos 2 sin 2 cos sin A3 sin 2 cos 2 sin cos cos 3 sin 3 sin 3 cos 3 cos 32 sin 32 0 1 Similarly A 32 sin 32 cos 32 1 0 cos 32 0 and sin32 1 32 4n 1 , n I 2 4n 1 , n I 64 for n 0 64
69.
70.
1 x f x loge , x 1 1 x 2x 1 1 2x 2 2x f n 2 1 x 1 2x 1 x2 x 12 1 x 2n n 2f x x 12 1 x 3 1 cos x sin x 2 Consider cot 1 2 1 3 sin x sin x 2 2 sin x 3 cot 1 cos x 3 cot 1 tan x tan1 tan x 3 2 3 x x ; 0 x 2 3 6 6 x 7 x ; x 2 3 2 6 6
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JEE-MAIN-2019 (8th Apr-First Shift)-PCM-37
2 x ; 0 x 6 6 2y 2 7 x ; 6x2 6
2 6 x . 1 ; 0 x 6 dy 2 dx 7 2 x . 1 ; x 6 6 2
x 2 x
71.
x 2
x
2
x 2
x 4 20 2
4 x 20
x 2 20
x 2 2 (not possible) or
x 2 1
x 2 1, 1 x 3,1 x 9, 1 Sum = 10 sin2 x 2 2 1 cos x x lim x 0 1 cos x x2
72.
73.
1
2
. 2 2 1 2
4
2
2. 20C0 5. 20 C1 8. 20C2 11. 20C3 ...... 62. 20C 20 2 20
3r 2 20Cr r 0
20
20
3 r . 20Cr 2 r 0
20
Cr
r 0
20 20 3 r 19Cr 1 2.220 r 0 r 60.219 2.220 225
74.
dy 2x 1 2 y 2 dx x 1 x2 1
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JEE-MAIN-2019 (8th Apr-First Shift)-PCM-38
I.F. e
x2 1
n x 2 1
So, general solution is y. x 2 1 tan1 x c As y 0 0
c 0
1
tan x x2 1 a, y 1 32 1 1 a a 4 16
y x
As,
75.
2 x.cos x g f x n f x n 2 x.cos x /4 2 x.cos x 2 x.cos x I n dx 2 x.cos x 2 x .cos x 0 /2
0 dx 0 log 1 e
0
76.
Required Area 1
x 2 3x dx Area of 0
rectangle PQRS 11 59 8 6 6
77.
Let 7 observations be x1, x 2 , x3 , x 4 , x 5 , x 6 , x 7 7
x 8 xi 56
………..(1)
i 1
Also 2 16 1 7 16 xi2 x 7 i1
2
16
1 7 2 xi 64 7 i1
7 xi2 560 ……..(2) i1 Now, x1 2, x 2 4, x 3 10, x 4 12, x 5 14
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JEE-MAIN-2019 (8th Apr-First Shift)-PCM-39
x6 x 7 14, (from (1)) and x 26 x 27 100 (from (2)) 2
x 26 x 72 x 6 x 7 2x 6 x 7 x 6 x 7 48
78.
Now, MP. 10iˆ 7ˆj kˆ 0
1 2 Length of perpendicular 1 49 PM 0 4 4 50 25 5 , 4 4 2 which is greater than 3 but less than 4.
79.
The contrapositive of a statement p q is ~ q ~ p Here, p : your are born in India q : you are citizen of India So, contrapositive of above statement is “If you are not a citizen of India, then you are not born in India”.
80.
S A sum of numbers between 100 and 200 which are divisible by 7. S A 105 112 ..... 196 14 SA 105 196 2107 2 SB Sum of numbers between 100 and 200 which are divisible by 13. 8 SB 104 117 ..... 195 104 195 1196 2 SC Sum of numbers between 100 and 200 which are divisible by 7 and 13.
SC 182
H.C. F. (91, n) > 1 S A SB SC 3121 81.
f x 9x 4 12x3 36x 2 25 f x 36x 3 36x 2 72x
36x x 2 x 2
36x x 1 x 2
Point of minima 2, 1 S1 Point of maxima 0 S2 82.
6
x x3 1
x x3 1
6
2 3 2 6 C0 x 6 6C2 x 4 x3 1 6C4 x 2 x 3 1 6C6 x3 1
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JEE-MAIN-2019 (8th Apr-First Shift)-PCM-40
2 6 C0 x 6 6C2 x7 6 C2 x 4 6C4 x 8 6C4 x 2 2 6C4 x5 x 9 1 3x 6 3x3 Sum of coefficient of even powers of x 2 1 15 15 15 1 3 24 15 3t t 5 15 3t 15 3t t or t 5 5 15 15 t or t 8 2 15 15 st So, P , 1 quadrant 8 8 15 15 nd or P , II Quadrant 2 2
83.
Now,
84.
4a2 b 2 8 ………..(1) dy 4x 2 dx 1,2 y
4a 1 b 2 b 8a b2 64a2
68a2 8 , a2
85.
86.
2 17
3 1 cos , tan 5 3 4 tan 3 4 1 9 tan 3 3 4 1 13 1 , 3 3 9 sin 5 10 9 sin1 5 10 The required plane is 2x y 4 y 2z 4 0 it passes through (1, 1, 0) 2 1 4 1 4 0 3 3 0 1 xyz0
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JEE-MAIN-2019 (8th Apr-First Shift)-PCM-41
87.
p
n n , but 4 n 1,2,3, 4,5 2 2
Length of chord AB 2 16
n2 2
64 2n2 (say) For n 1, 2 62
n 2, 2 56 n 3, 2 46 n 4, 2 32 n 5, 2 14 Required sum = 62 + 56 + 46 + 32 + 14 = 210 88.
For non – trivial solution D=0 1 c c c
1
c 0 2c 3 3c 2 1 0
c
c
1 2
c 1 2c 1 0 Greatest value of c is
89.
1 2
x f x f 2 x ' x f x f ' 2 x
………(1)
Since f " x 0 f x is increasing x 0,2 Case – I : When x 2 x x 1 ' x 0 x 1,2 x is increasing on 1, 2 Case – II : When x 2 x x 1 ' x 0 x 0, 1 x is decreasing on (0, 1)
90.
P A B
P A B P B
PA P B
(as A B P A B P A )
P A B P A
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FIITJEE Solutions to JEE(Main)-2019 Test Date: 8th April 2019 (Second Shift)
PHYSICS, CHEMISTRY & MATHEMATICS Paper - 1 Time Allotted: 3 Hours
Maximum Marks: 360
Please read the instructions carefully. You are allotted 5 minutes specifically for this purpose.
Important Instructions: 1.
The test is of 3 hours duration.
2.
This Test Paper consists of 90 questions. The maximum marks are 360.
3.
There are three parts in the question paper A, B, C consisting of Physics, Chemistry and Mathematics having 30 questions in each part of equal weightage. Each question is allotted 4 (four) marks for correct response.
4.
Out of the four options given for each question, only one option is the correct answer.
5.
For each incorrect response 1 mark i.e. ¼ (one-fourth) marks of the total marks allotted to the question will be deducted from the total score. No deduction from the total score, however, will be made if no response is indicated for an item in the Answer Box.
6.
Candidates will be awarded marks as stated above in instruction No.3 for correct response of each question. One mark will be deducted for indicating incorrect response of each question. No deduction from the total score will be made if no response is indicated for an item in the answer box.
7.
There is only one correct response for each question. Marked up more than one response in any question will be treated as wrong response and marked up for wrong response will be deducted accordingly as per instruction 6 above.
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JEE-MAIN-2019 (8th Apr-Second Shift)-PCM-2
PART – A (PHYSICS) 1.
A solid sphere and solid cylinder of identical radii approach an incline with the same linear velocity (see figure). Both roll without slipping all throughout. The two climb h maximum heights hsph and hcyl on the incline. The radio sph is given by: hcyl
(A) 1 (C) 2.
4 5 14 (D) 15
(B) 2 5
In the circuit shown, a four wire potentiometer is made of a 400 cm long wire, which extends between A and B. The resistance per unit length of the potentiometer wire is r = 0.01 /cm. If an ideal voltmeter is connected as shown with jockey J at 50 cm from end A, the expected reading of the voltmeter will be:
(A) 0.75V (C) 0.25V
(B) 0.20V (D) 0.50V
3.
A parallel plate capacitor has 1F capacitance. One of its two plates is given + 2C charge and the other plate, +4C charge. The potential difference developed across the capacitor is: (A) 3V (B) 1V (C) 5V (D) 2V
4.
If Surface tension (S), Moment of Inertia (I) and Planck’s constant (h), were to be taken as the fundamental units, the dimensional formula for linear momentum would be: (A) S1 / 2 I1 / 2 h0 (B) S1 / 2 I3 / 2 h1 (C) S3 / 2 I1 / 2 h0 (D) S1 / 2 I1 / 2 h1
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JEE-MAIN-2019 (8th Apr-Second Shift)-PCM-3 5.
Two magnetic dipoles X and Y are placed at a separation d, with their axes perpendicular to each other. The dipole moment of Y is twice that of X. A particle of charge q is passing through their mid-point P, at angle = 450 with the horizontal line as shown in the figure. What would be the magnitude of force on the particle at that instant? (d is much larger than the dimensions of the dipole)
M (A) 0 q 3 4 d / 2
(B) 0
2M (C) 0 q 3 4 d / 2
(D)
M 2 0 q 3 4 d / 2
6.
A damped harmonic oscillator has a frequency of 5 oscillations per second. The amplitude drops to half its value for every 10 oscillations. The time it will take to drop to 1 of the original amplitude is close to: 1000 (A) 10s (B) 100s (C) 50s (D) 20s
7.
A convex lens (of focal length 20 cm) and a concave mirror, having their principal axes along the same lines, are kept 80 cm apart from each other. The concave mirror is to the right of the convex lens. When an object is kept at a distance of 30 cm to the left o the convex lens, its image remains at the same position even if the concave mirror is removed. The maximum distance of the object for which this concave mirror, by itself would produce a virtual image would be: (A) 20 cm (B) 10 cm (C) 30 cm (D) 20 cm
8.
A cell of internal resistance r drives current through an external resistance R. The power delivered by the cell to the external resistance will be maximum when: (A) R = 0.001 r (B) R = 1000 r (C) R = 2r (D) R = r
9.
In a simple pendulum experiment for determination of acceleration due to gravity (g), time taken for 20 oscillations is measured by using a watch of 1 second least count. The mean value of time taken comes out to be 30s.. The length of pendulum is measured by using a meter scale of least count 1 mm and the value obtained is 55.0 cm. The percentage error in the determination of g is close to: (A) 0.7 % (B) 3.5% (C) 6.8% (D) 0.2%
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JEE-MAIN-2019 (8th Apr-Second Shift)-PCM-4 10.
The temperature, at which the root mean square velocity of hydrogen molecules equals their escape velocity from the earth is closest to: [Boltzmans Constant kB 1.38 1023 J / K Avogadro number NA 6.02 10 26 / kg Radius of Earth: 6.4 10 6 m Gravitation acceleration on Earth = 10 ms-2] (A) 800k (B) 104 K 5 (C) 3 10 (D) 650 K
11.
A particle starts from origin O from rest and moves with a uniform acceleration along the positive x – axis. Identify all figures that correctly represent the motion qualitatively. (a = acceleration, v = velocity, x = displacement, t = time) (a)
(b)
(c)
(d)
(A) a, b, c (C) b, c 12.
13.
(B) a (D) a, b, d
The election field in a region is given by E Ax B ˆi where E is in NC-1 and x in meters. The values of constants are A = 20 SI unit and B = 10 SI unit. If the potential at x =1 is V1 and that at x = – 5 is V2 then V1 – V2 is: (A) 320 V (B) – 48 V (C) – 520 V (D) 180 V
Young’s moduli of two wires A and B are in the ratio 7 : 4. Wire A is 2 m long and has radius R. Wire A is 2 m long and has radius R. Wire B is 1.5 m long and has radius 2 mm. If the two wires stretch by the same length for a given load, then the value of R is close to: (A) 1.3 mm (B) 1.5 mm (C) 1.7 mm (D) 1.9 mm
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JEE-MAIN-2019 (8th Apr-Second Shift)-PCM-5
14.
A body of mass m1 moving with an unknown velocity of v1ˆi undergoes a collinear collision with a body of mass m2 moving with a velocity v 2 ˆi . After collision m1 and m2 move with velocities of v ˆi and v ˆi respectively. 3
4
If m2 = 0.5 m1 and v3 = 0.5 v1 then v1 is: v (A) v 4 2 2 (C) v 4 v 2
v2 4 (D) v 4 v 2
(B) v 4
40
Ca and 16 O is close to: (B) 5 (D) 1
15.
The ratio of mass densities of nuclei of (A) 0.1 (C) 2
16.
A rectangles solid box of length 0.3 m is held horizontally, with one of its sides on the edge of a platform of height 5.. When released, it slips off the table in a very short time = 0.01 s, remaining essentially horizontal. The angle by which it would rotate when it hits the ground will be (in radians) close to:
(A) 0.5 (C) 0.02
(B) 0.6 (D) 0.28
17.
An electric dipole is formed by two equal and opposite charges q with separation d. The charges the same mass m. It is kept in a uniform electric field E. If it slightly rotated from its equilibrium orientation, then its angular frequency is: qE 2qE (A) (B) 2md md qE qE (C) (D) 2 md md
18.
The magnetic field of an electromagnetic wave is given by: Wb B 1.6 10 6 cos 2 107 z 6 1015 t 2iˆ ˆj 2 2 The associated electric field will be: V (A) E 4.8 102 cos 2 107 z 6 1015 t ˆi 2jˆ m V (B) E 4.8 102 cos 2 107 z 6 1015 t 2ˆj 2tˆ m V (C) E 4.8 102 cos 2 107 z 6 1015 t ˆi 2jˆ m V (D) E 4.8 102 cos 2 107 z 6 1015 t 2iˆ ˆj m
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JEE-MAIN-2019 (8th Apr-Second Shift)-PCM-6
19.
A common emitter amplifier circuit, built using an npn transistor, is shown in the figure. Its dc current gain is 250, RC = 1k and VCC = 10V. What is the minimum base current for VCE to reach saturation?
(A) 7A (C) 10 A 20.
21.
A positive point charge is released from rest at a distance r0 from a positive line charge with uniform density. The speed (v) of the point charge, as a function of instantaneous distance r from line charge, is proportional to
(A) v e r /ro
r (B) v In r0
r (C) v In r0
r (D) v r0
Let A 1 3, A 2 5, and A 1 A 2 5. The value of 2A 1 3A 2 3A 1 2A 2 is:
(A) –106.5 (C) –118.5 22.
(B) 40A (D) 100 A
(B) –112.5 (D) –99.5
In the figure shown, what is the current (in Ampere) drawn from the battery? You are given R1 15,R2 10, R3 20, R4 5,R5 25, R6 30, E 15V
(A) 13/24 (C) 9/32
(B) 7/18 (D) 20/3
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JEE-MAIN-2019 (8th Apr-Second Shift)-PCM-7 23.
In a line of sight radio communication, a distance of about 50 km is kept between the transmitting and receiving antennas. If the height of the receiving antenna is 70m, then the minimum height of the transmitting antenna should be: (Radius of the Earth = 6.4 10 6 m) (A) 32 m (B) 40 m (C) 51 m (D) 20 m
24.
A rocket has to be launched from each in such a way that it never returns. If E is the minimum energy delivered by the rocket launcher, what should be the minimum energy that the launcher should have if the same rocket is to be launched from the surface of the moon? Assume that the density of the earth and the moon are equal and that the earth’s volume is 64 times the volume of the moon. E E (A) (B) 32 16 E E (C) (D) 64 4
25.
A circuit connected to an ac source of emf e = e0 sin (1000t) with t in seconds, gives a phase difference of between the emf e and current i. Which of the following circuits 4 will exhibit this? (A) RC circuit with R = 1 k and C = 1 F (B) RL circuit with R = 1 k and L = 10 mH (C) RL circuit with R = 1 k and L = 1 mH (D) RC circuit with R = 1 k and C = 10 F
26.
A nucleus A, with a finite de – broglie wavelength A, undergoes spontaneous fission into two nuclei B and C of equal mass. B flies in the same direction as that of A, while C flies in the opposite direction with a velocity equal to half of that of F. The de – Broglie wavelength B and C of B and C are respectively : (A) A , 2 A (B) 2 A , A (C) A , A (D) A , A 2 2
27.
Calculate the limit of resolution of a telescope objective having a diameter of 200 cm, if it has to detect light of wavelength 500 nm coming from a star. (A) 457.5 10 9 radian (B) 610 10 9 radian (C) 305 109 radian
(D) 152.5 10 9 radian
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JEE-MAIN-2019 (8th Apr-Second Shift)-PCM-8 28.
Two very long, straight and insulated wires are kept at 900 angle from each other In xy – plane as shown in the figure.
These wires carry currently of equal magnitude I, whose directions are shown in the figure. The met magnetic field at point P will be: I 0I (A) 0 xˆ yˆ (B) zˆ 2d d I (C) Zero (D) 0 xˆ yˆ 2d 29.
The given diagram shows four processes i.e., isochoric, isothermal and adiabatic. The correct assignment of the processes, in the same order is given by:
(A) a d c b (C) d a c b 30.
(B) a d b c (D) d a b c
A uniform rectangular thin sheet ABCD of mass M has length a and breadth b, as shown in the figure. If the shaded portion HBGO is cut off, the coordinates of the centre of mass of the remaining portion will be:
5a 5b (A) , 3 3 3a 3b (C) , 4 4
2a 2b (B) , 3 3 5a 5b (D) , 12 12
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JEE-MAIN-2019 (8th Apr-Second Shift)-PCM-9
PART –B (CHEMISTRY) 31.
The statement that is INCORRECT about the interstitial compound is: (A) they have metallic conductivity (B) they have high melting points (C) they are chemically reactive (D) they are very hard
32.
The major product of the following reaction is:
33.
34.
(A)
(B)
(C)
(D)
The major product of the following reaction is:
(A)
(B)
(C)
(D)
If p is the momentum of the fastest electron ejected from a metal surface after the irradiation of light having wavelength , then for (A)5 p momentum of the photoelectron, the wavelength of the light should be: (Assume kinetic energy of ejected photoelectron to be very high in comparison to work function)
3 4 4 (C) 9 (A)
1 2 2 (D) 3 (B)
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JEE-MAIN-2019 (8th Apr-Second Shift)-PCM-10 35.
The ion that has sp3d2 hybridization for the central atom is: (A) IF6 – (B) ICI4 – (C) ICI2 –
36.
37.
(D) BrF2 –
The major product obtained in the following reaction is:
(A)
(B)
(C)
(D)
k
k
1 2 For a reaction scheme. A B C if the rate of formation of B is set to be zero then the concentration of B is given by
k1 A k 2
38.
(A)
(B) k1 k 2 A
(C) k1k 2 A
(D)
k1 k 2 A
For the following reactions, equilibrium constants are given: 52 S s O2 g SO2 g ;K1 10 129 2S s 3O2 g 2SO3 g;K 2 10
The equilibrium constant for the reaction 2SO2 g O2 g 2SO3 g is: (A) 1077 (C) 10181 39.
(B) 1025 (D) 10154
Calculate the standard cell potential (in V) of the cell in which following reaction takes place:
Fe2 aq Ag aq Fe3 aq Ag s Given that:
E 0 Ag'/ Ag xV E 0Fe2 /Fe yV E 0Fe3 /Fe zV (A) x – z (C) x – y
(B) x + y – z (D) x + 2y – 3z
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JEE-MAIN-2019 (8th Apr-Second Shift)-PCM-11
40.
41.
The covalent alkaline earth metal halide X CI,Br,I is: (A) BeX2 (C) SrX2
(B) CaX2 (D) MgX2
The structure of Nylon – 6 is: (A)
(B)
(C)
(D)
42.
The IUPAC symbols for the element with atomic number 119 would be: (A) uun (B) uue (C) unh (D) une
43.
Which of the following compounds will show the maximum ‘enol’ content? (A) CH3COCH3 (B) CH3COCH2CONH2 (C) CH3COCH2COCH3 (D) CH3COCH2COOC2H5
44.
0.27 g of a long chain fatty acid was dissolved in 100 cm3 of hexane. 10 mL of this solution was added dropwise to the surface of water in a round watch glass. Hexane evaporates and a monolayer is formed. The distance from edge to centre of the watch glass is 10 cm. What is the height of the monolayer? [Density of fatty acid = 0.9 g cm-3, = 3] (A) 10-4 m (B) 10-8 m -2 (C) 10 m (D) 10-6 m
45.
The percentage composition of carbon by mole in methane is: (A) 80% (B) 20% (C)75% (D) 25%
46.
The Mond process is used for the: (A) Extraction of Mo (C) Purification of Zr and Ti
(B) Extraction of Zn (D) Purification of Ni
47.
Which one of the following alkenes when treated with HCI yields majorly an anti Markovnikov product? (A) H2N – CH = CH2 (B) F3C – CH = CH2 (C) CH3 O – CH = CH2 (D) CI – CH = CH2
48.
5 moles of an ideal gas at 100 K are allowed to undergo reversible compression till its temperature becomes 200 K. If C V 28 JK 1 mol1 , calculate U and pV for this process. (R = 8.0 J K–1 mol-1) (A) U 2.8kJ; pV 0.8kJ (B) U 14 kJ; pV 4kJ (C) U 14kJ; pV 18kJ
(D) U 14kJ; pV 0.8J
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JEE-MAIN-2019 (8th Apr-Second Shift)-PCM-12 49.
The calculated spin only magnetic moments (BM) of the anionic and cationic species of
Fe H2O 6 and Fe CN 6 , respectively, are: 2 (A) 0 and 5.92 (C) 0 and 4.9
(B) 2.84 and 5.92 (D) 4.9 and 0
50.
Fructose and glucose can be distinguished by: (A) Benedict’s test (B) Fehling’s test (C) Barfoed’s test (D) Seliwanoff’s test
51.
The major product in the following reaction is:
52.
(A)
(B)
(C)
(D)
Polysubstitution is a major drawback in: (A) Friedel Craft’s alkylation (C) Acetylation of aniline
(B) Reimer Tiemann reaction (D) Friedel Craft’s acylation
53.
The strength of 11.2 volume solution of H2 O 2 is: [Given that molar mass of H = 1 g mol-1 and O = 16 g mol-1] (A) 3.4% (B) 1.7% (C) 13.6% (D) 34%
54.
For the solution of the gases w, x, y and z in water at 298 K, the Henrys law constants (KH) are 0.5, 2 35 and 40 kbar, respectively. The correct plot for the given data is:
(A)
(B)
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JEE-MAIN-2019 (8th Apr-Second Shift)-PCM-13
(C)
55.
56.
(D)
Among the following molecules / ions C 22 ,N22 ,O 22 ,O 2 which one is diamagnetic and has the shortest bond length? (A) O 22
(B) C 22
(C) O2
(D) N22
The major product obtained in the following reactions is”
(A)
(B)
(C)
(D)
57.
The maximum prescribed concentration of copper in drinking water is: (A) 0.5 ppm (B) 3 ppm (C) 5 ppm (D) 0.05 ppm
58.
The correct statement about ICI5 and ICI4 is: (A) ICI5 is trigonal bipyramidal and ICI4 is tetrahedral (B) ICI5 is square pyramidal and ICI4 is square planar (C)ICI5 is square pyramidal and ICI4 is tetrahedral (D) Both are isostructural
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JEE-MAIN-2019 (8th Apr-Second Shift)-PCM-14 59.
The compound that inhibits the growth of tumors is: (A) trans - Pd CI 2 NH3 2 (B) trans - Pt CI 2 NH3 2 (C) cis - Pd CI 2 NH3 2
60.
(D) cis - Pt CI 2 NH3 2
Consider the bcc unit cells of the solid 1 and 2 with the position of atoms as shown below. The radius of atom B is twice that of atom A. The unit cell edge length is 50% more in solid 2 than in 1. What is the approximate packing efficiency in solid 2?
(A) 90% (C) 65%
(B) 75% (D) 45%
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JEE-MAIN-2019 (8th Apr-Second Shift)-PCM-15
PART–C (MATHEMATICS) 61.
Let S
x, y : y
2
x,0 and A is area of the regions S . If for a
, 0 4, A : A 4 2 : 5 then equals: 1
1
2 3 (A) 4 5
2 3 (B) 2 5 1
1
4 3 (C) 4 25
4 3 (D) 2 25
x
62.
x
Let f x g t dt, were g is a non zero even function. If f(x+5) =g(x) , then 0
f t dt 0
equals 5
(A)
x5
g t dt
(B) 2
x 5 x 5
(C)
g t dt
g t dt 5 5
(D) 5
5
g t dt
x5
63.
Suppose that the points (h, k), (1, 2) and (–3, 4) lie on the line L1. If a line L2 passing k through the points (h, k) and (4, 3) is perpendicular to L1, then equals: h 1 1 (A) (B) 7 3 (C) 3 (D) 0
64.
Let f x a x a 0 be written as f x f1 x f2 x , where f1 x is an even function and f2 x is an odd function. Then f1 x y f1 x y equals
65.
(A) 2f1 x f2 y
(B) 2f1 x f1 y
(C) 2f1 x y f2 x y
(D) 2f1 x y f1 x y
Let f : 1,3 R be defined as x x , 1 x 1 f x x x , 1 x 2 x x, 2x3 where [t] denotes the greatest integer less than or equal to t. Then, ƒ is discontinuous at: (A) four or more points (B) only one point (C) only two points (D) only three points
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JEE-MAIN-2019 (8th Apr-Second Shift)-PCM-16
66.
3 i i 1 , then 1 iz z5 iz8 2 2
If z
9
(A) –1 (C) (–1 + 2i)9 67.
(B) 1 (D) 0
Which one of the following, statements is not a tautology: (A) p v q p v q (B) p v q p (D) p ^ q p v q
(C) p p v q 68.
2y . If the x2 curve passes through the centre of the circle x 2 y 2 2x 2y = 0, then its equation is:
Given that the slope of the tangent to a curve y = y(x) at any point (x, y) is
(B) x loge y 2 x 1
(A) x loge y x 1 2
(C) x loge y 2 x 1 69.
is equal to:
(D) x loge y 2 x 1
If f 1 1,f ' 1 3 , then the derivative of f f f x f x (A) 33 (C) 9
2
at x = 1 is:
(B) 15 (D) 12
70.
The number of four – digit numbers strictly greater than 4321 that can be formed using the digits 0, 1, 2, 3, 4, 5 (repetition of digits is allowed) is: (A) 360 (B) 288 (C) 310 (D) 306
71.
If a point R (4, y, z) lies on the line segment joining the points P (2, –3, 4) and Q (8, 0, 10), then the distance of R from the origin is: (A) 53 (B) 6 (C) 2 14
(D) 2 21
72.
If the system of linear equations x 2y kz 1 2x y z 2 3x y kz 3 Has a solution (x, y, z) 0, then (x, y) lies on the straight line whose equation is: (A) 3x – 4y – 1 = 0 (B) 4x – 3y – 4 = 0 (C) 4x – 3y – 1 = 0 (D) 3x – 4y – 4 = 0
73.
A student score the following marks in five tests : 45, 54, 41, 57, 43. His score is not known for the sixth test. If the mean score is 48 in the six tests, then the standard deviation of the marks in six tests is: 10 100 (A) (B) 3 3 100 10 (C) (D) 3 3
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JEE-MAIN-2019 (8th Apr-Second Shift)-PCM-17 74.
The vector equation of the plane through the line of intersection of the planes x + y + z = 1 and 2x + 3y + 4z = 5 which is perpendicular to the plane x – y + z = 0 is: (A) r ˆi kˆ 2 0 (B) r. ˆi kˆ 2 0 (C) r. ˆi kˆ 2 0 (D) x ˆi kˆ 2 0
75.
If the lengths of the sides of a triangle are in A.P. and the greatest angle is double the smallest, then a ratio of lengths of the sides of this triangle: (A) 4:5:6 (B) 5:6:7 (C) 3:4:5 (D) 5:9:13
76.
In an ellipse, with centre at the origin, if the difference of the lengths of major axis and minor axis is 10 and one of the foci is at (0, 5 3 ), then the length of its latus rectum is: (A) 6 (B) 5 (C) 8 (D) 10
77.
The minimum number of times one has to toss a fair coin so that the probability; of observing at least one head is at least 90% is: (A) 2 (B) 3 (C) 4 (D) 5
78.
If the eccentricity of the standard hyperbola passing, through the point (4, 6) is 2, then the equation of the tangent to the hyperbola at (4, 6) is: (A) 2x – 3y + 10 = 0 (B) x – 2y + 8 = 0 (C) 2x – y – 2 = 0 (D) 3x – 2y = 0
79.
The number of integral values of m for which the equation 1 m2 x 2 2 1 3m x 1 8m 0 has no real root is:
(A) infinitely many (C) 3 20
80.
Then sum
1
k 2
k
(B) 2 (D) 1 is equal to:
k 1
11 219 21 (C) 2 20 2
(A) 2
81.
11 220 3 (D) 2 17 2
(B) 1
ˆ for some real x. Then a b r is possible if: Let a 3iˆ 2ˆj xkˆ and b ˆi ˆj k,
(A) r 5 (C)
3 2
3 3 r3 2 2
3 3 r 5 2 2 3 (D) 0 r 2
(B) 3
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JEE-MAIN-2019 (8th Apr-Second Shift)-PCM-18
82.
Then tangent and the normal lines at the point
3,1 to the circle x 2 y 2 4 and the x
– axis form a triangle. The area of this triangle (in square units) is: 1 4 (A) (B) 3 3 1 2 (C) (D) 3 3 83.
The tangent to the parabola y2 = 4x at the point where it intersects the circle x2 + y2 = 5 in the first quadrant, passes through the point: 1 4 3 7 (A) , (B) , 3 3 4 4 1 1 1 3 (C) , (D) , 4 2 4 4
84.
The height of a right circular cylinder of maximum volume inscribed in a sphere of radius 3 is: (A) 3 (B) 6 2 (C) 2 3 (D) 3 3 1 1 1 Let the numbers 2, b, c be in an A.P and A 2 b c . If det A 2,16 then ce lies 4 b 2 c 2
85.
86.
in the interval (A) 3,2 22/4
(B) 2 23/4 , 4
(C) {2,3)
(D) [4, 6]
Let f : R R be a differentiable function satisfying f’’(3) + f’(2) = 0. Then 1
1 f 3 x f 3 x lim is equal to: x 1 f 2 x f 2 (A) e2 (C) e
87.
If
x
dx 3
6 2/3
1 x
xf x 1 x 6
function f(x) is equal to: 1 (A) 2 2x 1 (C) 3 2x
1 3
(B) 1 (D) e–1
C where C is a constant of integration, then the
1 2x 3 3 (D) 2 x
(B)
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JEE-MAIN-2019 (8th Apr-Second Shift)-PCM-19 88.
89.
90.
Two vertical poles of heights, 20 m and 80m stand a apart on a horizontal plane. The height (in meters) of the point of intersection of the lines joining the top of each pole to the foot of the other, from his horizontal plane is: (A) 18 (B) 12 (C) 16 (D) 15 1 1 If the fourth term in the binomial expansion of 1 log x x 12 x 10 1, then the value of x is: (A) 104 (B) 100 (C) 103 (D) 10
6
is equal to 200, and x >
If three distinct number a, b, c are in G.P. and the equations ax2 + 2bc + c = 0 and dx2 + 2ex + f = 0 have a common root, then which one of the following statements is correct? d e f (A) , , are in A.P (B) d, e, f are in A.P a b c d e f (C) , , are in G.P (D) d, e, f are in G.P a b c
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JEE-MAIN-2019 (8th Apr-Second Shift)-PCM-20
JEE (Main) – 2019 ANSWERS PART A – PHYSICS 1. 5. 9. 13. 17. 21. 25. 29.
D B C C B C D D
2. 6. 10. 14. 18. 22. 26. 30.
C D B C A C D D
3. 7. 11. 15. 19. 23. 27.
B B D D B A C
4. 8. 12. 16. 20. 24. 28.
A D D A C B C
34. 38. 42. 46. 50. 54. 58.
C B B D D D B
64. 68. 72. 76. 80. 84. 88.
B D B B A C C
PART B – CHEMISTRY 31. 35. 39. 43. 47. 51. 55. 59.
C B D C B B B D
32. 36. 40. 44. 48. 52. 56. 60.
B B A D B A D A
33. 37. 41. 45. 49. 53. 57.
C A C B C A B
PART C – MATHEMATICS 61. 65. 69. 73. 77. 81. 85. 89.
C 62. A A 66. A A 70. C D 74. C C 78. C A 82. D D 86. B All options are incorrect
63. 67. 71. 75. 79. 83. 87. 90.
B A C A A B B A
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JEE-MAIN-2019 (8th Apr-Second Shift)-PCM-21
HINTS AND SOLUTIONS PART A – PHYSICS 1.
For solid sphere 1 1 2 v2 mv 2 mR 2 2 mghsph 2 2 5 R For solid cylinder 1 1 1 v2 mv 2 mR 2 2 mghcyl 2 2 2 R hsph 7 / 5 14 hcyl 3 / 2 15
2.
Resistance of wire AB = 400 × 0.01 = 4 3 i 0.5 A 6 Now voltmeter reading = i (Resistance of 50 cm length) = (0.5 A) (0.01 × 50) = 0.25 volt
3.
4.
Charges at inner plates are 1 C and –1 C. Potential difference across capacitor q 1C 1 10 6 C 1V = c 1F 1 10 6 Farad
p = k saIbhc Where k is dimensionless constant MLT–1 = (MT–2)a(ML2)b(ML2T–1)c a+b+c=1 2b + 2c = 1 –2a – c = –1 1 1 a b c 0 2 2 S1/2II/2h0
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JEE-MAIN-2019 (8th Apr-Second Shift)-PCM-22 5.
M 2M B1 2 0 ; B2 0 3 3 4 (d / 2) 4 (d / 2) B1 = B2 Bnet is at 45° ( = 45°)
Velocity of charge and Bnet are parallel so by F q(b B) force on charge particle is zero. 6.
A = A0e–t A A 0 after 10 oscillations 2 After 2 seconds A0 A 0 e (2) ; 2 = e2 2 n2 n2 2 ; 2 –t A = A0e A n2 n 0 t ; n1000 t A 2 6n10 3n10 2 t ; t n2 n2 t = 19.931 sec t 20 sec
7.
Image formed by lens 1 1 1 1 1 1 ; v u f v 30 20 v = +60 cm If image position does not change even when mirror is removed it means image formed by lens is formed at centre of curvature of spherical mirror. Radius of curvature of mirror = 80 – 60 = 20 cm. Focal length of mirror f = 10 cm for virtual image, object is to be kept between focus and pole. Maximum distance of object from spherical mirror for which virtual image is formed, is 10 cm.
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JEE-MAIN-2019 (8th Apr-Second Shift)-PCM-23
8.
E r R Power generated in R P = i2R E2R P (r R)2 dP For maximum power 0 dR (r R)2 1 R 2(r R) E2 0 (r R)4 Current i =
r=R 9.
T
30 sec 1 T sec, 20 20 L = 55 cm L = 1 mm = 0.1 cm 2 4 L g= T2 percentage error in g is g L 2T 100 100% g T L 0.1 = 55
10.
v rms
3RT m
1 2 20 100% 6.8%. 30 20
v escape 2gRe
v rms v escape
3RT 2gRe m 3 1.38 10 23 6.02 10 26 10 103 104 k 2 11.
Given initial velocity u = 0 and acceleration is constant At time t v = 0 + at v = at 1 Also x = 0(t) + at 2 2 1 2 x at 2
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JEE-MAIN-2019 (8th Apr-Second Shift)-PCM-24
Graph (a), (b) and (d) are correct. 12.
E (20x 10)iˆ 1
V1 V2 (20x 10)dx 5
V1 V2 (10x 2 10x)15 V1 – V2 = 10(25 – 5 – 1 – 1) V1 – V2 = 180 V
13.
Given: YA 7 YB 4
LA = 2m
AA = R2
LB = 1.5 m
AB = (2mm))2
F Y A L given F and are same
AY is same L
A A YA A B YB LA LB 7 ( R2 ) YB 2 4 (2 mm) YB 2 1.5 R = 1.74 mm
14.
Applying linear momentum conservation m1v1ˆi m 2 v 2 ˆi m1v 3 ˆi m 2 v 4 ˆi m1v 1 0.5 m1v 2 m1(0.5 v1 ) 0.5 m1v 4 0.5 m1v1 = 0.5 m1(v4 – v2) v1 = v 4 – v2
15.
Mass densities of all nuclei are same so their ratio is 1.
16.
Angular impulse = change in angular momentum. t = L m 2 mg 0.01 2 3 3g 0.01 3 10 0.01 1 = = 0.5 rad/s 2 2 0.3 2
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JEE-MAIN-2019 (8th Apr-Second Shift)-PCM-25 Time taken by rod to hit the ground 2h 2 5 t = 1 sec. g 10 In this time angle rotate by rod = t = 0.5 × 1 = 0.5 radian 17.
Moment of inertia 2 md2 d (I) m 2 2 2 Now by = I md2 (qE) (d sin ) = 2 2qE sin for small md
2qE md Angular frequency 18.
19.
If we use that direction of light propagation will be along E B. Then (A) option is correct. Magnitude of E = CB E = 3 × 108 × 1.6 × 10–6 × 5 E = 4.8 × 102 5 E and B are perpendicular to each other E B = 0 Either direction of E is ˆi 2ˆj or ˆi 2ˆj from given option Also wave propagation direction is parallel to E B which is kˆ E is along ( ˆi 2ˆj) At saturation state, VCE becomes zero 10V iC = 10 mA 1000 i Now current gain factor C iB iB
20.
2qE md
10 mA = 40 A 250
1 mV 2 q(Vf Vi ) 2 E 20r FIITJEE Ltd., FIITJEE House, 29-A, Kalu Sarai, Sarvapriya Vihar, New Delhi -110016, Ph 46106000, 26569493, Fax 26513942 website: www.fiitjee.com.
JEE-MAIN-2019 (8th Apr-Second Shift)-PCM-26 r n 0 20 r 1 q r0 mv 2 n 2 2 0 r
V
r v n r0
21.
A 1 3, A 2 5, and A1 A 2 5. 2 2 A 1 A 2 A 1 A 2 2 A 1 A 2 cos
3 cos 10 2A 1 3A 2 3A 1 2A 2 2 2 = 6 A1 9A1 A 2 4A 1 A 2 6 A 2
= –118.5 22.
25 45 25 90 160 30 3 3 3 E 15 3 9 I amp. Req 160 32 Req 15
23.
Range =
2RhT 2RhR
50 103 2 6400 103 hT 2 6400 103 70 By solving hT = 32 m.
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JEE-MAIN-2019 (8th Apr-Second Shift)-PCM-27 24.
Minimum energy required (E) = – (Potential energy of object at surface of earth) GMm GMm R R Now Mearth = 64 Mmoon 4 4 R3e 64 Rm3 Re = 4Rm 3 3 E M R 1 4 Now moon moon earth E earth Mearth Rmoon 64 1
25.
Emoon
E 16
and = 100 rad/s 4 Reactance (X) = Resistance (R) Now by checking option. Given phase difference =
Option (A) R = 1000 and XC
1 = 104 10 100 6
Option (B) R = 103 and XL = 10 × 10–3 × 100 = 1 Option (C) R = 103 and XL = 10–3 100 = 10–1 Option (D) R = 103 and XC
1 = 103 10 10 6 100
26.
Let mass of B and C is m each. By momentum conservation mv 2mv 0 mv 2 v = 4v0 PA = 2mv0 pB = 4mv0 pc = 2mv0 h De-Broglie wavelength p h h h A ; B ; C 2mv 0 4mv 0 2mv 0
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JEE-MAIN-2019 (8th Apr-Second Shift)-PCM-28
27.
Limit of resolution of telescope =
1.22 D
1.22 500 10 9 = 305 × 10—9 radian 200 10 2
28.
Magnetic field at point P i ˆ i ˆ Bnet 0 ( k) 0 (k) =0 2d 2d
29.
isochoric Process d Isobaric Process a Adiabatic slope will be more than isothermal so Isothermal Process b Adiabatic Process c Order d a b c
30.
a M 3a M x 2 4 4 M M 4 a 3a 3a 5a 2 16 4 16 5a = 3 3 12 4 4 b M 3b M 5b y= 2 4 4 M 12 M 4
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JEE-MAIN-2019 (8th Apr-Second Shift)-PCM-29
PART B – CHEMISTRY 31.
In case of interstial compounds there is presence of small atoms (or impurity) in the lattice of metal. So interstial compounds are hard, high melting point and chemically inert.
32. O
O
t BuO t BuOH
Cl
Cl
O
H
OH
O
O-H
33.
Reaction involves free radical chlorination followed by hydrolysis. (a) h Cl2 Cl Cl (b)
Cl
CH3 Cl Cl
Cl
CH2 Cl Cl
(c)
(d)
34.
H O
Cl
2 Cl CH2Cl
Cl
2 2 CH2OH Cl
(e)
CH2 HCl
Cl /H O O
CH2Cl
CH2OH
CHO
1 1 m 2u2 1 P 2 mu2 2 2 m 2 m hc 1 P2 Wo K.E 1 Wo 1 2m
K.E
hc 1 1.5P Wo K.E 2 Wo 2 2 m
2
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JEE-MAIN-2019 (8th Apr-Second Shift)-PCM-30 On solving 4 2 1 9 Since the K.E is very high is comparison to work function, then we can assume that K.E + Wo = K.E 35.
[ICl4]– sp3d2 [IF6]– sp3d2 [ICl2]– sp3d [BrF2]– sp3d
36.
37.
Applying steady state of approximation d B K1 A K 2 B dt O K 1 A K 2 B K1 A B K2
38.
2SO2 g O2 g 2SO3 g 2
K eq
SO3 2 O2 SO2
K 2 10129 1025 K1 10104 Oxidation
39.
Fe
2
aq Ag aq Fe3 aq Ag s E0Cell 3 Reduction
Given:
E0 Ag'/ Ag x
1
E0Fe2 /Fe y
2
E0Fe3 /Fe z
3
Using equation:
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JEE-MAIN-2019 (8th Apr-Second Shift)-PCM-31 G0 nFE0 G01 Fx G02 2Fy G03 3Fz Fe2 2e Fe 2Fy Fe3 3e Fe 3Fz
Fe2 Fe3 e 2Fy 3Fz
Ag e Ag Fx 0 GTotal 2Fy 3Fz Fx FECell
E0Cell x 2y 3z 40.
The % covalent character can be predicted by using Fajan’s rule % of covalent character Be
2
Mg Ca Sr
1 cationic radius
2
2
2
Size
41.
The direct monomer of Nylon-6 is caprolactum which polymerises to give Nylon-6 as follows: H N
O 533 K N2 inert
Nylon-6 42
43.
The IUPAC name of element having atomic number 119 is “Ununennium”. So its symbol is uue. O
O
C
C
H3C CH2 CH3 Due to presence of active methylene group and stabilization of enol by intramoelcular H bond forming 6 membered ring structure.
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JEE-MAIN-2019 (8th Apr-Second Shift)-PCM-32 44.
Mass of fatty acid = 0.027 g Radius of plate = 10 cm Density of fatty acid = 0.9 g/cm3 0.027 Volume of fatty acid = 0.03 cm3 0.9 Area of plate = r2 = 3 102 = 300 cm2 Volume 0.03 Height of fatty acid = 104 cm 106 m Area 300
45.
In CH4 Mole of carbon nC = 1 Mole of hydrogen = nH = 4 % of nC =
46.
nC 1 100 100 20% nC nH 5
The mond’s process is used for refining of nickel as per following reaction.
Ni CO g Ni CO 4 Ni s 4CO g 330350K 450 470K
Impure
47.
Pure
In this case antimarkonikov product will be formed as major product because carbocation formed at a double bonded carbon having lesser number of H atom will be unstable due to presence of an electron withdrawing group (CF3) attached to it. HCl
F3C CH CH2 CF3 CH CH3 CF3 CH2 CH2 min or
major
Cl
CF3 CH2 CH2 CF3 CH2 CH2 Cl (major product)
48.
∆U = nCvm ∆T = 5 28 100 = 14 kJ ∆PV = nR∆T = 5 8 100 = 4 kJ
49.
[Fe(CN)6]4– Fe2 3d6 t 62geg0
[Fe(H2O)6]2+ Fe2 3d6 t 42geg2
n n 2
n n 2 24 4.9BM
= 0 50.
Barfoed’s test → Detecting presence of monosaccharides Fehling’s test → For aldehdyes Benedict’s test → For reducing sugars Seliwanoff’s test → Differentiate between aldose and ketose
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JEE-MAIN-2019 (8th Apr-Second Shift)-PCM-33 NH2
51. N
NH2 N
N
N
CH3 I
Base
N
N
N
H (Adenine)
N
CH3
The nitrogen atom marked with * contain high electron density. It is attached to one Hatom which can be removed by base OH– and the CH3 group gets attached there. 52.
Polysubstitution is a major drawback of Friedal craft’s alkylation. It is so because activating behaviour of benzene ring increases with increase in number of alkyl group on benzene ring.
53.
2H2O2 2H2O 68 g/L
O
2 22.4 of O 2 STP
11.2 L of O2 is given by 34 g of H2O2 present in 1 L of H2O2 solution. 34 So % strength of H2O2 = 100 3.4g / 100mL 3.4% 1000 54.
From Henry’s law Pgas = KH.Xgas Pgas = KH 1 XH O
2
Pgas = KH – KH XH O 2
Comparing this equation with straight line equation y = mx + c 55.
2 2 * 2 2 * 2 2 2 2 C 2 1s 1s 2s 2s 2p x 2Py 2Pz 10 4 B.O 3 diamagnetic 2 2 2 * 2 2 * 2 2 2 2 1 * 1 N2 1s 1s 2s 2s 2p x 2Py 2Pz 2p x 2Py 10 6 B.O 2 paramagnetic 2 2 2 * 2 2 * 2 2 2 2 2 * 2 O2 1s 1s 2s 2s 2Pz 2p x 2Py 2p x 2Py 10 8 B.O 1 diamagnetic 2 2 * 2 2 * 2 2 2 2 1 * 1 O2 1s 1s 2s 2s 2Pz 2p x 2Py 2p x 2Py 10 6 B.O 2 paramagnetic 2
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JEE-MAIN-2019 (8th Apr-Second Shift)-PCM-34 56.
The reaction involves carbylamine reaction of 1o amine with CHCl3/KOH followed by reduction with Pd/C/H2. NC NH2 CHCl3 /KOH
H CN
CN
O
O Pd/C/H2
H N - CH3
CH2NH2 OH 57.
Maximum prescribed concentration of copper in drinking water is 3 ppm. Above this concentration water becomes toxic.
58.
Cl
Cl
I
ICl4 sp3 d2
Cl
Cl Cl
Cl
Cl
I
ICl5 sp3d2 Cl
59.
Cl
cis–[PtCl2(NH3)2] is used in chemotherapy to inhibits the growth of tumors 4
60.
r 3 Vol. occupied by atom 3 % packing efficiency 100 3 100 Vol. of unit cell a Let radius of corner atom is r and radius of central atom is 2r So, 3a 2 2r 2r 6r 6r 2 3r 3 Now a
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JEE-MAIN-2019 (8th Apr-Second Shift)-PCM-35
% P.E =
4 3
4 3
4
3
r 3 3 2r
r 3 8r 3 8 3 3 r3
2 3r
3
100
100
4 9r 3 3 8 3 3 r3
100 90.6% 90%
PART C – MATHEMATICS
61.
S 2 x dx 0
S S 4
4 3/ 2 3
2 3/2 2 3/2 5 4 5 (, 0)
1/3
4 4 25
62.
Since g x is even with f 0 0 f x is odd function g x f x 5 g x f x 5 g x f x 5 Replace x by x 5 f x g x 5 x
x
f t dt g t 5 dt 0
0
x 5
g t dt 5
5
g t dt
x 5
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JEE-MAIN-2019 (8th Apr-Second Shift)-PCM-36 63.
Equation of L1 is x 2y 5 and equation of L2 is 2x y 5 Their point of intersection is (3, 1) k 1 h 3
(–3, 4) (h, k)
(1, 2) (4, 3)
L1 L2 x
64.
x
x
a a a a and f2 x 2 2 f1 x y f1 x y
x
f1 x
1 xy a a x y a x y a x y 2 1 ax ay a y a x a y a y 2 ax a x ay a y 2. 2 2
2f1 x f1 y
65.
x 1 x 1,0 x 0,1 x f x x 1,2 2x x 2 x 2, 3 f x is discontinuous at x 0,1 i
66.
3 i z e6 2
8 9
1 iz z iz 1 e e e 5
i /2
i /6
i5 /6
ei /2ei8 /6
11 i 1 ei 2 /6 ei5 /6 e 6
9
9
9
1 3 i /3 i e 2 2
9
ei3 1
67. (A) p q p ~ q ~ p q p ~ q ~ p ~ q p ~ q T FIITJEE Ltd., FIITJEE House, 29-A, Kalu Sarai, Sarvapriya Vihar, New Delhi -110016, Ph 46106000, 26569493, Fax 26513942 website: www.fiitjee.com.
JEE-MAIN-2019 (8th Apr-Second Shift)-PCM-37
(B) p q p ~ p q p ~ p ~ q p ~ p p ~ q =T (C) ~ p p q ~ p p q T
(D) ~ p q ~ p q ~ p ~ q ~ p q ~ p T T
68.
dy 2y dx x 2 2 nC x Passes through (1, 1) 0 2 n C 2 n y 2 x xn y 2 x 1 n y
69.
2
f f f f x f x
dy f ' f f x f ' f x f ' x 2f x f ' x dx Put x = 1 dy 27 6 33 dx
70.
Starting with 5 63 216 Starting with 44 62 36 Starting with 45 62 36 Starting with 43 18 (Not using 0, 1, 2) Starting with 432 = 4 Total = 310
71.
Equation of PQ is
x2 y3 z4 6 3 6 R (4, y, z) lies on this 1 y3 z4 3 3 6 R 4, 2, 6
QR 16 4 36 2 14
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JEE-MAIN-2019 (8th Apr-Second Shift)-PCM-38 72.
For infinitly many solution 1 2 k 2
1
1 0
3
1 k
1 2 Also consider x 2y k 1 and 2x y z 2 2x 4y z 2 2x y z 2 4x 3y 4 k
73.
41 45 54 57 43 x = 48 6 x 48 1 2 482 412 452 542 572 432 482 6 14024 2 2304 6 100 3
AM
74.
P1 : x y z 1 P2 :2x 3y 4z 5 Required plane is P1 P2 0
1 2 x 1 3 y 1 4 z 1 5 which is perpendicular to x y z 0
…..(i)
1 2 1 3 1 4 0
1 3 (i) x z 2 0 r . ˆi kˆ 2 0
75.
Given 2b a c Let A , B 3, C 2 2 sinB sin A sin C 2 sin3 sin sin 2
2 3 4 sin2 1 2cos 2
8 cos 2cos 3 0 3 cos 4 sin A :sinB :sin C
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JEE-MAIN-2019 (8th Apr-Second Shift)-PCM-39
1: 4 3 sin2 :2 cos 5 6 1: : 4 4 4:5:6 76.
be 5 3 b2 e2 75 b a b a 75 b a 15 b 10, a 5 LR
2a2 5 b
1 9 n 2 10 1 1 10 2n 2n 10 n4
77.
1
78.
Let equation of hyperbola be
x2 y2 1 a 2 b2
passes through (4, 6) 16 36 2 2 1 (i) a b b2 Also e2 1 2 b2 3a 2 a From (i) and (ii) a2 4, b2 12 Equation
(ii)
x2 y2 1 4 12
Tangent at (4, 6) is x
y 1 2
Or 2x y 2 79.
2
D 4 1 3m 4 1 m2 1 8m
2
4 1 9m 6m 1 8m m2 8m3
2
8m 4m 4m 1 2
8m 2m 1 0 Infinitely many values of m.
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JEE-MAIN-2019 (8th Apr-Second Shift)-PCM-40 20
k k 2 k 1 1 2 3 20 S 2 3 ..... 20 2 2 2 2 1 1 2 20 S 3 ....... 21 2 2 2 2 2 1 1 1 1 1 20 S 2 3 ...... 20 21 2 2 2 2 2 2 1 1 1 20 2 2 20 21 1 2 2 21 1 21 2 11 s 2 19 2
80.
S
81.
ˆi ab 3
ˆj
kˆ
2
x
1 1 1 2 x ˆi 3 x ˆj 5kˆ a b 2 x 2 x 19
2
82.
x 1/ 2
2
191/4
5 3 2
1 3 Equation of PQ is
Slope of OP
y 1 3 x 3
P
3, 1
y 3x 4 4 Q , 0 3 2 Area 3
O
Q
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JEE-MAIN-2019 (8th Apr-Second Shift)-PCM-41 83.
x 2 4x 5 x 5, x 1 P 1, 2 Tangent at P is y x 1
P (1, 2)
3 7 4 , 4 lies on this.
84.
h 2 3 cos 6 cos , r 3 sin
V r 2h 4 sin2 6 cos
54 sin2 cos dv 0 2 sin cos sin3 0 d 2 sin 3 sin3 0 sin cos
85.
3
h
2 3
1 3
h
6 3
1
1
1
2
b
c
2
c2
4 b
1 log xr
C2 C2 C1, C3 C3 C1 1 2
0
0
b2
c 2
2
4 b 4 c2 4 b 2 c 2
1
1
b2 c2
A b 2 c 2 c b 2,b,c are in AP 2,2 d, 2 2d A d 2d d 2d3 2, 16 d3 1, 8
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JEE-MAIN-2019 (8th Apr-Second Shift)-PCM-42 2d 2, 4 2 2d 4, 6
86.
1 Form f 3 x f 2x f 3 f 2 k lim x 0 x 1 f 2 x f 2 f ' 3 x f ' 2 x lim x 0 1 f 2 x f 2 x f ' 2 x 0 ek 1
87.
I
dx 3
x 1 x6
2/3
dx 1 x7 1 6 x
2/3
dx dt x7 6 1/3 1 dt 1 1 I 2/ 3 1 6 C 6 t 2 x
Put 1 x 6 t
1/3
6 1 1 x x f x 1 x6 2 2 x 1 f x 3 2x
88.
89.
1/3
h 20 h 80 , y x xy x h y h xy , 20 x 80 x h h 1 20 80 100 h 1 1600 h 16
C
80 x–y
h y
3 6 C3 x 1 log x . x1/4 200 2 1 3 1 logx 2
x4
10
1 3 1 log10 x .log10 x 1 4 2
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JEE-MAIN-2019 (8th Apr-Second Shift)-PCM-43 6t 2 5t 4 0, t log10 x D0 So no real solution All options are incorrect
90.
b2 ac Also roots of ax 2 2bx c 0 are equal b x , common root a 2 b b d 2e 0 a a db2 2aeb fa2 0, b2 ac dac 2eab fa2 0 dc 7eb fa 0 Dividing by ac d 2e f 0 a b c d f e 2. a c b
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FIITJEE Solutions to JEE(Main)-2019 Test Date: 9th April 2019 (First Shift)
PHYSICS, CHEMISTRY & MATHEMATICS Paper - 1 Time Allotted: 3 Hours
Maximum Marks: 360
Please read the instructions carefully. You are allotted 5 minutes specifically for this purpose.
Important Instructions: 1.
The test is of 3 hours duration.
2.
This Test Paper consists of 90 questions. The maximum marks are 360.
3.
There are three parts in the question paper A, B, C consisting of Physics, Chemistry and Mathematics having 30 questions in each part of equal weightage. Each question is allotted 4 (four) marks for correct response.
4.
Out of the four options given for each question, only one option is the correct answer.
5.
For each incorrect response 1 mark i.e. ¼ (one-fourth) marks of the total marks allotted to the question will be deducted from the total score. No deduction from the total score, however, will be made if no response is indicated for an item in the Answer Box.
6.
Candidates will be awarded marks as stated above in instruction No.3 for correct response of each question. One mark will be deducted for indicating incorrect response of each question. No deduction from the total score will be made if no response is indicated for an item in the answer box.
7.
There is only one correct response for each question. Marked up more than one response in any question will be treated as wrong response and marked up for wrong response will be deducted accordingly as per instruction 6 above.
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JEE-MAIN-2019 (9th Apr-First Shift)-PCM-2
PART –A (PHYSICS) 1.
In the density measurement of a cube, the mass and edge length are measured as (10.000.10) kg and (0.100.01)m, respectively. The error in the measurement of density is: (A) 0.10 kg/m3 (B) 0.31 kg/m3 3 (C) 0.07 kg/m (D) 0.01 kg/m3
2.
The total number of turns and cross-section area in solenoid is fixed. However, its length L is varied by adjusting the separation between windings. The inductance of solenoid will be proportional to: (A) 1/L (B) L (C) 1/L2 (D) L2
3.
The figure shows a Young’s double slit experimental setup. It is observed that when a thin transparent sheet of thickness t and refractive index is put in front of one of the slits, the central maximum gets shifted by a distance equal to n fringe widths. If the wavelength of light used is , t will be:
4.
(A)
2nD a 1
(B)
nD a 1
(C)
2D a 1
(D)
D a 1
A system of three charges are placed as shown in the figure:
If D >> d, the potential energy of the system is best given by: 1 q2 qQd 1 q2 qQd (A) (B) 4 o d D2 4 o d 2D2 (C)
5.
1 4 o
q2 2qQd D2 d
(D)
1 4 o
q2 qQd 2 D d
A moving coil galvanometer has resistance 50 and it indicates full deflection at 4mA current. A voltmeter is made using this galvanometer and a 5 k resistance. The maximum voltage, that can be measured using this voltamenter, will be close to: (A) 15 V (B) 20 V (C) 10 V (D) 40 V FIITJEE Ltd., FIITJEE House, 29-A, Kalu Sarai, Sarvapriya Vihar, New Delhi -110016, Ph 46106000, 26569493, Fax 26513942 website: www.fiitjee.com.
JEE-MAIN-2019 (9th Apr-First Shift)-PCM-3 6.
If ‘M’ is the mass of water that rises in a capillary tube of radius ‘r’, then mass of water which will rise in a capillary tube of radius ‘2r’ is: M (A) 4 M (B) 2 (C) M (D) 2 M
7.
An NPN transistor is used in common emitter configuration as an amplifier with 1 k load resistance. Signal voltage of 10 mV is applied across the base-emitter. This produces a 3 mA change in the collector current and 15 A change in the base current of the amplifier. The input resistance and voltage gain are: (A) 0.67 k, 300 (B) 0.67 k, 200 (C) 0.33 k, 1.5 (D) 0.33 k, 300
8.
An HCl molecule has rotational, translational and vibrational motions. If the rms velocity of HCl molecules in its gaseous phase is , m is its mass and kB is Boltzmann constant, then its temperature will be: 2
m (A) 3kB
2
m (C) 5kB
9.
10.
2
m (B) 7kB
2
m (D) 6k B
2x N The electric field of light wave is given as E 10 3 cos 2 6 1014 t x . This 7 5 10 C light falls on a metal plate of work function 2eV. The stopping potential of the photoelectrons is: (A) 0.48 V (B) 2.48 V (C) 0.72 V (D) 2.0 V
Following figure shows two processes A and B for a gas. If QA and QB are the amount of heat absorbed by the system in two cases, and UA and UB are changes in internal energies, respectively, then:
(A) QA = QB; UA = UB (C) QA < QB; UA < UB
(B) QA > QB; UA = UB (D) QA > QB; UA > UB
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JEE-MAIN-2019 (9th Apr-First Shift)-PCM-4 11.
A uniform cable of mass ‘M’ and length ‘L’ is placed on a horizontal surface such that its th 1 part is hanging below the edge of the surface. To lift the hanging part of the cable n upto the surface, the work done should be: MgL (A) nMgL (B) 2n2 2MgL MgL (C) (D) 2 n n2
12.
The following bodies are made to roll up (without slipping) the same inclined plane from R a horizontal place: (i) a ring of radius R, (ii) a solid cylinder of radius and (iii) a solid 2 R sphere of radius . If, in each case, the speed of the center of mass at bottom of the 4 incline is same, the ratio of the maximum heights they climb is: (A) 10 : 15 : 7 (C) 4 : 3 : 2
13.
(B) 14: 15: 20 (D) 2 : 3 : 4
A signal A cos t is transmitted using 0 sin 0 t as carrier wave. The correct amplitude modulated (AM) signal is: (A) 0 sin 0 1 0.01A sin t t A A sin 0 t sin 0 t 2 2 (C) 0 sin 0 t A cos t
(B) 0 sin 0 t
(D) 0 A cos t sin 0 t 14.
A concave mirror for face viewing has focal length of 0.4 m. The distance at which you hold the mirror from your face in order to see your image upright with a magnification of 5 is: (A) 0.16 m (B) 1.60 m (C) 0.24 m (D) 0.32 m
15.
Determine the charge on the capacitor in the following circuit:
(A) 200 C (C) 10 C 16.
(B) 60 C (D) 2 C
A rectangular coil (Dimension 5 cm 2 cm) with 100 100 turns, carrying a current of 3 A in the clock-wise direction, is kept centered at the origin and in the X-Z plane. A magnetic field of 1 T is applied along X-axis. If the coil is tilted through 45o about Z-axis, then the torque on the coil is: (A) 0.42 Nm (B) 0.55 Nm (C) 0.27 Nm (D) 0.38 Nm
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JEE-MAIN-2019 (9th Apr-First Shift)-PCM-5 17.
A stationary horizontal disc is free to rotate about its axis. When a torque is applied on it, its kinetic energy as a function of , where is the angle by which it has rotated, is given as k2. If its moment of inertia is then the angular acceleration of the disc is: k k (A) (B) 2 k 2k (C) (D) 4
18.
A simple pendulum oscillating in air has period T. The bob of the pendulum is completely 1 immersed in a non-viscous liquid. The density of the liquid is th of the material of the 16 bob. If the bob is inside liquid all the time, its period of oscillation in this liquid is: 1 1 (A) 2T (B) 2T 10 14 (C) 4T
19.
1 15
(D) 4T
For a given gas at 1 atm pressure, rms speed of the molecules is 200 m/s at 127oC. At 2 atm pressure and at 227oC, the rms speed of the molecules will be: (A) 80 m/s (B) 80 5 m / s (C) 100 m/s
20.
1 14
(D) 100 5 m / s
The magnetic field of a plane electromagnetic wave is given by: B B0 i cos kz t B1j cos kz t where B0 = 3 10–5T and B1 = 2 10–6T. The rms value of the force experienced by a stationary charge Q = 10–4=C at z = 0 is closet to: (A) 0.9 N (B) 0.6 N (C) 0.1 N (D) 3 10–2N
21.
The stream of a river is flowing with a speed of 2 km/h. A swimmer can swim at a speed of 4 km/h. What should be the direction of the swimmer with respect to the flow of the river to cross the river straight? (A) 60o (B) 90o o (C) 120 (D) 150o
22.
A rigid square loop of side ‘a’ and carrying current 2 is laying on a horizontal surface near a long current 1 wire in the same plane as shown in figure. The net force on the loop due to the wire will be:
(A) Repulsive and equal to (C) Zero
0 1 2 2
0 1 2 3 (D) Repulsive and equal to 0 1 2 4
(B) Attractive and equal to
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JEE-MAIN-2019 (9th Apr-First Shift)-PCM-6 23.
Taking the wavelength of first Balmer line in hydrogen spectrum ( n = 3 to n = 2) as 660 nm , the wavelength of the 2nd Balmer line (n =4 to n = 2) will be (A) 889.2 nm (B) 642.7 nm (C) 448.9 nm (D) 388.9 nm
24.
A solid sphere of mass ‘M’ and radius ‘a’ is surrounded by a uniform concentric spherical shell of thickness 2a and mass 2M. The gravitational field at distance ‘3a’ from the centre will be: 2GM 2GM (A) (B) 3a 2 9a 2 GM GM (C) (D) 2 9a 3a2
25.
A string is clamed at both the ends and it is vibrating in its 4th harmonic. The equation of the stationary wave is Y =0.3 sin(0.157x) cos(200t). The length of the string is: (all quantities are in S units) (A) 80 m (B) 60 m (C) 40 m (D) 20 m
26.
A ball is thrown vertically up (taken as +z-axis) from the ground. The correct momentumheight (p-h) diagram is:
(A)
(B)
(C)
(D)
27.
A capacitor with capacitance 5 F is charged to 5 C. If the plates are pulled apart to reduce the capacitance to 2 F, how much work is done? (A) 3.75 10–6 J (B) 2.55 10–6 J (C) 6.25 10–6 J (D) 2.16 10–6 J
28.
A body of mass 2 kg makes an elastic collision with a second body at rest and continues to move in the original direction but with one fourth of its original speed. What is the mass of the second body? (A) 1.5 kg (B) 1.2 kg (C) 1.8 kg (D) 1.0 kg
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JEE-MAIN-2019 (9th Apr-First Shift)-PCM-7 29.
The pressure wave, P = 0.01 sin[1000t – 3x]NM–2, corresponds to the sound produced by a vibrating blade on a day when atmospheric temperature is 0o C . On some other day when temperature is T, the speed of sound produced by the same blade and at the same frequency is found to be 336 ms–1. Approximate value of T is: (A) 12o C (B) 11o C o (C) 15 C (D) 4o C
30.
A wire of resistance R is bent to form a square ABCD as shown in the figure. The effective resistance between E and C is: (E is mid-point of arm CD)
1 R 16 3 (C) R 4
(A)
(B)
7 R 64
(D) R
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JEE-MAIN-2019 (9th Apr-First Shift)-PCM-8
PART –B (CHEMISTRY) 31.
Magnesium powder burns in air to give: (A) MgO and Mg(NO3)2 (C) MgO only
(B) MgO and Mg3N2 (D) Mg(NO3)2 and Mg3N2
32.
The number of water molecule(s) not coordinated to copper ion directly in CuSO45H2O, is: (A) 1 (B) 3 (C) 2 (D) 4
33.
The one that will show optical activity is: (en = ethane-1, 2-diamine)
34.
(A)
(B)
(C)
(D)
Consider the van der Waals constants, a and b, for the following gases. Gas a/(atm dm6mol–2) –2
3
Ar
Ne
Kr
Xe
1.3
0.2
5.1
4.1
–1
b/(10 dm mol ) 3.2 1.7 1.0 Which gas is expected to have the highest critical temperature? (A) Ar (B) Xe (C) Kr (D) Ne
5.0
35.
Among the following, the molecule expected to be stabilized by anion formation is: (A) F2 (B) C2 (C) O2 (D) NO
36.
Liquid ‘M’ and liquid ‘N’ form an ideal solution. The vapour pressures of pure liquids ‘M’ and ‘N’ are 450 and 700 mm Hg, respectively at the same temperature. Then correct statement is: (xM = mole fraction of ‘M’ in solution; xN = mole fraction of ‘N’ in solution; yM = mole fraction of ‘M’ in vapour phase; yN = mole fraction of ‘M’ in vapour phase) x y x y (A) M M (B) M M xN y N xN y N x y (C) xM yM xN yN) (D) M M xN y N FIITJEE Ltd., FIITJEE House, 29-A, Kalu Sarai, Sarvapriya Vihar, New Delhi -110016, Ph 46106000, 26569493, Fax 26513942 website: www.fiitjee.com.
JEE-MAIN-2019 (9th Apr-First Shift)-PCM-9 37.
Among the following the set of parameters that represents path functions, is: (a) q + w (b) q (c) w (d) H – TS (A) (b) and (c) (B) (b), (c) and (d) (C) (a), (b) and (c) (D) (a) and (d)
38.
The degenerate orbitals of Cr H2O 6 (A) dxz and dyz
39.
40.
are: (B) dx2 y 2 and dxy
(C) dyz and dz2
(D) dz2 and dxz
The aerosol is a kind of colloid in which: (A) solid is dispersed in gas (C) liquid is dispersed in water
(B) gas is dispersed in solid (D) gas is dispersed in liquid
The given plots represent the variation of the concentration of a reactant R with time for two different reactions (i) and (ii). The respective orders of the reactions are:
(A) 1, 1 (C) 0, 1 41.
3
(B) 0, 2 (D) 1, 0
The major product of the following reaction is: CH2CH3
i alkaline KMnO4 ii H3O CH2CHO
(A)
CH2COOH
(B)
COOH
(C)
42.
COCH3
(D)
The organic compound that gives following qualitative analysis is: Test Inference (a) Dil. HCl Insoluble (b) NaOH solution Soluble (c) Br2/water Decolourization
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JEE-MAIN-2019 (9th Apr-First Shift)-PCM-10 OH
OH
(A)
(B) NH2
NH2
(C)
43.
(D)
Aniline dissolved in dilute HCl is reacted with sodium nitrite at 0oC . This solution was added dropwise to a solution containing equimolar mixture of aniline and phenol in dil. HCl. The structure of the major product is: (A)
N
N
(C)
N
N
NH
NH2
(B)
N
N
(D)
N
N
44.
Excessive release of CO2 into the atmosphere results in: (A) formation of smog (B) depletion of ozone (C) global warming (D) polar vortex
45.
The ore that contains the metal in the form of fluoride is: (A) saphalerite (B) malachite (C) magnetite (D) cryolite
46.
The correct IUPAC name of the following compound is
OH
O
NO 2
Cl CH3
(A) 5-chloro-4-methyl-1-nitrobenzene (C) 3-chloro-4-methyl-1-nitrobenzene
(B) 2-methyl-5-nitro-1-chlorobenzene (D) 2-chloro-1-methyl-4-nitrobenzene
47.
The major product of the following reaction is : LiAlH4 CH3CH CHCO2CH3 (A) CH3CH2CH2CHO (B) CH3CH = CHCH2OH (C) CH3CH2CH2CO2CH3 (D) CH3CH2CH2CH2OH
48.
The element having greatest difference between its first and second ionization energies, is (A) Ca (B) K (C) Ba (D) Sc
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JEE-MAIN-2019 (9th Apr-First Shift)-PCM-11 49.
For a reaction, N2(g) + 3H2(g) 2NH3(g); identify dihydrogen (H2) as a limiting reagent in the following reaction mixtures. (A) 14g of N2 + 4g of H2 (B) 28g of N2 + 6g of H2 (C) 56g of N2 + 10g of H2 (D) 35g of N2 + 8g of H2
50.
The major product of the following reaction is : i DCI 1equiv. CH3C CH 2 DI (A) CH3CD(I)CHD(Cl) (C) CH3CD2CH(Cl)(I)
(B) CH3C(I)(Cl)CHD2 (D) CH3CD(Cl)CHD(I)
51.
The osmotic pressure of a dilute solution of an ionic compound XY in water is four times that of a solution of 0.01 M BaCl2 in water. Assuming complete dissociation of the given ionic compounds in water, the concentration of XY (in mol L–1) in solution is (A) 4 10–4 (B) 6 10–2 –4 (C) 16 10 (D) 4 10–2
52.
For any given series of spectral lines of atomic hydrogen, let V Vmax Vmin be the difference in maximum and minimum frequencies in cm–1. the ration VLymann / VBalmer is (A) 5 : 4 (C) 9 : 4
53.
(B) 4: 1 (D) 27 : 5
The major product of the following reaction is:
(A)
(B)
(C)
(D)
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JEE-MAIN-2019 (9th Apr-First Shift)-PCM-12 54.
55.
The increasing order of reactivity of the following compounds towards aromatic electrophilic substitution reaction is
(A) D < B < A < C (C) D < A < C < B
(B) A < B < C < D (D) B < C < A < D
C60, an allotrope of carbon contains: (A) 20 hexagons and 12 pentagons. (C) 18 hexagons and 14 pentagons.
(B) 12 hexagons and 20 pentagons. (D) 16 hexagons and 16 pentagons
56.
The correct order of the oxidation states of nitrogen in NO, N2O, NO2 and N2O3 is: (A) N2O < N2O3 < NO < NO2 (B) NO2 < NO < N2O3 < N2O (C) NO2 < N2O3 < NO < N2O (D) N2O < NO < N2O3 < NO2
57.
The standard Gibbs energy for the given cell reaction in kJ mol –1 at 298 K is: Zn s Cu2 aq Zn2 aq Cu s ,Eo 2 V at 298 K
[Faraday’s constant F = 96500 C mol–1] (A) -192 (C) -384 58.
(B) 384 (D) 192
The major product of the following reaction is: 1.PBr3 OH 2. KOH alc.
O OH
(A)
(B) O
O
(C)
(D) HO
59.
O
Which of the following statements is not true about sucrose? (A) The glycosidic linkage is present between C1 of -glucose and C1 of -fructose (B) It is a non-reducing sugar (C) It is also named as invert sugar (D) On hydrolysis it produces glucose and fructose
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JEE-MAIN-2019 (9th Apr-First Shift)-PCM-13 60.
Match the catalysis(Column – I) with products (Column-II) Column-I Column-II Catalyst Product (a) V2O5 (i) Polyethylene (b) TiCl4/Al(Me)3 (ii) Ethanal (c) PdCl2 (iii) H2SO4 (d) Iron oxide (iv) NH3 (A) (a)-(iii); (b)-(iv); (c)-(i); (d)-(ii) (B) (a)-(iv); (b)-(iii); (c)-(ii); (d)-(i) (C) (a)-(ii); (b)-(iii); (c)-(i); (d)-(iv) (D) (a)-(iii); (b)-(i); (c)-(ii); (d)-(iv)
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JEE-MAIN-2019 (9th Apr-First Shift)-PCM-14
PART–C (MATHEMATICS) 61.
62.
63.
(A) R 1,0
x2 , is surjective, then A is equal to: 1 x2 (B) R–(–1,0)
(C) R–{–1}
(D) [0, ]
If the function f :R–{1, –1}A defined by f x
Let p, q R. if 2 3 is a root of the quadratic equation, x2 + px + q = 0, then: (A) q2 + 4p + 14 = 0 (B) p2 – 4q + 12 = 0 2 (C) p – 4q – 12 = 0 (D) q2 – 4p – 16 = 0 Let 3i j and 2i j 3k . If 1 2 , where 1 is parallel to and 2 is perpendicular to , then 1 2 is equal to: 1 1 (A) (B) 3 i 9 j 5 k 3 i 9 j 5 k 2 2 (C) 3 i 9 j 5 k (D) 3 i 9 j 5 k
64.
The integral
sec
2/3
(A) 3 tan1/3 x C (C) 3 cot 1/3 x C 65.
x cos ec 4/3 x dx is equal to: (Here C is a constant of integration) 3 tan 4/3 x C 4 (D) 3 tan1/3 x C
(B)
A plane passing through the points (0, –1, 0) and (0, 0, 1) and making an angle
with 4
plane y – z + 5 = 0, also passes through the point: (A) 2,1,4 (B) 2, 1, 4
(C)
2, 1, 4
(D) 2, 1, 4
66.
If the tangent to the curve, y = x3 + ax – b at the point (1, –5) is perpendicular to the line, –x + y + 4 = 0, then which one of the following, points lies on the curve? (A) (2, –2) (B) (–2, 2) (C) (–2, 1) (D) (2, –1)
67.
If the standard deviation of the numbers –1, 0, 1, k is (A) 4
5 3
(C) 2 6 68.
(B) (D) 2
5 where k > 0, then k is equal to:
6 10 3
Let f(x) = 15–|x – 10|; x R. then the set of all values of x, at which the function, g(x) = f(f(x)) is not differentiable, is: (A) {5, 10, 15} (B) {10} (C) {5, 10, 15, 20} (D) {10, 15} FIITJEE Ltd., FIITJEE House, 29-A, Kalu Sarai, Sarvapriya Vihar, New Delhi -110016, Ph 46106000, 26569493, Fax 26513942 website: www.fiitjee.com.
JEE-MAIN-2019 (9th Apr-First Shift)-PCM-15 6
69.
2 If the fourth term in the Binomial expansion of xlog8 x x 0 is 20 87, then a value x of x is: (A) 83 (B) 8–2 (C) 8 (D) 82
70.
If f(x) is a non-zero polynomial of degree four, having local extreme points at x = –1, 0, 1; then the set S = {x R; f(x) = f(0)} contains exactly: (A) four irrational numbers (B) four rational numbers (C) two irrational and one rational number (D) two irrational and two rational numbers
71.
For any two statements p and q, the negation of the expression p p q is: (A) p q (C) p q
(B) p q (D) p q
72.
A committee of 11 members is to be formed from 8 males and 5 females. If m is the number of ways the committee is formed with at least 6 males and n is the number of ways the committee is formed with at least 3 females, then: (A) n = m – 8 (B) m + n = 68 (C) m = n = 78 (D) m = n = 68
73.
If the line,
74.
The value of cos2 10o cos10o cos 50o cos2 50o is: 3 3 (A) 1 cos 20o (B) 2 4 3 3 (C) (D) cos 20o 2 4
x 1 y 1 z 2 meets the plane, x + 2y + 3z = 15 at a point P, then the 2 3 4 distance of P from the origin is: 5 (A) (B) 2 5 2 9 7 (C) (D) 2 2
75.
76.
1 1 1 2 1 3 1 n 1 1 78 If ........... , then the inverse of 1 0 1 0 1 0 1 0 1 0 1 12 1 0 (A) (B) 0 1 13 1 1 0 1 13 (C) (D) 1 12 1 0
1 n 0 1 is:
i All the points in the set S : R i 1 lie on a: i (A) straight line whose slope is 1 (B) circle whose radius is 2 (C) straight line whose slope is –1 (D) circle whose radius is 1
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JEE-MAIN-2019 (9th Apr-First Shift)-PCM-16 77.
Let the sum of the first n terms of a non-constant A.P., a1, a2, a3, …… be n n 7 50n A , where A is a constant. If d is the common difference of this A.P., then 2 the ordered pair (d,a50) is equal to: (A) (A, 50 + 46A) (B) (A, 50 + 45A) (C) (50, 50 + 45A) (D)(50, 50 + 46A)
78.
The value of
/2
0
sin3 x dx is: sin x cos x
2 (A) 4 1 (C) 4
79.
If the line y mx 7 3 is normal to the hyperbola (A) (C)
80.
1 2 2 (D) 8
(B)
2
(B)
5 15 2
(D)
The solution of the differential equation x x3 1 2 5 5x 4 3 1 (C) y x 2 5 5x
(A) y
x2 y2 1 , then a value of m is: 24 18
5 2 3 5
dy 2y x 2 x 0 with y(1) = 1, is: dx x2 3 (B) y 2 4 4x 3 2 1 (D) y x 2 4 4x
81.
If one end of a focal chord of the parabola, y2 = 16x is at (1, 4), then the length of this focal chord is: (A) 25 (B) 24 (C) 22 (D) 20
82.
Let S be the set of all values of x for which the tangent to the curve y = f(x) = x3 – x2 – 2x at (x, y) is parallel to the line segment joining the points 1, f 1 and –1, f –1 , then S is equal to: 1 (A) , 1 3 1 (C) ,1 3
1 (B) , 1 3 1 (D) ,1 3
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JEE-MAIN-2019 (9th Apr-First Shift)-PCM-17 83.
Slope of a line passing through P(2, 3) and intersecting the line, x + y = 7 at a distance of 4 units from P, is: 5 1 1 5 (A) (B) 5 1 1 5 7 1 1 7 (C) (D) 7 1 1 7
84.
Let
10
f a k 16 2
10
1 , where the function f satisfies f(x + y) = f(x) f(y) for all natural
k 1
numbers x, y and f(1) = 2. Then the natural number ‘a’ is: (A) 4 (B) 16 (C) 2 (D) 3 85.
Let S = {[–2, 2]: 2 cos2 + 3 sin = 0}. Then the sum of the elements of S is: 13 (A) (B) 2 6 5 (C) (D) 3
86.
If a tangent to the circle x2 + y2 = 1 intersects the coordinate axes at distinct points P and Q, then the locus of the mid-point of PQ is: (A) x2 + y2 – 16x2y2 = 0 (B) x2 + y2 – 2x2y2 = 0 2 2 2 2 (C) x + y – 4x y = 0 (D) x2 + y2 – 2xy = 0
87.
Four persons can hit a target correctly with probabilities
88.
The area (in sq. units) of the region A = {(x, y): x2 y x + 2} is: 31 13 (A) (B) 6 6 9 10 (C) (D) 2 3
89.
Let and be the roots of the equation x2 + x + 1 = 0. Then for y 0 in R, y 1
1 1 1 1 , , and respectively. If all 2 3 4 8 hit at the target independently, then the probability that the target would be hit, is: 25 25 (A) (B) 32 192 7 1 (C) (D) 32 192
y
1
1
y
(A) y(y2 – 3) (C) y3
is equal to:
(B) y3 – 1 (D) y(y2 – 1)
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JEE-MAIN-2019 (9th Apr-First Shift)-PCM-18
90.
2 cos x 1 , x 4 is continuous, then k is If the function f defined on , by f x cot x 1 6 3 k, x 4 equal to: (A) 1 (B) 2 1 1 (C) (D) 2 2
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JEE-MAIN-2019 (9th Apr-First Shift)-PCM-19
JEE (Main) – 2019 ANSWERS PART A – PHYSICS 1. 5. 9. 13. 17. 21. 25. 29.
Bonus B A B D C B D
2. 6. 10. 14. 18. 22. 26. 30.
A D B D C B A B
3. 7. 11. 15. 19. 23. 27.
D A B A B C A
4. 8. 12. 16. 20. 24. 28.
D A B B D A B
34. 38. 42. 46. 50. 54. 58.
C A B D B C B
64. 68. 72. 76. 80. 84. 88.
D A C D B D C
PART B – CHEMISTRY 31. 35. 39. 43. 47. 51. 55. 59.
B B A C B B A A
32. 36. 40. 44. 48. 52. 56. 60.
A A D C B C D D
33. 37. 41. 45. 49. 53. 57.
A A C D B C C
PART C – MATHEMATICS 61. 65. 69. 73. 77. 81. 85. 89.
A A D C A A B C
62. 66. 70. 74. 78. 82. 86. 90.
B or Bonus A C B C D C C
63. 67. 71. 75. 79. 83. 87.
A C C D A D A
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JEE-MAIN-2019 (9th Apr-First Shift)-PCM-20
HINTS AND SOLUTIONS PART A – PHYSICS 1.
2.
m v Maximum % error in will be given by p m L 100% 100% 3 100% …(i) p m L This is not applicable as error is big. m 9.9 min min = 7438 kg/m3 v max (0.11)3 m 10.1 & max max = 13854.6 kg/m3 v min (0.09)3 p = 6416.6 kg/m3 No option is matching. Therefore this question should be awarded bonus. = NBA = LI N 0 nIR2 = LI N N 0 IR 2 LI N and R constant Self inductance (L)
1 1 length
3.
Path difference at central maxima x = ( – 1)t, whole pattern will shift by same amount which will be given by D D n ( 1)t n , according to eh question t d d ( 1) No option is matching, therefore question should be award bonus. Correct option should be (Bonus)
4.
Utotal = Uself of dipole + Uinteraction kq2 kQ = qd d D2
q2 qQd = k 2 D d 5.
G = 50 S = 5000 Ig = 4 × 10–3 V = ig (G + S) V = 4 × 10–3 (50 + 5000) = 4 × 10–3 (5050) = 20.2 volt
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JEE-MAIN-2019 (9th Apr-First Shift)-PCM-21
6.
Height of liquid rise in capillary tube h
2T cos C rg
1 r When radius becomes double height become half h h 2 Now, M = r2h × and M = (2r)2 (h/2) × = 2M. h
7.
Input current = 15 × 10–6 Output current = 3 × 10–3 Resistance out put = 1000 Vinput = 10 × 10–3 Now Vinput = rinput × iinput 10 × 10–3 = rinput × 15 × 10–6 2000 rinput = 0.67 K. 3 Voutput 1000 3 10 3 Voltage gain = = 300 Vinput 10 10 3
8.
According to equipartion energy theorem 1 1 2 m(v rms ) 3 K bT 2 2 2
mv rms T 3k 9.
= 6 × 1014 × 2 f = 6 × 1014 C=f C 3 108 = 5000 Å f 6 1014 12375 Energy of photon = 2.475 eV 5000 From Einstein’s equation KEmax = E – eVs = E – eVs = 2.475 – 2 eVo = 0.475 eV Vo = 0.48 V
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JEE-MAIN-2019 (9th Apr-First Shift)-PCM-22 10.
Initial and final states for both the processes are same, UA = UB Work done during process A is greater than in process B. Because area is more By First law of thermodynamics Q = U + W QA > QB
11.
Mass of the hanging part =
hCOM
M n
L 2n
M L MgL Work done W = mghCOM g 2 n 2n 2n 12.
1 1 m 2 v 2 = mgh 2 R If radius of gyration is k, then k2 2 1 2 v k k solid cylinder R 1 ring h ; 1, 2g Rring R solid cylinder 2
k solid sphere Rsolid sphere
2 5
1 2 H1 : h2 : h3 :: (1 + 1) : 1 : 1 :: 20 :15 : 14 2 5 Therefore most appropriate option is (B) Although which is not in correct sequence. 13.
14.
f f u 40 5 ; u = –32 cm 40 u
m
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JEE-MAIN-2019 (9th Apr-First Shift)-PCM-23 15.
Different potential is shown at different points.
q = eV q = 10F × 20 = 200 C 16.
t MB
NI A B sin 45o = 0.27 Nm 17.
18.
1 2 I k2 2 2k2 2k 2 I I Differentiate (A) wrt time d 2k d dt I dt 2k 2k by (1) I I 2k I Kinetic energy KE =
For a simple pendulum T = 2
…(A)
L gerr
Situation 1: when pendulum is in air geff = g Situation 2:when pendulum is in liquid liquid 1 15g geff g 1 g 1 body 16 16 L 2 15g / 16 T So, T L 2 g T
19.
Vrms
4T 15
3RT Mw
v rms
T
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JEE-MAIN-2019 (9th Apr-First Shift)-PCM-24
v 500 v 5 200 400 200 2 v 100 5 m / s Now,
20.
Maximum electric field E = (B) (C) E0 (3 105 )c ˆj E1 (2 10 6 )c ˆi
Maximum force Fnet 10 4 3 10 8 (3 10 5 )2 (2 10 6 )2 0.9 N Frms
F0 2
= 0.6 N (approx)
21.
For swimmer to cross the river straight 4 sin = 2 1 sin = 30o 2 So, angle with direction of river flow = 90° + = 120°.
22.
F3 & F4 cancel each other. Force on PQ will be F1 = 2B1 a I = I2 0 1 a 2a 0Ii = a= 2a 0I1I2 2 Force on RS will be F2 = I2 B2 a I = I2 0 1 a 22a 0I1I2 = 4 II Net force = F1 – F2 = 0 1 2 repulsion 4
23.
1 1 5R 1 R 2 2 660 2 3 36 1 1 3R 1 R 2 2 4 16 2 Divide equation (1) with (B) 5 16 660 36 3
…(1) …(B)
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JEE-MAIN-2019 (9th Apr-First Shift)-PCM-25
4400 = 488.88 = 488.9 nm 9
24.
We use Gauss’s Law for gravitation g4r2 = (Mass enclosed) 4G 3M4 G GM g= 4 (3a)2 3a2
25.
4th harmonic
; 2 2 2 From equation = 0.157 = 40 ; 2 = 80 m 4
26.
Momentum p = mv
…(1) 2
and for motion under gravity h h
27.
u v 2g
2
…(2)
u2 p 2 / m 2g
Work done = U = Uf – Ui q2 q2 = 2Cr 2Ci
(5 10 6 )2 1 1 6 2 5 10 6 2 10 15 = 106 4 = 3.75 × 10–6 J =
28.
By conservation of linear momentum: v v 2v 0 2 0 mv 2v 0 0 mv 2 2 4 3v 0 mv …(1) 2 Since collision is elastic Vseparation = vapproch v 6 v 0 v 0 m = 1.2 kg 4 5
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JEE-MAIN-2019 (9th Apr-First Shift)-PCM-26 29.
Speed of wave from wave equation (coeffecient of t) v (coeffecient of x) 1000 1000 v ( 3) 3 Since speed of wave 1000 273 So = 3 T 336 T = 277.41 K T = 4.41°C
T
30.
1 8 8 Req 7R R
7R 1 8 56 ; Reql Req 7R 64
PART B – CHEMISTRY 31.
Mg O 2 N2 MgO Mg2N3 Air
32.
[Cu(H2O)4]SO4.H2O
33.
No plane of symmetry 34.
TC
8a 27Rb
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JEE-MAIN-2019 (9th Apr-First Shift)-PCM-27 35.
C2 configuration
1s2 * 1s2 2s2 * 2s2
2py 2 2pz2
2Px
C 2 configuration
1s2 * 1s2 2s2 * 2s2 36.
2py 2 2pz2
2Px1
PMo 450 PNo 700 PM = xM 450 PN = xN 700 YMPT = PM YNPT = PN YMPT xM 450 YNPT xN 700 YM xM 0 : 64 YN xN xM YM xN YN
37.
U = q + w G = H - TS
38.
39.
Fact based (Given in NCERT)
40.
For first order reaction n[R]t = -Kt + n[R]o For zero order reaction [R]t = -Kt + [R]o Where ‘R’ is reactant.
41.
COO
COOH H3O
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JEE-MAIN-2019 (9th Apr-First Shift)-PCM-28
42.
Phenol being acidic insoluble in HCl OH
O Na
NaOH
H2O Soluble in NaOH
OH
OH Br
Br
Br2 /Water
Decolourises Bromine water
Br
43. o
NaNO2 , 0 C NH2
N
NCl
Ph - OH, Ph - NH 2, dil HCl N =N
44.
Fact based (Given in NCERT)
45.
Cryolite is Na3[AlF6]
46.
NH2
NO 2 4 3 2 1
Cl
CH3
47.
LiAlH4 does not reduce double bond it reduces ester in alcohol.
48.
After losing one electron ‘K’ acquires noble gas configuration.
49.
56 g of N2 means 2 mole 2 mole of N 2 requires 6 mole of H2 for complete reaction but available H 2 is 10 g i.e. 5 mole hence H 2 is limiting reagent.
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JEE-MAIN-2019 (9th Apr-First Shift)-PCM-29 Cl
50.
DCI CH3C CH CH3 C CHD
DI Cl
CH3 C CHD2 I
51.
xy 4BaCl2 xy iCRT xy 2 CRT BaCl2 3 0.01 RT RT 12 0.01RT 12 0.01 0.06 2
52.
RH 4 RH 9
VLymann
VBalmer
VLymann VBalmer
9 4
Formula 1 1 V RH 2 2 n1 n2
53. KOH alc
Cl Cl
Cl Free radical polymerisation
n Cl
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JEE-MAIN-2019 (9th Apr-First Shift)-PCM-30 54.
R effect
Hyperconjugation
Me
O - Me
R effect
I effect
CN
Cl
55.
Fact based (Given in NCERT)
56.
N O, N2 O, N O2 , N2 O3
57.
G = -2 96000 2 G = -nFEo G = -2 96000 2 J G = -384 kJ mol–1
2
58.
1
4
3
1.PBr3 OH
Br
2. KOH alc.
O
alc KOH
O
O
59.
C1 of -glucose and C2 of -fructose
60.
V2O5 is used in contact process for H2SO4 TiCl4/Al(Me)3 Ziegler natal catalyst used fin polymerization PdCl2 is used in ethanol formation Iron oxide is used in Haber process for NH3
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JEE-MAIN-2019 (9th Apr-First Shift)-PCM-31
PART C – MATHEMATICS 2
61.
62.
x 1 x2 Range of y :R 1, 0 for surjective function, A must be same as above range. y
In given question p,q R . If we take other root as any real number , then quadratic
equation will be x 2 2 3 x . 2 3 0 Now, we can have none or any of the options can be correct depending upon ' ' . Instead of p,q R it should be p,q Q then other root will be 2 3
p 2 3 2 3 4 and q 2 3 2 3 1 2
p 2 4q 12 4 4 12 16 16 0 Option (B) is correct. 63.
3iˆ ˆj 2iˆ ˆj 3kˆ 1 2 1 3iˆ ˆj , 2 3iˆ ˆj 2iˆ j 3kˆ 2 . 0
3 2 .3 1 0 9 6 1 0 1 2 3 1 1 ˆi ˆj 2 2 1 3 2 ˆi ˆj 3kˆ 2 2
ˆi
ˆj
kˆ
3 1 0 Now, 1 2 2 2 1 3 3 2 2 3 9 9 1 ˆi 0 ˆj 0 kˆ 2 2 4 4 3ˆ 9ˆ 5ˆ i j k 2 2 2 1 3iˆ 9ˆj 5kˆ 2
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JEE-MAIN-2019 (9th Apr-First Shift)-PCM-32
64.
I
dx
sin x
4/3
. cos x
2/3
dx
I
4/3
sin x 2 .cos x cos x sec 2 x I dx 4/3 tan x
put tan x t sec 2 x dx dt dt 3 I 4/3 I 1/3 c t t 3 I c 1/3 tan x 65.
Let ax by cz 1 be the equation of the plane 0 b 0 1 b 1 0 0 c 1 c 1 ab cos a b
1 2
0 1 1
a
2
1 1 0 1 1
2
a 24 a 2 2x y z 1 Now for – sign 2, 2 1 4 1 Hence option (A) 66.
y x3 ax b
1, 5
lies on the curve
5 1 a b a b 6 Also, y ' 3x 2 a
…….(i)
y ' 1, 5 3 a (slope of tangent) this tangent is to x y 4 0 3 a 1 1 a 4 ………(ii) By (i) and (ii) : a 4, b 2
y x3 4x 2 , (2, –2) lies on this curve.
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JEE-MAIN-2019 (9th Apr-First Shift)-PCM-33
2
67.
S.D.
x x
n x 1 0 1 k k x 4 4 4 2
Now
2
2
k k k k 1 4 0 4 1 4 k 4 5 4
2
2
k 5k 2 5 4 2 1 8 16 2 3k 18 4 2 k 24 k 2 6
68.
f x 15 x 10 , x R
f f x 15 f x 10 15 15 x 10 10 15 5 x 10 x 5, 10, 15 are differentiability
point 3
69.
6 2 T4 T31 . xlog8 x 3 x 160 20 87 3 . x 3log8 x x 6 log2 x 8 x 3 log2 x 3 18 2 x 18 log2 x 3 log2 x Let log2 x t
of
non
3
t 2 3t 18 0 t 6 t 3 0 t 6, 3
log2 x 6 x 26 82 log2 x 3 x 23 81 70.
f ' x x 1 x 0 x 1 x3 x
x4 x2 f x 2 4 Now f x f 0
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JEE-MAIN-2019 (9th Apr-First Shift)-PCM-34
x4 x2 2 4 x 0,0, 2 Two irrational and one rational number 71.
~ p ~ p q ~ p ~ ~ p q ~ p p q ~ p p ~ p ~ q c ~ p ~ q ~ p ~ q
72.
73.
Since there are 8 males and 5 females. Out of these 13, if we select 11 persons, then there will be at least 6 males and at least 3 females in the selection. 13 13 13 12 m n 78 2 11 2 Any point on the given line can be 1 2, 1 3, 2 4 ; R Put in plane 1 2 2 6 6 12 15 20 5 15 20 10 1 2 1 Point 2, , 4 2 Distance from origin 4
74.
1 16 1 64 81 16 4 2 2
9 2
1 2 cos2 10o 2cos10o cos 50o 2cos2 50o 2 1 1 cos 20o cos 60o cos 40o 1 cos100o 2 13 cos 20o 2 sin 70o sin 30o 22 13 cos 20o sin70o 22 3 4
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JEE-MAIN-2019 (9th Apr-First Shift)-PCM-35
75.
1 1 1 2 1 2 1 3 1 n 1 1 78 ...... 1 0 1 0 1 0 1 0 1 0 1 0 1 1 2 3 ...... n 1 1 78 1 0 0 1 n n 1 78 n 13, 12 (reject) 2 1 13 we have to find inverse of 0 1 1 13 1 0
i z 1 i z i 1 z circle of radius 1
76.
Let
77.
Sn 50n
n n 7 2
A
Tn Sn Sn 1
50n
n n 7 2
A 50 n 1
n 1n 8 2
A
A 2 n 7n n2 9n 8 2 50 A n 4 d Tn Tn 1 50
50 A n 4 50 A n 5 A T50 50 46A
d, A 50 A,50 46A /2
78.
I
0
sin3 x dx sin x cos x / 4
I
0
sin3 x cos3 x dx sin x cos x
/ 4
1 sin x cos x dx 0
/4
sin2 x x 2 0
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JEE-MAIN-2019 (9th Apr-First Shift)-PCM-36
1 4 4 1 4
79.
x 2 y2 1 a 24 :b 18 24 18 Parametric normal: 24 cos . x 18. y cot 42 42 At x 0; y tan 7 3 (from given equation) 18
3 3 sin 2 5
tan
slope of parametric normal m
80.
24 cos 18 cot
m
4 2 2 sin or 3 5 5
dy 2y x 2 : y 1 1 dx dy 2 y x (LDE in y) dx x
x
2
x dx e2 n x x 2 x4 y. x 2 x. x 2dx C 4 y 1 1
IF e
1 1 3 C C 1 4 4 4 4 x 3 yx 2 4 4 x2 3 y 2 4 4x
1
81.
y 2 4ax 16x a 4 A 1, 4 2. 4. t1 4 t1
1 2
1 length of focal chord a t t
2
2
25 1 4 2 4. 25 2 4
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JEE-MAIN-2019 (9th Apr-First Shift)-PCM-37 82.
f 1 1 1 2 2 f 1 1 1 2 0 m
f 1 f 1
2 0 1 2
1 1 dy 2 3x 2x 2 dx 3x 2 2x 2 1 3x 2 2x 1 0 x 1 3x 1 0
x 1, 83.
1 3
x 2 r cos y 3 r sin 2 r cos 3 r sin 7 r cos sin 2
2 2 1 r 4 2
sin cos 1 sin 2
1 4
3 4 2m 3 1 m2 4 2 3m 8m 3 0 4 7 m 1 7 sin 2
2
1 7 1 7
1 7 1 7
8 2 7 4 7 6 3
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JEE-MAIN-2019 (9th Apr-First Shift)-PCM-38 84.
From the given functional equation: f x 2x x N
16 2
1
2a 1 2a 2 .... 2a10 16 210 1 a
2
10
2 2 2 ..... 2 2a.
16
2. 210 1
1 a 1 2 16 24 a=3
85.
10
2
10
1
2 1 sin2 3 sin 0 2
2sin 3 sin 2 0 2 sin 1 sin 2 0
1 sin ;sin 2 (reject) 2 roots : , 2 , , 6 6 6 6 sum of values 2 86.
Let the mid point be S (h, k) P 2h,0 and Q 0,2k equation of
x y 1 2h 2k PQ is tangent to circle at R (say) PQ :
1
OR 1
2
1 1 2h 2k
2
1
1 1 2 1 2 4h 4k 2 2 x y 4x 2 y 2 0
87.
Let persons be A, B, C, D P (Hit) = 1 – P (none of them hits)
1 P A .P B .P C .P D 1 P A B C D
1 2 3 7 1 . . . 2 3 4 8 25 32
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JEE-MAIN-2019 (9th Apr-First Shift)-PCM-39
88.
x2 y x 2 x 2 y; y x 2 x2 x 2 x2 x 2 0 x 2 x 1 0 x 2, 1 2
Area
9
2
x 2 x dx 2
1
89.
Roots of the equation x 2 x 1 0 are and 2 where , 2 are complex cube roots of unity y 1 2
2
y 2
1
1
y
R1 R1 R2 R3 1 y 2
1
1 2
y
1
1
y
Expanding along R1 , we get
y . y2 D y3 Or If , we get same value or on expansion using 1 , 1 we get 2
value y 3 . 90.
function should be continuous at x
4
lim f x f x 4 4 lim
x 4
lim
x 4
2 cos x 1 k cot x 1 2 sin x k (Using L Hospital Rule) cos ec 2 x
lim 2 sin3 x k x
4
3
1 1 k 2 2 2
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FIITJEE Solutions to JEE(Main)-2019 Test Date: 9th April 2019 (Second Shift)
PHYSICS, CHEMISTRY & MATHEMATICS Paper - 1 Time Allotted: 3 Hours
Maximum Marks: 360
Please read the instructions carefully. You are allotted 5 minutes specifically for this purpose.
Important Instructions: 1.
The test is of 3 hours duration.
2.
This Test Paper consists of 90 questions. The maximum marks are 360.
3.
There are three parts in the question paper A, B, C consisting of Physics, Chemistry and Mathematics having 30 questions in each part of equal weightage. Each question is allotted 4 (four) marks for correct response.
4.
Out of the four options given for each question, only one option is the correct answer.
5.
For each incorrect response 1 mark i.e. ¼ (one-fourth) marks of the total marks allotted to the question will be deducted from the total score. No deduction from the total score, however, will be made if no response is indicated for an item in the Answer Box.
6.
Candidates will be awarded marks as stated above in instruction No.3 for correct response of each question. One mark will be deducted for indicating incorrect response of each question. No deduction from the total score will be made if no response is indicated for an item in the answer box.
7.
There is only one correct response for each question. Marked up more than one response in any question will be treated as wrong response and marked up for wrong response will be deducted accordingly as per instruction 6 above.
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JEE-MAIN-2019 (9th Apr-Second Shift)-PCM-2
PART –A (PHYSICS) 1.
Two coils P and Q are separated by some distance. When a current of 3 A flows through coil P a magnetic flux of 10–3 Wb passes through Q. No current is passed through Q. When no current passes through P and a current of 2 A passes through Q, the flux through P is: (A) 6.67 103 Wb (B) 6.67 10 4 Wb (C) 3.67 10 4 Wb
(D) 3.67 10 3 Wb
2.
A metal wire of resistance 3 is elongated to make a uniform wire of double its previous length. The new wire is now bent and the ends joined to make a circle. If two points on this circle make an angle 600 at the centre, the equivalent resistance between these two points will be: 12 5 (A) (B) 5 3 5 7 (C) (D) 2 2
3.
The resistance of a galvanometer is 50 ohm and the maximum current which can be passed through it is 0.002 A. What resistance must be connected to it in order to convert it into an ammeter of range 0 – 0.5 A? (A) 0.2 ohm (B) 0.002 ohm (C) 0.02 ohm (D) 0.5 ohm
4.
The position of a particle as a function of time t, is given by x t at bt 2 ct 3 where a, b and c are constants. When the particle attains zero acceleration, then its velocity will be: b2 b2 (A) a (B) a 4c c b2 b2 (C) a (D) a 2c 3c
5.
A thin convex lens L (refractive index = 15) is placed on a plane mirror M. When a pin is placed at A, such that OA = 18 cm, its real inverted image is formed at A itself, as shown in figure. When a liquid of refractive index 1 is put between the lens and the mirror, the pin has to be moved to A’. such that OA’ = 27cm, to get its inverted real image at A itself. The value of 1 will be:
(A)
2
(C)
3
4 3 3 (D) 2
(B)
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JEE-MAIN-2019 (9th Apr-Second Shift)-PCM-3 6.
A moving coil galvanometer has a coil with 175 turns and area 1 cm2 It uses a torsion band of torsion constant 10–6 N – m/rad. The coil is placed in a magnetic field B parallel to its plane. The coil deflects by 10 for a current of 1 mA. The value of B (in Tesla) is approximately:(A) 10-3 (B) 10-1 -4 (C) 10 (D) 10-2
7.
A very long solenoid of radius R is carrying current I t kte t k 0 , as a function of time (t 0). counter clockwise current is taken to be positive. A circular conducting coil of radius 2R is placed in the equatorial plane of the solenoid and concentric with the solenoid. The current induced in the outer coil is correctly depicted, as a function of time, by:(A) (B)
(C)
(D)
8.
A massless spring (k = 800 N/m), attached with a mass (500 g) is completely immersed in 1 kg of water. The spring is stretched by 2 cm and released so that it starts vibrating. What would be the order of magnitude of the change in the temperature of water when the vibrations stop completely? (Assume that the water container and spring receive negligible heat and specific heat of mass = 400 J/kg K, specific heat of water = 4184 J/kg K) (A) 10-3 K (B) 10-4 -1 (C) 10 K (D) 10-5 K
9.
A particle P is formed due to a completely inelastic collision of particles x and y having de – Broglie wavelengths x and y respectively. If x and y were moving in opposite directions, then the de – Broglie wavelength of P is: xy (A) x y (B) x y (C)
10.
xy x y
(D) x y
A convex lens of focal length 20 cm produces images of the same magnification 2 when an object is kept at two distances x1 and x2 (x1 > x2) from the lens. The ratio of x1 and x2 is: (A) 5 : 3 (B) 2 : 1 (C) 4 : 3 (D) 3 : 1
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JEE-MAIN-2019 (9th Apr-Second Shift)-PCM-4
11.
Diameter of the objective lens of a telescope is 250 cm. For light of wavelength 600 nm. Coming from a distant object, the limit of resolution of the telescope is close to:(A) 1.5 10 7 rad (B) 2.0 10 7 rad (C) 3.0 10 7 rad
(D) 4.5 10 7 rad
12.
Moment of inertia of a body about a given axis is 1.5 kg m2 Initially the body is at rest. In order to produce a rotational kinetic energy of 1200 J, the angular acceleration of 20 rad/s2 must be applied about the axis of rotation for a duration of: (A) 2 s (B) 5 s (C) 2.5 s (D) 3 s
13.
The logic gate equivalent to the given logic circuit is:
(A) OR (C) NOR 14.
(B) AND (D) NAND
50 W/m2 energy density of sunlight is normally incident on the surface of a solar panel. Some part of incident energy (25%) is reflected from the surface and the rest is absorbed. The force exerted on 1 m2 surface area will be close to c 3 108 m / s (A) 15 10 8 N (C) 10 10 8 N
(B) 35 10 8 N (D) 20 10 8 N
15.
The area of a square is 5.29 cm2 The area of 7 such squares taking into account the significant figures is: (A) 37 cm2 (B) 37.0 cm2 2 (C) 37.03 cm (D) 37.030 cm2
16.
The physical sizes of the transmitter and receiver antenna in a communication system are: (A) Proportional of carrier frequency (B) Inversely proportional to modulation frequency (C) Inversely proportional to carrier frequency (D) Independent of both carrier and modulation frequency
17.
Four point charges –q, +q, +q and –q are placed on y axis at y = -2d, y = -d, y = +d and y = +2d, respectively. The magnitude of the electric field E at a point on the x – axis at x = D, with D > > d, will vary as: 1 1 (A) E (B) E 3 D D 1 1 (C) E 2 (D) E 4 D D
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JEE-MAIN-2019 (9th Apr-Second Shift)-PCM-5 18.
The specific heats, CP and CV of a gas of diatomic molecules, A, are given (in units of J mol-1 K-1) by 29 and 22, respectively. Another gas of diatomic molecules B, has the corresponding values 30 and 21. If they are treated as ideal gases, then:(A) A has one vibrational mode and B has two (B) Both A and B have a vibrational mode each (C) A is rigid but B has a vibrational mode (D) A has a vibrational mode but B has none
19.
The position vector of a particle changes with time according to the relation r t 15t 2 ˆi 4 20t 2 ˆj . What is the magnitude of the acceleration at t = 1? (A) 40 (C) 25
(B) 100 (D) 50
20.
A test particle is moving in a circular orbit in the gravitational field produced by a mass K density r 2 . Identify the correct relation between the radius R of the particle’s orbit r and its period T: (A) T/R2 is a constant (B) TR is constant (C) T2/R3 is a constant (D) T/R is a constant
21.
A particle of mass m is moving with speed 2 v collides with a mass 2m moving with speed v in the same direction. After collision, the first mass is stopped completely while the second one splits into two particles each of mass m, which move at angle 450 with respect to the original direction. The speed of each of the moving particle will be: (A) v / 2 2 (B) 2 2v
(C)
2v
(D) v / 2
4 of its volume submerged. When 5 certain amount of an oil is poured into the bucket, it is found that the block is just under the oil surface with half of its volume under water and half in oil. The density of oil relative to that of water is:(A) 0.5 (B) 0.7 (C) 0.6 (D) 0.8
22.
A wooden block floating in a bucket of water has
23.
Two cars A and B are moving away from each other in opposite directions. Both the cars are moving with a speed of 20 ms-1 with respect to the ground. If an observer in car A detects a frequency 2000 Hz of the sound coming from car B, what is the natural frequency of the sound source of car B? (speed of sound in air = 340 ms-1) (A) 2250 Hz (B) 2060 Hz (C) 2150 Hz (D) 2300 Hz
24.
A wedge of mass M = 4m lies on a frictionless plane. A particle of mass m approaches the wedge with speed v. There is no friction between the particle and the plane or between the particle and the wedge. The maximum height climbed by the particle on the wedge is given by: 2v 2 v2 (A) (B) 7g g (C)
2v 2 5g
(D)
v2 2g
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JEE-MAIN-2019 (9th Apr-Second Shift)-PCM-6
25.
A He+ ion is in its first excited state. Its ionization energy is: (A) 6.04 eV (B) 13.60 eV (C) 54.40 eV (D) 48.36 eV
26.
Two materials having coefficients of thermal conductivity 3K and K and thickness d and 3d, respectively, are joined to form a slab as shown in the figure. The temperatures of the outer surfaces are 2 and 1 respectively 2 1 . The temperature at the interface is:
2 1 2 (C) 1 2 3 3
(A)
1 92 10 10 5 (D) 1 2 6 6
(B)
27.
A thin smooth rod of length L and mass M is rotating freely with angular speed 0 about an axis perpendicular to the rod and passing through its centre. Two beads of mass m and negligible size are at the centre of the rod initially. The beads are free to slide along the rod. The angular speed of the system, when the beads reach the opposite ends of the rod will be:M0 M0 (A) (B) M 3m Mm M0 M0 (C) (D) M 2m M 6m
28.
The parallel combination of two air filled parallel plate capacitors of capacitance C and nC is connected to a battery of voltage, V. When the capacitor are fully charged, the battery is removed and after that a dielectric material of dielectric constant K is placed between the two plates of the first capacitor. The new potential difference of the combined system is: V (A) (B) V Kn n 1 V nV (C) (D) Kn K n
29.
In a conductor, if the number of conduction electrons per unit volume is 8.5 x 1028 m–3 and mean free time is 25fs (femto second), its approximate resistivity is: me 9.1 1031kg (A) 10-5 m (C) 10 7 m
30.
(B) 10-6 m (D) 10-8 m
A string 2.0 m long and fixed at its end is driven by a 240 Hz vibrator. The string vibrates in its third harmonic mode. The speed of the wave and its fundamental frequency is: (A) 320 m/s, 120 Hz (B) 180 m/s, 80 Hz (C) 180m/s, 120 Hz (D) 320 m/s, 80 Hz FIITJEE Ltd., FIITJEE House, 29-A, Kalu Sarai, Sarvapriya Vihar, New Delhi -110016, Ph 46106000, 26569493, Fax 26513942 website: www.fiitjee.com.
JEE-MAIN-2019 (9th Apr-Second Shift)-PCM-7
PART –B (CHEMISTRY) 31.
Increasing order of reactivity of the following compounds for SN1 substitution is: (a) (b)
(c)
32.
(d)
(A) (b) < (c) < (d) < (a) (C) (b) < (a) < (d) < (c)
(B) (a) < (b) < (d) < (c) (D) (b) < (c) < (a) < (d)
The one that is not a carbonate is: (A) bauxite (C) calamine
(B) siderite (D) malachite
33.
During compression of a spring the work done is 10kJ and 2kJ escaped to the surroundings as heat. The change in internal energy, U (in kJ) is: (A) 8 (B) 12 (C) -12 (D) -8
34.
The amorphous form of silica is: (A) quartz (C) cristobalite
35.
(B) kieselguhr (D) tridymite
The major products A and B for the following reactions are, respectively
(A)
(B)
(C)
(D)
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JEE-MAIN-2019 (9th Apr-Second Shift)-PCM-8 36.
Which one of the following about an electron occupying the 1s orbital in a hydrogen atom is incorrect? (The Bohr radius is represented by a0) (A) The electron can be found at a distance 2a0 from the nucleus (B) The probability density of finding the electron is maximum at the nucleus (C) The magnitude of potential energy is double that of its kinetic energy on an average (D) The total energy of the electron is maximum when it is at a distance a0 from the nucleus
37.
The maximum possible denticities of a ligand given below towards a common transition and inner – transition metal ion, respectively, are:
(A) 6 and 8 (C) 8 and 8
(B) 8 and 6 (D) 6 and 6
38.
The correct statements among I to III regarding group 13 element oxides are, I. Boron trioxide is acidic II. Oxides of aluminium and gallium are amphoteric III. Oxides of indium and thallium are basic (A) I, II and III (B) II and III only (C) I and III only (D) I and II only
39.
Among the following species, the diamagnetic molecule is (A) O2 (B) NO (C) B2 (D) CO
40.
The peptide that gives positive ceric ammonium nitrate and carbylamine tests is: (A) Lys - Asp (B) Ser – Lys (C)Gln - Asp (D) Asp – Gln
41.
Assertion : For the extraction of iron, haematite ore is used Reason : Haematite is a carbonate ore of iron (A) Only the reason is correct. (B) Both the assertion and reason are correct and the reason is the correct explanation for the assertion. (C) Only the assertion is correct (D) Both the assertion and reason are correct, but the reason is not the correct explanation for the assertion
42.
10 mL of 1mM surfactant solution forms a monolayer covering 0.24 cm2 on a polar substrate. If the polar head is approximated as cube, what is its edge length? (A) 2.0 pm (B) 2.0 nm (C)1.0 pm (D) 0.1 nm
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JEE-MAIN-2019 (9th Apr-Second Shift)-PCM-9 43.
Consider the given plot of enthalpy of the following reaction between A and B. A+BC+D Identify the incorrect statement
(A) C is the thermodynamically stable product (B) Formation of A and B from C has highest enthalpy of activation (C) D is kinetically stable product (D) Activation enthalpy to form C is 5kJ mol-1 less than that to form D 44.
At a given temperature T, gases Ne, Ar, Xe and Kr are found to deviate from ideal gas RT behaviour. Their equation of state is given as P at T. Here, b is the van der V b Waals constant. Which gas will exhibit steeper increase in the plot of Z (compression factor) versus P? (A) Ne (B) Ar (C) Xe (D) Kr
45.
A solution of Ni(NO3)2 is electrolysed between platinum electrodes using 0.1 Faraday electricity. How many mole of Ni will be deposited at the cathode? (A) 0.20 (B) 0.05 (C). 0.10 (D) 0.15
46.
In the following reaction
Carbonyl compound + MeOH acetal Rate of the reaction is the highest for: (A) Acetone as substrate and methanol in stoichiometric amount (B) Proponal as substrate and methanol in stoichiometric amount (C) Acetone as substrate and methanol in excess (D) Propanal as substrate and methanol in excess HCI
47.
The structures of beryllium chloride in the solid state and vapour, phase, respectively, are: (A) chain and dimeric (B) chain and chain (C)dimeric and dimeric (D) dimeric and chain
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JEE-MAIN-2019 (9th Apr-Second Shift)-PCM-10 48.
49.
Which of the following potential energy (PE) diagrams represents the SN1 reaction?
(A)
(B)
(C)
(D)
The major product of the following reaction is:
(A)
(B)
(C)
(D)
50.
The maximum number of possible oxidation states of actinoids are shown by (A) berkelium (Bk) and californium (Cf) (B) nobelium (No) and lawrencium (Lr) (C) actinium (Ac) and thorium (Th) (D) neptunium (Np) and plutonium (Pu)
51.
Molal depression constant for a solvent is 4.0 K Kg mol-1. The depression in the freezing point of the solvent for 0.03 mol kg–1 solution of K2SO4 is: (Assume complete dissociation of the electrolyte) (A) 0.12 K (B) 0.36 K (C) 0.18 K (D) 0.24 K
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JEE-MAIN-2019 (9th Apr-Second Shift)-PCM-11 52.
53.
Noradrenaline is a/an (A) Neurotransmitter (C) Antihistamine
(B) Antidepresessant (D) Antacid
Hinsberg’s reagent is: (A) C6H5SO2CI
(B) C6H5COCI
(C) SOCI2
(D) COCI2
54.
The layer of atmosphere between 10 km to 50 km above the sea level is called as: (A) troposphere (B) mesosphere (C)stratosphere (D) thermosphere
55.
HF has highest boiling point among hydrogen halides, because it has: (A) lowest dissociation enthalpy (B) strongest van der Waals’ interactions (C) strongest hydrogen bonding (D) lowest ionic character
56.
What would be the molality of 20% (mass/mass) aqueous solution of KI? (molar mass of KI = 166 g mol–1) (A) 1.08 (B) 1.48 (C)1.51 (D) 1.35
57.
Which of the following compounds is a constituent of the polymer?
(A) Formaldehyde (C) Methylamine 58.
(B) Ammonia (D) N – methyl urea
The correct statements among I to III are: I. Valence bond theory cannot explain the color exhibited by transition metal complexes II. Valence bond theory can predict quantitatively the magnetic properties of transition metal complexes III. Valence bond theory cannot distinguish ligands as weak and strong field ones (A) I and II only (B) I, II and III (C) I and III only (D) II and III only
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JEE-MAIN-2019 (9th Apr-Second Shift)-PCM-12 59.
In an acid – base titration, 0.1 M HCI solution was added to the NaOH solution of unknown strength. Which of the following correctly shows the change of pH of the titration mixture in this experiment?
(A) (a) (C) (d) 60.
(B) (c) (D) (b)
p-Hydroxybenzophenone upon reaction with bromine in carbon tetrachloride gives:
(A)
(B)
(C)
(D)
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JEE-MAIN-2019 (9th Apr-Second Shift)-PCM-13
PART–C (MATHEMATICS) 61.
If the tangent to the parabola y 2 x at a point , , ( > 0) is also a tangent to the ellipse, x 2 2y 2 1, then a is equal to (A) 2 2 1 (C) 2 1
62.
(B) 2 1 (D) 2 2 1
Some identical balls are arranged in rows to form an equilateral triangle. The first row consists of one ball, the second row consists of two balls and so on. If 99 more identical balls are added to the total number of balls used in forming the equilateral triangle, then all these balls can be arranged in a square whose each side contains exactly 2 balls less than the number of balls each side of the triangle contains. Then the number of balls used to form the equilateral triangle is (A) 190 (B) 262 (C) 225 (D) 157 f x
63.
If f : R R is a differentiable function and f 2 6, then lim
x2
2 tdt
x 2
is:
6
(B) 2f ' 2 (D) 24f’ (2)
(A) 0 (C) 12f’ (2) 64.
If the system of equations 2x 3y z 0, x ky 2z 0 and 2x y z 0 has a non – x y z trivial solution (x, y, z), then k is equal to y z x 3 (A) (B) –4 4 1 1 (C) (D) 2 4
65.
The common tangent to the circles x 2 y 2 4 and x 2 y 2 6x 8y 24 0 also passes through the point (A) (–4, 6) (B) (6. –2) (C) (–6, 4) (D) (4, –2)
66.
If the sum and product of the first three term in an A.P. are 33 and 1155, respectively, then a value of its 11th tern is (A) –25 (B) 25 (C) –36 (D) –35
67.
The value of the integral
1
x cot 1 x 1
2
x 4 dx is
0
1 loge 2 4 2 1 (C) loge 2 2 2
(A)
loge 2 2 (D) loge 2 4
(B)
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JEE-MAIN-2019 (9th Apr-Second Shift)-PCM-14 68.
The value of sin10o sin 30o sin50o sin70o is 1 1 (A) (B) 36 32 1 1 (C) (D) 18 16
69.
Let z C be such that z 1 . If w
5 3z , then 5 1 z
(A) 5Im w 1
(B) 4Im w 5
(C) 5 Re w 1
(D) 5 Re w 4 n
70.
If some three consecutive in the binomial expansion of x 1 in powers of x are in the ratio 2 : 15 : 70, then the average of these three coefficient is: (A) 964 (B) 625 (C) 227 (D) 232
71.
If cos x
72.
If the two lines x a 1 y 1 and 2x a 2 y 1 a R 0,1 are perpendicular, then the
dy y sin x 6x, 0 x and y 0, then y is equal to: dx 2 3 6 2 2 (A) (B) 2 4 3 2 2 (C) (D) 2 3 2 3
distance of their point of intersection from the origin is: 2 2 (A) (B) 5 5 2 2 (C) (D) 5 5 73.
A water tank has the shape of an inverted right circular cone, whose semi vertical angle 1 is tan 1 .Water is poured in at a constant rage of 5 cubic meter per minute. Then the 2 rate (in m/min) at which the level of water is rising at the instant when the depth of water in the tank is 10 m is: 2 1 (A) (B) 5 1 1 (C) (D) 10 15
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JEE-MAIN-2019 (9th Apr-Second Shift)-PCM-15 74.
Two poles standing on a horizontal ground are of heights 5m and 10 m respectively. The line joining their tops makes an angle of 15o with ground. Then the distance (in m) between the poles, is 5 (A) 2 3 (B) 5 3 1 2 (C) 5 2 3 (D) 10 3 1
75.
77.
x 2 y 1 z such that BC = 5 units. 3 0 4 Then the area (in sq. units) of this triangle, given that the point A (1, –1, 2) is: (A) 2 34 (B) 34
(D) 5 17
0 2x 2x The total number of matrices A 2y y y ; x, y R, x y for which A T A 3I3 1 1 1 (A) 6 (B) 2 (C) 3 (D) 4
The area in sq. units) of the smaller of the two circles that touch the parabola, y 2 4x at the points (1, 2) and the axis is (A) 4 2 2 (B) 8 3 2 2
(C) 4 3 2 78.
79.
The vertices B and C of a ABC lie on the line,
(C) 6
76.
(D) 8 2 2
a x 1, x 5 If the function f x is continuous at x = 5, then the value of a – b is b x 3, x 5 2 2 (A) (B) 5 5 2 2 (C) (D) 5 5 x If f x x , x R , where [x] denotes the greatest integer function, then: 4 (A) Both lim f x and lim f x exist but are not equal x 4
x 4
(B) lim f x exists but lim f x does not exist x 4
x 4
(C) lim f x exists but lim f x does not exist x 4
x 4
(D) f is continuous at x = 4
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JEE-MAIN-2019 (9th Apr-Second Shift)-PCM-16
80.
If
e
sec x
sec x tan x f x sec x tan x sec x dx e 2
sec x
f x C , then a possible
choice of f x is (A) sec x tan x
1 2
(C) sec x x tan x 81.
(B) x sec x tan x 1 2
(D) sec x tan x
1 2
1 2 2
If m is chosen in the quadratic equation m2 1 x 2 3x m2 1 0 such that the sum of its roots is greatest, then the absolute difference of the cubes of its roots is: (A) 8 3 (B) 4 3 (C) 10 5
(D) 8 5
82.
Two newspaper A and B are published in a city. It is known that 25% of the city populations reads A and 20% reads B while 8% reads both A and B. Further, 30% of those who read A but not B look into advertisements and 40% of those who read B but not A also look into advertisements, while 50% of those who read both A and B look into advertisements. Then the percentage of the population who look into advertisement is: (A) 12.8 (B) 13.5 (C) 13.9 (D) 13
83.
Let P be the plane, which contains the line of intersection of the planes, x y z 6 0 and 2x 3y z 5 0 and it is perpendicular to the xy – plane. Then the distance of the point (0, 0, 256) from P is equal to: (A) 63 5 (B) 205 5 17 11 (C) (D) 5 5
84.
If P q r is false, then the truth values of p, q, r are respectively (A) F, T, T (B) T, F, F (C) T, T, F (D) F, F, F 1 The domain of the definition of the function f x log x3 x is 2 4x (A) 1,2 2, (B) 1,0 1,2 3,
85.
(C) 1,0 1,2 2,
(D) 2, 1 1,0 2,
86.
The sum of the series 1 2 3 3 5 4 7 .... upto 11th term is (A) 915 (B) 946 (C) 945 (D) 916
87.
The mean and the median of the following ten numbers in increasing order 10, 22, 26, y 29, 34, x, 42, 67, 70, y are 42 and 35 respectively, then is equal to x 7 9 (A) (B) 3 4 7 8 (C) (D) 2 3 FIITJEE Ltd., FIITJEE House, 29-A, Kalu Sarai, Sarvapriya Vihar, New Delhi -110016, Ph 46106000, 26569493, Fax 26513942 website: www.fiitjee.com.
JEE-MAIN-2019 (9th Apr-Second Shift)-PCM-17
88.
89.
90.
y2 The area (in sq. units) of the region A x, y : x y 4 is: 2 53 (A) (B) 18 3 (C) 30 (D) 16
If a unit vector r makes angles with ˆi, with ˆj and 0, with kˆ , then a value of 3 4 is 5 5 (A) (B) 12 6 2 (C) (D) 3 4 A rectangle is inscribed in a circle with a diameter lying along the line 3y x 7 . If the two adjacent vertices of the rectangle are (–8, 5) and (6, 5) then the area of the rectangle (in sq. units) is (A) 72 (B) 84 (C) 98 (D) 56
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JEE-MAIN-2019 (9th Apr-Second Shift)-PCM-18
JEE (Main) – 2019 ANSWERS PART A – PHYSICS 1. 5. 9. 13. 17. 21. 25. 29.
B B C A D B B D
2. 6. 10. 14. 18. 22. 26. 30.
B A D D D C B D
3. 7. 11. 15. 19. 23. 27.
A D C C D A D
4. 8. 12. 16. 20. 24. 28.
D D A C D C C
34. 38. 42. 46. 50. 54. 58.
B A A D D C C
64. 68. 72. 76. 80. 84. 88.
C D D D D B B
PART B – CHEMISTRY 31. 35. 39. 43. 47. 51. 55. 59.
C B D D A B C A
32. 36. 40. 44. 48. 52. 56. 60.
A D B C D A C D
33. 37. 41. 45. 49. 53. 57.
A A C B C A A
PART C – MATHEMATICS 61. 65. 69. 73. 77. 81. 85. 89.
C B C B B D C C
62. 66. 70. 74. 78. 82. 86. 90.
A A D C A C B B
63. 67. 71. 75. 79. 83. 87.
C A D B D D A
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JEE-MAIN-2019 (9th Apr-Second Shift)-PCM-19
HINTS AND SOLUTIONS PART A – PHYSICS 1.
Mutual induction q MIp 10–3 = M(3) 1 M 10 3 3 p = MIq p = 6.67 × 10–4 Wb
2.
2 (V Volume of wire) A (V / ) V Final resistance = 3 × (B)2 = 12 Req = 2 || 10 5 = 3 R
3.
Since shunt resistance is connected in parallel with galvanometer, both will have same voltage drop. (0.002) (RG) = (0.5 – 0.002) (rs) rs 0.2
4.
x = at + bt2 – ct3 dx V = a + 2bt – 3ct2 dt dv a 2b 6ct dt Put acceleration = 0 b t 3c b Find V at t 3c b2 V= a 3c
5.
For image to form at object itself, says must retrace their path back to object. Hence must incident on mirror normally. Case 1: Object will be at focus of lens 1 1 1 1 ( 1) f R R 18 R = 18 cm Case 2: Retraction at 1st surface:
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JEE-MAIN-2019 (9th Apr-Second Shift)-PCM-20
1 1.5 1 1.5 27 V1 R
…(i)
2nd retraction:
1.5 1.5 u V1 R
…(ii)
From (i) and (ii) = 6.
7.
4 . 3
MB C = i N A B 106 103 104 175 B 180 B = 10–3 Tesla
d d = (onI) ( R2 ) dt dt dI = ( on R 2 ) dt d (kt e t ) = on R2 dt = –C (t(e–t) (–) + e–t) = –C (e–t) (1 – t)
(C constant)
Clearly, at t = 0 Induced current 0 Also, apply Lenz law to find correct option. 8.
By law of conservation of energy 1 2 kx = (m1s1 + m2s2)T 2 16 10 2 T = 3.65 × 10–5 4384
9.
Conservation of momentum p x p y p final mxvx – myvy = (mx + my) V h h h x y
xy x y
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JEE-MAIN-2019 (9th Apr-Second Shift)-PCM-21 10.
Magnification is 2 If image is real, x1
3f 2
If image is virtual, x2 =
f 2
x1 =3:1 x2
1.22 d 1.22 600 10 9 = 250 10 2 = 2.9 × 10–7 rad.
11.
Limit of resolution =
12.
KE =
13.
Truth table can be formed as
1 2 I 1200 (given) 2 = 40 rad/s = o + t 40 = 0 + (20) t t = 2 sec.
A 0 0 1 1
B 0 1 0 1
Equivalent 0 1 1 1
Hence the equivalent is “OR” gate. 14.
Radiation pressure for 100% reflection = Radiation pressure for 0% reflection =
2I C
I C
2I I Hence, in given case, radiation pressure = (0.25) (0.75) C C I = (1.25) C Force = P × (Area) = 20.83 × 10–8 N 15.
Total Area = A1 + A2 + ……… A7 = A + A + ……….. 7 times = 37.03 m2 Addition of 7 terms all having 2 terms beyond decimal, so final answer must have 2 terms beyond decimal (as per rules of significant digits)
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JEE-MAIN-2019 (9th Apr-Second Shift)-PCM-22
16.
The physical size of antenna of receiver and transmitter both are inversely proportional to carrier frequency.
17.
Electric field at p = 2E1cos1 –2E2cos2 2Kq D 2Kq D = 2 2 2 2 1/2 2 2 2 (d D ) (d D ) [(2d) D ] [(2d) D 2 ]1/ 2 = 2KqD (d2 D2 )3 /2 (4d2 D2 )3/ 2 3/ 2 3/ 2 d2 4d2 1 2 1 2 D D Applying binomial approximation d << D 2KqD 3 d2 3 4d2 = 1 1 D3 2 D2 2D2 2KqD 12 d2 3 d2 = D3 2 D2 2 D2
2KqD = D3
= 18.
9kqd2 D4
For A: Cp 2 29 1 Cv f 22 It gives f = 6.3 6
(3 translational, 2 rotational and 1 vibrational)
For B: Cp 2 30 1 Cv f 21 f = 4.67 5 (3 translational, 2 rotational, no vibrational)
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JEE-MAIN-2019 (9th Apr-Second Shift)-PCM-23
19.
20.
r (15t 2 )iˆ (4 20t 2 )ˆj dr v (30t)iˆ (40t)jˆ dt dv a (30)iˆ (40)jˆ dt a 50
For circular motion of particle: mV 2 mE r GM = m 2 r r
k Where M = (4 x 2 dx) 2 x 0 = 4kr 2 mV G(4 k) m r r V = constant 2R T= V T Constant R
21.
Linear momentum conservation v m 2v 2m v m 0 m 2 2 v 2 2 v. 22.
In 1st situation Vbbg = Vswg Vs b 4 Vb w 5
…(i)
Here Vb is volume of block Vs is submerged volume of block b is density of block w is density of water & Let o is density of oil Finally in equilibrium condition V V Vb b g b o g b w g 2 2 FIITJEE Ltd., FIITJEE House, 29-A, Kalu Sarai, Sarvapriya Vihar, New Delhi -110016, Ph 46106000, 26569493, Fax 26513942 website: www.fiitjee.com.
JEE-MAIN-2019 (9th Apr-Second Shift)-PCM-24 2b = 0 + w 3 o 0.6 w 5 23.
Doppler effect: v uo f (fo ) v us 340 ( 20) 2000 (fo ) 340 ( 20) fo = 2250 Hz
24.
Let mass attains height ‘h’ on wedge and at that time, both attain velocity vf. Conservation of momentum: mv = (m + 4m)Vf …(A) COE: 1 1 mv 2 (m 4m)Vf2 mgh 2 2 From (A) and (B) 2V 2 h 5g
25.
26.
27.
…(B)
z2 T.E. = (13.6) 2 eV n z=n=2 Ionisation energy = –T.E. = 13.6 eV At steady state: q q t 1 t 2 3kA( 2 ) kA( 1 ) d 3d 1 92 10 Conservation of angular momentum about rotation axes: Li = Lf M 2 2 M 2 2 m f o 2 12 12
M f o M 6m
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JEE-MAIN-2019 (9th Apr-Second Shift)-PCM-25 28.
After fully charging, battery is disconnected.
Total chare of the system = CV + nCV = (n + 1)CV After the insertion of dielectric of constant K New potential (common) Total charge VC Total capacitance (n 1)CV (n 1)V = KC nC K n
2m ne2 = 3.34 × 10–8 m
29.
30.
We have: nv f 2 3v 240 = 22 v = 320 m/s Fundamental frequency =
v = 80 Hz. 2
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JEE-MAIN-2019 (9th Apr-Second Shift)-PCM-26
PART B – CHEMISTRY 31.
Stability of carbonium ions involved in the reactions follow the order:
CH2
CH2 H3C >
>
CH - CH2 > CH3CH2
H3C OCH3 Reactivity Stability of carbonium ions 32.
Malachite = CuCO3.Cu(OH)2 Bauxite = Al2O3.2H2O Calamine = ZnCO3 Siderite = FeCO3
33.
Work done on system = +10 kJ Heat escaped = -2kJ U = q + w = 10 – 2 = 8 kJ
34.
Kieselguhr is amorphous form of silica.
35.
O
O I
CN KCN/DMSO SN 2
H2/Pd OH CH2NH2
36.
37.
PE 2 The energy of electrons increases as they are away from the nucleus. The probability of finding the 1s electron may be higher at a0 but the energy is not. The probability density of finding the electron is not zero at any place in the atom. So, the energy may be higher when it is far from a0. T.E K.E
General coordination number of CN– in transition element is 6 and in inner transition element is 8-12. because inner transition metal ions can make available more number of vacant orbitals of nearly same energy than transition metal ions. The high effective nuclear charge of inner-transition metal ions make them form complex with high coordination number.
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JEE-MAIN-2019 (9th Apr-Second Shift)-PCM-27 38.
B2O3 is acidic in nature Al2O3 and Ga2O3 are amphoteric Oxides of In and Tl are basic in nature. Because the metallic character of the elements increases on moving down the group.
39.
CO is diamagnetic in nature due to absence of any unpaired electron.
40.
Due to –OH group of serine it give cerric ammonium nitrate test whereas due to –NH2 group lysine give +ve carbylamine test.
41.
Extraction of Fe is done form haematite ore this is true but reason is wrong as haematite is Fe2O3.
42.
MVml 103 10 105 mole 1000 1000 10–5 NA molecules covering area = 0.24 cm2 0.24 1 ---------------------------------------- = cm2 10 5 NA Moles
0.24 a2 23 10 6 10 a2 = 4 10–20 cm2 a = 2 10–10 cm a = 2 10–12 m a = 2 pm 5
43.
Ea = (D C) = 15 – 0 = 15 kJ mol–1 Ea = (A + B) C = 15 kJ mol–1 Ea = (A + B) D = 10 kJ mol–1 Ea = C (A + B) = 20 kJ mol–1 (high activation enthalpy in the reaction) Activation enthalpy to form C is 5 kJ mol–1 more than that of form D.
44.
P
RT V b
P(V – b) = RT a P 2 V b RT V At high pressure. P(V - b) = RT PV – Pb = RT PV Pb 1 RT RT Pb Z=1+ RT Z > 1, Z b The value of ‘b’ for the gases follows the order Ne < Ar < Kr < Xe
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JEE-MAIN-2019 (9th Apr-Second Shift)-PCM-28
45.
1F deposits 1 g equivalent of Ni or ½ mole of Ni 1 0.1 F will deposit 0.05 moles of Ni 20
46.
Ketones < Aldehyde Rate of Nucleophilic addition reaction Only aldehydes are responsible for formation of acetals. Excess of MeOH is used to drive the reaction towards forward direction.
47.
Solid state – chain Cl Cl
- Be
Be
Be
-
Cl Cl In vapour state dimeric Cl
-
Cl Be
Be - Cl Cl
48.
For SN1 carbocation is formed.
The first transition state should have higher energy than the second transition state. Reaction intermediate is also formed. 49.
Chloroform is used as a solvent in this reaction
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JEE-MAIN-2019 (9th Apr-Second Shift)-PCM-29 50.
Ac +3
Th
Pu +3 +4 +5
U +3 +4 +5
Np Pu Am Cm Bk Cf Es Fm Md +3 +3 +3 +3 +3 +3 +3 +3 +3 +4 +4 +4 +4 +4 +4 +5 +5 +5 +6 +6 +6 +6 +7 +7 Np and Pu has maximum no. of possible oxidation states
No Lr +2 +3
51.
Kf = 4 K Kg mol–1 i=3 Molality = 0.03 Tf = i Kfm = 3(4) (0.03) Tf = 0.36 K
52.
It is an organic chemical in the catecholamine family that functions in the brain and body as hormone and neurotransmitter.
53.
Benzene sulphonyl chloride C6H5SO2Cl
54.
Fact based.
55.
HF has strong hydrogen bond. It has highest boiling point among hydrogen halides. The strong hydrogen bond is due to more difference in electronegativity between F and H atoms.
56.
20% w/w KI Mass of solute(KI) = 20 g Mass of solvent = 100 – 20 = 80 g Molar mass of KI = 38 + 128 = 166 gm solute 20 1000 Molality 1.506 1.51 mw Kg solvent 166 80
57.
58.
(i) VBT does not explain colour exhibited by complex of transition metal because splitting of d-orbitals in explained by CFT. (ii) VBT does not distinguish between strong and weak field complex. (iii) VBT does not give quantitative interpretation of magnetic properties. According to question statement I and III are correct.
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JEE-MAIN-2019 (9th Apr-Second Shift)-PCM-30 59.
The nature of the curve is such that the pH change of the reaction becomes larger for very small change in volume of titrant. O
60.
O Br Br2 /CCl4
HO
HO O C
– R effect, deactivate the ring for EAS. OH +R effect activate the
ring towards EAS
PART C – MATHEMATICS 61.
Equation of tangent to the parabola y 2 x At , is T 0
x 2 x 2 y 2 2 1 y x 2 2 y
1 ,c m 2 2 This is also a tangent to ellipse x 2 2y 2 1
C a 2m2 b 2
1 1 2 2 4 2
2 1 1 2 4 4 2
4 22 1 0
2
2 1 2 2 1 2 2 2 1 2 2 1
62.
n n 1
99 n 2
2
2 n2 n 198 2n2 8n 8 n2 9n 190 0
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JEE-MAIN-2019 (9th Apr-Second Shift)-PCM-31 n 19 n 10 0 n 19 19 20 Number of balls is 190 2
63.
lim
f x
x2 6
2t dt dx {given that f 2 6 } x 2
0 from, so we use L – Hospital Rule 0 f ' x .2f x lim x 2 1 f ' 2 .2f 2 12f ' 2
64.
system of equations has non trival solution 2 3 1 D 0 1
k
2 1
2 0 1
9 2 So equation are 2x 3y z 0 ……….(1) 9 x y 2z 0 ……….(2) 2 2x y z 0 ……….(3) (1) – (3) 4y 2z 0 2y z ………….(4) y 1 z 2 From equation (1) and (4) 2x 3y 2y 0 2x y 0 x 1 z or 4 y 2 x x y z 1 k y z x 2 k
65.
Circle x 2 y2 4 c1 0,0 ; r1 2
and circle x 2 y 2 6x 8y 24 0 c 2 3, 4 ; r2 7 d c 1c 2 5
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JEE-MAIN-2019 (9th Apr-Second Shift)-PCM-32 also d r1 r2 circles touch internally Equation of common tangent S1 S2 0 6x 8y 20 0 3x 4y 10 0 Point (6, –2) satisfy it. 66.
Let the three numbers in A.P. are a – d, a, a + d Given that a d a a d 33 a 11 and a d a a d 1155
a a2 d2 1155
11 121 d2 1155 2
d 16 d 4 If d = 4 then first term a – d = 7 If d = –4 then first term a – d = 15 T11 7 10 4 47 , T11 15 10 4 25 67.
1
I x cot 1 1 x 2 x 4 dx 0
1 1 I x tan1 dx 2 4 0 1 x x x 2 x 2 1 1 1 I x tan dx 2 2 0 1 x x 1
1
I x tan1 x 2 tan 1 x 2 1 dx 0
It x 2 t 2x dx dt 1 1 I tan1 t tan1 t 1 dt 2 0 1 1 1 1 tan1 t dt tan1 t 1 dt 2 0 2 0 1 1 1 1 tan1 t dt tan1 t dt 0 2 2 0 1 1 1 1 tan1 t dt tan1 t dt 2 0 2 1
1
tan1 t dt 0
t .tan1 t
1 t t 1
1
0
0
2
dt
1 1 log 1 t2 0 4 2 1 loge 2 4 2
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JEE-MAIN-2019 (9th Apr-Second Shift)-PCM-33
68.
sin10o sin30o sin50o sin70o sin10o sin30o sin50o sin70o
sin30o sin10o sin 60o 10o sin 60o 10o
1 sin30o sin3 10o 4 1 1 1 2 4 2 1 16
69.
W
5 3z 5 1 z
5w 5wz 5 3z 5w 5 z 3 5w given z 1
R w
5w 5 1 3 5w 5w 5 3 5w
w 1
3 5
1 5
1 5
1
3 w 5
n
Cr 1 2 r 2 Cr 15 n r 1 15 15r 2n 2r 2 ………..(1) 17r 2n 2 n C 15 r 1 3 also given n r Cr 1 70 n r 14 3n 3r 14r 14 ………..(2) 17r 3n 14 Solving (1) and (2) n = 16, r = 2 16 C1 16 C2 16 C3 Average of coefficient 3 16 120 560 3 = 232
70.
Given:
71.
cos x
n
dy y sin x 6x dx
dy y tan x 6x sec x dx
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JEE-MAIN-2019 (9th Apr-Second Shift)-PCM-34
tan x dx e e loge sec x
1 sec x
: solution of equation 1 1 y. 6x sec x. dx sec x sec x
y 3x 2 c sec x
…….(1)
given y 0 3 2 So, 0 3 C 9 2 C 3 Now from (1) y 2 3x 2 sec x 3 At x 6 3y 32 2 2 36 3 2 y 2 3 72.
Two lines are perpendicular m1m2 1
1 2 2 1 a 1 a a3 a 2 2 0 a 1 a 2 2a 2 0
a 1 So lines are
L1 : x 2y 1 0 L 2 :2x y 1 0
Solving these equation we get point of intersection 1 3 P , 5 6 Now distance of P from origin 1 9 2 OP 25 25 5
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JEE-MAIN-2019 (9th Apr-Second Shift)-PCM-35
73.
1 Given tan1 2 1 r tan 2 h h r 2 1 2 V r h 3 1 h3 V 3 4 dv dh 3h2 dt 12 dt dh dh 1 5 100 4 dt dt 5
74.
r
In ABC tan15o
5 x x 5 2 3
h = 10 m
B
5 x
2 3
5 cm
A
15o
10 cm C
5 x
E
D
A (1, –1, 2)
75.
Let any point on given line is D 3 2, 1, 4 Now AD BC D.R. of BC a1 3, b1 0, c1 y D.R. of AD a2 3 3, b2 2, c 2 4 2 a1a 2 b1b 2 c 1c 2 0
B
D
C
x 2 y 1 z 3 0 4
5
3 3 3 0 4 4 2 0 25 17 17 25 Co – ordinate of point D
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JEE-MAIN-2019 (9th Apr-Second Shift)-PCM-36
68 1 25 ,1, 25 576 324 2 AD 4 34 625 25 5 1 Area of ABC BC AD 2 1 2 5 34 2 6 34 76.
A T A 3I3
0 2x 2x 0 2y 2y y y 2x y 1 1 1 2x y 8x 2 0 0 3 0 0 6y 2 0 0 3 0 0 3 0 0 3 8x 2 3 x 8
1 3 0 0 1 0 3 0 1 0 0 3 0 0 3
1 2 4 matrices are possible 6y 2 3 y
77.
Equation of tangent parabola y 2 4x at
to
the
x 1 (1, 2) is 2y 4 2 y x 1 Equation of normal y x 3 Let centre be C (3 – r, r) Now PC2 r 2 2
2
3 r 1 r 2 r
y=x+1
y P (1, 2)
C r
2
(3 – r, r) x
2
2 2 r r2 r 2 8r 8 0 r 42 2 For r 4 2 2 possible So r 4 2 2
(3 – r, 0)
3 r 0
Area r 2 16 8 16 2
8 3 2 2
Not
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JEE-MAIN-2019 (9th Apr-Second Shift)-PCM-37
78.
a x 1, x 5 f x b x 3, x 5 Contributes at x = 5 L.H.L. = R.H.L = f (5) b 5 3 a 5 1
b 5 3 a 5 1 a b 5 2
ab 79.
2 2 5 5
x f x x 4 x lim f x lim x 4 1 3 x 4 x4 4 x lim f x lim x 3 0 3 x4 x 4 4 f 4 3 Continuous at x = 4
80.
e
sec x
sec x tan x f x sec x tan x sec x dx e 2
sec x
f x C
Diff. both side w.r.t.x
esec x sec x tan x f x sec x tan x sec 2 x
esec x .sec x tan xf x esec x f n f ' x sec 2 x tan x sec x f x tan x sec x C
81.
m
2
2
1 x 2 3x m 1 0
3 m 1 2 m 1
2
m2 1 is maximum
m2 1 is minimum m0 3 and 1
3 3 2 2
2
2
4
9 4 9 1
8 5
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JEE-MAIN-2019 (9th Apr-Second Shift)-PCM-38 82.
Let population = 100 n A 25 n B 20 n A B 8
B
A
n A B 12 n A B 17
17
8
12
Now % of the population who look advertisement 30 40 50 17 12 8 100 100 100 13.9 83.
Equation of plane P1 P2 0
x y z 6 2x 3y z 5 0 x 1 2 y 1 3 2 1 6 5 0 This plane is to xy – plane 1 0 So, equation of plane x 2y 11 0 x 2y 11 0 distance of the point (0, 0, 256) from this plane. 0 0 11 11 1 4 5 84.
p q r is false
T F F So, p T, q F and r F 85.
1 log10 4 x2 1 Let f1 and 4 x2 4 x2 0 x 2 f x
x
3
x
f2 log10 x 3 x
x3 x 0 x x 1 x 1 0
x 1, 0 1, 2 x 1, 0 1,2 2,
86.
S 1 2 3 3 5 4 7 ....... upto 11 terms nth term of the series is Tn n 2n 1 11
11
S Tn 2n2 n n 1
n 1
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JEE-MAIN-2019 (9th Apr-Second Shift)-PCM-39
Sn
2n n 1 2n 1
n n 1
6 2 Put n 1 2 1112 23 1112 S11 6 2 S11 946
87.
mean = 42 10 22 26 29 34 x 42 67 70 y 45 10 x y 120 …………(i) and median = 35 34 x 35 x 36 2 from (i) y = 84 y 84 7 x 36 3
88.
y 2 2x …..(i) and x y 4 0 ………(ii) solving (1) and (2) 2 x y 2x x 2 10x 16 0 x 8,2 and y 4, 2 4
A
y2 y 4 dy 2 2 4
y2 y3 A 4y 6 2 4 64 8 A 4 16 1 8 18 6 6 89.
cos2 cos2 cos2 1 1 1 cos2 1 4 2 3 1 cos2 1 4 4 1 cos 2 2 or 3 3
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JEE-MAIN-2019 (9th Apr-Second Shift)-PCM-40 90.
Let vertex C is 6,k then centre of circle
5k 1, 2 It lies on diameter 3y x 7 5k 3 1 7 2 k 1 So, AB = 14 and BC = 6 Area = 14 6 84
D
C (6, k)
A (–8, 5)
B (6, 5)
3y = x + 7
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FIITJEE Solutions to JEE(Main)-2019 Test Date: 10th April 2019 (First Shift)
PHYSICS, CHEMISTRY & MATHEMATICS Paper - 1 Time Allotted: 3 Hours
Maximum Marks: 360
Please read the instructions carefully. You are allotted 5 minutes specifically for this purpose.
Important Instructions: 1.
The test is of 3 hours duration.
2.
This Test Paper consists of 90 questions. The maximum marks are 360.
3.
There are three parts in the question paper A, B, C consisting of Physics, Chemistry and Mathematics having 30 questions in each part of equal weightage. Each question is allotted 4 (four) marks for correct response.
4.
Out of the four options given for each question, only one option is the correct answer.
5.
For each incorrect response 1 mark i.e. ¼ (one-fourth) marks of the total marks allotted to the question will be deducted from the total score. No deduction from the total score, however, will be made if no response is indicated for an item in the Answer Box.
6.
Candidates will be awarded marks as stated above in instruction No.3 for correct response of each question. One mark will be deducted for indicating incorrect response of each question. No deduction from the total score will be made if no response is indicated for an item in the answer box.
7.
There is only one correct response for each question. Marked up more than one response in any question will be treated as wrong response and marked up for wrong response will be deducted accordingly as per instruction 6 above.
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JEE-MAIN-2019 (10th Apr-First Shift)-PCM-2
PART –A (PHYSICS) 1.
A particle of mass m is moving along a trajectory given by x = x0 + a cos 1t y = y0 + b sin 2t The torque, acing on the particle about the origin, at t = 0 is: (A) my 0a12kˆ (B) m( x0b y0 a)12kˆ (C) m( x b2 y a2 )kˆ (D) Zero 0
2
0
1
2.
A stationary source emits sound waves of frequency 500 Hz. Two observers moving along a line passing through the source detect sound to be of frequencies 480 Hz and 530 Hz. Their respective speeds are, in ms–1 (Given speed of sound = 300 m/s) (A) 16, 14 (B) 12, 16 (C) 8, 18 (D) 12, 18
3.
Two particles, of masses M and 2M, moving as shown, with speeds of 10 m/s and 5 m/s, collide elastically at the origin. After the collision, they move along the indicated directions with speeds v1 and v2 respectively. The value of v1 and v2 are nearly: (A) 3.2 m/s and 12.6 m/s (B) 6.5 m/s and 6.3 m/s (C) 6.5 m/s and 3.2 m/s (D) 3.2 m/s and 6.3 m/s
4.
Given below in the left column are different modes of communication using the kinds of waves given in the right column. A. Optical Fibre Communication P. Ultrasound B. Radar Q. Infrared Light C. Sonar R. Microwaves D. Mobile Phones S. Radio Waves From the options given below, find the most appropriate match between entries in the left and the right column. (A) A – S, B – Q, C – R, D – P (B) A – Q, B – S, C – P, D – R (C) A – R, B – P, C – S, D – Q (D) A – Q, B – S, C – R, D – P
5.
In an experiment, the resistance of a material is plotted as a function of temperature (in some range). As shown in the figure, it is a straight line. One may conclude that: 2 2 (A) R(T) R0 e T / T0 R0 T2 2 2 (C) R(T) R0 e T / T0
(B) R(T)
2
(D) R(T) R0 e T0 / T
2
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JEE-MAIN-2019 (10th Apr-First Shift)-PCM-3 6.
A cylinder with fixed capacity of 67.2 lit contains helium gas at STP. The amount of heat needed to raise the temperature of the gas by 20°C is: [Given that R = 8.31 J mol–1 K–1] (A) 350 J (B) 700 J (C) 748 J (D) 374 J
7.
A ball is thrown upward with an initial velocity V0 from the surface of the earth. The motion of the ball is affected by a drag force equal to m2 (where m is mass of the ball, is its Instantneous velocity and is a constant). Time taken by the ball to rise to its zenith is: 1 1 (A) ln 1 V0 (B) ta n1 V0 g g g g (C)
sin1 V0 g g
1
(D)
2 tan1 V0 2 g g 1
8.
A thin disc of mass M and radius R has mass per unit area (r) = kr2 where r is the distance from its centre. Its moment of inertia about an axis going through its centre of mass and perpendicular to its plane is: MR2 MR2 (A) (B) 2 3 MR2 2MR2 (C) (D) 6 3
9.
Two coaxial discs, having moments of inertia I1 and
I1 , area rotating with respectively 2
1 , about their common axes. They are brought in contact 2 with each other and thereafter they rotate with a common angular velocity. If Ef and Ei are the final and initial total energies, then (Ef – Ei) is: I 2 3 (A) 1 1 (B) I112 6 8 2 I I 2 (C) 1 1 (D) 1 1 12 24
angular velocities 1 and
10.
The ratio of surface tensions of mercury and water is given to be 7.5 while the ratio of their densities is 13.6. Their contact angles, with glass, are close to 135° and 0°, respectively. It is observed that mercury gets depressed by an amount h in a capillary tube of radius r1, while water rises by the same amount h in a capillary tube of radius r2. The ratio, (r1/r2), is then close to: (A) 3/5 (B) 4/5 (C) 2/3 (D) 2/5
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JEE-MAIN-2019 (10th Apr-First Shift)-PCM-4 11.
Figure shows charge (q) versus voltage (V) graph for series and parallel combination of to given capacitors. The capacitances are: (A) 40 F and 10 F (B) 50 F and 30 F (C) 60 F and 40 F (D) 20 F and 30 F
12.
Two wires A and B are carrying currents I1 and I2 as shown in the figure. The separation between them is d. A third wire C carrying a current I is to be kept parallel to them at a distance x from A such that the net force acting on it is zero. The possible values of x are: I I2 (A) x 1 d and x d (I1 I2 ) I1 I2 I I (C) x 2 d and x 2 d I1 I2 I1 I2
(B) x
I2d (I1 I2 )
I (D) x 1 I1 I2
I2 d and x d I1 I2
13.
A proton, an electron, and a Helium nucleus, have the same energy. They are in circular orbitals in a plane due to magnetic field perpendicular to the plane. Let rp, re and rHe be their respective radii, then, (A) re > rp = rHe (B) re > rp > rHe (C) re < rp < rHe (D) re < rp = rHe
14.
The value of acceleration due to gravity at Earth’s surface is 9.8 ms–2. The altitude above its surface at which the acceleration due to gravity decreases to 4.9 ms–2, is close to: (Radius of earth = 6.4 × 106 m) (A) 6.4 × 106 m (B) 9.0 × 106 m 6 (C) 2.6 × 10 m (D) 1.6 × 106 m
15.
In the given circuit, an ideal voltmeter connected across the 10 resistance reads 2 V. The internal resistance r, of each cell is: (A) 1 (B) 0.5 (C) 1.5 (D) 0
16.
The electric field of a plane electromagnetic wave is given by E E0 ˆi cos(kz) cos( t) The corresponding magnetic field B is then given by: E E (A) B 0 ˆj sin(kz) cos( t) (B) B 0 kˆ sin(kz) cos( t) C C E E (C) B 0 ˆj cos(kz) sin( t) (D) B 0 ˆj sin(kz) sin( t) C C
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JEE-MAIN-2019 (10th Apr-First Shift)-PCM-5 17.
An npn transistor operates as a common emitter amplifier, with a power gain of 60 dB. The input circuit resistance Is 100 and the output load resistance is 10 k. the common emitter current gain is: (A) 6 × 102 (B) 102 (C) 60 (D) 104
18.
In a meter bridge experiment, the circuit diagram and the corresponding observation table are shown in figure.
Sl. No. 1. 2. 3. 4.
R () 1000 100 10 1
l (cm) 60 13 1.5 1.0
Which of the readings is inconsistent? (A) 4 (C) 2
(B) 3 (D) 1
19.
A transformer consisting of 300 turns in the primary and 150 turns in the secondary gives output power of 2.2 kW. If the current in the secondary coils is 10 A, then the input voltage and current in the primary coil are: (A) 440 V and 5 A (B) 440 and 20 A (C) 220 V and 20 A (D) 220 V and 10 A
20.
A uniformly charged ring of radius 3a and total charge q is placed in xy-plane centered at origin. A point charge q is moving towards the ring along the z-axis and has speed v at z = 4a. The minimum value of v such that it crosses the origin is:
21.
1/2
(A)
2 m
1 q2 5 4 0 a
(C)
2 m
4 q2 15 4 0 a
1/2
(B)
2 m
1 q2 15 4 0 a
(D)
2 m
2 q2 15 4 0 a
1/2
1/2
A current of 5 A passes through a copper conductor (resistivity = 1.7 × 10–8 m) of radius of cross-section 5 mm. Find the mobility of the charges if their drift velocity is 1.1 × 10–3 m/s. (A) 1.8 m2/Vs (B) 1.0 m2/Vs 2 (C) 1.3 m /Vs (D) 1.5 m2/Vs
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JEE-MAIN-2019 (10th Apr-First Shift)-PCM-6 22.
n-moles of an ideal gas with constant volume heat capacity CV undergo an isobaric expansion by certain volume. The ratio of the work done in the process, to the heat supplied is nR nR (A) (B) CV nR CV nR 4nR 4nR (C) (D) CV nR CV nR
23.
A message signal of frequency 100 MHz and peak voltage 100 V is used to execute amplitude modulation on a carrier wave of frequency 300 GHz and peak voltage 400 V. The modulation index and difference between the two side band frequencies are: (A) 4 ; 2 × 108 Hz (B) 4 ; 1 × 108 Hz 8 (C) 0.25 ; 1 × 10 Hz (D) 0.25 ; 2 × 108 Hz
24.
A ray of light AO in vacuum is incident on a glass slab at angle 60° and refracted at angle 30° along OB as shown in the figure. The optical path length of light ray from A to B is : 2b 2b (A) 2a (B) 2a 3 3 (C)
25.
2 3 2b a
(D) 2a + 2b
The displacement of a damped harmonic oscillator is given by x(t) = e–0.1 t cos(10t + ). Here t is in seconds. The time taken for its amplitude of vibration to drop to half of its initial value is close to: (A) 13 s (B) 27 S (C) 4 s (D) 7 s
26.
Two radioactive materials A and B have decay constants 10 and , respectively. If initially they have the same number of nuclei, then the ratio of the number of nuclei of a to that of B will be 1/e after a time: 1 1 (A) (B) 11 10 1 11 (C) (D) 9 10
27.
A 25 × 10–3 m3 volume cylinder is filled with 1 mol of O2 gas at room temperature (300 K). The molecular diameter of O2, and its root mean square speed, are found to be 0.3 nm and 200 m/s, respectively. What is the average collision rate (per second) for an O2 molecule? (A) ~1010 (B) ~1011 12 (C) ~10 (D) ~1013
28.
In a photoelectric effect experiment the threshold wavelength of incident light is 260 nm, 1237 the maximum kinetic energy of emitted electrons will be: Given E (in eV) = (in nm) (A) 15.1 eV
(B) 1.5 eV
(C) 4.5 eV
(D) 3.0 eV
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JEE-MAIN-2019 (10th Apr-First Shift)-PCM-7 29.
One plano-convex and one plano-concave lens of same radius of curvature ‘R’ but of different materials are joined side by side as shown in the figure. If the refractive index of the material of 1 is 1 and that of 2 is 2, then the focal length of the combination is R 2R (A) (B) 2(1 2 ) 1 2 R R (C) (D) 1 2 2 (1 2 )
30.
A moving coil galvanometer allows a full scale current of 10–4 A. A series resistance of 2 M is required to convert the above galvanometer into a voltmeter of range 0 – 5 V. Therefore the value of shunt resistance required to convert the above galvanometer into a ammeter of range 0–10 mA is: (A) 200 (B) 100 (C) 10 (D) 500
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JEE-MAIN-2019 (10th Apr-First Shift)-PCM-8
PART –B (CHEMISTRY) 31.
The species that can have a trans-isomer is: (en = ethane-1,2-diamine, ox = oxalate) (A) [Pt(en)Cl2] (B) [Pt(en)2Cl2]2+ (C) [Cr(en)2(ox)]+ (D) [Zn(en)Cl2]
32.
The major product of the following reaction is
(A)
(B)
(C) CH3CH = CH – CH2NH2
(D)
33.
The synonym for water gas when used in the production of methanol is (A) syn gas (B) natural gas (C) fuel gas (D) laughing gas
34.
The regions of the atmosphere, where clouds form and where we live, respectively, are: (A) Troposphere and Stratosphere (B) Troposphere and Troposphere (C) Stratosphere and Stratosphere (D) Stratosphere and Troposphere
35.
Consider the hydrated ions of Ti2+, V2+, Ti3+ and Sc3+. The correct order of their spin-only magnetic moments is: (A) Ti3+ < T2+ < Sc3+ < V2+ (B) Sc3+ < Ti3+ < Ti2 < V2+ 2+ 2+ 3+ 3+ (C) V < Ti < Ti < Sc (D) Sc3+ < Ti3+ < V2+ < Ti2+
36.
Amylopectin is composed of (A) -D-glucose, C1–C4 and C2–C6 linkages (B) -D-glucose, C1–C4 and C2–C6 linkages (C) -D-glucose, C1–C4 and C1–C6 linkages (D) -D-glucose, C1–C4 and C1–C6 linkages
37.
A gas undergoes physical adsorption on a surface and follows the given Freundlich adsorption isotherm equation x kp0.5 m Adsorption of the gas increases with: (A) Increase in p and decrease in T (B) Increase in p and increase in T (C) Decrease in p and increase in T (D) Decrease in p and decrease in T
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JEE-MAIN-2019 (10th Apr-First Shift)-PCM-9 38.
39.
A bacterial infection in an internal would grows as N(t) = N0 exp(t), where the time t is in hours. A dose of antibiotic, taken orally, needs 1 hour to reach the wound. Once it dN reaches there, the bacterial population goes down as 5N2 . What will be the plot of dt N0 vs, t after 1 hour? N
(A)
(B)
(C)
(D)
2
The graph between and r(radial distance) is shown below. This represents:
(A) 1s orbital (C) 2s orbital
(B) 3s orbital (D) 2p orbital
40.
Consider the following table: Gas a/ (k Pa dm6 mol–1) b/ (dm3 mol–1) A 642.32 0.05196 B 155.21 0.04136 C 431.91 0.05196 D 155.21 0.4382 a and b a van der Waals constants. The correct statement about the gases is: (A) Gas C will occupy more volume than gas A ; gas B will be lesser compressible than gas D (B) Gas C will occupy lesser volume than gas A ; gas B will be more compressible than gas D. (C) Gas C will occupy lesser volume than gas A ; gas B will be lesser compressible than gas D (D) Gas C will occupy more volume than gas A; gas B will be more compressible than gas D.
41.
During the change of O2 to O 2 , the incoming electron goes to the orbital: (A) 2p y (B) 2p x (C) * 2p z
(D) * 2p x
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JEE-MAIN-2019 (10th Apr-First Shift)-PCM-10 42.
The principle of column chromatography is (A) Differential absorption of the substances on the solid phase. (B) Differential adsorption of the substances on the solid phase. (C) Gravitational force. (D) Capillary action.
43.
At 300 K and 1 atmospheric pressure, 10 mL of a hydrocarbon required 55 mL of O2 for complete combustion, and 40 mL of CO2 is formed. The formula of the hydrocarbon is: (A) C4H10 (B) C4H6 (C) C4H7Cl (D) C4H8
44.
Three complexes, [CoCl(NH3)5]2+ (I), [co(NH3)5H2O]3+ (II) and [Co(NH3)6]3+ (III) Absorb light in the visible region. The correct order of the wavelength of light absorbed by them is (A) (II) > (I) > (III) (B) (III) > (II) > (I) (C) (I) > (II) > (III) (D) (III) > (I) > (II)
45.
Major products of the following reaction are:
(A)
CH3OH and HCO2H
(C)
46.
(B)
(D)
The major product of the following reaction is:
(A)
(B)
(C)
(D)
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JEE-MAIN-2019 (10th Apr-First Shift)-PCM-11 47.
Match the refining methods (Column I) with metals (Column II). Column I Column II (Refining methods) (Metals) (I) Liquation (a) Zr (II) zone Refining (b) Ni (III) Mond process (C) Sn (IV) Van Arkel Method (d) Ga (A) (I) – (b); (II) – (c); (III) – (d); (IV) – (a) (B) (I) – (b); (II) – (d); (III) – (a); (IV) – (c) (C) (I) – (c); (II) – (a); (III) – (b); (IV) – (d) (D) (I) – (c); (II) – (d); (III) – (b); (IV) – (a)
48.
The increasing order of the reactivity of the following compounds towards electrophilic aromatic substitution reaction is:
(A) (III) < (I) < (II) (C) (II) < (I) < (III)
(B) (III) < (II) < (I) (D) (I) < (III) < (II)
49.
A process will be spontaneous at all temperatures if: (A) H < 0 and S < 0 (B) H < 0 and S > 0 (C) H > 0 and S > 0 (D) H > 0 and S < 0
50.
Consider the following statements (a) The pH of a mixture containing 400 mL of 0.1 M H2SO4 and 400 mL of 0.1 M NaOH will be approximately 1.3. (b) Ionic product of water is temperature dependent. (c) A monobasic acid with Ka = 10–5 has a pH = 5. the degree of dissociation of this acid is 50%. (d) The Le Chatelier’s principle is not applicable to common-ion effect. The correct statements are: (A) (a) and (b) (B) (a), (b) and (c) (C) (b) and (c) (D) (a), (b) and (d)
51.
Ethylamine (C2H5NH2) can be obtained from N-ethylphthalimide on treatment with: (A) CaH2 (B) H2O (C) NaBH4 (D) NH2NH2
52.
The correct order of catenation is: (A) C > Sn > Si Ge (C) Si > Sn > C > Ge
(B) Ge > Sn > Si > C (D) C > Si > Ge Sn
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JEE-MAIN-2019 (10th Apr-First Shift)-PCM-12 53.
The major product of the following reaction is
(A)
(B)
(C)
(D)
54.
Which of the following is a condensation polymer? (A) Nylon 6, 6 (B) Neoprene (C) Buna–S (D) Teflon
55.
Consider the statements S1 and S2: S1: Conductivity always increases with decrease in the concentration of electrolyte. S2: Molar conductivity always increases with decrease in the concentration of electrolyte. The correct option among the following is: (A) S1 is wrong and S2 is correct (B) Both S1 and S2 are wrong (C) S1 is correct and S2 is wrong (D) Both S1 and S2 are correct
56.
Increasing rate of SN1 reaction in the following compounds is
(A) (A) < (B) < (C) < (D) (C) (B) < (A) < (D) < (C)
(B) (A) < (B) < (D) < (C) (D) (B) < (A) < (C) < (D)
57.
The oxoacid of sulphur that does not contain bond between sulphur atoms is: (A) H2S2O3 (B) H2S2O4 (C) H2S2O7 (D) H2S4O6
58.
The isoelectronic set of ions is (A) Li+, Na+, O2– and F– (C) N3–, O2–, F– and Na+
(B) F–, Li+, Na+ and Mg2+ (D) N3–, Li+, Mg2+ and O2–
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JEE-MAIN-2019 (10th Apr-First Shift)-PCM-13
59.
The alloy used in the construction of aircrafts is (A) Mg–Mn (B) Mg-Zn (C) Mg-Al (D) Mg-Sn
60.
At room temperature, a dilute solution of urea is prepared by dissolving 0.60 g of urea in 360 g of water. If the vapour pressure of pure water at this temperature is 35 mm Hg, lowering of vapour pressure will be: (molar mass of urea = 60 g mol–1) (A) 0.027 mmHg (B) 0.031 mmHg (C) 0.028 mmHg (D) 0.017 mmHg
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JEE-MAIN-2019 (10th Apr-First Shift)-PCM-14
PART–C (MATHEMATICS) 61.
If Q(0, –1, –3) is the image of the point P in the plane 3x – y, 1.4z = 2 and R is the point (3, –1, –2), then the area (in square units) of PQR is: 91 (A) (B) 2 13 2 65 91 (C) (D) 2 4
62.
If the coefficients of x2 and x3 are both zero, in the expansion of the expression (1 + ax + bx2) (1 – 3x)t5 in powers of x, then the ordered pair (a, b) is equal to (A) (–54, 315) (B) (28, 861) (C) (28, 315) (D) (–21, 714)
63.
The sum 3 1 5 (13 23 ) 7 (13 23 33 ) ........ 12 12 22 12 22 3 2 Upto 10th term, is (A) 620 (B) 660 (C) 680 (D) 600
64.
If the length of the perpendicular from the point (, 0, ) ( = 0) to the line, x y 1 z 1 3 is , then is equal to: 1 0 1 2 (A) 1 (B) –1 (C) 2 (D) –2
65.
66.
sin (p 1)x sin x , x0 x If f(x) = q , x0 2 xx x , x0 x/2 Is continuous at x = 0, then the ordered pair (p, d) is equal to 3 1 5 1 (A) , (B) , 2 2 2 2 1 3 3 1 (C) , (D) , 2 2 2 2
Which one of the following Boolean expressions is a tautology? (A) (p q) (p~q) (B) (p q) (p~q) (C) (p q) (~p~q) (D) (p q) (p~q)
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JEE-MAIN-2019 (10th Apr-First Shift)-PCM-15
67.
If and are the roots of the quadratic equation, x2 + x sin – 2sin = 0, 0, then 2 12 12 is equal to: 12 ( 12 ) ( )24 212 (sin 8)12 212 (C) (sin 8)6
212 (sin 4)12 26 (D) (sin 8)12
(A)
68.
(B)
ABC is a triangular park with AB = AC = 100 metres. A vertical tower is situated at the mid-point of BC. If the angles of elevation of the top of the tower at A and B are cot– 1 1 (3 2 ) and cos ec (2 2 ) respectively, then the height of the tower (in metres) is: (A) 25 100 (C) 3 3
69.
(B) 10 5 (D) 20
If a > 0 and z
(1 i)2 , has magnitude ai
3 1 i 5 5 1 3 (C) i 5 5
(A)
x
70.
sin cos
If 1 sin
x
1
cos
1
x
x
and
sin 2 cos 2
2 sin 2
x
1
cos 2
1
x
, x 0 ; then
for all 0, : 2 (A) 1 – 2 = –2x3 (C) 1 – 2 = x(cos2 – cos4)
71.
2 , then z is equal to: 5 1 3 (B) i 5 5 1 3 (D) i 5 5
(B) 1 + 2 = –2 (x3 + x –1) (D) 1 + 2 = –2x3
If y = 1f(x) is the solution of the differential equation
dy (tan x y) sec 2 x, dx
x , , such that y(0) = 0, then y is equal to 2 2 4 1 1 (A) e (B) 2 2 e 1 (C) e – 2 (D) 2 e
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JEE-MAIN-2019 (10th Apr-First Shift)-PCM-16 72.
Let f: R R be differentiable at c R and f(c) = 0. If g(x) = f(x) , then at x =c, g is: (A) differentiable if f(c) = 0 (C) not differentiable
(B) differentiable if f(c) 0 (D) not differentiable if f(c) = 0
73.
Let f(x) = x2, x R. For any A R, define g(A) = {x R : f(x) A}. If S = [0, 4], then which one of the following statements is not true? (A) f(g(S) f(S) (B) f(g(S)) = S (C) g(f(S)) S (D) g(f(S)) = g(S)
74.
If the system of linear equations x+y+z=5 x = 2y + 2z = 6 x + 3y + z = u ( R), has infinitely many solutions then the value of + is: (A) 12 (B) 7 (C) 10 (D) 9
75.
The number of 6 digit numbers hat can be formed using the digits 0, 1, 2, 5, 7 and 9 which are divisible by 11 and no digits is repeated, is (A) 36 (B) 60 (C) 72 (D) 48
76.
The line x = y touches a circle at the point (1, 1). If the circle also passes through the point (1, –3), then its radius is (A) 3 2 (B) 3 (C) 2 (D) 2 2
77.
Let f(x) = ex – x and g(x) = x2 – x, x R. Then the set h(x) = (fog) (x) is increasing is: 1 1 (A) 0, 1, (B) 1, 2 2 1 (C) , 0 1, (D) 0, 2
of all x R, where the function 1 2 ,
2
78.
The value of
[sin 2x(1 cos 3x)] dx where [t] denotes the greatest integer function, is: 0
(A) (C) 2
(B) –2 (D) –
79.
If the circles x2 + y2 + 5Kx + 2y + K = 0 and 2(x2 + y2) + 2Kx + 3y – 1 = 0, (KR), intersect at the points P and Q, then the line 4x + 5y – K = 0 passes through P and Q for: (A) exactly one value of K (B) not value of K (C) infinitely many values of K (D) exactly two values of K
80.
If the lines x – 2y = 12 is tangent to the ellipse length of the latus rectum of the ellipse is (A) 12 2 (C) 8 3
x2 y2 9 2 1 at the point 3, , then the 2 a b 2
(B) 9 (D) 5
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JEE-MAIN-2019 (10th Apr-First Shift)-PCM-17 81.
If a1, a2, a3 ………… an are in A.P and a1 + a4 + a7 + …………… + a16 = 114, then a1 + a6 + a11 + a16 is equal to (A) 76 (B) 64 (C) 98 (D) 38
82.
If for some x R, the frequency distribution of the marks obtained by 20 students in a test is Marks 2 3 5 7 2 2 Frequency (x+1) 2x – 5 x – 3x x Then the mean of the marks is: (A) 2.8 (B) 3.2 (C) 2.5 (D) 3.0
83.
If lim x 1
x4 1 x3 k 3 lim 2 , then k is x 1 x k x k 2
3 8 4 (C) 3
8 3 3 (D) 2
(A)
84.
85.
(B)
All the pairs (x, y) that satisfy the inequality 2 1 2 sin x 2 sin x 5 sin2 y 1 also 4 Satisfy the equation: (A) 2 sin x 3 sin y (B) sin x sin y (C) 2sin x = sin y (D) sin x = 2 sin y (n 1)1/3 (n 2)1/3 (2n)1/3 lim .... 4/3 n n4/3 n 4/3 n is equal to: 3 3 (A) (2)4/3 4 4 3 4 (C) (2)4/3 4 3
4 3/4 (2) 3 4 (D) (2)4/3 3
(B)
86.
If a directrix of a hyperbola centered at the origin and passing through the point (4, 2 3 ) is 5x 4 5 and its eccentricity is e, then (A) 4e4 + 8e2 – 35 = 0 (B) 4e4 – 24e2 + 35 = 0 4 2 (C) 4e – 12e – 27 = 0 (D) 4e4 – 24e2 + 27 = 0
87.
The region represented by x y 2 and x y 2 is bounded by a: (A) rhombus of area 8 2 sq. units (C) rhombus of side length 2 units
(B) square of area 16 sq. units (D) square of side length 2 2 units
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JEE-MAIN-2019 (10th Apr-First Shift)-PCM-18 88.
Assume that each born child is equally likely to be a boy or a girl. If two families have two children each, then the conditional probability that all children are girls given that at least two are girls is: 1 1 (A) (B) 10 17 1 1 (C) (D) 12 11
89.
Let A(3, 0, –1), B(2, 10, 6) and C(1, 2, 1) be the vertices of a triangle and M be the midpoint of AC. If G divides BM in the ratio, 2 : 1, then cos(GOA) (O being he origin) is equal to: 1 1 (A) (B) 30 2 15 1 1 (C) (D) 6 10 15
90.
If
(x
2
dx 2x 10)2
f(x) x 1 = A tan1 C 2 3 x 2x 10 Where C is a constant of integration, then: 1 1 (A) A and f(x) (x 1) (B) A and f(x) 9(x 1)2 27 54 1 1 (C) A and f(x) 3(x 1) (D) A and f(x) 3(x 1) 54 81
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JEE-MAIN-2019 (10th Apr-First Shift)-PCM-19
JEE (Mains) – 2019 ANSWERS PART A – PHYSICS 1. 5. 9. 13. 17. 21. 25. 29.
A C D D B B D C
2. 6. 10. 14. 18. 22. 26. 30.
D C D C A B C Bonus
3. 7. 11. 15. 19. 23. 27.
B B A B A D A
4. 8. 12. 16. 20. 24. 28.
B D B D D D B
34. 38. 42. 46. 50. 54. 58.
B D A C B A C
64. 68. 72. 76. 80. 84. 88.
B D A D B B D
PART B – CHEMISTRY 31. 35. 39. 43. 47. 51. 55. 59.
B B C B D D A C
32. 36. 40. 44. 48. 52. 56. 60.
D C D C A D D D
33. 37. 41. 45. 49. 53. 57.
A B C D B A C
PART C – MATHEMATICS 61. 65. 69. 73. 77. 81. 85. 89.
A D B C A A A D
62. 66. 70. 74. 78. 82. 86. 90.
C D D C D A B C
63. 67. 71. 75. 79. 83. 87.
B A C B B B D
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JEE-MAIN-2019 (10th Apr-First Shift)-PCM-20
HINTS AND SOLUTIONS PART A – PHYSICS 1.
2.
F ma m[ a12 cos , tiˆ b22 sin 2 tjˆ ft 0 = ma12 ˆi rt 0 (x0 a)iˆ y ˆj r F my0 a12kˆ
v vo o v vo 1 v 0 530 vo 1 300 = 18 m/s 500
480 vo 1 300 = 12 m/s 500
3.
2MV, cos 30° + Mv2 cos 45° = 10 M cos 30° + 10 cos 45° v v1 3 2 5 3 5 2 …(i) 2 2MV, sin 30° – MV2 sin 45° = –10 M sin 30° + 10 M sin 45° V V1 2 5 5 2 …(ii) 2 5( 3 1) 10 2 17.5 V1 = 6.5 m/s 2.7 3 1 V2 = 6.3 m/s
4.
Factual
5.
1 T 2 ln R(T) 1 1 ln R(To ) T02 T2 ln R(T) = ln R(To) 1 o2 T
R(T) = Ro e
T2 02 T
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JEE-MAIN-2019 (10th Apr-First Shift)-PCM-21
6.
7.
3 Q = nCv T = n RT 2 67.2 3 = 8.31 (20) 22.4 2 748 J dv dt
(g v 2 )
g dv gdt g 2 v
Integrating 0 t and V0 0: V g gt tan1 0 g 1 t tan1 V g 0 g R
8.
R
IDisc (dm)r 2 IDisc ( 2rdr)r 2 0
0
R
IDisc (kr 2 2rdr)r 2
Mass of disc
0
R
R
IDisc 2k r 2dr 0
R
IDisc
r6 2k 6 0 6
IDisc IDisc IDisc
R 2k 6
M 2rdr kr 2 0
R
M 2k r 3 dr 0
r4 M 2k 4
R
0
kR 6 kR 4 R 2 2 r4 M 2k 3 4 2 3 2 M2R 2 ; IDisc MR2 3 3
R
0
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JEE-MAIN-2019 (10th Apr-First Shift)-PCM-22
9.
1 1 I 12 II 12 2 22 4 I 2 9 9 = 1 1 I112 2 8 16 I 3I 5 3I I11 1 1 1 ; I11 1 4 2 4 2 5 1 3I1 25 2 1 ; Ef 1 6 2 2 36 25 2 = I11 48 25 2 2 Ef Ei I112 I2 1 49 48 25 2 = I11 48 25 9 2 2 E f Ei I112 I11 48 16 48 Ei
=
10.
I112 24
2S1 cos r11g 2s cos 2 h 2 r22 g r 2 1 r2 5 h
11.
As q = CV Hence slope of graph will give capacitance. Slope will be more in parallel combination. Hence capacitance in parallel should be 50 F and in series combination must be 8 F. Only in option 40 f and 10 F. Cparallel = 40 + 10 = 50 F 40 10 Cseries = = 8 F 40 10
12.
Net force on wire carrying current I per unit length is 0I1I 0I2I 0 2x 2(d x) I1 I 2 x xd Id x 1 I1 I2
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JEE-MAIN-2019 (10th Apr-First Shift)-PCM-23
13.
r rHe
14.
mv 2mK qB qB rp re
GM GM 2 (R h) 2R2
R + h = 2R h ( 2 1)R 2.6 × 106 m 15.
15 10 2 2r 25 = 8 + 2r 3 i 8 2r 3 2 i R eq 6 8 2r 16 + 4r = 18 r = 0.5 Req
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JEE-MAIN-2019 (10th Apr-First Shift)-PCM-24
16.
17.
E B || v
Given that wave is propagating along positive z-axis and E along positive x-axis. Hence B along y-axis. From Maxwell equation B V E t E E B i.e. and B0 0 Z dt C
Av × = Pgain P 60 10 log10 P0
P 10 6 2
Rout Rin
104 100 = 100
= 2 2 = 104 ;
18.
as x
R(100 )
for (A)
x
1000 (100 60) 667 40
for (B)
x
100 (100 13) 669 13
for (C)
x
10 (100 1.5) 656 98.5
for (D)
x
1 (100 1) 95 1
So reading in serial no. (4) is completely different hence correct answer (A). 19.
Given NP = 300, Ns = 150, P0 = 2200W Is = 10 A P0 = V0I0 2200 = V0 × 10 V0 = 220 V Vi NP Vi = 2 × 220 = 440 V V0 Ns Also, P0 = ViIi 2200 Ii =5A 440
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JEE-MAIN-2019 (10th Apr-First Shift)-PCM-25 20.
Ui + Ki = Uf + Kf kq2 1 kq2 mv 2 3a 16a 2 9a 2 2
1 kq2 1 1 2kq2 mv 2 2 a 3 5 15a v
21.
=
4kq2 15ma
Vd E
E = J 1.1 103
5 25 10 6 1.1 103 25 10 6 = 1.01 m2 / Vs 1.7 10 8 5 1.7 10 8
22.
w = nRT H = (Cv + nR)T nR H Cv nR
23
fm = 100 MHz = 108 Hz, (Vm)0 = 100 V fc = 300 GHz, (Vc)0 = 400 V UBF – LBF = 2fm = 2 × 108 Hz
24.
From Snell’s law 1 sin 60° = sin 30° 3 Optical path = AO + (OB) a b 3 = o cos 60 cos 30o = 2a + 2b
25.
A = A0e–0.1 t =
A0 2
ln2 = 0.1 t t = 10 ln2 = 6.93 7 sec. 26.
N1 = N0e–10t ; 1 N1 e 9 t e N2 9t = 1 1 t= 9
N2 = N0e–t
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JEE-MAIN-2019 (10th Apr-First Shift)-PCM-26
27.
Vav RT
2 2NAP = 2 × 0.3 × 10–9 RT P V V 2 2NA
8 Vrms 3
Vav
200 2 2NA 8 v 3 25 10 3 = 17.68 × 108/sec = 0.1768 × 1010/sec ~ 1010
This answer does not match with JEE – Answer key 28.
K max
hc hc o
K max hc 0 0
380 260 K max (1237) 380 260 = 1.5 eV
29.
For 1st lens
1 1 1 1 1 1 1 f1 1 R R
For 2nd lens
1 1 2 1 1 0 2 f2 1 R R
1 1 1 feq f1 f2 1 1 1 ( 2 1) R feq feq R R 1 2 30.
200 + 10–4 G = 5 G = –ve So, answer is Bonus.
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JEE-MAIN-2019 (10th Apr-First Shift)-PCM-27
PART B – CHEMISTRY 31.
The trans-isomer of [Pt(en)2Cl2]2+
32.
33.
The synonoyms for water gas is syn gas.
34.
Fact based.
35.
36.
Amylopetcin is a homopolymer of a-D-glucose where C1–C4 linkage and C1 – C6 linkage are present.
37.
Increase in pressure leads to the increase in adsorption capacity and the physical adsorption is an exothermic process with the increase in temperature adsorption decrease.
38.
Initially N > N0 and ‘’N” is increasing through first-order kinetics. So N0/N in initial time decreases.
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JEE-MAIN-2019 (10th Apr-First Shift)-PCM-28
But after 1 hr the value of N decreases with a faster rate. So N0/N will increases. 39.
As we know that for s-orbital graph starts from top and no. of radial node = n - -1 For 2s orbital it will = 2 - 0 - 1 = 1.
The graph
40.
is of 2s
Gas A and C have same value of 'b' but different value of 'a' so gas having higher value of 'a' have more force of attraction so molecules will be more closer hence occupy less volume. Gas B and D have same value of 'a' but different value of 'b' so gas having lesser value of 'b' will be more compressible.
41.
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JEE-MAIN-2019 (10th Apr-First Shift)-PCM-29 42.
The principle of column chromatography is differential adsorption of substance and hence option 1 is correct.
43.
44.
As we known that Strong ligand CFSE Eabsorbed
1
absorbed We have [CoCl(NH3)5]2+, [Co(NH3)5H2O]3+ and [CO(NH3)6]3+ (I) (II) (III) III > II> I (as per Eabsorbed) absorbed I > II > III
45.
46. O
O H
OH
H OH
NC
NC O
O
NO I
H
OH
NC
I
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JEE-MAIN-2019 (10th Apr-First Shift)-PCM-30 47.
48.
49.
Liquation is used for Sn. Zone refining is used for Ga. Mond's process is used for Ni. Van arkel process is used for Zr.
Rate of aromatic electrophilic substitution is
At constant P and T and for the process to be spontaneous we should have G = -ve and we know that G = H - TS If H = -ve and S = +ve then at all the temperature the process will be spontaneous.
50.
51.
It is the final step of Gabriel pthalimide synthesis reaction. O
O
C
C - OH
C2H5 NH2
H3O N - C2H5
C O
Product
C - OH O
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JEE-MAIN-2019 (10th Apr-First Shift)-PCM-31 52.
In this order of catenation is asked. Catenation is a self-linking property here and for group 14: Self-linking is through covalent bonding C > Si > Ge Sn In C there is 2p – 2p overlapping further 3p – 3p, 4p-4p and so on and the extent of overlapping is more in 2p-2p > 3p-3p > 4p-4p 5p-5p
53.
It is SN1 reaction mechanism is favourable hence reaction complete via SN1 mechanism
54.
Nylon-6,6 is a condensation polymer of hexamethylene diamine and adipic acid. Buna-S, Teflon and Neoprene are addition polymer.
55.
We know that K Here m = molar conductivity m C K = conductivity C = concentration When concentration increase conductivity always increases. The molar conductivity always increase with the decrease in the concentration.
56.
Rate of SN1 reaction stability of C – I.M
57.
H2S2O7
58.
In this we have to choose isoelectronic set of ions: Isoelectronic species are those which have same no. of electron in total, so option 3 is correct.
59.
For aircraft construction aluminium and its alloys are used, because they are lighter.
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JEE-MAIN-2019 (10th Apr-First Shift)-PCM-32 60.
PART C – MATHEMATICS 61.
MQ
1 12 2 9 1 16
13
26
Q
13 2
PM 26 RQ 9 1 10 13 7 2 2 1 7 91 Ar PQR 26 2 2 2
M
RM 10
R
P
62.
Coefficient of x 2 15C2 9 3a
15C2 9 45a b 0 3
15
C1 b 0
(1) 15
Coefficient of x 27 C3 9a 15C2 3b 15 C1 0 (2) 273 21a b 0 (1) + (2) give a = 28, b = 315 24a 672 0
3 n 1 2 1
3
63.
Tn
1 2
2
23 ..... n3
.... n2
3 n n 1 2 Sn Tn
3 .n. n 1 2 On solving Sn
n n 1n 2 2
S10 660
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JEE-MAIN-2019 (10th Apr-First Shift)-PCM-33
64.
x y 1 z 1 1 0 1 A point on this line is A 0,1, 1 AB.BC 0 1 We get 2 1 1 C , 1, 2 2 2 BC 3
B ,0,
,1, , 1
2
2
1 2 1 2 2 1 2 3 0, 1
1
65.
C
A 0,1, 1
0
x x2 x x 0 x3/2 1 x 1 1 lim x 0 x 2 sin 1 x sin x LHL lim x 0 x p 1 1 p2 For function to be continuous LHL RHL f 0 RHL lim
3 1 p,q , 2 2 66.
From options p q ~ p ~ q p q ~ p q Not a tautology
p q p q p q ~ q p q p ~ q p q ~ q p q p ~ q p q ~ q 67.
tautology Not a tautology Not a tautology
x 2 x sin 2sin 0 sin 2 sin 12
12 12 Now, 24 1 1 24 12 12 FIITJEE Ltd., FIITJEE House, 29-A, Kalu Sarai, Sarvapriya Vihar, New Delhi -110016, Ph 46106000, 26569493, Fax 26513942 website: www.fiitjee.com.
JEE-MAIN-2019 (10th Apr-First Shift)-PCM-34 12
12 2 4 2 4
12
12
2 sin 2 sin 8 sin 212 12 sin 8
68.
Q
cos ec 2 2 cot 3 2 x 3 2 ….(i) h 1 So ….(ii) 4 2 7 10 x From (i) and (ii) h = 20
B
h
P
100
A
69.
z
1 i
2
100
C
2i a i
a2 1 2 z a3 2 5 a 1 2i 3 i z 10 1 3i 5 a i 2
x
70.
x
sin cos
1 sin
x
1
cos
1
x
2
x x 1 sin x sin cos cos sin x cos
x
3
x
sin 2 cos 2
2 sin 2
x
1
cos 2
1
x
x3 1 2 2x 3
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JEE-MAIN-2019 (10th Apr-First Shift)-PCM-35
71.
dy tan x y sec 2 x dx dy y sec 2 x tan x sec 2 x dx dt Let tan x t sec 2 x dx dy t y dt dy y t (Linear differential equation) dt After solving we get ye t e t t 1 c y tan x 1 ce tan x y 0 0 c 1
y tan x 1 e tan x So, y e2 4 72.
g' c lim
f c h f c h
h 0
lim
f c 0
f c h
h f c h f c h lim . h 0 h h h 0
h
0 if f ' c 0 h i.e. g x is differentiable at x c if f ' c 0 lim f ' c h 0
73.
g s 2, 2
f g s 0,4 5 f S 0, 16 f g s f s g f s 4, 4 g s therefore, g f s S 74.
x 3y z u a x y z 5 b x 2y 2z 6 Comparing coefficients we get a b 1 and a 2b 3 a,b 1, 2 So, x 3y z u x 3y 3z u 7, 3
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JEE-MAIN-2019 (10th Apr-First Shift)-PCM-36 75.
Let the six digit number be abcdef for this number to be divisible by 11 a c e b d f must be multiple of 11 possibility is a c c b d f 12 Case I a,c,e 7,5,0 and b,d,f 9,2,1 So, number of numbers 2 2! 3! 24
Case II a,c,e 3,2,1 b,d,f 7,5,0 So number of numbers 3! 3! 36 Total 24 36 60 76.
Equation of circle is given as S L 0 2
2
x 1
y 1 x y 0 passes through (1, –3) 16 4 0 4 2
(1, 1)
2
x 1 y 1 4 x y 0 r 2 2
77.
h x f g x h' x f ' g x g' x and f ' x e x 1
1 g' x h ' x e 1 2x 1 0 h ' x e
g x
x2 x
Case : 1
ex
2
x
1 and 2x 1 0 1 x 0, …..(i) 2 Case : 2 2 ex x 1 and 2x 1 0 x 1, …….(ii) from (i) and (ii) 1 x 0, 1, 2 2
78.
I
sin 2x 1 cos 3x dx 0
2
2I
sin 2x 1 cos 3x sin 2x sin 2x cos 3x dx 0
2
2I
dx 0
2I 2 dx 0
I dx 0
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JEE-MAIN-2019 (10th Apr-First Shift)-PCM-37
79.
80.
81.
1 1 yk 0 2 2 and given line 4x 5y k 0 …..(ii) 1 k 1 2 on comparing (i) and (ii) we get k 10 k No real value of k exist. Equation of common chord 4kx
……..(i)
9 Tangent at 3, 2 3x 9y 1 a2 2b2 Comparing with x 2y 12 3 9 1 2 2 a 4b 12 and b3 3 a6 2b2 Length of latus rectum 9 a
a1 a4 a7 a10 a13 a16 114 6 a1 a16 114 2 a1 a16 38
So a1 a6 a11 a16
4 a1 a16 2
2 38 76
82.
Mean x
xf f
i i i
2
fi x 1 2x 5 x 2 3x x 20
x 3, 4 (rejected) x
x f f
i i
2.8
i
83.
x4 1 lim x 1 x 2 1 ……..(1) x 1 x 1 x 1 3 3 x k k 2 k 2 k2 lim 2 ………….(2) x k x k 2 2k (1) = (2) 8 k 3 lim
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JEE-MAIN-2019 (10th Apr-First Shift)-PCM-38
84.
2
sin2 x 2sin x 5
2
sin x 1
2
. 4 sin
4
2
y
4sin
1
2
y
2
sin x 1 4 2 sin2 y
0
2
2
this is possible only if sin x 1 and sin y 1 1/ 3
85.
n 1n r lim n r 1 n n 1
1/ 3
1 x
dx
0
86.
3 4/3 2 1 4
Let hyperbola be
x2 y2 1 and passes through 4, 2 3 therefore a 2 b2
16 12 1 ……….(i) a2 b2 b2 a 2 e 2 1
4 5 a 16 2 a2 e 5 e 5 On solving (i) and (ii) 4e4 24e2 35 0 x
87.
………(ii)
y
Shown figure is square with side length 2 2
(0, 2) x – y = –2 x+y=2 x
(–2, 0) x + y = –2
(2, 0) x–y=2 (0, –2)
88.
P (Boy) = P (girl)
1 2
Required probability
all four girls Atleast two girls
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JEE-MAIN-2019 (10th Apr-First Shift)-PCM-39 4
1 2 4 4 4 1 1 1 4 4 2 C 3 2 C2 2 1 11
89.
G will be centroid of ABC G 2,4,2 OG 2iˆ 4ˆj 2kˆ OA 3i kˆ OG.OA 1 cos GOA 15 OG OA
90.
dx
x
2
2x 10
2
dx
x 1
21
9
2
Let x 1 3 tan
dx 3 sec 2 d 1 1 1 sin2 cos2 d 1 cos 2 d 27 54 54 2 3 x 1 1 1 x 1 tan C 2 54 3 x 2x 10
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FIITJEE Solutions to JEE(Main)-2019 Test Date: 10th April 2019 (Second Shift)
PHYSICS, CHEMISTRY & MATHEMATICS Paper - 1 Time Allotted: 3 Hours
Maximum Marks: 360
Please read the instructions carefully. You are allotted 5 minutes specifically for this purpose.
Important Instructions: 1.
The test is of 3 hours duration.
2.
This Test Paper consists of 90 questions. The maximum marks are 360.
3.
There are three parts in the question paper A, B, C consisting of Physics, Chemistry and Mathematics having 30 questions in each part of equal weightage. Each question is allotted 4 (four) marks for correct response.
4.
Out of the four options given for each question, only one option is the correct answer.
5.
For each incorrect response 1 mark i.e. ¼ (one-fourth) marks of the total marks allotted to the question will be deducted from the total score. No deduction from the total score, however, will be made if no response is indicated for an item in the Answer Box.
6.
Candidates will be awarded marks as stated above in instruction No.3 for correct response of each question. One mark will be deducted for indicating incorrect response of each question. No deduction from the total score will be made if no response is indicated for an item in the answer box.
7.
There is only one correct response for each question. Marked up more than one response in any question will be treated as wrong response and marked up for wrong response will be deducted accordingly as per instruction 6 above.
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JEE-MAIN-2019 (10th Apr-Second Shift)-PCM-2
PART –A (PHYSICS) 1.
One mole of an ideal gas passes through a process where pressure and volume obey 1 V 2 the relation P Po 1 o . Here Po and Vo are constants. Calculate the change in 2 V the temperature of the gas if its volume change from Vo to 2Vo 1 Po Vo 1 Po Vo (A) (B) 4 R 2 R 5 Po Vo 3 Po Vo (C) (D) 4 R 4 R
2.
A solid sphere of mass M and radius R is divided into two unequal parts. The first part 7M has a mass of and is converted into a uniform disc of radius 2R. The second part is 8 converted into a uniform solid sphere. Let I1 be the moment of inertia of the disc about its axis and I2 be the moment of inertia of the new sphere about its axis. The ratio of I1/I2 is given by: (A) 285 (B) 185 (C) 65 (D) 140
3.
The correct figure that shows, schematically, the wave pattern produced by superposition of two waves of frequencies 9 Hz and 11 Hz,
4.
(A)
(B)
(C)
(D)
In an experiment, brass and steel wires of length 1 m each with areas of cross section 1 mm2 are used. The wires are connected in series and one end of the combined wire is connected to a rigid support and other end is subjected to elongation. The stress requires to produced a new elongation of 0.2 mm is [Given, the Young’s Modulus for steel and brass are respectively 120 109 N/m2 and 60 109 N/m2] (A) 1.8 106 N/m2 (B) 0.2 106 N/m2 6 2 (C) 1.2 10 N/m (D) 4.0 106 N/m2
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JEE-MAIN-2019 (10th Apr-Second Shift)-PCM-3 5.
When heat Q is supplied to a diatomic gas of rigid molecules at constant volume its temperature increases by T. The heat required to produce the same change in temperature, at constant pressure is 3 5 (A) Q (B) Q 2 3 7 2 (C) Q (D) Q 5 3
6.
A bullet of mass 20 g has an initial speed of 1 ms–1 just before it starts penetrating a mud wall of thickness 20 cm. If the wall offers a mean resistances of 2.5 10–2 N, the speed of the bullet after emerging from the other side of the wall is close to (A) 0.7 ms–1 (B) 0.3 ms–1 –1 (C) 0.1 ms (D) 0.4 ms–1
7.
The elastic limit of brass is 379 MPa. What should be the minimum diameter of a brass rod if it is to support a 400 N load without exceeding its elastic limit? (A) 1.00 mm (B) 1.16 mm (C) 0.90 mm (D) 1.36 mm
8.
A plane is inclined at an angle = 30o with respect to the horizontal. A particle is projected with a speed u = 2 ms–1, from the base of the plant, making an angle = 15o with respect to the plane as shown in the figure. The distance from the base at which the particle hits the plane is close to (Take g = 10 ms2)
(A) 18 cm (C) 26 cm
(B) 14 cm (D) 20 cm
9.
The magnitude of the magnetic field at the centre of an equilateral triangular loop of side 1 m which is carrying a current of 10 A is: [Take o = 4 10–7 NA–2] (A) 9T (B) 1T (C) 3T (D) 18T
10.
Two radioactive substances A and B have decay constants 5 and respectively. At t = 0, a sample has the same number of the two nuclei. The time taken for the ratio of the 2
1 number of nuclei to become will be e (A) 1/ (C) 2/
(B) 1/4 (D) 1/2
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JEE-MAIN-2019 (10th Apr-Second Shift)-PCM-4 11.
Two blocks A and B of masses mA = 1 kg and mB = 3 kg are kept on the table as shown in figure. The coefficient of friction between A and B is 0.2 and between B and the surface of the table is also 0.2. The maximum force F that can be applied on B horizontal, so that the block A does not slide over the block B is: [Take g = 10 m/s2]
(A) 8 N (C) 12 N
(B) 16 N (D) 40 N
12.
The formula X = 5YZ2 X and Z have dimensions of capacitance and magnetic field respectively. What are the dimensions of Y in SI units? (A) [M-2 L0 T–4 A–2] (B) [M-3 L-2 T8 A–1] -2 -2 6 3 (C) [M L T A ] (D) [M-1 L-2 T4 A2]
13.
In Li++, electron in first Bohr orbit is excited to a level by a radiation of wavelength . When the ion gets deexcited to the ground state in all possible ways(including intermediate emission) a total of six spectral lines are observed. What is the value of ? (Given: h = 6.63 1034 js; e = 3 108 ms–1) (A) 10.8 nm (B) 11.4 nm (C) 9.4 nm (D) 12.3 nm
14.
The graph shows how the magnification m produced by a thin lens varies with image distance . What is the focal length of the lens used?
b2 ac b 2c (C) a
(A)
15.
a c b (D) c
(B)
A spaceship orbits around a planet at a height of 20 km from its surface. Assuming that only gravitational field of the plant acts on the spaceship. What will be the number of complete revolutions made by the spaceship in 24 hours around the plane? [Given: Mass of plane = 8 1022 kg, Radius of planet = 2 106 m, Gravitational constant G = 6.67 10–11 Mn2/kg2] (A) 9 (B) 11 (C) 13 (D) 17 FIITJEE Ltd., FIITJEE House, 29-A, Kalu Sarai, Sarvapriya Vihar, New Delhi -110016, Ph 46106000, 26569493, Fax 26513942 website: www.fiitjee.com.
JEE-MAIN-2019 (10th Apr-Second Shift)-PCM-5 16.
Light is incident normally on a completely absorbing surface with an energy flux of 25 W cm–2. If the surface has an area of 25 cm2, the maximum transferred to the surface in 40 min time duration will be (A) 6.3 10–4 Ns (B) 3.5 10–6 Ns –3 (C) 5.0 10 Ns (D) 1.4 10–6 Ns
17.
The time dependence of the position of a particle of mass m = 2 is given by r t 2t ˆi 3t 2 ˆj Its angular momentum with respect to the origin at time t = 2 is . (A) -48 kˆ (B) 48( ˆi + ˆj ) (C) 36 kˆ
(D) -34( kˆ - ˆi )
18.
Water from a tap emerges vertically downwards with an initial speed of 1.0 ms–1. The cross-sectional area of the tap is 10–4m2. Assume that the pressure is constant throughout the stream of water and that flow is streamlined. The cross-sectional area of the stream, 0.15 m below the tap would be: (take g = 110 ms–2) (A) 5 10–4 m2 (B) 5 10–5 m2 –5 2 (C) 1 10 m (D) 2 10–5 m2
19.
Space between two concentric conducting spheres of radii a and b (b >a) is filled with a medium of resistivity . The resistance between the two spheres will be 1 1 1 1 (A) (B) 2 a b 2 a b 1 1 1 1 (C) (D) 4 a b 4 a b
20.
A metal coin of mass 5 g and radius 1 cm is fixed to a thin stick AB of negligible mass as shown in the figure. The system is initially at rest. The constant torque, that will make the system rotate about AB at 25 rotations per second is 5 s is close to
(A) 2.0 10–5 Nm (C) 1.6 10–5 Nm
(B) 4.0 10–6 Nm (D) 7.9 10–6 Nm
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JEE-MAIN-2019 (10th Apr-Second Shift)-PCM-6 21.
The figure represents a voltage regulator circuit using a Zener diode. The breakdown voltage of the Zener diode is 6 V and the load resistance is RL= 4k. The series resistance of the circuit is Ri = 1k. If the battery voltage V8 varies from 8 V to 16 V, what are the minimum and maximum values of the current through Zener diode?
(A) 0.5 mA; 0.6 mA (C) 1.5 mA; 8.5 mA
22.
(B) 1 mA; 8.5 mA (D) 0.5 mA; 8.5 mA
A simple pendulum of length L is placed between the plates of a parallel plate capacitor having electric field E, as shown in figure. Its bob has mass m and charge q. the time period of the pendulum is given by
(A) 2
L qE g2 m
(C) 2
L qE g m
2
(B) 2
(D) 2
L qE g m
L g2
q2E 2 m2
23.
In free space, a particle A of charge 1 C is held fixed at a point P. Another particle B of the same charge and mass 4 g is kept at a distance of 1 mm from P. If B is released, then its velocity at a distance of 9 mm from P is 1 9 109 Nm2C2 Take 4o (A) 1.5 102 m/s (B) 2.0 103 m/s (C) 1.0 m/s (D) 3.0 104 m/s
24.
A 2 mW laser operates at a wavelength of 500 nm. The number of photons that will be emitted per second is [Given Planck’s constant h = 6.6 10–34 Js, speed of light c = 3.0 108 m/s] (A) 1 1016 (B) 1.5 1016 16 (C) 2 10 (D) 5 1015
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JEE-MAIN-2019 (10th Apr-Second Shift)-PCM-7 25.
A coil of self inductance 10 mH and resistance 0.1 is connected through a switch to a battery of internal resistance 0.9 . After the switch is closed the time taken for the current to attain 80% of the saturation values is: [take ln 5 = 1.6] (A) 0.016 s (B) 0.324 s (C) 0.002 s (D) 0.103 s
26.
A submarine experiences a pressure of 5.05 106 Pa at a depth of d1 in a sea. When it goes further to a depth of d2, it experiences a pressure of 8.08 106 Pa. Then d2 – d1 is approximately (density of water = 103 kg/m3 and acceleration due to gravity = 10 ms–2) (A) 400 m (B) 500 m (C) 600 m (D) 300 m
27.
A cubical block of side 0.5 m floats on water with 30% of its volume under water. What is the maximum weight that can be put on the block without fully submerging it under water? [Take density of water = 103 kg/m3] (A) 46.3 kg (B) 65.4 kg (C) 30.1 kg (D) 87.5 kg
28.
In a Young’s double slit experiment the ratio of the slit’s width is 4 : 1. The ratio of the intensity of maxima to minima, close to central fringe on the screen will be (A)
4
3 1 : 16
(C) 9 : 1
(B) 25 : 9 (D) 4:1
29.
A source of sound S is moving with the velocity of 50 m/s towards a stationary observer. The observer measures the frequency of the sound as 1000 Hz. What will be the apparent frequency of the source when it is moving away from the observer after crossing him? (Take velocity of sound in air is 350 m/s) (A) 1143 Hz (B) 857 Hz (C) 750 Hz (D) 807 Hz
30.
A square loop is carrying a steady current I and the magnitude of its magnetic dipole moment is m. If this square loop is changed to a circular loop and it carries the same current, the magnitude of the magnetic dipole moment of circular loop will be m 3m (A) (B) 4m 2m (C) (D)
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JEE-MAIN-2019 (10th Apr-Second Shift)-PCM-8
PART –B (CHEMISTRY) 31.
The difference between H and U (H - U), when the combustion of one mole of heptane(I) is carried out a temperature T is equal to (A) -4 RT (B) -3 RT (C) 3 RT (D) 4 RT
32.
The major product obtained in the given reaction is
(A)
(B)
(C)
(D)
33.
The ratio of the shortest wavelength of two spectral series of hydrogen spectrum is found to be about 9. The spectral series are: (A) Lyman and Paschen (B) Brackett and Pfund (C) Paschen and Pfund (D) Balmer and Brackett
34.
The correct order of the first ionization enthalpies is (A) Mn < Ti < Zn < Ni (B) Zn < Ni < Mn < Ti (C) Ti < Mn < Zn < Ni (D) Ti < Mn < Ni < Zn
35.
The correct statements among (a) to (b) are: (a) saline hydrides produce H2 gas when reacted with H2O. (b) reaction of LiAH4 with BF3 leads to B2H6. (c) PH3 and CH4 are electron - rich and electron-precise hydrides, respectively. (d) HF and CH4 are called as molecular hydrides. (A) (c) and (d) only (B) (a), (b) and (c) only (C) (a), (b), (c) and (d) (D) (a), (c) and (d) only
36.
Air pollution that occurs in sunlight is: (A) oxidising smog (C) reducing smog
37.
(B) acid rain (D) fog
For the re action of H2 w i t h I2, the rate constant is 2.5 10-4 dm3 mol–1 s–1 at 327°C and 1.0 dm3 mol–1 s–1 at 527°C. The activation energy for the reaction, in kJ mol–1 is: (R = 8.314 J K–1 mol–1) (A) 72 (B) 166 (C) 150 (D) 59 FIITJEE Ltd., FIITJEE House, 29-A, Kalu Sarai, Sarvapriya Vihar, New Delhi -110016, Ph 46106000, 26569493, Fax 26513942 website: www.fiitjee.com.
JEE-MAIN-2019 (10th Apr-Second Shift)-PCM-9 38.
In chromatography, which of the following statements is INCORRECT for Rf? (A) Rf value depends on the type of chromatography. (B) The value of Rf can not be more than one. (C) Higher Rf value means higher adsorption. (D) Rf value is dependent on the mobile phase.
39.
A hydrated solid X on heating initially gives a monohydrated compound Y. Y upon heating above 373K leads to an anhydrous white powder Z. X and Z, respectively, are: (A) Washing soda and soda ash. (B) Washing soda and dead burnt plaster. (C) Baking soda and dead burnt plaster. (D) Baking soda and soda ash.
40.
The INCORRECT statement is: (A) the spin-only magnetic moments of [Fe(H2O)6]2+ and [Cr(H2O)6]2+ are nearly similar. (B) the spin-only magnetic moment of [Ni(NH3)4(H2O)2]2+ is 2.83 BM. (C) the gemstone, ruby, has Cr3+ ions occupying the octahedral sites of beryl. (D) the color of [CoCl(NH3)5]2+ is violet as it absorbs the yellow light.
41.
Which of these factors does not govern the stability of a conformation in acyclic compounds? (A) Torsional strain (B) Angle strain (C) Steric interactions (D) Electrostatic forces of interaction
42.
The correct statement is: (A) zincite is a carbonate ore (B) aniline is a froth stabilizer (C) zone refining process is used for the refining of titanium (D) sodium cyanide cannot be used in the metallurgy of silver
43.
For the reaction, 2 SO2 g O2 g 2 SO3 g H = –57.2 kJ mol–1 and KC = 1.7 1016 Which of the following statement is INCORRECT? (A) The equilibrium constant is large suggestive of reaction going to completion and so no catalyst is required. (B) The equilibrium will shift in forward direction as the pressure increase. (C) The equilibrium constant decreases as the temperature increases. (D) The addition of inert gas at constant volume will not affect the equilibrium constant.
44.
The increasing order of nucleophilicity of the following nucleophiles is : (a) CH3CO2 (b) H2O (c) CH3SO3 (A) (b) < (c) < (a) < (d) (C) (d) < (a) < (c) < (b)
(d) OH (B) (a) < (d) < (c) < (b) (D) (b) < (c) < (d) < (a)
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JEE-MAIN-2019 (10th Apr-Second Shift)-PCM-10 45.
The correct match between Item-I and Item-II is:
(A) (a) (III), (b)(I), (c)(II), (d)(IV) (B) (a)(IV), (b)(II), (c)(I), (d)(III) (C) (a)(II), (b)(IV), (c)(I), (d)(III) (D) (a)(III), (b)(I), (c)(IV), (d)(II) 46.
The major product 'Y' in the following reaction is:Ph
CH3
i SOCl
NaOCl 2 X ii Aniline Y
O
47.
(A)
(B)
(C)
(D)
Points I, II and III in the following plot respectively correspond to (Vmp : most probable velocity)
(A) Vmp of N2 (B) Vmp of H2 (C) Vmp of O2 (D) Vmp of N2
(300K); Vmp of H2(300K); V mp (300K); Vmp of N2(300K); Vmp (400K); Vmp of N2(300K); Vmp (300K); Vmp of O2(400K); Vmp
of O2(400K) of O2(400K) of H2(300K) of H2(300K)
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JEE-MAIN-2019 (10th Apr-Second Shift)-PCM-11 48.
The highest possible oxidation states of uranium and plutonium, respectively, are (A) 6 and 4 (B) 7 and 6 (C) 4 and 6 (D) 6 and 7
49.
Compound A (C9H10O) shows positive iodoform test. Oxidation of A with KMnO4/KOH gives acid B(C8H6O4). Anhydride of B is used for the preparation of phenolphthalein. Compound A is:-
50.
The noble gas that does NOT occur in the atmosphere is: (A) He (B) Ra (C) Ne (D) Kr
51.
The pH of a 0.02 M NH4Cl solution will be [given Kb (NH4OH) = 10–5 and log 2 = 0.301] (A) 2.65 (B) 5.35 (C) 4.35 (D) 4.65
52.
The crystal field stabilization energy (CFSE) of [Fe(H 2 O) 6 ]Cl 2 and K 2 [NiCl 4 ], respectively, are :(A) –0.4o and –0.8t (B) –0.4o and –1.2t (C) –2.4o and –1.2t (D) –0.6o and –0.8t
53.
The correct option among the following is: (A) Colloidal particles in lyophobic sols can be precipitated by electrophoresis. (B) Brownian motion in colloidal solution is faster the viscosity of the solution is very high. (C) Colloidal medicines are more effective because they have small surface area. (D) Addition of alum to water makes it unfit for drinking.
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JEE-MAIN-2019 (10th Apr-Second Shift)-PCM-12
54.
Which one of the following graphs between molar conductivity (m) correct? (A) (B)
(C)
55.
versus
C is
(D)
1 g of non-volatile non-electrolyte solute is dissolved in 100g of two different solvents A and B whose ebullioscopic constants are in the ratio of 1 : 5. The ratio of the elevation Tb A in their boiling points, is : Tb B (A) 5 : 1 (C) 1 : 5
(B) 10 : 1 (D) 1 : 0.2
56.
Which of the following is NOT a correct method of the preparation of benzylamine from cyanobenzene? (A) (i) HCl/H2O (ii) NaBH4 (B) (i) LiAIH4 (ii) H3O+ (C) (i) SnCl2 + HCl(gas) (ii) NaBH4 (D) H2/Ni
57.
The number of pentagons in C60 and trigons (triangles) in white phosphorus, respectively, are: (A) 12 and 3 (B) 20 and 4 (C) 12 and 4 (D) 20 and 3
58.
The minimum amount of O2(g) consumed per gram of reactant is for the reaction : (Given atomic mass : Fe = 56, O = 16, Mg = 24, P = 31, C = 12, H = 1) (A) C3H8(g) + 5O2(g) 3 CO2(g) + 4 H2O(l) (B) P4(s) + 5O2(g) P4O10(s) (C) 4Fe(s) + 3O2(g) 2 Fe2O3(s) (D) 2 Mg(s) + O2(g) 2 MgO(s)
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JEE-MAIN-2019 (10th Apr-Second Shift)-PCM-13 59.
Number of stereo centers present in linear and cyclic structures of glucose are respectively (A) 4 & 5 (B) 5 & 5 (C) 4 & 4 (D) 5 & 4
60.
The major product 'Y' in the following reaction is:
(A)
(B)
(C)
(D)
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JEE-MAIN-2019 (10th Apr-Second Shift)-PCM-14
PART–C (MATHEMATICS) x 2 ax b 3 , then a + b is equal to x 1 x 1 (A) 5 (B) 1 (C) -4 (D) -7
61.
If lim
62.
The sum of the real roots of the equation x 6 1 2
3x
x 3 0 is equal to
3
2x
x2
(A) -4 (C) 6 63.
(B) 0 (D) 1
Lines are drawn parallel to the line 4x – 3y + 2 = 0 at a distance
3 from the origin. Then 5
which one of the following points lies on any of these lines? 1 2 1 1 (A) , (B) , 4 3 4 3 1 1 1 2 (C) , (D) , 4 3 4 3 x , x R, x 3 at a point (, ) (0, 0) on it is x 3 parallel to the line 2x + 6y – 11 = 0 then (A) |2 + 6| = 11 (B) |2 + 6| = 19 (C) |6 + 2| = 19 (D) |6 + 2| = 9
64.
If the tangent to the curve y
65.
The distance of the point having position vector ˆi 2ˆj 6kˆ from the straight line passing through the point (2, 3, -4) and parallel to the vector 6iˆ 3ˆj 4kˆ is
2
(A) 7
(B) 4 3
(C) 2 13
(D) 6
66.
If the line ax + y = c, touches both the curves x2 + y2 = 1 and y 2 4 2 x , then |c| is equal to 1 (A) (B) 2 2 1 (C) (D) 2 2
67.
Let f(x) = loge(sinx), (0 < x < ) and g(x) = sin–1(e–x), (x 0). If is a positive real number such that a = (fog)’() and b = (fog)(), then (A) a2 + b – a = 22 (B) a2 – b – a = 0 (C) a2 – b – a = 1 (D) a2 + b + a = 0 FIITJEE Ltd., FIITJEE House, 29-A, Kalu Sarai, Sarvapriya Vihar, New Delhi -110016, Ph 46106000, 26569493, Fax 26513942 website: www.fiitjee.com.
JEE-MAIN-2019 (10th Apr-Second Shift)-PCM-15 68.
If 5x + 9 = 0 is the directrix of the hyperbola 16x2 – 9y2 = 144, then its corresponding focus is 5 (A) (5, 0) (B) ,0 3 (C) (-5, 0)
69.
y y If cos–1x – cos–1 2 , where -1 x 1, – 2 y 2, x 2 , then for all x, y, 4x2 – 4xy
cos + y2 is equal to (A) 4 sin2 - 2x2y2 (C) 2 sin2 70.
5 (D) ,0 3
(B) 4 cos2 + 2x2y2 (D) 4 sin2
The locus of the centres of the circles, which touch the circle, x2 + y2 = 1 externally, also touch the y-axis and lie in the first quadrant is (A) x 1 2y, y 0 (B) y 1 4x, x 0 (C) x 1 4y, y 0
(D) y 1 2x, x 0
71.
Let a1, a2, a3, …… be and A.P with a6 = 2. Then the common difference of this A.P., which maximizes the product a1a4a5 is 3 8 (A) (B) 2 5 2 6 (C) (D) 3 5
72.
The smallest natural number n, such that the coefficient of x in the expansion of n
2 1 n x 3 is C23 is x (A) 38 (C) 23
(B) 58 (D) 35
73.
Suppose that 20 pillars of the same height have been erected along the boundary of a circular stadium. If the top of each pillar has been connected by beams with the top of all its non-adjacent pillars, then the total number of beams is (A) 210 (B) 180 (C) 170 (D) 190
74.
A spherical iron ball of radius 10 cm is coated with a layer of ice of uniform thickness that melts at a rate of 50 cm3/min. When the thickness of the ice is 5 cm, then the rate at which the thickness (in cm/min) of ice decreases is 1 5 (A) (B) 36 6 1 1 (C) (D) 9 18
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JEE-MAIN-2019 (10th Apr-Second Shift)-PCM-16 75.
If both the means and the standard deviation of 50 observations x1, x2, ………, x50 are equal to 16, then the mean of (x1 - 4)2, (x2 - 4)2, …., (x50 - 4)2 is (A) 400 (B) 380 (C) 525 (D) 480
76.
The number of real roots of the equation 5 + |2x - 1| = 2 x(2 x - 2) is (A) 4 (C) 2
(B) 3 (D) 1
77.
The tangent and normal to the ellipse 3x2 + 5y2 = 32 at the point P(2, 2) meet the x-axis at Q and R, respectively. Then the area(in sq. units) of the triangle PQR is 34 68 (A) (B) 15 15 14 16 (C) (D) 3 3
78.
A perpendicular is drawn from a point on the line
79.
The integral
x 1 y 1 z to the plane x + y + z 2 1 1 = 3 such that the foot of the perpendicular Q also lies on the plane x – y + z = 3. Then the co-ordinates of Q are (A) (2, 0, 1) (B) (-1, 0, 4) (C) (1, 0, 2) (D) (4, 0, -1)
/3
/ 6
sec 2/3 x cos ec 4/3 x dx is equal to
(A) 35/6 – 32/3 (C) 37/6 – 35/6
(B) 35/3 – 31/3 (D) 34/3 – 31/3
80.
The angles A, B and C of a triangle ABC are in A.P and a : b = 1 : the area (in sq. cm) of this triangle is 4 (A) 2 3 (B) 3 2 (C) 4 3 (D) 3
3 . If c = 4 cm, then
81.
Let a, b and c be in G.P with common ratio r, where a 0 and 0 < r
1 . If 3a, 7b and 2
15c are the first three terms of an A.P., then the 4th term of this A.P is 2 7 (A) a (B) a 3 3 (C) 5a (D) a 82.
Let y = y(x) be the solution of the differential equation dy y tan x 2x x 2 tan x, x , , such that y(0) = 1. Then dx 2 2 (A) y ' y ' (B) y ' y ' 2 2 4 4 4 4 (C) y y 2 4 4
2 (D) y y 2 4 4 2
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JEE-MAIN-2019 (10th Apr-Second Shift)-PCM-17 1 2 and units from the planes 4x – 2y 3 3 + 4z + = 0 and 2x – y + 2z + = 0, respectively, then the maximum value of + us equal to (A) 15 (B) 13 (C) 5 (D) 9
83.
If the plane 2x – y + 2z + 3 = 0 has the distances
84.
The area(in sq. units) of the region bounded by the curves y = 2x and y = |x +1| in the first quadrant is 3 3 (A) (B) loge 2 2 2 3 1 1 (C) (D) 2 loge 2 2
85.
Let be a real number for which the system of linear equations x+y+z=6 4x + y - z = - 2 3x + 2y – 4z = -5 Has indefinitely many solutions. Then is a root of the quadratic equation (A) 2 - – 6 = 0 (B) 2 - 3 – 4 = 0 2 (C) + 3 – 4 = 0 (D) 2 + – 6 = 0
86.
The sum 3 3 3 3 3 3 3 3 3 1 2 1 2 3 1 2 3 ..... 15 1 1 ...... 1 2 3 ...... 15 is equal to 1 2 1 2 3 1 2 3 ..... 15 2 (A) 620 (B) 1860 (C) 1240 (D) 660
87.
Minimum number of times a fair coin must be tossed so that the probability of getting at least one head is more than 99% is (A) 8 (B) 6 (C) 7 (D) 5
88.
The negation of the Boolean expression ~ s(~rs) is equivalent to (A) sr (B) ~s ~r (C) r (D) s r
89.
If
5 x2
x e
2
dx g x e x c , where c is a constant of integration, then g(-1) is equal to
(A) -1 (C) 90.
(B) 1 5 2
(D)
1 2
If z and w are two complex numbers such that |zw| = 1 and arg(z) – arg(w) = (A) zw i (C) zw
(B) zw
1 i 2
, then 2
1 i 2
(D) zw i
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JEE-MAIN-2019 (10th Apr-Second Shift)-PCM-18
JEE (Mains) – 2019 ANSWERS PART A – PHYSICS 1. 5. 9. 13. 17. 21. 25. 29.
D C C A A D A C
2. 6. 10. 14. 18. 22. 26. 30.
D A D D B D D C
3. 7. 11. 15. 19. 23. 27.
A B B B D None D
4. 8. 12. 16. 20. 24. 28.
None D B C A D C
34. 38. 42. 46. 50. 54. 58.
D C B A C B C
64. 68. 72. 76. 80. 84. 88.
C C A D A C D
PART B – CHEMISTRY 31. 35. 39. 43. 47. 51. 55. 59.
A C A A D B C A
32. 36. 40. 44. 48. 52. 56. 60.
C A C A D A A C
33. 37. 41. 45. 49. 53. 57.
A B B D A A C
PART C – MATHEMATICS 61. 65. 69. 73. 77. 81. 85. 89.
D A D C B D A C
62. 66. 70. 74. 78. 82. 86. 90.
B B D D A B A D
63. 67. 71. 75. 79. 83. 87.
A C B A C B C
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JEE-MAIN-2019 (10th Apr-Second Shift)-PCM-19
HINTS AND SOLUTIONS PART A – PHYSICS 1.
n = 1 mole 1 V 2 P = Po 1 o ; PV = nRT = RT 2 V RT P V V2 RT Po 1 o 2 V 2V PV Vo2 V 2 Po T o 1 = V R 2V 2 R 2V 2 P V2 1 1 T o (2Vo Vo ) o R 2 2Vo Vo Vo2 V o 2 P V2 1 1 T o (2Vo Vo ) o R 2 2Vo Vo
=
Po R
Po Vo2 (1 2) Vo R 2 2Vo P V 3 Po Vo = o Vo o = R 4 4 R
=
2.
7M 2 (ZR) 7M 4R 2 7MR2 8 I1 2 28 4 2 2 2 2 MR 2M R MR I2 5 82 58 4 80 2 I1 7MR 80 140 I2 4MR 2
3.
By looking into graph.
4.
=1M A = 10–6 M2 F Stress = A Stress Stress = Y
Ys = 120 × 109
Ys = 120 × 109
F 9
YB = 60 × 10
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JEE-MAIN-2019 (10th Apr-Second Shift)-PCM-20
F AY
1 2
1F 2F 0.2 10 3 AY1 AY2
F 0.2 10 3 A Y1 Y2 0.2 10 3 0.2 10 3 109 120 1 1 1 2 9 120 10 60 109 0.2 10 6 120 = 8 106 3 =
5.
Q = CVT Q = CPT C 2 7 Q = P Q 1 Q Q CV 5 5
6.
2.5 × 10–2 × 0.2 =
7.
8.
1 20 10 3 V 2 12 2 5 × 10–3 = 10 × 10–3 (1 –V2) 1 1 1 1 – V2 = ; V2 = ; V= = 0.7 2 2 2
400 379 106 2 d 4 4 400 106 d2 = 0.336 × 10–6 × 4 379 d 2 0.336 10 3 M 1.16 mm
T
2 u sin g cos
1 g sin T 2 2 ucos 2u sin gsin 4u2 sin2 = gcos 2 g2 cos2
R ucos T
=
u2 sin2 u2 sin 1 cos 2 g cos g cos2
1 u2 sin 3 2 = 1 2 2 3 g cos 10 2 4
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JEE-MAIN-2019 (10th Apr-Second Shift)-PCM-21
= = 9.
4 10 3
4 5 3
8 3 1 30 2
8 8 3 8 8( 3 1) = 20 cm 30 30 30
i = 10 A =1m o = 4 × 10–7 B=
N A2
i C
o i
3 4 3 2 i 3 4 10 7 10 3 = o = 20 3 107 2 2 1 = 3 T
10.
11.
1 e t 5 t 2 e 1 t 2 MA = 1 kg, MB = 3 kg AB 0.2 B = 0.2 Fmax = (MA + MB) × 0.2 × 10 + (MA + MB) × 0.2 ×
A B
F
10 = 4 × 2 + 4 × 2 = 16 12.
13.
X = 5YZ2 X Y= = M–3 L–2 T8A4 2 5Z
hc 1 13.6 eV(g) 1 16 1240 eV 15 9 13.6 eV 16 1240 16 = 10.8 nm 15 9 13.6
h=4
h=1
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JEE-MAIN-2019 (10th Apr-Second Shift)-PCM-22
14.
f
m
b c C
c b
a
15.
b
V
mV 2 GMm 2 r r GM V r
n
VT GM T 2r r 2r
GM T 6.67 1011 8 1022 T = r 3 2 (202 104 )3 2 =
16.
17.
24 3600 2 3.14
24 3600 6.67 8 1011 24 3600 = 11 3 12 (202) 10 2 3.14 1242.8 78.51
W 25 10 4 W / m2 cm2 P = 25 × 25 ; W = 625 W hc dn P dt h dn P 625 F dt C 3 108 625 40 60 Momentum = = 5 × 10–3 Ns 3 108 I 25
v 2iˆ 6 ˆj At t = 2 v 2iˆ 12ˆj P mv 4i 24 ˆj At t = 2 r 4iˆ 12ˆj
ˆi ˆj kˆ L r P 4 12 0 4 24 0
= 4( 24) 4 12 kˆ = –96 48 kˆ = ( ) 48 kˆ
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JEE-MAIN-2019 (10th Apr-Second Shift)-PCM-23
18.
10 4 1 (1)2 2 10 0.15 A A
10 4 = 5 × 10–5 2 b
19.
R a
=
dx 4 x 2
1 1 4 a b
20.
m = 5 × 10–3 kg, r = 10–2 m = 25 × 2 rad/5 = 50 rad/sec t I I 5mr 2 t 4 t 3 5 5 10 10 4 50 = 45 25 = 106 = 2 × 10–5 4
21.
Vbreakdwon = 6V, RL = 4k, Ri = 1 k 6 iL = 103 1.5 10 3 = 1.5 mA 4 ii = 2 × 10–3 i = i1 – iL = 0.5 mA – minimum current
Ri ii
i RL
VB iL
ii = 10 × 10–3 = 10 mA imax = 8.5 mA 22.
T 2
L q2E 2 g 2 M 2
+ + + + + + + + + + + +
– – – – – E – – – – – – m – q qE – – – – Mg –
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JEE-MAIN-2019 (10th Apr-Second Shift)-PCM-24 23.
qA = 1 c ; qB = 1 c,
mB = 4 × 10–9 kg, rAB = 10–3 m
1 1 1 MB V 2 k qA qB 3 2 9 10 3 10
1 8 4 109 V 2 9 109 10 6 10 6 103 2 9 8 V 2 109 4 109 2 hc dn dt dn 2 103 dt hc 2 10 3 500 10 9 = 6.6 10 34 3 108 1000 = 1014 = 5 × 1015 6.6 3
24.
2 103
25.
L = 10 × 10–3 H, r1 = 0.1 i = 1 e t/2
L = 10
–2
r1 = 0.1
isaturation = 80% isaturation = 0.8 0.8 = 1 e t/2
t/2
0.8 = 1 e ; e t/2 0.2 et/L = 5 t = L ln 5 = 10 × 10–3 × 1.6 = 16 × 10–3
r2 = 0.9
26.
P1 = 5.05 × 106 ; P2 = 8.08 × 106 P2 – P1 = g(d2 – d1) 3.03 106 d2 d1 = 3.03 × 102 = 303 103 10
27.
0.3 3 = 3
××
kg m3 m + 3 = 3
0.3 3
= 300
M = 3(w – ) = (5)3 {1000 – 300} = 700 × (5)3 = 87.5 kg 28.
IMax 9 IMin 1
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JEE-MAIN-2019 (10th Apr-Second Shift)-PCM-25
29.
fa
V fo = 1000 Hz V Vs
s
V = 50 m/s
V fo V Vs fa V Vs 350 50 300 3 fa V Vs 350 50 400 4 3 fa 1000 = 750 Hz 4 fa
30.
m = I2 I4 2 m 4 m 4 m m
m =
2r = 4
r
2
r 2
4 2 42 = 2
PART B – CHEMISTRY 31.
32.
33.
1 1 1 RH 2 2 Z2 2 n1 n2 1 1 1 RH 1 1 Z2 1 n1 n2 As for shortest wavelength both n1 and n12 are
1 9 n11 2 1 n12
Now if n11 = 3 and n1 is 1 it will justify the statement hence Lyman and Paschen is correct.
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JEE-MAIN-2019 (10th Apr-Second Shift)-PCM-26
34.
As Zn is fully filled and left to right in group IP increases.
35.
36.
Fact based
37.
38.
Rf value can’t measure the extent of adsorption.
39.
40.
In gemstone, ruby has Cr3+ ion occupying the octahedral sites of aluminium oxide (Al2O3) normally occupied by Al3+ion.
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JEE-MAIN-2019 (10th Apr-Second Shift)-PCM-27 41.
Angle strain govern stability in cyclic compound.
42.
Fact based.
43.
In option (B)- ng is –ve therefore increase in pressure will bring reaction in forward direction. In option (C)- as the reaction is exothermic therefore increase in temperature will decrease the equilibrium constant. In option (D)- Equilibrium constant changes only with temperature. Hence, option (B), (C) and (D) are correct therefore option (1) is incorrect choice.
44.
45
(a) (b) (c) (d)
High density polythene Polyacrylonitrile Novolac Nylon 6
(III) (I) (IV) (II)
Ziegler-Natta Catalyst Peroxide catalys t Acid or base catalyst Condensation at high temperature & pressure
46.
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JEE-MAIN-2019 (10th Apr-Second Shift)-PCM-28 47.
48.
The highest oxidation state of U and Pu is 6+ and 7+ respectively.
49.
50.
Fact based.
51.
52.
CFSE = [-0.4n t2g + 0.6 n eg] o
53.
In electrophoresis precipitation occurs at the electrode which is oppositely charged therefore (A) is correct. FIITJEE Ltd., FIITJEE House, 29-A, Kalu Sarai, Sarvapriya Vihar, New Delhi -110016, Ph 46106000, 26569493, Fax 26513942 website: www.fiitjee.com.
JEE-MAIN-2019 (10th Apr-Second Shift)-PCM-29
54.
Both NaCl and KCl are strong electrolytes and as Na+(aq.) has less conductance than K+ (aq.) due to more hydration therefore the graph of option (B) is correct.
55.
56.
Benzylamine will not give cyanobenzene with HCl/H2O & NaBH4.
57.
Refer structure of C60 & P4 P
P
P P
58.
4 mol of Fe require 3 32 gram 1 3 32 1 mol of Fe require 0.428 g 56 4 56
59.
60.
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JEE-MAIN-2019 (10th Apr-Second Shift)-PCM-30
PART C – MATHEMATICS 2
61.
62.
63.
x ax b 5 x 1 ……..(i) 1 a b 0 . ……..(ii) 2a5 a b 7 lim
x 1
By expansion, we get 5x 3 30x 30 5x 0 5x3 35x 30 0 x3 7x 6 0 , All roots area real So, sum of roots = 0 Y
Required line is 4x 3y 0
3 5 5 3 So, required equation of line 4x 3y 3 0 and 4x 3y 3 0
is
(–2, 0) X
1 2 (1) 4 3 3 0 4 3 4x – 3y + 2 = 0
64.
dy 2 3 2 dx , 2 3
Given that : 2 3 1 2 2 3 3
0, 3
1 . 2 6 2 19
65.
0 0
AP.n AD 61 n
P (–1, 2, 6)
PD AP 2 AD 2 110 61 7
A (2, 3, –4)
D
n 6iˆ 3 ˆj 4kˆ
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JEE-MAIN-2019 (10th Apr-Second Shift)-PCM-31
66.
Tangent to y 2 4 2 x is y mx
2 /m 1 m2
2 it is also tangent to x 2 y 2 1 m
1 m 1
Tangent will be y x 2 or y x 2 compare with y ax C 67.
a 1 and C 2 fog x x fg b
fg x ' 1 fg ' 1 a 68.
x2 y2 1 9 16
16 5 9 3 corresponding focus will be (–ae, 0) i.e. 5, 0 .
a 3, b 4 and e 1
69.
y 2 y cos cos1 x cos1 cos 2 cos1 x cos1
x
y y2 1 x2 1 cos 2 4 2
xy y2 2 cos 1 x 1 2 4 y2 x2 xy cos 1 cos2 sin2 4
70.
h2 k 2 h 1
x 2 y 2 x 2 1 2x y 2 1 2x y 1 2x; x 0
71.
Let a is first term and d is common difference then, a 5d 2 (given) ……(1) f d 2 5d 2 2d 2 d
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JEE-MAIN-2019 (10th Apr-Second Shift)-PCM-32
2 8 , 3 5 8 f " d 0 at d 5 8 d 5 f ' x 0 d
n
72.
Tr 1 nCr x 2n2r . x 3r r 0
2n 5r 1 2n 5r 1 for r 15, n 38 smallest value of n is 38.
73.
74.
75.
Total cases = number of diagonals in 20 sided polygon. 20C2 20 170
4 3 10 h 103 3 dV 2 dh 4 10 h dt dt 2 dh 50 4 10 5 dt dh 1 cm dt 18 min V
Mean
x 50
i
16
Standard deviation
256 2
e
x
2 i
50
2
16
2 i
50
New mean
x
i
4
2
x
2 i
16 50 8 x i
50 50 256 2 16 8 16 400
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JEE-MAIN-2019 (10th Apr-Second Shift)-PCM-33 76.
Let 2x t 5 t 1 t 2 2t
t 1 t 2 2t 5
gt f t From the graph So, number of real root is 1.
77.
3x 2 5y 2 32 dy 3 dx 2,2 5 3 16 x 2 Q , 0 5 3 5 4 Normal : y 2 x 2 R , 0 3 5 1 68 Area is QR 2 QR 2 15 Tangent : y 2
78.
79.
Let point P on the line is 2 1, 1, foot of perpendicular Q is given by x 2 1 y 1 z 2 3 1 1 1 3 Q lies on x y z 3 and x y z 3 x z 3 and y 0 2 3 y 0 1 0 3 Q is (2, 0, 1) 1 I dx 2/3 cos x sin1/3 x.sin x
tan2/3 x .sec 2 x.dx tan2 x sec 2 x .dx tan x t, sec 2 x dx dt 4/3 tan x dt t 1/3 3 t 1/3 tan4/3 1/ 3
1 3 tan x 3 1/3
tan x
1/3
/3
I
/6
3
1
1/3
1/3
3
3
37/6 3 5/6
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JEE-MAIN-2019 (10th Apr-Second Shift)-PCM-34
80.
81.
, by sine Rule 3 1 sin A 2 A 30o ,a 2, b 2 3, c 4 1 2 3 2 2 3 sq.cm 2 B
b ar c ar 2 3a,7b and 15c are in A.P. 14b 3a 15c 14 ar 3a 15 ar 2
14r 3 15r 2 15r 2 14r 3 0
r
3r 1 5r 3 0
1 3 , 3 5
1 1 Only acceptable value is r , because r 0, 3 2 7 2 c .d 7b 3a 7ar 3a a 3a a 3 3 2 15 2 4th term 15c a a aa 3 9 3 82.
dy y tan x 2x x 2 tan x dx tan x dx I.F. e eln.sec x sec x
y.sec x 2x x 2 tan x sec x.dx 2
2x sec x dx x sec x .tan x dx
y sec x x 2 sec x y x 2 cos x y 0 0 1
1
2
y x cos x 1 2 y 2 4 16 2 1 y 2 4 16 y ' x 2x sin x 1 y ' 2 4 2
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JEE-MAIN-2019 (10th Apr-Second Shift)-PCM-35
1 y ' 2 4 2 y ' y' 2 4 4 83.
4x 2y 4z 6 0 6 16 4 16 6 2 8, 4 3 4 4 1 3 2 5, 1
6 1 6 3
2 3
Maximum value of 13 .
84.
Required Area 1
x 1 2 dx x
0
1
x2 2x x ln 2 0 2 1 2 1 1 0 0 ln 2 ln 2 2 3 1 2 ln 2 85.
D0 1 1
1
4
0 3
3 2 4
13 23 .....n3 1 15.16 . 2 2 n 1 1 2 ..... n 15 n n 1 60 2 n 1 15
86.
Sum
15
n 1
n n 1 n 2 n 1 6
60
15.16.17 60 620 6
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JEE-MAIN-2019 (10th Apr-Second Shift)-PCM-36 n
87.
99 1 1 100 2 n
1 1 2 100 n7
88.
~ ~ s ~ s s r ~ s
s r s ~ s s r s r 89.
Let x 2 t 2xdx dt 1 2 t 1 2 t t .e dt t .e 2t.e t , dt 2 2 1 t 2 .e t t.e t 1.e t .dt 2 t2 t 2e t te t e t t 1 e t 2 2
x4 2 x 2 1 e x C 2 for k 0 1 5 g 1 1 1 2 2 90.
z . w 1
z. w e
i 2
z re
i 2
i 2
z.w e
.ei e
1 and w ei r
i 2
i
i
.e i e 2 i
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FIITJEE Solutions to JEE(Main)-2019 Test Date: 12th April 2019 (First Shift)
PHYSICS, CHEMISTRY & MATHEMATICS Paper - 1 Time Allotted: 3 Hours
Maximum Marks: 360
Please read the instructions carefully. You are allotted 5 minutes specifically for this purpose.
Important Instructions: 1.
The test is of 3 hours duration.
2.
This Test Paper consists of 90 questions. The maximum marks are 360.
3.
There are three parts in the question paper A, B, C consisting of Physics, Chemistry and Mathematics having 30 questions in each part of equal weightage. Each question is allotted 4 (four) marks for correct response.
4.
Out of the four options given for each question, only one option is the correct answer.
5.
For each incorrect response 1 mark i.e. ¼ (one-fourth) marks of the total marks allotted to the question will be deducted from the total score. No deduction from the total score, however, will be made if no response is indicated for an item in the Answer Box.
6.
Candidates will be awarded marks as stated above in instruction No.3 for correct response of each question. One mark will be deducted for indicating incorrect response of each question. No deduction from the total score will be made if no response is indicated for an item in the answer box.
7.
There is only one correct response for each question. Marked up more than one response in any question will be treated as wrong response and marked up for wrong response will be deducted accordingly as per instruction 6 above.
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JEE-MAIN-2019 (12th Apr-First Shift)-PCM-2
PART –A (PHYSICS) 1.
The value of numerical aperature of the objective lens of a microscope is 1.25. If light of
wavelength 5000 A is used, the minimum separation between two points, to be seen as distinct, will be : (A) 0.48 m (B) 0.38 m (C) 0.24 m (D) 0.12 m 2.
The resistive network shown below is connected to a D.C. source of 16 V. The power consumed by the network is 4 Watt. The value of R is:
(A) 8 (B) 16 3.
The figure shows a square loop L of side 5 cm which is connected to a network of resistances. The whole set up is moving towards right with a constant speed of 1 cms–1. At some instant, a part of L is in a uniform magnetic field of 1 T, perpendicular to the plane of the loop. If the resistance of L is 1.7 , the current in the loop at that instant will be close to :
(A) 115 A (C) 60 A 4.
(B) 6 (D) 1
(B) 170 A (D) 150 A
A circular disc of radius b has a hole of radius a at its centre (see figure). If the mass per unit area of the disc varies as 0 , then the radius of gyration of the disc about its axis r passing through the centre is : (A)
ab 3
(B)
a 2 b 2 ab 3
(C)
ab 2
(D)
a 2 b 2 ab 2
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JEE-MAIN-2019 (12th Apr-First Shift)-PCM-3 5.
A man (mass = 50 kg) and his son (mass = 20 kg) are standing on a frictionless surface facing each other. The man pushes his son so that he starts moving at a speed of 0.70 ms–1 with respect to the man. The speed of the man with respect to the surface is : (A) 0.47 ms–1 (B) 0.28 ms–1 –1 (C) 0.14 ms (D) 0.20 ms–1
6.
Two identical parallel plate capacitors, of capacitance C each, have plates of area A, separated by a distance d. The space between the plates of the two capacitors, is filled with three dielectrics, of equal thickness and dielectric constants K1 , K 2 and K 3 . The first capacitor is filled as shown in fig. I, and the second one is filled as shown in fig. II. If these two modified capacitors are charged by the same potential V, the ratio of the energy stored in the two, would be ( E1 refers to capacitor (I) and E2 to capacitor (II)) :
E1 K1K 2K 3 E 9K 1K 2K 3 (B) 1 E 2 (K1 K 2 K 3 )(K 2K 3 K 3K 1 K1K 2 ) E 2 (K1 K 2 K 3 )(K 2K 3 K 3K 1 K1K 2 ) E (K K 2 K 3 )(K 2K 3 K 3K1 K1K 2 ) E (K K 2 K 3 )(K 2K 3 K 3K1 K1K 2 ) (C) 1 1 (D) 1 1 E2 9K 1K 2K 3 E2 K1K 2K 3
(A)
7.
Two moles of helium gas is mixed with three moles of hydrogen molecules (taken to be rigid). What is the molar specific heat of mixture at constant volume? (R = 8.3 J/mol K) (A) 17.4 J/mol K (B) 15.7 J/mol K (C) 19.7 J/mol K (D) 21.6 J/mol K
8.
The stopping potential V0 (in volt) as a function of frequency () for a sodium emitter, is shown in the figure. The work function of sodium, from the data plotted in the figure, will be : (Given : Planck's constant (h) = 6.63 × 10–34 Js, electron charge e = 1.6 × 10–19 C) (A) 1.82 eV (B) 1.66 eV (C) 2.12 eV (D) 1.95 eV
9.
At 40°C, a brass wire of 1 mm is hung from the ceiling. A small mass, M is hung from the free end of the wire. When the wire is cooled down from 40ºC to 20ºC it regains its original length of 0.2 m. The value of M is close to : (Coefficient of linear expansion and Young's modulus of brass are 10–5/ºC and 1011 N/m2, respectively; g = 10 ms–2) (A) 0.5 kg (B) 9 kg (C) 0.9 kg (C) 1.5 kg
10.
A magnetic compass needle oscillates 30 times per minute at a place where the dip is 45°, and 40 times per minute where the dip is 30º. If B1 and B2 are respectively the total magnetic field due to the earth at the two places, then the ratio B1/B2 is best given by : (A) 3.6 (B) 1.8 (C) 2.2 (D) 0.7 FIITJEE Ltd., FIITJEE House, 29-A, Kalu Sarai, Sarvapriya Vihar, New Delhi -110016, Ph 46106000, 26569493, Fax 26513942 website: www.fiitjee.com.
JEE-MAIN-2019 (12th Apr-First Shift)-PCM-4 11.
An excited He+ ion emits two photons in succession, with wavelengths 108.5 nm and 30.4 nm, in making a transition to ground state. The quantum number n, corresponding 1240 eV to its initial excited state is (for photon of wavelength , energy E :) (in nm) (A) n = 4 (B) n = 6 (C) n = 5 (D) n = 7
12.
When M1 gram of ice at –10ºC (specific heat = 0.5 cal g–1°C–1) is added to M2 gram of water at 50ºC, finally no ice is left and the water is at 0ºC. The value of latent heat of ice, in cal g–1 is: 50M2 5M2 (A) 5 (B) 5 M1 M1 50M2 5M1 (C) (D) 50 M1 M2
13.
A shell is fired from a fixed artillery gun with an initial speed u such that it hits the target on the ground at a distance R from it. If t1 and t2 are the values of the time taken by it to hit the target in two possible ways, the product t1t2 is: (A) 2R / g (B) R / 4g (C) R / g (D) R/2g
14.
The transfer characteristic curve of a transistor, having input and output resistance 100 and 100 k respectively, is shown in the figure. The Voltage and Power gain, are respectively :
15.
(A) 5 10 4 , 2.5 106
(B) 5 10 4 , 5 106
(C) 5 10 4 , 5 105
(D) 2.5 104 , 2.5 106
Shown in the figure is a shell made of a conductor. It has inner radius a and outer radius b, and carries charge Q. At its centre is a dipole P as shown. In this case : (A) Surface charge density on the inner surface of the shell is zero everywhere. (B) Electric field outside the shell is the same as that of a point charge at the centre of the shell. (C) Surface charge density on the inner surface is uniform and (Q / 2) equal to 4a2 (D) Surface charge density on the outer surface depends on P
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JEE-MAIN-2019 (12th Apr-First Shift)-PCM-5 16.
Which of the following combinations has the dimension of electrical resistance ( 0 is the permittivity of vacuum and 0 is the permeability of vacuum)? 0 (A) (B) 0 0 0 (C)
17.
0 0
0 0
(D)
A galvanometer of resistance 100 has 50 divisions on its scale and has sensitivity of 20 A / division. It is to be converted to a voltmeter with three ranges, of 0–2 V, 0–10 V and 0–20 V. The appropriate circuit to do so is : (A)
(B)
(C)
(D)
18.
In a double slit experiment, when a thin film of thickness t having refractive index is introduced in front of one of the slits, the maximum at the centre of the fringe pattern shifts by one fringe width. The value of t is ( is the wavelength of the light used) : (A) (B) 2( 1) ( 1) 2 (C) (D) (2 1) ( 1)
19.
A progressive wave travelling along the positive x-direction is represented by y(x, t) = A sin (kx – t + ). Its snapshot at t = 0 is given in the figure:
For this wave, the phase is : (A) (C)
(B) 2
2
(D) 0
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JEE-MAIN-2019 (12th Apr-First Shift)-PCM-6 20.
21.
A uniform rod of length is being rotated in a horizontal plane with a constant angular speed about an axis passing through one of its ends. If the tension generated in the rod due to rotation is T(x) at a distance x from the axis, then which of the following graphs depicts it most closely?
(A)
(B)
(C)
(D)
A point dipole p p0 x is kept at the origin. The potential and electric field due to this dipole on the y-axis at a distance d are, respectively : (Take V = 0 at infinity)
p
p (A) , 4 0 d2 4 0 d3 p p (C) , 4 0 d2 4 0 d3
22.
p (B) 0, 4 0 d3 p (D) 0, 4 0 d3
An electromagnetic wave is represented by the electric field E E0 n sin[t (6y 8z)]. Taking unit vectors in x, y and z directions to be i, j, k , the direction of propogation s , is
: 3 j 4k (A) S 5 4k 3 j (C) S 5
23.
4 j 3k (B) S 5 3i 4j (D) S 5
A submarine (A) travelling at 18 km/hr is being chased along the line of its velocity by another submarine (B) travelling at 27 km/hr. B sends a sonar signal of 500 Hz to detect A and receives a reflected sound of frequency . The value of is close to : (Speed of sound in water = 1500 ms–1) (A) 499 Hz (B) 502 Hz (C) 504 Hz (D) 507 Hz
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JEE-MAIN-2019 (12th Apr-First Shift)-PCM-7
24.
The trajectory of a projectile near the surface of the earth is given as y = 2x – 9x2. If it were launched at an angle 00 with speed v0 then (g = 10 ms–2) : 1 5 2 3 1 1 (A) 0 cos1 (B) 0 cos1 and v 0 ms and v 0 ms 3 5 5 5 2 3 1 (C) 0 sin1 and v 0 ms 5 5
1 5 1 (D) 0 sin1 and v 0 ms 3 5
25.
A thin ring of 10 cm radius carries a uniformly distributed charge. The ring rotates at a constant angular speed of 40 rad s–1 about its axis, perpendicular to its plane. If the magnetic field at its centre is 3.8 × 10–9 T, then the charge carried by the ring is close to ( 0 4 107 N / A 2 ). (A) 2 × 10–6 C (B) 7 × 10–6 C –5 (C) 4 × 10 C (D) 3 × 10–5 C
26.
The truth table for the circuit given in the fig is :
(A)
(C)
27.
A B Y
A B Y
0
0
1
0
0
0
0
1
0
0
1
0
1 1
0 1
0 0
1 1
0 1
1 1
(B)
A B Y
A B Y
0
0
1
0
0
1
0
1
1
0
1
1
1 1
0 1
1 1
1 1
0 1
0 0
(D)
A sample of an ideal gas is taken through the cyclic process abca as shown in the figure. The change in the internal energy of the gas along the path ca is –180 J. The gas absorbs 250 J of heat along the path ab and 60 J along the path bc. The work done by the gas along the path abc is : (A) 120 J (B) 100 J (C) 140 J (D) 130 J
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JEE-MAIN-2019 (12th Apr-First Shift)-PCM-8 28.
A concave mirror has radius of curvature of 40 cm. It is at the bottom of a glass that has water filled up to 5 cm (see figure). If a small particle is floating on the surface of water, its image as seen, from directly above the glass, is at a distance d from the surface of water. The value of d is close to : (Refractive index of water = 1.33) (A) 13.4 cm (B) 8.8 cm (C) 6.7 cm (D) 11.7 cm
29.
To verify Ohm's law, a student connects the voltmeter across the battery as, shown in the figure. The measured voltage is plotted as a function of the current, and the following graph is obtained: If V0 is almost zero, identify the correct statement: (A) The potential difference across the battery is 1.5 V when it sends a current of 1000 mA. (B) The emf of the battery is 1.5 V and the value of R is 1.5 (C) The emf of the battery is 1.5 V and its internal resistance is 1.5 (D) The value of the resistance R is 1.5
30.
A person of mass M is, sitting on a swing of length L and swinging with an angular amplitude 0. If the person stands up when the swing passes through its lowest point, the work done by him, assuming that his centre of mass moves by a distance ( < < L), is close to : (A) Mg(1 0 2 ) (B) Mg(1 02 ) (C) Mg
2 (D) Mg 1 0 2
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JEE-MAIN-2019 (12th Apr-First Shift)-PCM-9
PART –B (CHEMISTRY) 31.
The mole fraction of a solvent in aqueous solution of a solute is 0.8.The molality (in mol kg–1) of the aqueous solution is (A) 13.88 10–2 (B) 13.88 10–1 (C) 13.88 (D) 13.88 10–3
32.
The major products of the following reaction are:
(A)
(B)
(C)
(D)
33.
The correct statement among the following is (A) (SiH3 )3 N is planar and less basic than (CH3 )3 N. (B) (SiH3 )3 N is planar and more basic than (CH3 )3 N. (C) (SiH3 )3 N is pyramidal and less basic than (CH3 )3 N. (D) (SiH3 )3 N is pyramidal and more basic than (CH3 )3 N.
34.
The complex ion that will lose its crystal field stabilization energy upon oxidation of its metal to +3 state is
2
(A) Ni(phen)3
2
(C) Co(phen)3
2
(B) Zn(phen)3
2
(D) Fe(phen)3
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JEE-MAIN-2019 (12th Apr-First Shift)-PCM-10 35.
The correct set of species responsible for the photochemical smog is : (A) NO, NO2, O3 and hydrocarbons (B) CO2, NO2 SO2 and hydrocarbons (C) N2, NO2 and hydrocarbons (D) N2, O2, O3 and hydrocarbons
36.
The major product(s) obtained in the following reaction is/ are:
37.
38.
(A)
(B)
(C)
(D)
The major product of the following addition reaction is: Cl2 /H2O H3 C CH CH2 (A)
(B)
(C)
(D)
The correct sequence of thermal stability of the following carbonates is : (A) BaCO3 SrCO3 CaCO3 MgCO3 (B) BaCO3 CaCO3 SrCO3 MgCO3 (C) MgCO3 CaCO3 SrCO3 BaCO3
(D) MgCO3 SrCO3 CaCO3 BaCO3
39.
The group number, number of valence electrons and valency of an element with atomic number 15, respectively, are: (A) 15, 5 and 3 (B) 15, 6 and 2 (C) 16, 5 and 2 (D) 16, 6 and 3
40.
Peptization is a : (A) process of converting soluble particles to form colloidal solution (B) process of converting precipitate into colloidal solution (C) process of converting a colloidal solution into precipitate (D) process of bringing colloidal molecule into solution
41.
The metal that gives hydrogen gas upon treatment with both acid as well as base is: (A) iron (B) magnesium (C) zinc (D) mercury
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JEE-MAIN-2019 (12th Apr-First Shift)-PCM-11 42.
The electrons are more likely to be found:
(A) in the region a and b (C) only in the region a 43.
(B) in the region a and c (D) only in the region c
An example of a disproportionation reaction is : (A) 2KMnO 4 K 2MnO 4 MnO2 O2 (B) 2NaBr Cl2 2naCl Br2 (C) 2 CuBr CuBr2 Cu (D) 2MnO 4 10 16H 2Mn2 5 2 8 H2O
44.
5 moles of AB2 weigh 125 10–3 kg and 10 moles of A2B2 weigh 300 10–3 kg. The molar mass of A(MA) and molar mass of B(MB) in kg mol–1 are : (A) MA = 50 10–3 and MB = 25 10–3 (B) MA = 10 10–3 and MB = 5 10–3 –3 –3 (C) MA = 5 10 and MB = 10 10 (D) MA = 25 10–3 and MB = 50 10–3
45.
The basic structural unit of feldspar, zeolites, mica and asbestos is :
(A)
(C)
SiO2
(B)
(SiO3 )2
(D)
(SiO 4 )4
46.
Enthalpy of sublimation of iodine is 24 cal g–1 at 200°C. If specific heat of I2(s) and I2 (vap) are 0.055 and 0.031 cal g–1K–1 respectively, then enthalpy of sublimation of iodine at 250°C in cal g–1 is : (A) 2.85 (B) 11.4 (C) 5.7 (D) 22.8
47.
Which of the following is a thermosetting polymer? (A) PVC (B) Bakelite (C) Buna-N (D) Nylon 6
48.
An element has a face-centred cubic (fcc) structure with a cell edge of a. The distance between the centres of two nearest tetrahedral voids in the lattice is: a (A) (B) 2a 2 3 (C) a (D) a 2
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JEE-MAIN-2019 (12th Apr-First Shift)-PCM-12 49.
An organic compound 'A' is oxidized with Na2O2 followed by boiling with HNO3.The resultant solution is then treated with ammonium molybdate to yield a yellow precipitate. Based on above observation, the element present in the given compound is : (A) Phosphorus (B) Sulphur (C) Nitrogen (D) Fluorine
50.
The idea of froth floatation method came from a person X and this method is related to the process Y of ores. X and Y, respectively, are: (A) washer woman and concentration (B) fisher woman and concentration (C) fisher man and reduction (D) washer man and reduction
51.
In the following reaction; xA yB d[A] d[B] log10 log10 0.3010 dt dt ‘A’ and ‘B’ respectively can be : (A) C2H2 and C6H6 (C) N2O4 and NO2
(B) n-Butane and Iso-butane (D) C2H4 and C4H8
52.
Glucose and Galactose are having identical configuration in all the positions except position. (A) C–4 (B) C–3 (C) C–5 (D) C–2
53.
Which of the following statements is not true about RNA? (A) It has always double stranded -helix structure (B) It is present in the nucleus of the cell (C) It controls the synthesis of protein (D) It usually does not replicate
54.
What is the molar solubility of Al(OH)3 in 0.2 M NaOH solution? Given that, solubility product of Al(OH)3 = 2.4 10–24 : (A) 3 10 19 (B) 12 10 21 23 (C) 12 10 (D) 3 10 22
55.
The increasing order of the pKb of the following compound is :
(A)
(B)
(C)
(D)
(A) (c) < (a) < (d) < (b) (C) (a) < (c) < (d) < (b)
(B) (b) < (d) < (a) < (c) (D) (b) < (d) < (c) < (a)
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JEE-MAIN-2019 (12th Apr-First Shift)-PCM-13 56.
Given : Co3 e Co 2 ; E 1.81V Pb4 2e Pb2 ; E 1.67 V Ce 4 e Ce3 ; E 1.61V Bi3 3e Bi ; E 0.20V Oxidizing power of the species will increase in the order (A) Ce4 Pb 4 Bi3 Co 3 (B) Co3 Pb 4 Ce 4 Bi3 3 4 4 3 (C) Bi Ce Pb Co (D) Co3 Ce4 Bi3 Pb4
57.
But-2-ene on reaction with alkaline KMnO4 at elevated temperature followed by acidification will give : (A) One molecule of CH3CHO and one molecule of CH3COOH (B)
2 molecules of CH3CHO
(C)
2 molecules of CH3COOH
(D) 58.
The major product of the following reaction is :
(A)
(B)
(C)
(D)
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JEE-MAIN-2019 (12th Apr-First Shift)-PCM-14 59.
60.
Complete removal of both the axial ligands (along the z-axis) from an octahedral complex leads to which of the following splitting patterns? (relative orbital energies not on scale).
(A)
(B)
(C)
(D)
An ideal gas is allowed to expand from 1 L to 10 L against a constant external pressure of 1bar. The work done in kJ is: (A) – 9.0 (B) – 0.9 (C) – 2.0 (D) + 10.0
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JEE-MAIN-2019 (12th Apr-First Shift)-PCM-15
PART–C (MATHEMATICS) 61.
Let P be the point of intersection of the common tangents to the parabola y2 = 12x and the hyperbola 8x2 – y2 = 8. If S and S' denote the foci of the hyperbola where S lies on the positive x-axis then P divides SS' in a ratio: (A) 2:1 (B) 13:11 (C) 5:4 (D) 14:13
62.
If three of the six vertices of a regular hexagon are chosen at random, then the probability that the triangle formed with these chosen vertices is equilateral is : 3 1 (A) (B) 10 5 1 3 (C) (D) 10 20
63.
Let f : R R be a continuously differentiable function such that f(2) = 6 and f '(2)
f (x)
6
4t 3 dt (x 2)g(x), then lim g(x) is equal to : x2
(A) 24 (C) 12 64.
1 . If 48
(B) 18 (D) 36
2x3 1 x 4 x dx is equal to : (Here C is a constant of integration)
The integral
x3 1 1 (A) loge C 2 x2
(C) loge
x3 1 x
C
2
x3 1 1 (B) loge C 2 x3
(D) loge
x3 1 x2
C
65.
The coefficient of x18 in the product (1 x)(1 x)10 (1 x x 2 )9 is : (A) 84 (B) 126 (C) –126 (D) –84
66.
The number of ways of choosing 10 objects out of 31 objects of which 10 are identical and the remaining 21 are distinct, is : (A) 220 (B) 220 + 1 21 (C) 2 (D) 220 – 1
67.
Let a random variable X have a binomial distribution with mean 8 and variance 4. If k P(X 2) 16 , then k is equal to : 2 (A) 17 (B) 137 (C) 1 (D) 121
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JEE-MAIN-2019 (12th Apr-First Shift)-PCM-16
68.
The equation z i z 1 ,i 1, represents : 1 2 (C) a circle of radius 1
(A) a circle of radius
69.
(B) the line through the origin with slope 1 (D) the line through the origin with slope –1
If m is the minimum value of k for which the function f(x) x kx x 2 is increasing in the interval [0,3] and M is the maximum value of f in [0, 3] when k = m, then the ordered pair (m, M) is equal to : (A) (5, 3 6 ) (B) (4, 3 2) (C) (3, 3 3)
(D) (4, 3 3 )
70.
If the data x1, x2, ...., x10 is such that the mean of first four of these is 11, the mean of the remaining six is 16 and the sum of squares of all of these is 2,000; then the standard deviation of this data is : (A) 2 2 (B) 2 (C) 4 (D) 2
71.
If the area (in sq. units) of the region then a – b is equal to : 10 (A) 3 8 (C) 3
72.
73.
(x, y) : y
2
4x, x y 1, x 0, y 0 is a 2 b,
(B) 6 (D)
2 3
1 Consider the differential equation, y 2 dx x dy 0 . If value of y is 1 when x = 1, y then the value of x for which y = 2, is : 3 1 1 (A) e (B) 2 2 e 3 1 5 1 (C) (D) 2 2 e e
If and are the roots of the equation 375x2 – 25x – 2 = 0, then n
n
lim r lim r is equal to :
n
r 1
1 12 7 (C) 116
(A)
n
r 1
29 358 21 (D) 346
(B)
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JEE-MAIN-2019 (12th Apr-First Shift)-PCM-17 74.
75.
76.
77.
78.
79.
A 2 m ladder leans against a vertical wall. If the top of the ladder begins to slide down the wall at the rate 25 cm/ sec., then the rate (in cm/sec.) at which the bottom of the ladder slides away from the wall on the horizontal ground when the top of the ladder is 1 m above the ground is : 25 (A) 25 (B) 3 25 (C) 25 3 (D) 3 5 5 The number of solutions of the equation 1 + sin4x = cos23x, x , is : 2 2 (A) 3 (B) 4 (C) 5 (D) 7
dy d2 y If ey + xy = e, the ordered pair , 2 at x = 0 is equal to : dx dx 1 1 1 1 (A) , 2 (B) , 2 e e e e 1 1 1 1 (C) , 2 (D) , 2 e e e e 12 3 The value of sin1 sin 1 is equal to : 13 5 33 63 (A) cos1 (B) sin1 65 65 9 56 (C) cos1 (D) sin1 2 2 65 65
If the normal to the ellipse 3x2 + 4y2 = 12 at a point P on it is parallel to the line, 2x + y = 4 and the tangent to the ellipse at P passes through Q(4, 4) then PQ is equal to : 157 5 5 (A) (B) 2 2 221 61 (C) (D) 2 2 Let a 3i 2j 2k and b i 2j 2k be two vectors. If a vector perpendicular to both the vectors a b and a b has the magnitude 12 then one such vector is: (A) 4 2i 2j k (B) 4 2i 2j k
(C) 4 2i 2j k 80.
(D) 4 2i 2j k
The equation y = sinx sin(x + 2) – sin2(x+1) represents a straight line lying in : (A) first, third and fourth quadrants (B) first, second and fourth quadrants (C) third and fourth quadrants only (D) second and third quadrants only
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JEE-MAIN-2019 (12th Apr-First Shift)-PCM-18
81.
1 x2 3 For x 0, , let f(x) x, g(x) tan x and h(x) = . If (x) ((hof )og)(x), then 1 x2 2 is equal to : 3 11 (A) tan (B) tan 12 12 5 7 (C) tan (D) tan 12 12
82.
If the angle of intersection at a point where the two circles with radii 5 cm and 12 cm intersect is 90°, then the length (in cm) of their common chord is : 13 120 (A) (B) 2 13 13 60 (C) (D) 5 13
83.
If
84.
Let Sn denote the sum of the first n terms of an A.P.. If S4 = 16 and S6 = –48, then S10 is equal to : (A) –410 (B) –260 (C) –320 (D) –380
85.
86.
87.
2 0
cot x dx m( n),thenm n is equal to cot x cos ecx 1 (A) 1 (B) 2 1 (C) (D) –1 2
5 2 1 If B = 0 2 1 is the inverse of a 3 3 matrix A, then the sum of all values of for 3 1 which det (A) + 1 = 0, is : (A) 0 (B) –1 (C) 1 (D) 2 x 2 y 1 z 1 intersects the plane 2x + 3y – z + 13 = 0 at a point P and 3 2 1 the plane 3x + y + 4z = 16 at a point Q, then PQ is equal to : (A) 2 14 (B) 14 (C) 2 7 (D) 14
If the line
j k and i k is If the volume of parallelepiped formed by the vectors i j k, minimum, then is equal to : 1 (A) 3 (B) 3 1 (C) (D) 3 3 FIITJEE Ltd., FIITJEE House, 29-A, Kalu Sarai, Sarvapriya Vihar, New Delhi -110016, Ph 46106000, 26569493, Fax 26513942 website: www.fiitjee.com.
JEE-MAIN-2019 (12th Apr-First Shift)-PCM-19
88.
2 3 If A is a symmetric matrix and B is a skew-symmetrix matrix such that A + B = , 5 1 then AB is equal to : 4 2 4 2 (A) (B) 1 4 1 4 4 2 (C) 1 4
4 2 (D) 1 4
89.
If the truth value of the statement p (~ q r) is false (F), then the truth values of the statement p, q, r are respectively : (A) T, T, F (B) F, T, T (C) T, F, T (D) T, F, F
90.
For x R,let [x] denote the greatest integer x, then the sum of the series 1 1 2 1 1 1 99 3 3 100 3 100 3 100 (A) –135 (B) –153 (C) –133 (D) –131
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JEE-MAIN-2019 (12th Apr-First Shift)-PCM-20
JEE (Mains) – 2019 ANSWERS PART A – PHYSICS 1. 5. 9. 13. 17. 21. 25. 29.
C D B A C D D C
2. 6. 10. 14. 18. 22. 26. 30.
A B D A B A D A
3. 7. 11. 15. 19. 23. 27.
B A C B A B D
4. 8. 12. 16. 20. 24. 28.
B B A D D A B
34. 38. 42. 46. 50. 54. 58.
D C B D A D C
64. 68. 72. 76. 80. 84. 88.
C B C C C C B
PART B – CHEMISTRY 31. 35. 39. 43. 47. 51. 55. 59.
C A A C B D C B
32. 36. 40. 44. 48. 52. 56. 60.
C C B C A A C B
33. 37. 41. 45. 49. 53. 57.
A B C D A A C
PART C – MATHEMATICS 61. 65. 69. 73. 77. 81. 85. 89.
C A D A D A C A
62. 66. 70. 74. 78. 82. 86. 90.
C A B D B B A C
63. 67. 71. 75. 79. 83. 87.
B B B C A D Bonus
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JEE-MAIN-2019 (12th Apr-First Shift)-PCM-21
HINTS AND SOLUTIONS PART A – PHYSICS 1.
Numerical aperature of the microscope is given as 0.61 NA d Where d = minimum sparaton between two points to be seen as distinct d
0.61 (0.61) (5000 10 m10 ) = 2.4 × 10–7 m NA 1.25 = 0.24 m
2.
162 4 8R R = 8 P
3.
Since it is a balanced wheatstone bridge, its equivalent resistance =
4 3
= Bv = 5 × 10–4 V So total resistance 4 R 1.7 3 3 i 166 A 170 A R 4.
dI = (dm)r2 = ( dA)r2 = 0 2dr r 2 = (0 20r2dr r b
I = DI 0 2r 2 dr a
b3 a3 = 0 2 3 m = dm dA b
= 0 2 dr a
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JEE-MAIN-2019 (12th Apr-First Shift)-PCM-22
5.
0 = 50V1 – 20V2 and V1 + V2 = 0.7 V1 = 0.2 6.
30 AK1 d 3 AK C2 0 2 d 30 AK 3 C3 d C1
1 1 1 1 Ceq C1 C2 C3
Ceq
30 AK 1K 2K 3 d(K1K 2 K 2K 3 K 3K1 )
…(i)
0K1A 3d 0K 2 A C2 3d K A C3 0 3 3d C1
Ceq C1 C2 C3
=
0 A (K1 K 2 K 3 ) 3d
…(ii)
Now, 1 C V2 9K1K 2K 3 E1 2 eq 1 E2 (K1 K 2 K 3 ) (K 1K 2 K 2K 3 K 3K1 ) Ceq V 2 2
7.
8.
n1f1 n2 f2 2 3 3 5 21 n1 n2 5 5 fR 21 R Cv = = 17.4 J/mol K 5 5 2 fmix
hv = + ev0 h v0 e e v0 is zero for = 4 × 1014 Hz
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JEE-MAIN-2019 (12th Apr-First Shift)-PCM-23
hv e e = h 6.63 10 34 4 1014 = = 1.66 eV 1.6 10 19 0
9.
Ay Mg Mg = (Ay)T = 2 It is closest to 9.
10.
1 B1 cos 45o 2 I o f1 B1 cos 45 f2 B2 cos 30o
f1
11.
12.
1 1 1 R 2 2 z2 n m 1 1 1 R 2 2 22 1085 n m m=2 n=5 Heat lost = Heat gain
M2 × 1 × 50 = M1 × 0.5 × 10 + M1Lf
Lf
= 13.
1 B 2 cos 30o 2 I B1 0.7 B2
f2
50M2 5M1 M1 50M2 5 M1
Range will be same for time t1 and t2, so angles of projection will be ‘’ & ‘90° – ’
t1
2u sin 2u sin(90o ) u2 sin 2 t2 and R g g g
t1t 2
4u2 sin cos 2 2u2 sin cos g2 g g 2R = g
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JEE-MAIN-2019 (12th Apr-First Shift)-PCM-24
14.
I R 5 10 3 Vgain C out = 103 6 100 10 IB Rin 1 108 5 10 4 20 I Pgain C (Vgain ) Ib 5 103 = (5 10 4 ) 6 100 10 = 2.5 × 106
15.
Total charge of dipole = 0, so charge induced on outside surface = 0. But due to non uniform electric field of dipole, the charge induced on inner surface is non zero and non uniform. So, for any observer outside the shell, the resultant electric field is due to Q uniformly distributed on outer surface only and it is equal to. E
16.
KQ r2
[0] = M–1 L–3 T4 A2 [0] = M L T–2 A–2 [R] = M L2 T–3 A–2
[R] 0 0 17.
20 × 50 × 10–6 = 10–3 Amp. 2 V1 3 100 R1 10 1900 = R1 10 V2 3 (2000 R 2 ) 10 R2 = 8000 20 V3 3 10 103 R3 = 10 × 103 R3 10
18.
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JEE-MAIN-2019 (12th Apr-First Shift)-PCM-25 x = ( – 1)t = 1 for one maximum shift t 1 19.
y = A sin (kx – wt + ) at x = 0, t = 0 and slope is negative = 20.
T
x
x x
dm 2 x
x
xx
m dx2 x T
m2 2 ( x2 ) 2 m 2 2 T ( x2 ) 2
=
21.
22.
V=0
KP E 3 r p = 40 d3
E E0nˆ sin( t (6y 8z) = E0nˆ sin(t k r ) where r xiˆ yjˆ zkˆ and k r = 6y – 8z k 6 ˆj 8kˆ direction of propagation sˆ kˆ 3ˆj 4kˆ = 5
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JEE-MAIN-2019 (12th Apr-First Shift)-PCM-26
23.
f0 = 500 Hz frequency received by B again = 1500 (B) (A) & 7.5 m/s 5 m/sec 1500 7.5 1500 5 f2 f0 = 502 Hz 1500 5 1500 7.5 24.
25.
26.
Equation of trajectory is given as y = 2x – 9x2 …(A) Comparing with equation: g y = x tan – x2 …(B) 2 2 2u cos We get, tan = 2 1 cos 5 g Also, =9 2 2u cos2 10 25 u 2 ; u2 2 9 1 29 5 5 u m/s 3
0i 0 q 2R 2R 2 q = 3 × 10–5 C B
C=A+B and y = A.C
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JEE-MAIN-2019 (12th Apr-First Shift)-PCM-27 27.
28.
Light incident from particle P will be reflected at mirror.
R 20 cm 2 1 1 1 20 ; v1 cm v u f 3 This image will act as object for light getting refracted at water surface. 20 35 So, object distance d 5 cm 3 3 Below water surface. After refraction, final image is at 35 1 d d 2 1 3 4 / 3 35 8.75 cm 4 8.8 cm u 5cm, f m
29.
V = E – Ir When V = V0 = 0 0 = E – Ir E=r When I = 0, V = E = 1.5 V r = 1.5 30.
Angular momentum conservation MV0L = MV1(L – )
L V1 V0 L wg + wp = KE
mg w p
1 m V12 V02 2
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JEE-MAIN-2019 (12th Apr-First Shift)-PCM-28
w p mg
L 2 1 L
1 mV02 2
1 = mg mV02 2
2 L 1 1 L
Now, << L By, Binomial approximation 1 = mg mV02 2
= mg
2 L 1 1 L
1 2 mV02 2 L
W P = mg mV02
L
g Here, V0 = maximum velocity = × A = ( L) L 0
So, w p mg m 0 gL
2
= mg 1 0
2
L
PART B – CHEMISTRY
31.
Xsolvent = 0.8 = 8/10 NTotal= 10, nsolutent = 8, nsolute = 2 Wt of solvent = 8 18 2 1000 Molality 8 18
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JEE-MAIN-2019 (12th Apr-First Shift)-PCM-29 32.
33. 34.
p-d bonding in N(SiH3)3. When Fe2+ oxidizes to Fe3+ 3d X
35.
Smog photochemical is NO, NO2, O3.
36.
37.
38.
Smaller the size of cation, more polarisability high thermal decomposition of carbonates.
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JEE-MAIN-2019 (12th Apr-First Shift)-PCM-30
39.
15 1s2, 2s2, 2p6, 3s2, 3p3 Period- III, Group-V
40.
Peptization is process to form collidal solution by adding peptizing agent.
41.
HCl Zn ZnCl2 H2 NaOH Na 2 ZnO2 H2
42.
At a & c, probability of finding electron is maximum 43.
2 Cu Cu2 Cu
44.
Mol. wt is of 1 mol AB2 A + 2B = 25 A2B2 2A + 2B = 30
45.
Basic unit is silicate (SiO 4 )4
46.
CP
47.
Thermosetting are cross-linked polymer.
48.
Two nearest tetrahedral voids are along one body-diagonal.
49.
Canary yellow ppt comes in test of PO34 ion.
50.
Concentration(on dressing), i.e. washing of ore.
51.
H2 H1 T2 T1
1 dA 1 dB x dt y dt d A x dB log10 log10 log y dt dt
x
log y log 0.3
x 2 y
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JEE-MAIN-2019 (12th Apr-First Shift)-PCM-31 6
52.
CH2OH O OH
5
Glucose:
4
OH
1 2
OH 3
OH 6
They differ at C-4
CH2OH
OH Galactose:
4
O OH
OH 3
1 2
OH 53.
Not always double stranded -helix.
54.
3 Al OH3 Al 3OH
2.4 10
-24
s (3s 0.2) = s(3s + 0.2)3
55.
CH3O–, CH3–, H is electron donating, But –NO2 is electron withdrawing group.
56.
Lower the standard reduction potential, more the ability to get reduced higher the oxidizing power.
57.
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JEE-MAIN-2019 (12th Apr-First Shift)-PCM-32 58.
59.
Ligand filed exert mass repulsion along x, y axis as compared to Z-axis so dx 2 y 2 and dxx will have increase in energy y
60.
W = - Pext(V2 – V1)
PART C – MATHEMATICS 61.
Tangents y 2 12x y 2x
3 m
x2 y2 1 y mx m2 8 1 8 Common tangent gives 3 m m2 8 m4 8m2 9 0 m 3 1 y 3x 1 P ,0 3 y 3x 1
x2 y2 1 1 8 e3 ae 3 S 3,0
S ' 3, 0
P divides SS’ in 5 : 4. 62.
Only two equilateral triangle are possible A1A 3 A 5 and A 2 A 5 A 6
2 2 1 6 C3 20 10
A4 A5
A3
A6
A2 A1
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JEE-MAIN-2019 (12th Apr-First Shift)-PCM-33 f x
4t 3 dt
63.
lim
6
x2
lim
x2 4.f 3 x .f ' x
1 4f 2 f ' 2 18 x2
3
64.
65.
2x 3 1 x 4 x dx 1 2x 2 2 x1 dx x x 1 x2 t x 1 2x x 2 dx dt dt t n t C 1 n x 2 C x 9
10
1 x 1 x 1 x x 2 3 9
1 x 1 x 2
9
66.
C6 84
Since
21
C0 ..... 21C10 21C11 ....... 21C21 221
but we have to select 10 objects and
67.
21
21
C0 ..... 21C10 21 C11 ....... 21C21
C0 ..... 21C10 220
np 8 npq 4 1 1 q p 2 2 n 16 16
1 p x r 16 Cr 2 16 C0 16 C1 16 C 2 p x 2 216 137 16 2
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JEE-MAIN-2019 (12th Apr-First Shift)-PCM-34 68.
z i z 1 gives y x
69.
f x x kx x 2 f ' x
3kx 4x 2
2 kx x 2 For increasing f x 0
3kx 4x 2 0 4x 2 3kx 0 3k 4x x 0 4 3k 3 0 4 k4
kx x 2 0 x 2 kx 0 x x k 0 so x 0, 3 ve x 3 minimum value of k is m 4
f x x kx x 2
3 4 3 32 3 3, M 3 3 70.
x1 ...... x 4 44 x 5 ...... x10 96
x 14, xi 140
x Variance
2 i
2
x 4
n Standard deviation = 2 71.
x, y : y
2
4x, x y 1, x 0, y 0
32 2
A
2 x dx
0
2 x 3/2
1 1 3 2 2 2
1 3 2 2
y2 = 4x
32 2
1 0 2 22 2 22 3/2 2 8 2 10 3 3 8 10 a ,b 3 3 a b 6 72.
1
y 2 dx xdy
O
32 2
1
dy y
dx x 1 2 3 dy y y
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JEE-MAIN-2019 (12th Apr-First Shift)-PCM-35 1
IF e
y 2 dy
1
e
1
e y .x e y . 1 y
1 y
xe e
C x
73.
1 y
1 dy C y3
1 y
e C y
1 e
3 1 when y = 2 2 e
375x 2 25x 2 0 25 2 , 375 375 2 ...upto inf inite terms 2 ..........upto inf inite terms
74.
1 1 1 12
dy x2 y2 4 25 dt dx dy x y 0 dt dt dx 3 1 25 0 dt dx 25 cm / sec dt 3 4
A 25 cm/s 2
y
B O
x
2
75.
1 sin x cos 3x sin x 0 and cos3x 1 0,2, 2, ,
76.
ey xy e differentiate w.r.t. x dy dy ey x y0 dx dx dy dy ey x y0 dx dx dy dy 1 x e y y, again differentiate w.r.t.x dx dx 0,1 e d2 y dy y dy d2 y dy dy .e . x. 0 dx 2 dx dx dx 2 dx dx 2 d2 y dy y dy x ey 0 .e 2 dx 2 dx dx
ey .
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JEE-MAIN-2019 (12th Apr-First Shift)-PCM-36
d2 y 1 1 e 2 0 dx 2 e1 e 2 dy 1 2 2 dx e e
77.
12 3 sin1 sin1 13 5
sin1 x 1 y 2 y 1 x 2
33 56 56 sin1 cos1 sin1 65 65 2 65 78.
3x 2 4y 2 12 x2 y2 1 4 3 x 2 cos , y 3 sin
Equation of normal is
a2 x b2 y a 2 b2 x1 y1
2x sin 3 cos y sin cos 2 Slope tan 2 tan 3 3 Equation of tangent is it passes through (4, 4) 3x cos 2 3 sin y 6 12 cos 8 3 sin 6 2
1 3 cos ,sin 120o 2 2
Hence point P is 2cos120o , 3 sin120o
79.
2 P 1, , Q 4, 4 2 5 5 PQ 2 ab ab
ˆi
ˆj
3
1
kˆ
2 1 2 2 3 2
2
2 8iˆ 8ˆj 4kˆ
2iˆ 2ˆj kˆ Required vector 12 3
4 2iˆ 2ˆj kˆ
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JEE-MAIN-2019 (12th Apr-First Shift)-PCM-37
80.
2y 2 sin x sin x 2 2 sin2 x 1
2y cos 2 cos 2x 2 1 cos 2x 2 cos 2 1
2y 2 sin2 y sin2
81.
1 2
1 0 2
f x x, g x tan x,h x
1 x2 1 x2
fog x tan x
hofog x h
tan x
1 tan x 1 tan x
tan x 4 x tan x 4 tan tan tan 12 3 4 3 12 11 tan tan 12 12 82.
Let length of common chord 2x 25 x 2 144 x 2 13 12 5 After solving x 13 120 2x 13 / 2
83.
0
/2
0
/ 2
12
5 x x 5
12
cot x dx cot x cos ec x
cos x cos x 1
1
1 2 sec 0
2
x 1 2 x 2cos2 2
2cos2
x dx 2
x2 x tan 2 0
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JEE-MAIN-2019 (12th Apr-First Shift)-PCM-38
1 2 2 mn 1 84.
m
1 , n 2 2
2 2a 3d 16 3 2a 5d 48 2a 3d 8 2a 5d 16 d 12 S10 5 44 9 12 320
85.
B 5 5 2 2
2 2 2 25 1 A 0 2 12 0 Sum = 1 86.
x 2 y 1 z 1 3 2 1 x 3 2, y 2 1, z 1 Intersection with plane 2x 3y z 13 0 2 3 2 3 2 1 1 13 0 13 13 0 1 P 1, 3, 2 Intersection with plane 3x y 4z 16 3 3 2 2 1 4 1 16 1 Q 5, 1,0 PQ 62 42 22 56 2 14
1 1 87.
Volume of paralleopiped 0 1
0 1
f 3 1 Its graphs as follows
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JEE-MAIN-2019 (12th Apr-First Shift)-PCM-39
2.0 1.5 1.0 0.5
1
1 3
3 where 1.32 For minimum value of volume of paralelopiped and corresponding value of ; the minimum value is zero, cubic always has at least one real root. Hence answer to the question must be root of cubic 3 1 0 . None of the options satisfies the cubic. Hence Question must be Bonus.
88.
A A ',B B ' 2 3 A B ……….(1) 5 1 2 5 A ' B ' 3 1 2 5 A B ……….(2) 3 1 After adding equation (1) and (2) 2 4 0 1 A , B 4 1 1 0 4 2 AB 1 4
89.
P ~ q r ~ p ~ q r
~ p F p T ~ q F q T ~ r F r F
90.
1 1 1 1 66 1 67 1 99 3 3 100 .... 3 100 3 100 ... 3 100 133 1 67
2 33
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FIITJEE Solutions to JEE(Main)-2019 Test Date: 12th April 2019 (Second Shift)
PHYSICS, CHEMISTRY & MATHEMATICS Paper - 1 Time Allotted: 3 Hours
Maximum Marks: 360
Please read the instructions carefully. You are allotted 5 minutes specifically for this purpose.
Important Instructions: 1.
The test is of 3 hours duration.
2.
This Test Paper consists of 90 questions. The maximum marks are 360.
3.
There are three parts in the question paper A, B, C consisting of Physics, Chemistry and Mathematics having 30 questions in each part of equal weightage. Each question is allotted 4 (four) marks for correct response.
4.
Out of the four options given for each question, only one option is the correct answer.
5.
For each incorrect response 1 mark i.e. ¼ (one-fourth) marks of the total marks allotted to the question will be deducted from the total score. No deduction from the total score, however, will be made if no response is indicated for an item in the Answer Box.
6.
Candidates will be awarded marks as stated above in instruction No.3 for correct response of each question. One mark will be deducted for indicating incorrect response of each question. No deduction from the total score will be made if no response is indicated for an item in the answer box.
7.
There is only one correct response for each question. Marked up more than one response in any question will be treated as wrong response and marked up for wrong response will be deducted accordingly as per instruction 6 above.
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JEE-MAIN-2019 (12th Apr-Second Shift)-PCM-2
PART – A (PHYSICS) 1.
Figure shown a DC voltage regulator circuit, with a Zener diode of breakdown voltage = 6V. If the unregulated input voltage varies between 10 V to 16 V, then what is the maximum Zener current?
(A) 2.5 mA (C) 7.5 mA 2.
Three particles of masses 50 g, 100 g and 150 g are placed at the vertices of an equilateral triangle of side 1 m (as shown in the figure). The (x, y) coordinates of the centre of mass will be :
(A) (C)
3.
(B) 3.5 mA (D) 1.5 mA
3 7 m, m 7 12 3 5 m, m 4 12
7 (B) m, 12 7 (D) m, 12
3 m 8 3 m 4
The ratio of the weights of a body on the Earth's surface to that on the surface of a 1 planet is 9 : 4. The mass of the planet is th of that of the Earth. If 'R' is the radius of the 9 Earth, what is the radius of the planet ? (Take the planets to have the same mass density) R R (A) (B) 3 4 R R (C) (D) 9 2
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JEE-MAIN-2019 (12th Apr-Second Shift)-PCM-3 4.
A block of mass 5 kg is (i) pushed in case (A) and (ii) pulled in case (B), by a force F = 20 N, making an angle of 30º with the horizontal, as shown in the figures. The coefficient of friction between the block and floor is m = 0.2. The difference between the accelerations of the block, in case (B) and case (A) will be : (g = 10 ms–2)
(A) 0.4 ms2 (C) 0 ms2
(B) 3.2 ms2 (D) 0.8 ms2
5.
In the given circuit, the charge on 4 F capacitor will be : (A) 13.4 C (B) 24 C (C) 9.6 C (D) 5.4 C
6.
A diatomic gas with rigid molecules does 10 J of work when expanded at constant pressure. What would be the heat energy absorbed by the gas, in this process? (A) 40 J (B) 30 J (C) 35 J (D) 25 J
7.
A Carnot engine has an efficiency of 1/6. When the temperature of the sink is reduced by 62ºC, its efficiency is doubled. The temperatures of the source and the sink are, respectively (A) 62ºC, 124ºC (B) 99ºC, 37ºC (C) 37ºC, 99ºC (D) 124ºC, 62ºC
8.
Consider an electron in a hydrogen atom, revolving in its second excited state (having radius 4.65 Å). The de-Broglie wavelength of this electron is : (A) 12.9 Å (B) 9.7 Å (C) 6.6 Å (D) 3.5 Å
9.
A small speaker delivers 2 W of audio output. At what distance from the speaker will one detect 120 dB intensity sound? [Given reference intensity of sound as 10–12W/m2] (A) 30 cm (B) 10 cm (C) 40 cm (D) 20 cm
10.
A system of three polarizers P1, P2, P3 is set up such that the pass axis of P3 is crossed with respect to that of P1. The pass axis of P2 is inclined at 60º to the pass axis of P3. When a beam of unpolarized light of intensity I0 is incident on P1, the intensity of light transmitted by the three polarizers is I. The ratio (I0/I) equals (nearly) : (A) 10.67 (B) 1.80 (C) 5.33 (D) 16.00
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JEE-MAIN-2019 (12th Apr-Second Shift)-PCM-4 11.
Let a total charge 2Q be distributed in a sphere of radius R, with the charge density given by r(r) = kr, where r is the distance from the centre. Two charges A and B, of –Q each, are placed on diametrically opposite points, at equal distance, a, from the centre. If A and B do not experience any force, then : 3R 1 (A) a 1 (B) a 2 4 R 2 4 (C) a 8
12.
1
4
(D) a R / 3
R
An electron, moving along the x-axis with an initial energy of 100 eV, enters a region of magnetic field B 1.5 10 3 T kˆ at S (See figure). The field extends between x = 0 and
x = 2 cm. The electron is detected at the point Q on a screen placed 8 cm away from the point S. The distance d between P and Q (on the screen) is : (electron's charge = 1.6 × 10–19 C, mass of electron = 9.1 × 10–31 kg)
(A) 12.87 cm (C) 1.22 cm
(B) 2.25 cm (D) 11.65 cm
13.
Two particles are projected from the same point with the same speed u such that they have the same range R, but different maximum heights, h1 and h2. Which of the following is correct? (A) R2 = 4 h1h2 (B) R2 = 2 h1h2 2 (C) R = 16 h1h2 (D) R2 = h1h2
14.
A particle is moving with speed v x along positive x-axis. Calculate the speed of the particle at time t = t(assume that the particle is at origin at t = 0). b2 (A) b 2 (B) 2 b2 b2 (C) (D) 4 2
15.
One kg of water, at 20ºC, is heated in an electric kettle whose heating element has a mean (temperature averaged) resistance of 20 . The rms voltage in the mains is 200 V. Ignoring heat loss from the kettle, time taken for water to evaporate fully, is close to : [Specific heat of water = 4200 J/kg ºC), Latent heat of water = 2260 kJ/kg] (A) 3 minutes (B) 10 minutes (C) 22 minutes (D) 16 minutes
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JEE-MAIN-2019 (12th Apr-Second Shift)-PCM-5 16.
Half lives of two radioactive nuclei A and B are 10 minutes and 20 minutes, respectively. If, initially a sample has equal number of nuclei, then after 60 minutes, the ratio of decayed numbers of nuclei A and B will be : (A) 9 : 8 (B) 1 : 8 (C) 8 : 1 (D) 3 : 8
17.
In an amplitude modulator circuit, the carrier wave is given by, C(t) = 4 sin (20000 t) while modulating signal is given by, m(t) = 2 sin (2000 t). The values of modulation index and lower side band frequency are: (A) 0.5 and 9 kHz (B) 0.3 and 9 kHz (C) 0.5 and 10 kHz (D) 0.4 and 10 kHz
18.
A uniform cylindrical rod of length L and radius r, is made from a material whose Young's modulus of Elasticity equals Y. When this rod is heated by temperature T and simultaneously subjected to a net longitudinal compressional force F, its length remains unchanged. The coefficient of volume expansion, of the material of the rod, is (nearly) equals to : (A) 9F/(r2YT) (B) F/(3r2YT) (C) 3F/(r2YT) (D) 6F/(r2YT)
19.
Consider the LR circuit shown in the figure. If the switch S is closed at t = 0 then the L amount of charge that passes through the battery between t = 0 and t is : R
EL 7.3R2 7.3 EL (C) R2
EL 2.7R 2 2.7 EL (D) R2
(A)
(B)
20.
Two sources of sound S1 and S2 produce sound waves of same frequency 660 Hz. A listener is moving from source S1 towards S2 with a constant speed u m/s and he hears 10 beats/s. The velocity of sound is 330 m/s. Then, u equals: (A) 15.0 m/s (B) 10.0 m/s (C) 5.5 m/s (D) 2.5 m/s
21.
A plane electromagnetic wave having a frequency n = 23.9 GHz propagates along the positive z-direction in free space. The peak value of the electric field is 60 V/m. Which among the following is the acceptable magnetic field component in the electromagnetic wave? (A) B 2 10 7 sin 1.5 10 2 x 0.5 1011 t ˆj (B) B 60 sin 0.5 103 x 1.5 1011 t kˆ (C) B 2 10 7 sin 0.5 10 2 z 1.5 1011 t ˆi (D) B 2 107 sin 0.5 103 z 1.5 1011 t ˆi
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JEE-MAIN-2019 (12th Apr-Second Shift)-PCM-6 22.
Find the magnetic field at point P due to a straight line segment AB of length 6 cm carrying a current of 5 A. (See figure) (0 = 4p × 10–7 N-A–2)
(A) 2.0 × 105 T (C) 2.5 × 105 T
(B) 3.0 × 105 T (D) 1.5 × 105 T
23.
The electron in a hydrogen atom first jumps from the third excited state to the second excited state and subsequently to the first excited state. The ratio of the respective wavelengths, 1/2, of the photons emitted in this process is : (A) 20/7 (B) 7/5 (C) 9/7 (D) 27/5
24.
A smooth wire of length 2r is bent into a circle and kept in a vertical plane. A bead can slide smoothly on the wire. When the circle is rotating with angular speed w about the vertical diameter AB, as shown in figure, the bead is at rest with respect to the circular ring at position P as shown. Then the value of 2 is equal to:
3g 2r
(B) g 3 / r
(C) 2g / r
(D) 2g / r 3
(A)
25.
A solid sphere, of radius R acquires a terminal velocity 1 when falling (due to gravity) through a viscous fluid having a coefficient of viscosity . The sphere is broken into 27 identical solid spheres. If each of these spheres acquires a terminal velocity, 2, when falling through the same fluid, the ratio (1/2) equals : (A) 27 (B) 1/27 (C) 9 (D) 1/9
26.
The number density of molecules of a gas depends on their distance r from the origin as, 4 n r n0 e r . Then the total number of molecules is proportional to : (A) n03/4 (C) n01/4
(B) n03 (D) n0 1/2
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JEE-MAIN-2019 (12th Apr-Second Shift)-PCM-7
27.
A transparent cube of side d, made of a material of refractive index 2, is immersed in a liquid of refractive index 1(1 < 2). A ray is incident on the face AB at an angle q(shown in the figure). Total internal reflection takes place at point E on the face BC.
The q must satisfy : (A) sin1
1 2
(B) sin1
22 1 12
(C) sin 1
1 2
(D) sin1
22 1 12
28.
A tuning fork of frequency 480 Hz is used in an experiment for measuring speed of sound () in air by resonance tube method. Resonance is observed to occur at two successive lengths of the air column, 1 30 cm and 2 70 cm. Then n is equal to : (A) 332 ms1 (B) 338 ms1 1 (C) 384 ms (D) 379 ms1
29.
A spring whose unstretched length is has a force constant k. The spring is cut into two pieces of unstretched lengths 1 and 2 where, 1 = n 2 and n is an integer. The ratio k1/k2 of the corresponding force constants, k1 and k2 will be: 1 (A) n (B) 2 n 1 (C) n2 (D) n
30.
A moving coil galvanometer, having a resistance G, produces full scale deflection when a current Ig flows through it. This galvanometer can be converted into (i) an ammeter of range 0 to I0 (I0 > Ig) by connecting a shunt resistance RA to it and (ii) into a voltmeter of range 0 to V(V = GI0) by connecting a series resistance RV to it. Then, Ig R (A) R AR V G2 and A R V I0 Ig Ig R (B) R AR V G and A R V I0 Ig
2
2
Ig I0 Ig R (C) R AR V G and A I I R V Ig 0 g
2
2
Ig I0 Ig R (D) R A R V G2 and A I R I I V g 0 g
2
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JEE-MAIN-2019 (12th Apr-Second Shift)-PCM-8
PART –B (CHEMISTRY) 31.
What will be the major product when m-cresol is reacted with propargyl bromide (HC C – CH2Br) in presence of K2CO3 in acetone
(A)
(B)
(C)
(D)
32.
Which one of the following is likely to give a precipitate with AgNO3 solution? (A) CH2 = CH – Cl (B) CHCl3 (C) (CH3)3CCl (D) CCl4
33.
The INCORRECT match in the following is: (A) G° < 0, K > 1 (B) G° < 0, K < 1 (C) G° = 0, K = 1 (D) G° > 0, K < 1
34.
Benzene diazonium chloride on reaction with aniline in the presence of dilute hydrochloric acid gives : (A)
(B)
(C)
(D)
35.
The INCORRECT statement is: (A) LiNO3 decomposes on heating to give LiNO2 and O2. (B) Lithium is least reactive with water among the alkali metals. (C) LiCl crystallises from aqueous solution as LiCl.2H2O. (D) Lithium is the strongest reducing agent among the alkali metals.
36.
The C–C bond length is maximum in (A) graphite (C) diamond
(B) C70 (D) C60
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JEE-MAIN-2019 (12th Apr-Second Shift)-PCM-9 37.
Consider the following reactions :
‘A’ is: (A) CH2 = CH2 (C) CH CH 38.
(B) CH3 – C CH (D) CH3 – C C – CH3
In which one of the following equilibria, Kp Kc? (A) 2NO g (B) 2C s O2 g N2 g O2 g 2CO g
(C) NO2 g SO2 g NO g SO3 g (D) 2HI g H2 g I2 g 39.
The coordination numbers of Co and Al in [Co(Cl)(en)2]Cl and K3[Al(C2O4)3], respectively, are (en = ethane-1,2-diamine) (A) 6 and 6 (B) 5 and 3 (C) 3 and 3 (D) 5 and 6
40.
Which of the given statements is INCORRECT about glycogen? (A) It is present in some yeast and fungi (B) It is present in animal cells (C) Only -linkages are present in the molecule (D) It is a straight chain polymer similar to amylase
41.
The primary pollutant that leads to photochemical smog is: (A) acrolein (B) nitrogen oxides (C) ozone (D) sulphur dioxide
42.
The ratio of number of atoms present in a simple cubic, body centered cubic and face centered cubic structure are, respectively : (A) 1 : 2 : 4 (B) 4 : 2 : 3 (C) 4 : 2 : 1 (D) 8 : 1 : 6
43.
Thermal decomposition of a Mn compound (X) at 513 K results in compound Y, MnO2 and a gaseous product. MnO2 reacts with NaCl and concentrated H2SO4 to give a pungent gas Z. X, Y and Z, respectively. (A) K2MnO4, KMnO4 and SO2 (B) K3MnO4, K2MnO4 and Cl2 (C) K2MnO4, KMnO4 and Cl2 (D) KMnO4, K2MnO4 and Cl2
44.
An ' Assertion' and a 'Reason' are given below. Choose the correct answer from the following options. Assertion (A): Vinyl halides do not undergo nucleophilic substitution easily. Reason (R): Even though the intermediate carbocation is stabilized by loosely held p-electrons, the cleavage is difficult because of strong bonding. (A) Both (A) and (R) are correct statements but (R) is not the correct explanation of (A) (B) Both (A) and (R) are wrong statements (C) Both (A) and (R) are correct statements and (R) is the correct explanation of (A) (D) (A) is a correct statement but (R) is a wrong statement. FIITJEE Ltd., FIITJEE House, 29-A, Kalu Sarai, Sarvapriya Vihar, New Delhi -110016, Ph 46106000, 26569493, Fax 26513942 website: www.fiitjee.com.
JEE-MAIN-2019 (12th Apr-Second Shift)-PCM-10 45.
The molar solubility of Cd(OH)2 is 1.84 × 10–5 M in water. The expected solubility of Cd(OH)2 in a buffer solution of pH = 12 is : (A) 6.23 × 1011 M (B) 1.84 × 109 M 2.49 (C) 10 9 M (D) 2.49 × 1010 M 1.84
46.
The compound used in the treatment of lead poisoning is: (A) EDTA (B) Cis-platin (C) D-penicillamine (D) desferrioxime B
47.
The pair that has similar atomic radii is: (A) Ti and Hf (C) Sc and Ni
(B) Mn and Re (D) Mo and W
48.
25 g of an unknown hydrocarbon upon burning produces 88 g of CO2 and 9 g of H2O. This unknown hydrocarbon contains. (A) 24g of carbon and 1 g of hydrogen (B) 22g of carbon and 3 g of hydrogen (C) 18g of carbon and 7 g of hydrogen (D) 20g of carbon and 5 g of hydrogen
49.
The decreasing order of electrical conductivity of the following aqueous solutions is: 0.1 M Formic acid (a) 0.1 M Acetic acid (b) 0.1 M Benzoic acid (c) (A) a > c > b (B) c > a > b (C) c > b > a (D) a > b > c
50.
Among the following, the energy of 2s orbital is lowest in: (A) K (B) Na (C) H (D) Li
51.
In the following skew conformation of ethane, H'–C–C–H" dihedral angle is :
(A) 58° (C) 149° 52.
(B) 120° (D) 151°
The correct statement is: (A) leaching of bauxite using concentrated NaOH solution gives sodium aluminate and sodium silicate (B) the Hall-Heroult process is used for the production of aluminium and iron (C) the blistered appearance of copper during the metallurgical process is due to the evolution of CO2 (D) pig iron is obtained from cast iron
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JEE-MAIN-2019 (12th Apr-Second Shift)-PCM-11 53.
NO2 required for a reaction is produced by the decomposition of N2O5 in CCl4 as per the equation 2N2O5 (g) 4NO 2 g O 2 g The initial concentration of N2O5 is 3.00 mol L–1 and it is 2.75 mol L–1 after 30 minutes. The rate of formation of NO2 is : (A) 1.667 × 102 mol L1 min1 (B) 4.167 × 103 mol L1 min1 (C) 8.333 × 103 mol L1 min1 (D) 2.083 × 103 mol L1 min1
54.
The correct name of the following polymer is:
(A) Polyisoprene (C) Polyisobutane 55.
56.
(B) Polytert-butylene (D) Polyisobutylene
Heating of 2-chloro-1-phenylbutane with EtOK/EtOH gives X as the major product. Reaction of X with Hg(OAc)2/H2O followed by NaBH4 gives Y as the major product. Y is: (A)
(B)
(C)
(D)
The IUPAC name of the following compound is :
(A) 3, 5-dimethyl-4-propylhept-1-en-6-yne (B) 3-methyl-4-(1-methylprop-2-ynyl)-1-heptene (C) 3-methyl-4-(3-methylprop-1-enyl)-1-heptyne (D) 3, 5-dimethyl-4-propylhept-6-en-1-yne 57.
Among the following, the INCORRECT statement about colloids is : (A) The range of diameters of colloidal particles is between 1 and 1000 nm (B) The osmotic pressure of a colloidal solution is of higher order than the true solution at the same concentration (C) They can scatter light (D) They are larger than small molecules and have high molar mass
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JEE-MAIN-2019 (12th Apr-Second Shift)-PCM-12 58.
In comparison to boron, berylium has: (A) greater nuclear charge and greater first ionisation enthalpy (B) lesser nuclear charge and lesser first ionisation enthalpy (C) greater nuclear charge and lesser first ionisation enthalpy (D) lesser nuclear charge and greater first ionisation enthalpy
59.
A solution is prepared by dissolving 0.6 g of urea (molar mass = 60 g mol–1) and 1.8 g of glucose (molar mass = 180 g mol–1) in 100 mL of water at 27ºC. The osmotic pressure of the solution is : (R = 0.08206 L atm K–1 mol–1) (A) 8.2 atm (B) 1.64 atm (C) 4.92 atm (D) 2.46 atm
60.
The temporary hardness of a water sample is due to compound X. Boiling this sample converts X to compound Y. X and Y, respectively, are : (A) Mg(HCO3)2 and MgCO3 (B) Ca(HCO3)2 and CaO (C) Mg(HCO3)2 and Mg(OH)2 (D) Ca(HCO3)2 and Ca(OH)2
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JEE-MAIN-2019 (12th Apr-Second Shift)-PCM-13
PART–C (MATHEMATICS) 61.
62.
The general solution of the differential equation (y2 – x3) dx – xydy = 0 (x 0) is : (where c is a constant of integration) (A) y2 + 2x3 + cx2 = 0 (B) y2 – 2x3 + cx2 = 0 2 2 3 (C) y + 2x + cx = 0 (D) y2 – 2x2 + cx3 = 0 ˆ b 2iˆ ˆj kˆ and c ˆi 2ˆj 3kˆ . Then Let R and the three vectors a ˆi ˆj 3k, the set S :a,b and c are coplanar
(A) Contains exactly two numbers only one of which is positive (B) is empty (C) Contains exactly two positive numbers (D) is singleton 63.
Let (0, /2) be fixed. If the integral
tan x tan
tan x tan dx
A(x) cos2 + B(x) sin2 + C, where C is a constant of integration, then the functions A(x) and B(x) are respectively: (A) x and loge sin x (B) x and loge cos x (C) x and loge sin x
(D) x and loge sin x
64.
Let S be the set of all R such that the equation, cos2 x + sin x = 2 – 7 has a solution. Then S is equal to : (A) [3, 7] (B) R (C) [2, 6] (D) [1, 4]
65.
Let f(x) 5 x 2 and g(x) x 1 , x R . If f(x) attains maximum value at and g(x) attains minimum value at , then lim
x 1 x 2 5x 6
x
3 2 1 (C) 2
(A)
66.
Let z C with Im(z) = 10 and it satisfies (A) n = 40 and Re(z) = 10 (C) n = 40 and Re(z) = – 10
x 2 6x 8 3 (B) 2 1 (D) 2
is equal to :
2z n 2i 1 for some natural number n. Then : 2z n (B) n = 20 and Re(z) = 10 (D) n = 20 and Re(z) = – 10 6
67.
1 x8 2 3 The term independent of x in the expansion of 2x 2 is equal to: x 60 81 (A) 36 (B) 36 (C) 108 (D) 72
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JEE-MAIN-2019 (12th Apr-Second Shift)-PCM-14
68.
If the area (in sq. units) bounded by the parabola y2 =4x and the line y = x, > 0, is then is equal to : (A) 48
(B) 4 3
(C) 2 6
(D) 24
1 , 9
69.
If [x] denotes the greatest integer x, then the system of linear equations [sin ] x + [– cos ] y = 0 [cot ] x + y = 0 2 7 (A) have infinitely many solutions if , and has a unique solution if , 2 3 6 2 7 (B) have infinitely many solutions if , , 2 3 6 2 7 (C) has a unique solution if , and have infinitely many solutions if , 2 3 6 2 7 (D) has a unique solution if , , 2 3 6
70.
For an initial screening of an admission test, a candidate is given fifty problems to solve. 4 If the probability that the candidate can solve any problem is , then the probability that 5 he is unable to solve less than two problems is : 164 1 (A) 25 5
(C)
54 4 5 5
48
201 1 (B) 5 5
49
(D)
49
316 4 25 5
48
71.
If a1, a2, a3, ……. are in A.P. such that a1 + a7 + a16 = 40, then the sum of the first 15 terms of this A.P. is: (A) 200 (B) 280 (C) 150 (D) 120
72.
An ellipse, with foci at (0, 2) and (0, –2) and minor axis of length 4, passes through which of the following points? (A) 2, 2 (B) 2,2 2
(C) 1,2 2 73.
(D)
2,2
The length of the perpendicular drawn from the point (2, 1, 4) to the plane containing the lines r ˆi ˆj ˆi 2jˆ kˆ and r ˆi ˆj ˆi ˆj 2kˆ is:
(A) (C)
1 3 1 3
(B)
3
(D) 3
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JEE-MAIN-2019 (12th Apr-Second Shift)-PCM-15
74.
Let A, B and C be sets such that A B C. Then which of the following statements is not true? (A) If (A – C) B, then A B (B) If (A – B) C, then A C (C) (C A) (C B) = C (D) B C
75.
If , and are three consecutive terms of a non-constant G.P. such that the equations x2 + 2x + = 0 and x2 + x – 1 = 0 have a common root, then ( + ) is equal to : (A) (B) 0 (C) (D)
76.
The tangents to the curve y = (x – 2)2 –1 at its points of intersection with the line x – y = 3, intersect at the point : 5 5 (A) ,1 (B) , 1 3 2 5 5 (C) ,1 (D) , 1 2 2
77.
The angle of elevation of the top of vertical tower standing on a horizontal plane is observed to be 45° from a point A on the plane. Let B be the point 30 m vertically above the point A. If the angle of elevation of the top of the tower from B be 30°, then the distance (in m) of the foot of the tower from the point A is: (A) 15 1 3 (B) 15 3 3
(C) 15 3 3
(D) 15 5 3
78.
A triangle has a vertex at (1, 2) and the mid points of the two sides through it are (–1, 1) and (2,3). Then the centroid of this triangle is : 7 1 (A) 1, (B) ,1 3 3 1 1 5 (C) ,2 (D) , 3 3 3
79.
A person throws two fair dice. He wins Rs. 15 for throwing a doublet (same numbers on the two dice), wins Rs.12 when the throw results in the sum of 9, and loses Rs. 6 for any other outcome on the throw. Then the expected gain/loss (in Rs.) of the person is : 1 (A) loss (B) 2 gain 4 1 1 (C) gain (D) loss 2 2
80.
If
20
C1 22
20
C3 32
(A, ) is equal to: (A) (420, 18) (C) (420, 19)
20
C3 22 ....... 202
20
C20 A 2 , then the ordered pair
(B) (380, 18) (D) (380, 19)
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JEE-MAIN-2019 (12th Apr-Second Shift)-PCM-16
81.
x sin x cos x The derivative of tan 1 , with respect to , where x 0, is: 2 sin x cos x 2 1 (A) 2 (B) 2 2 (C) 1 (D) 3
82.
A circle touching the x-axis at (3, 0) and making an intercept of length 8 on the y-axis passes through the point : (A) (3, 5) (B) (1, 5) (C) (3, 10) (D) (2, 3)
83.
The equation of a common tangent to the curves, y2 = 16x and xy = –4 is : (A) x – 2y + 16 = 0 (B) 2x – y + 2 = 0 (C) x + y + 4 = 0 (D) x – y + 4 = 0
84.
A plane which bisects the angle between the two given planes 2x – y + 2z – 4 = 0 and x + 2y + 2z – 2 = 0, passes through the point: (A) (1, 4, 1) (B) (2, 4, 1) (C) (2, 4, 1) (D) (1, 4, 1)
85.
A value of such that
1
dx
x x 1 log
e
(A) (C) 86.
1 2
(B) 2
1 2
(D) 2
A group of students comprises of 5 boys and n girls. If the number of ways, in which a team of 3 students can randomly be selected from this group such that there is at least one boy and at least one girl in each team, is 1750, then n is equal to: (A) 24 (B) 28 (C) 27 (D) 25 1 cos2
87.
A value of (0, /3), for which
2
cos 2
cos
18 7 (C) 36
sin2
4 cos 6
2
4 cos 6
1 sin 2
sin
0, is:
1 4 cos 6
9 7 (D) 24
(A)
88.
9 8 is:
(B)
A straight line L at a distance of 4 units from the origin makes positive intercepts on the coordinate axes and the perpendicular from the origin to this line makes an angle of 60° with the line x + y = 0. Then an equation of the line L is : (A) 3 1 x 3 1 y 8 2 (B) 3x y 8
(C) x 3y 8
(D)
3 1 x
3 1 y 8 2
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JEE-MAIN-2019 (12th Apr-Second Shift)-PCM-17
89.
lim x 0
(A) 2 (C) 3 90.
x 2 sin x x 2 2 sin x 1 sin2 x x 1
is: (B) 6 (D) 1
The Boolean expression (p (q)) is equivalent to: (A) (p) q (B) p q (C) p q (D) q p
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JEE-MAIN-2019 (12th Apr-Second Shift)-PCM-18
JEE (Mains) – 2019 ANSWERS PART A – PHYSICS 1. 5. 9. 13. 17. 21. 25. 29.
B B C C A C C D
2. 6. 10. 14. 18. 22. 26. 30.
D C A B C D A B
3. 7. 11. 15. 19. 23. 27.
D B C C B A D
4. 8. 12. 16. 20. 24. 28.
D B C A D D C
34. 38. 42. 46. 50. 54. 58.
B B A A A D D
64. 68. 72. 76. 80. 84. 88.
C D D D A B D
PART B – CHEMISTRY 31. 35. 39. 43. 47. 51. 55. 59.
C A D D D C A C
32. 36. 40. 44. 48. 52. 56. 60.
C C D D A A A C
33. 37. 41. 45. 49. 53. 57.
B B B D A A B
PART C – MATHEMATICS 61. 65. 69. 73. 77. 81. 85. 89.
B C A B C A B A
62. 66. 70. 74. 78. 82. 86. 90.
B C C A C C D C
63. 67. 71. 75. 79. 83. 87.
C B A D D D B
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JEE-MAIN-2019 (12th Apr-Second Shift)-PCM-19
HINTS AND SOLUTIONS PART A – PHYSICS 1.
Maximum current will flow from zener if input voltage is maximum. When zener diode works in breakdown state, voltage across the zener will remain same. Vacross 4k 6V
6 6 A mA 4000 4 Since input voltage = 16 V Potential difference across 2K = 10V 10 Current through 2k = 5mA 2000 Current through zener diode = (Is – IL) = 3.5 mA Current through 4K =
2.
The co-ordinates of the centre of mass 1 3 ˆ 0 150 i j 100 ˆi 2 2 rcm 300 7 ˆ 3ˆ rcm i j 12 4 7 3 Co-ordinate , m 12 4
3.
Since mass of the object remains same Weight of object will be proportional to ‘g’ (acceleration due to gravity) Given: Wearth 9 gearth Wplanet 4 gplanet Also, gsurface
GM (M is mass planet, G is universal gravitational constant, R is radius of R2
planet)
2 2 2 Rplanet 9 GMearth Rplanet Mearth Rplanet 9 2 2 2 4 GMplanet Rearth Mplanet Rearth R earth
Rplanet
Rearth R 2 2
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JEE-MAIN-2019 (12th Apr-Second Shift)-PCM-20 4.
N1 = 60 10 3 0.2 60 a1 5 a1 – a2 = 0.8
N2 = 40 10 3 0.2 40 a2 5
5.
So total charge flow = q = 5.4 F × 10V = 54 F The charge will be distributed in the ratio of capacitance q1 2.4 4 q2 3 5 9X = 54 C X = 6 C Charge on 4 F capacitor will be = 4X = 4 × 6 C = 24 C 6.
7 R 2 Since gas undergoes isobaric process For a diatomic gas, Cp =
7 7 Q = n RT (nRT) = 35 J 2 2 7.
Efficiency of Carnot engine = 1
Tsink Tsource
Given,
T Tsink 1 5 1 sink 6 Tsource Tsource 6
…(i)
Also,
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JEE-MAIN-2019 (12th Apr-Second Shift)-PCM-21
T 62 2 62 1 1 sink 6 Tsource Tsource 6
…(ii)
Tsource = 372 K = 99°C
5 372 = 310 K = 37°C 6 (Note: Temperature of source is more than temperature of sink) Also, Tsink
8.
2rn = nn 2(4.65 10 10 ) 3 3 3 = 9.7 Å
9.
Loudness of sound is given by I dB 10log (I is intensity of sound, I0 is reference intensity of sound) I0 I 120 10 log I0 2 I = 1 W/m P 2 Also, I 2 4r 4r 2 2 I r m = 0.399 m 4 2 40 cm
10.
Since unpolarised light falls on P1
I0 . 2 Pass axis of P2 will be at an angle of 30° with P1 Intensity of light transmitted from I 3I P2 0 cos2 30o 0 2 8 Pass axis of P3 is at an angle of 60° with P2 Intensity of light transmitted from 3I 3I P3 0 cos2 60o 0 8 32 I0 32 10.67 I 3 Intensity of light transmitted from P1
11.
E 4 a
E
2
0
kr 4r 2 dr e0
k 4 a 4 4 4 0
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JEE-MAIN-2019 (12th Apr-Second Shift)-PCM-22 R
2Q kr 4r 2 dr 0
2Q k R 4 1 QQ 40 (2a)2 R = a81/4 QE
12.
mv qB
R
2m (K.E.) qB
=
2 9.1 10 31 (100 1.6 10 19 ) 1.6 1019 1.5 10 3 R = 2.248 cm 2 QU sin ; tan 2.248 TU 2 QU 1.026 6 QU = 11.69 R=
PU = R(1 – cos ) = 1.22 d = QU + PU 13.
For same range angle of projection will be will be & 90 – u2 2 sin cos R g h1
u2 sin2 g
h2
u2 sin2 (90 ) g
R2 16 h1h2 14.
v b x dv b dx bv ; a dt 2 x dt 2 x
b b x a
2 x
;
dv b2 a ; dt 2
v
b2 2
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JEE-MAIN-2019 (12th Apr-Second Shift)-PCM-23 15.
Q=P×t Q = mcT + mL V2 P = rms R 4200 × 80 + 2260 × 103 =
(200)2 t 20
t = 1298 sec t 22 min 16.
NOA NOA 6 2 2t/t1/ 2 Number of nuclei decayed N 63NOA = NOA OA 6 2 64 NOB NOB NB NOBet t/t1/ 2 3 2 2 Number of nuclei decayed N 7N = NOB OB OB 3 2 8 Since, NOA = NOB Ratio of decayed numbers of nuclei 63 NOA 8 9 A&B= 64 7NOB 8 NA NOA et
17.
Modulation index is given by A 2 m m 0.5 Ac 4 & (a) carrier wave frequency is given by = 2fc = 2 × 104 fc = 1 kHz lower side band frequency fc – fm 10 kHz – 1 kHz = 9 kHz
18.
Length of cylinder remains unchanged F F so A Compressive A Thermal F YT ( is linear coefficient of expansion) r 2 F YTr 2 The coefficient of volume expansion = 3 F 3 YTr 2
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JEE-MAIN-2019 (12th Apr-Second Shift)-PCM-24
19.
q Idt L /R
Rt L 1 e dt 0 EL 1 EL q 2 ; q R e 2.7R2
q
20.
E R
f = 660 Hz, v = 330 m/s
v u v u f1 f ; f2 f v v f f2 f1 [v u (v u)] v f 10 = f2 – f1 = [2u] v u = 2.5 m/s 21.
22.
23.
Magnetic field when electromagnetic wave propagates in +z direction. B = B0 sin (kz – t) where 60 B0 2 10 7 8 3 10 2 k 0.5 103 = 2f = 1.5 × 1011
0I 2 sin 4d d = 4 cm 3 sin = 5 B
1 1 1 1 1 1 R 2 2 R 2 2 ; 1 4 3 nf ni 1 1 1 7 1 R R 2 2 ; 1 2 3 9 16 2 5 = R 49 5 1 20 36 7 2 7 9 16
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JEE-MAIN-2019 (12th Apr-Second Shift)-PCM-25
24.
r N sin = m 2 2 N cos = mg r2 tan 2g r 2
25.
3r 2
r2 2g
;
…(i) …(ii)
2
2g 3r
We have
VT
2 r2 (0 )g 9
VT r2 Since mass of the sphere will be same 4 4 R3 27 r 3 3 3 r
26.
R 3
v1 R2 9 v2 r2
Given number density of molecules of gas as a function of r is n(r) = n0 er
4
Total number of molecule =
4
r 2 n(r)dV = n0 e 4r dr 0
0
Number of molecules is proportional to n0–3/4 27.
1 2 1 sin = 2 sin (90° – C) 2 2 1 12 2 sin 1 sin c =
sin1
12 22 12
For TIR sin1
u22 1 12
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JEE-MAIN-2019 (12th Apr-Second Shift)-PCM-26 28.
v = 2f(l2 – l1) v = 2 × 480 × (70 – 30) × 10–2 v = 960 × 40 × 10–2 v = 38400 × 10–2 m/s v = 384 m/s
29.
k1
30.
When galvanometer is used an ammeter shunt is used in parallel with galvanometer.
C 1 C k2 2 k1 C 2 1 2 k 2 1C n 2 n
IgG = (I0 – Ig)RA Ig RA G I0 Ig When galvanometer is used as a voltmeter, resistance is used in series with galvanometer.
Ig(G + Rv) = V = GI0 (given V = GI0) (I0 Ig )G RV Ig Ig R R AR V G & A R V I0 Ig
2
2
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JEE-MAIN-2019 (12th Apr-Second Shift)-PCM-27
PART B – CHEMISTRY 31.
O
CH3 Above anion act as nucleophile for SN2 attack on CH C – CH2 – Br and acetone acting as polar aprotic solvent. 32.
(CH3)3CCl will give Cl– and most stable carbocation. Hence (CH3)3CCl likely to give a precipitate with AgNO3 solution.
33.
G° = -2.303 RT log Keq
34.
According to NCERT C – N coupling take place, when diazonium ion is treated with aniline.
35.
1 2LiNO 3 Li2O 2NO 2 O2 2
36.
Since carbon in diamond is sp3 hybridized and its C – C bond order is 1. In graphite and fullerene there is both C – C and C = C in conjugation, hence there is partial double bond character between carbon atoms.
37.
Ag2O CH3 C C H CH3 C C Ag
(A) Hg2+/H+
White PPt.
O H3C
C
CH3
(B) NaBH4 OH ZnCl2 CH3 C CH3 Turbidity within 5 minutes Conc.HCl
(C) H
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JEE-MAIN-2019 (12th Apr-Second Shift)-PCM-28 38.
If ng = 0 KP = KC If ng 0, KP KC Hence (B) is correct answer.
39.
en and C2 O24 are a bidentate ligand. So coordination number of [Co(Cl)(en)2]Cl is 5 and K3[Al(C2O4)3] is 6
40.
Glycogen is a multibranched polysaccharide.
41.
NO2 and Hydrocarbons are primary precursors of photochemical smog.
42.
No. of atoms in simple cubic = 1, bcc = 2 & fcc = 4
43.
44.
Due to resonance there is partial double bond character between carbon and chlorine, hence it do not undergoes nucleophilic substitution reaction.
45.
Ksp of Cd(OH)2 = 4s3 = 4 × (1.84 × 10–5)3 If pH = 12 pOH = 2 [OH–] = 10–2 M Ksp = [Cd2+] [OH–]2 4 × (1.84 ×10–5)3 = [Cd2+] [OH–]2 3 4 1.84 10 15 2+ [Cd ] = 10 4 2+ Cd = 4 × 6.22 × 10–11 = 2.49 × 1010 M
46.
(A) EDTA (ethylene diamine tetra acetate) is used for lead poisoning (B) Cis platin is used as a anti cancer drug (C) D-penicillamine is used for copper poisoning (D) desferrioxime B is used for iron poisoning
47.
Due to lanthanoide contraction Mo & W have similar atomic radii.
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JEE-MAIN-2019 (12th Apr-Second Shift)-PCM-29 48.
49.
Stronger the acidic strength greater will be its electrical conductivity. Ka value of formic acid > benzoic acid > acetic acid.
50.
Greater the nuclear charge, stronger will be the attraction, hence lower will be energy of 2s
51.
Dihedral angle is then angle between bond pairs present on adjacent atoms.
52.
Since NaOH is a strong base hence it reacts with Al2O3 and SiO2 to form salts.
53.
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JEE-MAIN-2019 (12th Apr-Second Shift)-PCM-30 54.
Isobutylene on polymerization will form given polymer. OH
55. Ph
alc.KOH
Ph X
Cl
Hg OAc 2 ,H2O NaBH4
Ph
OMDM follow Markonikovs addition rule. 56.
According to IUPAC rules, select the largest chain including functional group, if alkene and alkyne are present at equivalent position then priority is given to alkene.
57.
Definitions and property of colloidal will explain & solve above question..
58.
Since boron has higher nuclear charge because it has greater atomic number and lower 1st I.E. then beryllium due to fully filled s-orbital.
59.
60.
18 6 CRT .0821 300 60 180 = 0.2 .082 300 = 4.926 atm.
Mg HCO3 2 aq Mg OH2 2CO2
PART C – MATHEMATICS 61.
y
2
x3 dx xy dy 0
y 2 x 3 xy
x0
dy 0 dx
dy y2 x3 dx dy 1 2 y y x 2 ……..(i) dx x Let y 2 dy d 2y dx dx Putting this value in equation (i) 1 d 1 x2 2 dx x d 2 2x2 (ii) dx x 2 dx 1 I.F. e x e 2 n x 2 x Sol. of equation (ii) 1 1 3 2x 2 dx C x 2 2x C x2 or, xy
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JEE-MAIN-2019 (12th Apr-Second Shift)-PCM-31
y 2 2x3 cx 2 y 2 2x 3 cx 2 0 Hence correct answer is option B. 62.
a.b.c 0 3 1 2 1 0 2 3
3 2 1 2 6 3 4 0 2
2
3 2 6 12 3 0 32 18 0 2 6 0 not possible for real S is empty set 63.
tan x tan
sin x
tan x tan dx sin x dx Let, x t sin t 2 dt cos 2 dt cot t sin 2 dt sin t t.cos 2 n sin t .sin 2 C
x cos 2 ln sin x .sin 2 C Hence the correct answer is option (C) 64.
Given, cos 2x 2 sin x 2 7
1 2sin2 x sin x 2 7 2sin2 x sin x 2 8 0 sin x
2 8 2 8
4 8 sin x 4 8 8 sin x , 4 4 sin x 2 (Not possible) For solution 2 8 1 1 4 4 2 8 4 4 2 12 2, 6
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JEE-MAIN-2019 (12th Apr-Second Shift)-PCM-32 65.
f x 5 x 2 f x attains maximum value when x 2 0 x 2 g x x 1 g x attains minimum value of x 1 Lim
x 1 x 2 5x 6
x
Lim
x 2 6x 8 x 1 x 2 x 3
x 2 x 4 2 1 2 3 1 2 2 4 x 2
66.
Let z x 10i 2z n given 2i 1 2z n 2 x 10i n 2i 1 2 x 10i n
2x n 20i 2i 1 2x n 20i Comparing real and imaginary part 2x n 2 20 2x n and 20 2 2x n 20 2x n 40 2x n and 20 4x 2n 20 4x 40 and 4x 2n 40 x 10 and 40 2n 40 n 40 n 40 and Re z 10
67.
1 x8 2 3 2x 2 x 60 81
6
1 independent of x in 60 6 6 1 2 3 2 3 8 2x Term of x in 2x x2 8 x3
Term independent of x will be
6
Tr 1
3 in 2x 2 2 will be Tr 1 6Cr 2x 2 x
6 r
3 2 x
r
r
6Cr 26 r 1 3r x122r 2r Case I : For term independent of x, 12 4r 0 r 3 T4 6C3 23 33 6 20 23 33 Case II : For term of x 8 12 4r 8 4r 20 r 5
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JEE-MAIN-2019 (12th Apr-Second Shift)-PCM-33 T6 6C5 .21 1 .35 . x 8
Required Answer
1 1 20 23 33 6 2 1 35 60 81
72 36 36 Hence the correct answer is option (B).
68.
y 2 4x and y x 2 x 2 4x x 0 and x
4
y
4/
Area
4x x dx
0
4/
y x
1 9 4/
x3/2 x2 1 2 9 3 / 2 0 2 0 3/4 x 4 x 16 1 22 3/2 3 2 9 32 8 1 3 9 8 1 3 9 24
x
69.
sin x cos y 0 cot x y 0
……..(1) ………(2)
Case I
2 When , 2 3 3 sin , 1 2 1 1 cos , 0 cos 0, 2 2 1 cot , 0 3 sin 0 cos 0 cot 1 Equation (1) and (2) will 0x 0y 0 system will have infinitely many solution x y 0 Case II When
7 1 , sin , 0 6 2
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JEE-MAIN-2019 (12th Apr-Second Shift)-PCM-34
3 cos 1, 2
cot
3,
sin 1, cos 1 cot 1,2,3,...... x y 0 Ix y 0 I 1,2,..... It will have unique solution in all cases x 0, y 0 70.
Total problems = 50 4 P (Solving) 5 1 P (Not solving) 5 P (unable to solve less than two problems) = P (not solving one problem) + P (not solving zero problem) 0 50 1 49 1 4 1 4 50 C0 50 C1 5 5 5 5 50 49 4 4 50 50. 5 5.5 49 50 49 4 4 10. 5 5 49
4 4 10 5 5
71.
54 4 . 5 5
49
a1,a 2 ,......an are in A.P. a1 a7 a16 40 a a 6d a 15d 40 3a 21d 40 40 a 7d 3 15 515 2a 14d 2 15 a 7d
15
40 3
= 200
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JEE-MAIN-2019 (12th Apr-Second Shift)-PCM-35 72.
Given 2a 4 and 2be 4 a 2, be 2
b2e2 4 b2 a2 4 b2 8 equation of ellipse x2 y2 1 4 8 Clearly option (D) satisfy the given curve. 73.
Equation of plane containing both lines is x 1 y 1 z 1
2
1 0
1
1
2
x 1 4 1 y 11 2 z 1 2 0 3 x 1 3 y 1 3z 0 x 1 y 1 z 0 x y z 0 distance from point (2, 1, 4) is
74.
2 1 4 12 12 12
3
For A C, A C B But A B option A is NOT true Let x Cx C A C B x C A and x C B x C or x A
x C or
and
x B
x C or x A B xC or x C A B C) xC C A C B C Now
x C x C A
(as A=C
B
(1) and
x C B x C A C B C C A C B from (1) and (2) C C A C B option B is true
(2)
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JEE-MAIN-2019 (12th Apr-Second Shift)-PCM-36 Let x A and x B x A B
x C (as A B C ) Let x A and x B x A B x C (as A B C ) Hence x A x C AC Option C is true As C A B B C A B As A B B C Hence the correct answer is option (A)
75.
, , are in G.P.
x 2 2 x 0 and x 2 x 1 0 have a common roots. Both roots will be common. 2 1 1 1 2 2 2 76.
xy3 0 ……..(i) will be chord of contact of parabola Let the required point is P x1,y1 chord
of
contact
for
point
P
x x1 y y1 xx1 4 3 2 2 y y1 2x1x 4x 4x1 6 As equation (i) and (ii) are same line 2x1 4 1 4x1 y1 6 1 1 3 2x1 4 1
x1
5 2
4x1 y1 6 3 10 y1 9 0 y 1 1
5 Hence correct answer is , 1 which is option (D). 2
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is
JEE-MAIN-2019 (12th Apr-Second Shift)-PCM-37 77.
AB 30m NP
In ANM tan 45o
MN 1 AN M
MN AN PM MN 30 AN 30
P
PM AN 30 In BPM tan 30 PB AN
30o
B
o
1 AN 30 AN 3
45o N
AN 3AN 30 3
AN
30 3 3 1
30 3
A
15 3 3
3 1
2
78. A (1, 2)
E (–1, 1) F (2, 3)
B (x2, y2)
C (x3, y3)
2 1 y 2 1, 2 1 2 2
x3 1 y 2 2 and 3 3 2 2
x 2 3, y 2 0
x 3 3, y3 4
B 3, 0
C 3, 4
x x 2 x 3 y1 y 2 y 3 , Centroid 1 3 3 1 3 3 2 D 4 1 , 3 , 2 3 3
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JEE-MAIN-2019 (12th Apr-Second Shift)-PCM-38 79. Coin Probability
+15 6 36
+12 4 36
–6 26 36
6 36 4 Probability of sum of 9 36 26 Other probability 36 6 4 26 Expected gain/loss 15 12 6 36 36 36 90 48 156 1 36 36 36 2 Hence correct answer is option (D). Probability of doublet
80.
20
20C0 20C1 20C2 x 2 ....... 20 C20 x 20 ……….(i) Differential equation w.r.t. x 19 20 1 x 20 C1.1 2. 20C2 x ....... 20 20C20 x19 …………..(ii) Multiply equation (2) by x 19 20x 1 x 20 C1x 2. 20C2 x 2 ...... 20 20C20 x 20 ……….(iii)
1 x
Differential equation 3 w.r.t. x 19 18 20 1 x 19x 1 x 1. 20C1 22 . 20C 2 x ...... 20 2 Put x 1 in equation (iv) 20 219 19.218 12 20C1 22 20 C2 ...... 20 2 20 C20
18
20 2
20
C 20 x19 …….(iv)
18
2 19 20 21 2
420 218 A 420, 18 Hence correct Option is (A). 81.
sin x cos x Given, y tan1 sin x cos x tan x 1 y tan1 tan x 1 1 tan x y tan1 1 tan x y tan1 tan x 4 0 x 2
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JEE-MAIN-2019 (12th Apr-Second Shift)-PCM-39
x 0 2 x0 4 4 y x tan1 tan x x x , 4 2 2 dy 1 yx 2 4 x 1 d 2 2
82.
2
2
Equation of required circle will be x 3 y r r 2
x 2 6x 9 y 2 2r y r 2 r 2 x 2 y 2 6x 2ry 9 0
………(1) 2
r f
Length of y intercept 2 f c 2
82 r 9 16 r 2 9 r=5 So equation of required circle will be x 2 y 2 6x 10y 9 0 two circles
x 2 y 2 6x 10y 9 0 ………..(2) x 2 y 2 6x 10y 9 0 ………..(3) Given option (C) i.e. (3, 10) satisfy equation (3). 83.
4 ……(i) is always tangent to y 2 16x m If it is tangent to the xy 4 y mx
4 x mx 4 m 2 2 m x 4x 4m m2 x 2 4x 4m m2 x 2 4x 4m 0 for tangent D = 0 16 16m3 0 m 1 put in equation (i) yx4 So the correct answer is option (D) 84.
Equation of angle bisectors x 2y 2z 2 2x y 2z 4 2 2 2 2 1 2 2 22 1 22
……..(1)
Case I: take positive sign 2x y 2z 4 x 2y 2z 2
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JEE-MAIN-2019 (12th Apr-Second Shift)-PCM-40
x 3y 2 0 ………(2) Case II : take negative sign 2x y 2z 4 x 2y 2z 2 2x y 2z 4 x 2y 2z 2 3x y 4z 6 0 ………(3) Option (B) satisfy equation (3) 1
85.
dx
x x 1 log
e
1
9 8
x 1 x dx log
x x 1
e
1
dx x
1
9 8
dx 9 loge x 1 8 1
x 9 loge loge x 1 8
2 1 2 9 loge log loge 2 2 2 1 8 2 1 2 1 9 log loge 8 2 2 2 2
2 1 9 4 1 8 8 42 4 1 9 4 2 4
322 32 8 36 2 36 4 2 4 8 0 2 2 0 2 1 0 1, 2 Hence the correct answer is option (B). 86.
Given 5 boys and n girls Total ways of farming team of 3 Members under given condition 5 C1 . nC2 5C2 . nC1
5C1. nC2 5C2 . nC1 1750
5n n 1 2 n n 1
10n 1750
2n 350 2 n2 3n 700 n2 3n 700 0 n 25
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JEE-MAIN-2019 (12th Apr-Second Shift)-PCM-41
87.
0, 3 1 cos2 sin2 4 cos 6 2 2 cos 1 sin 4 cos 6 0 2 2 cos sin 1 4 cos 6
R2 R 2 R1,R3 R3 R1 1 cos2 sin2 4 cos 6
1
1
0
1
0
1
0
C1 C1 C2
sin2 4 cos 6
2 0 1
1
0
0
1
0
expanding along first column 2 1 0 1 4 cos 6 0
2 4 cos 6 0 1 cos 6 2 2 6 3 9 88.
OP = 4 given OP makes 60o with x y 0 Let slope of OP = m m1 tan60o 1 m m 1 3 or 3 m 1 m 1 3m 3 or m 1 3 3m
m
m
3 1 3 1 or m 1 3 3 1 3 1 3 1
tan
or m
3 1
3 1 3 1
or tan
3 1
3 1 3 1 equation of line x cos y sin P
3 1 x
3 1 y 8 2
or
3 1 x
3 1 y 8 2
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JEE-MAIN-2019 (12th Apr-Second Shift)-PCM-42
89.
Lim x 0
x 2 sin x x 2 2 sin x 1 sin2 x x 1
x 2 sin x x 2 sin x 1 sin2 x x 1 x 2 sin x Lim 2 .2 x 0 x 2 sin x sin2 x x Applying L’H Rule 2. 1 2 cos x Lim x 0 2x 2 cos x 2 sin x cos x 1 Lim x 0
2
x 2 2 sin x 1 sin2 x x 1
2 3
2 2 1 Hence the correct answer is option (A).
90.
~ p ~ q p q Hence the correct answer is option (C).
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PHYSICS, CHEMISTRY & MATHEMATICS Paper - 1 Time Allotted: 3 Hours
Maximum Marks: 360
Please read the instructions carefully. You are allotted 5 minutes specifically for this purpose.
Important Instructions: 1.
The test is of 3 hours duration.
2.
This Test Paper consists of 90 questions. The maximum marks are 360.
3.
There are three parts in the question paper A, B, C consisting of Physics, Chemistry and Mathematics having 30 questions in each part of equal weightage. Each question is allotted 4 (four) marks for correct response.
4.
Out of the four options given for each question, only one option is the correct answer.
5.
For each incorrect response 1 mark i.e. ¼ (one-fourth) marks of the total marks allotted to the question will be deducted from the total score. No deduction from the total score, however, will be made if no response is indicated for an item in the Answer Box.
6.
Candidates will be awarded marks as stated above in instruction No.3 for correct response of each question. One mark will be deducted for indicating incorrect response of each question. No deduction from the total score will be made if no response is indicated for an item in the answer box.
7.
There is only one correct response for each question. Marked up more than one response in any question will be treated as wrong response and marked up for wrong response will be deducted accordingly as per instruction 6 above.
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JEE-MAIN-2019 (9th Jan-First Shift)-PCM-2
PART – A (PHYSICS) 1.
A convex lens is put 10 cm from a light source and it makes a sharp image on a screen, kept 10 cm from the lens. Now a glass block (refractive index 1.5) of 1.5 cm thickness is placed in contact with the light source. To get the sharp image again, the screen is shifted by a distance d. Then d is: (A) 1.1 cm away fro the lens (B) 0 (C) 0.55 cm towards the lens (D) 0.55 cm away from the lens
2.
A resistance is shown in the figure. Its value and tolerance are given respectively by: (A) 270 , 10% (B) 27 k, 10% (C) 27 k, 20% (D) 270 , 5%
3.
Drift speed of electrons, when 1.5 A of current flows in a copper wire of cross section 5 mm2, is v. If the electron density in copper is 9 × 1028 /m3 the value of v in mm/s is close to (Take charge of electron to be = 1.6 × 10–19 C) (A) 0.02 (B) 3 (C) 2 (D) 0.2
4.
An L-shaped object, made of thin rods of uniform mass density, is suspended with a string as shown in figure. If AB = BC, and the angle made by AB with downward vertical is , then: 1 1 (A) tan (B) tan 2 2 3 2 1 (C) tan (D) tan 3 3
5.
6.
ˆ , where K is a constant. The general A particle is moving with a velocity v K(yiˆ xj) equation for its path is: (A) y = x2 + constant (B) y2 = x + constant 2 2 (C) y = x + constant (D) xy = constant
A mixture of 2 moles of helium gas (atomic mass = 4 u), and 1 mole of argon gas (atomic mass = 40 u) is kept at 300 K in a container. The ratio of their rms speeds Vrms (helium) , is close to: Vrms (arg on) (A) 3.16 (C) 0.45
(B) 0.32 (D) 2.24
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JEE-MAIN-2019 (9th Jan-First Shift)-PCM-3
7.
A current loop, having two circular arcs joined by two radial lines is shown in the figure. It carries a current of 10 A. The magnetic field at point O will be close to: (A) 1.0 × 10–7 T (B) 1.5 × 10–7 T (C) 1.5 × 10–5 T (D) 1.0 × 10–5 T
8.
A block of mass m, lying on a smooth horizontal surface, is attached to a spring (of negligible mass) of spring constant k. The other end of the spring is fixed, as shown in the figure. The block is initially at rest in a equilibrium position. If now the block is pulled with a constant force F, the maximum speed of the block is 2F F (A) (B) mk mk F F (C) (D) mk mk
9.
For a uniformly charged ring of radius R, the electric field on its axis has the largest magnitude at a distance h from its centre. Then value of h is: R R (A) (B) 5 2 (C) R (D) R 2
10.
Two coherent sources produce waves of different intensities which interfere. After interference, the ratio of the maximum intensity to the minimum intensity is 16. The intensity of the waves are in the ratio: (A) 16 : 9 (B) 25 : 9 (C) 4 : 1 (D) 5 : 3
11.
Surface of certain metal is first illuminated with light of wavelength 1 = 350 nm and then, the light of wavelength 2 = 540 nm. It is found that the maximum speed of the photo electrons in the two cases differ by a factor of 2. The work function of the metal (in eV) is 1240 close to: (Energy of photon = eV) (in nm) (A) 1.8 (B) 2.5 (C) 5.6 (D) 1.4
12.
Temperature difference of 120°C is maintained between two ends of a uniform rod AB of length 2L. Another bent rod PQ, of same cross-section as AB 3L and length , is connected across AB (See 2 figure). In steady state, temperature difference between P and Q will be close to: (A) 45°C (B) 75°C (C) 60°C (D) 35°C FIITJEE Ltd., FIITJEE House, 29-A, Kalu Sarai, Sarvapriya Vihar, New Delhi -110016, Ph 46106000, 26569493, Fax 26513942 website: www.fiitjee.com.
JEE-MAIN-2019 (9th Jan-First Shift)-PCM-4 13.
A gas can be taken from A to B via two different processes ACB and ADB. When path ACB is used 60 J of heat flows into the system and 30 J of work is done by the system. If path ADB is used work down by the system is 10 J. the heat flow into the system in path ADB is (A) 40 J (B) 80 J (C) 100 J (D) 20 J
14.
Consider a tank made of glass (refractive index 1.5) with a thick bottom. It is filled with a liquid of refractive index . A student finds that, irrespective of what the incident angle I (see figure) is for a beam of light entering the liquid, the light reflected from the liquid glass interface is never completely polarized. For this to happen, the minimum value of is 5 3 (A) (B) 3 5 5 4 (C) (D) 3 3
15.
Mobility of electron in a semiconductor is defined as the ratio of their drift velocity to the applied electric field. If, for an n-type semiconductor, the density of electrons is 1019 m–3 and their mobility is 1.6 m2/(V.s) then the resistivity of the semiconductor(since it is an ntype semiconductor contribution of holes is ignored) is close to: (A) 2 m (B) 4 m (C) 0.4 m (D) 0.2 m
16.
A plane electromagnetic wave of frequency 50 MHz travels in free space along the positive x-direction. At a particular point in space and time, E 6.3ˆj V / m. . The corresponding magnetic field B, at that point will be ˆ ˆ (A) 18.9 × 10–8 kT (B) 2.1 × 10–8 kT ˆ ˆ (C) 6.3 × 10–8 kT (D) 18.9 × 108 kT
17.
Three charges +Q, q, +Q are placed respectively, at distance, 0, d/2 and d from the origin, on the x-axis. If the net force experienced by +Q, placed at x = 0, is zero, then value of q is (A) –Q/4 (B) +Q/2 (C) +Q/4 (D) –Q/2
18.
A copper wire is stretched to make it 0.5% longer. The percentage change in its electric resistance if its volume remains unchanged is (A) 2.0% (B) 2.5% (C) 1.0% (D) 0.5%
19.
A sample of radioactive material A, that has an activity of 10 mCi (1 Ci = 3.7 × 1010 decays/s), has twice the number of nuclei as another sample of different radioactive material B which has an activity of 20 mCi. The correct choices for half-lives of A and B would then be respectively: (A) 5 days and 10 days (B) 10 days and 40 days (C) 20 days and 5 days (D) 20 days and 10 days
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JEE-MAIN-2019 (9th Jan-First Shift)-PCM-5 20.
A heavy ball of mass M is suspended from the ceiling of car by a light string of mass m (m << M). When the car is at rest, the speed of transverse waves in the string is 60 ms–1. When the car has acceleration a, the wave-speed increases to 60.5 ms–1. The value of a, in terms of gravitational acceleration g is closest to: g g (A) (B) 30 5 g g (C) (D) 10 20
21.
A conducting circular loop made of a thin wire, has area 3.5 × 10–3 m2 and resistance 10 . It is placed perpendicular to a time dependent magnetic field B(t) = (0.4T) sin (50 t). The field is uniform in space. Then the net charge flowing through the loop during t = 0 s and t = 10 ms is close to: (A) 14 mC (B) 7 mC (C) 21 mC (D) 6 mC
22.
An infinitely long current carrying wire and a small current carrying loop are in the plane of the paper as shown. the radius of the loop is a and distance of its centre from the wire is d (d >> a). If the loop applies a force F on the wire then: a (A) F = 0 (B) F d a2 (C) F 3 d
a (D) F d
2
23.
A block of mass 10 kg is kept on a rough inclined plane as shown in the figure. A force of 3 N is applied on the block. The coefficient of static friction between the plane and the block is 0.6. What should be the minimum value of force P, such that the block does not move downward? (take g = 10 ms–2) (A) 32 N (B) 18 N (C) 23 N (D) 25 N
24.
A parallel plate capacitor is made of two square plates of side ‘a’, separated by a distance d(d <
K 0 a 2 ln K d(K 1)
K0 a2 ln K d 1 K0 a2 (D) 2 d
(C)
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JEE-MAIN-2019 (9th Jan-First Shift)-PCM-6 25.
A rod of length L at room temperature and uniform area of cross section A, is made of a metal having coefficient of linear expansion /°C. It is observed that an external compressive force F, is applied on each of its ends, prevents any change in the length of the rod, when it temperature rises by TK. Young’s modulus, Y, for this metal is F F (A) (B) A T A( T 273) F 2F (C) (D) 2A T A T
26.
A bar magnet is demagnetized by inserting it inside a solenoid of length 0.2 m, 100 turns, and carrying a current of 5.2 A. The coercivity of the bar magnet is (A) 285 A/m (B) 2600 A/m (C) 520 A/m (D) 1200 A/m
27.
Three blocks A, B and C are lying on a smooth horizontal surface, as shown in the figure. A and B have equal masses, m while C has mass M. Block A is given an initial speed towards B due to which it collides with B perfectly inelastically. The combined mass collides with C, also perfectly inelastically. The combined mass 5 collides with C, also perfectly inelastically if th of the 6 initial kinetic energy is lost in whole process. What is value of M/m? (A) 5 (B) 2 (C) 4 (D) 3
28.
When the switch S, in the circuit shown, is closed, then the value of current i will be (A) 3 A (B) 5 A (C) 4 A (D) 2 A
29.
If the angular momentum of a planet of mass m, moving a round the Sun in a circular orbit its L, about the center of the Sun, its areal velocity is: L 4L (A) (B) m m L 2L (C) (D) 2m m
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JEE-MAIN-2019 (9th Jan-First Shift)-PCM-7
30.
m are connected at the two ends of a 2 massless rigid rod of length . The rod is suspended by a thin
Two masses m and
wire of torsional constant k at the centre of mass of the rodmass system (see figure). Because of torsional constant k, the restoring torque is = k for angular displacement . If the rod is rotated by 0 and released, the tension in it when it passes through its mean position will be 3k02 2k02 (A) (B) 2 k0 k 0 2 (C) (D) 2
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JEE-MAIN-2019 (9th Jan-First Shift)-PCM-8
PART –B (CHEMISTRY) 31.
Two complexes [Cr(H2O)6]Cl3 (A) and [Cr(NH3)6]Cl3 (B) are violet and yellow coloured respectively. The incorrect statement regarding them is (A) 0 values of (A) and (B) are calculated from the energies of violet and yellow light, respectively. (B) both are paramagnetic with three unpaired electrons (C) both absorbs energies corresponding to their complementary colours (D) 0 value for (A) is less than that of (B)
32.
The correct decreasing order for acidic strength is (A) NO2CH2COOH rel="nofollow"> FCH2COOH > CNCH2COOH > ClCH2COOH (B) FCH2COOH > NCCH2COOH > NO2CH2COOH > ClCH2COOH (C) CNCH2COOH > O2NCH2COOH > FCH2COOH > ClCH2COOH (D) NO2CH2COOH > NCCH2COOH > FCH2COOH > ClCH2COOH
33.
The major product of the following reaction is 1 AlH iBu2 R C N ? 2 H2O (A) RCOOH (B) RCONH2 (C) RCHO (D) RCH2NH2
34.
The highest value of the calculated spin-only magnetic moment (in BM) among all the transition metal complexes is (A) 5.92 (B) 6.93 (C) 3.87 (D) 4.90
35.
0.5 moles of gas A and x moles of gas B exert a pressure of 200 Pa in a container of volume 10 m3 at 1000 K. Given R is the gas constant in JK–1 mol–1, x is 2R 2R (A) (B) 4 R 4 R 4 R 4 R (C) (D) 2R 2R
36.
The one that is extensively used as a piezoelectric material is (A) tridymite (B) amorphous silica (C) quartz (D) mica
37.
Correct statements among a to d regarding silicones are (i) they are polymers with hydrophobic character (ii) they are biocompatible (iii) in general, they have high thermal stability and low dielectric strength (iv) usually they are resistant to oxidation and used as greases. (A) (i), (ii), (iii) and (iv) (B) (i), (ii) and (iii) (C) (i) and (ii) only (D) (i), (ii) and (iv)
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JEE-MAIN-2019 (9th Jan-First Shift)-PCM-9 38.
The major product of the following reaction is
(A)
(B)
(C)
(D)
39.
In general, the properties that decrease and increase down a group in the periodic table respectively are (A) atomic radius and electronegativity (B) electron gain enthalpy and electronegativity (C) electronegativity and atomic radius (D) electronegativity and electron gain enthalpy
40.
A solution of sodium sulphate contains 92 g of Na+ ions per kilogram of water. The Molality of Na+ ions in that solution in mol kg–1 is (A) 12 (B) 4 (C) 8 (D) 16
41.
The correct match between Item-I and Item-II is: Item – I Item –II (drug) (test) (a) Chloroxylenol (p) Carbylamine test (b) Norethindrone (q) Sodium hydrogen carbonate test (c) Sulpha pyridine (r) Ferric chloride test (d) Penicillin (s) Bayer’s test (A) a r, b p, c s, d q (B) a q, b s, c p, d r (C) a r, b s, c p, d q (D) a q, b p, c s, d r
42.
A water sample has ppm level concentration of the following metals: Fe = 0.2, Mn = 5.0, Cu = 3.0, Zn = 5.0. The metal that makes the water sample unsuitable for drinking is (A) Cu (B) Mn (C) Fe (D) Zn
43.
The anodic half-cell of lead-acid battery is recharged using electricity of 0.05 Faraday. The amount of PbSO4 electrolyzed in g during the process is: (Molar mass of PbSO4 = 303 g mol–1) (A) 22.8 (B) 15.2 (C) 7.6 (D) 11.4 FIITJEE Ltd., FIITJEE House, 29-A, Kalu Sarai, Sarvapriya Vihar, New Delhi -110016, Ph 46106000, 26569493, Fax 26513942 website: www.fiitjee.com.
JEE-MAIN-2019 (9th Jan-First Shift)-PCM-10 44.
Which one of the following statements regarding Henry’s law is not correct? (A) Higher the value of KH at a given pressure, higher is the solubility of the gas in the liquids (B) Different gases have different KH(Henry’s law constant) values at the same temperature (C) The partial pressure of the gas in vapour phase is proportional to the mole fraction of the gas in the solution (D) The value of KH increases with increase of temperature and KH is function of the nature of the gas.
45.
The following results were obtained during kinetic studies of the reaction 2A B products Experiment
[A] [B] Initial rate of reaction (in mol L–1) (in mol L–1) (in mol L–1 min-1) I 0.10 0.20 6.93 10–3 II 0.10 0.25 6.93 10–3 III 0.20 0.30 1.386 10–2 The time(in minutes) required to consume half of A is (A) 5 (B) 10 (C) 1 (D) 100 46.
Major product of the following reaction is
(A)
(B)
(C)
(D)
47.
The alkaline earth metal nitrate that does not crystallise with water molecules is (A) Mg(NO3)2 (B) Sr(NO3)2 (C) Ca(NO3)2 (D) Ba(NO3)2
48.
20 mL of 0.1 M H2SO4 is added to 30 mL of 0.2 M NH4OH solution. The pH of the resultant mixture is [pkb of NH4OH = 4.7] (A) 5.2 (B) 9.0 (C) 5.0 (D) 9.4
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JEE-MAIN-2019 (9th Jan-First Shift)-PCM-11 49.
Adsorption of a gas follows Freundlich adsorption isotherm. In the given plot, x is the x mass of gas adsorbed on mass m of the adsorbent at pressure p. is proportional to m
(A) p2 (C) p1/2
(B) p1/4 (D) p
50.
Which amongst the following is the strongest acid? (A) CHBr3 (B) CHI3 (C) CH(CN)3 (D) CHCl3
51.
The ore that contains both iron and copper is (A) copper pyrites (B) malachite (C) dolomite (D) azurite
52.
For emission line of atomic hydrogen from n1= 8 to nf = n, the plot of wave number 1 against 2 will be(The Rydberg constant, RH is in wave number unit) n (A) Linear with intercept - RH (B) Non linear (C) Linear with slope RH (D) Linear with slope – RH
53.
54.
The isotopes of hydrogen are (A) tritium and protium only (C) protium, deuterium and tritium
According to molecular orbital theory, which of the following is true with respect to Li2 and Li2 ? (A) Li2 is unstable and Li2 is stable (C) Both are stable
55.
(B) protium and deuterium only (D) deuterium and tritium only
(B) Li2 is stable and Li2 is unstable (D) Both are unstable
The major product of the following reaction is:
(A)
(B)
(C)
(D)
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JEE-MAIN-2019 (9th Jan-First Shift)-PCM-12 56.
Aluminium is usually found in +3 oxidation state. In contrast, thallium exists in +1 and +3 oxidation states. This is due to (A) inert pair effect (B) diagonal relationship (C) lattice effect (D) lanthanoid contraction
57.
The increasing order of pKa of following amino acids in aqueous solution is Gly, Asp, Lys, Arg (A) Asp < Gly < Arg < Lys (B) Gly < Asp < Arg < Lys (C) Asp < Gly < Lys < Arg (D) Arg < Lys < Gly < Asp
58.
Consider the reversible isothermal expansion of an ideal gas in a closed system at two different temperatures T1 & T2 (T1 < T2). The correct graphical depiction of the dependence of work done(w) on the final volume(v).
59.
(A)
(B)
(C)
(D)
Arrange the following amine in the decreasing order of basicity:
(A) I > II > III (C) III > I > I 60.
(B) III > I > II (D) I > III > II
The compounds A and B in the following reaction are, respectively
(A) A = Benzyl alcohol, B = Benzyl cyanide (B) A = Benzyl chloride, B = Benzyl cyanide (C) A = Benzyl alcohol, B = Benzyl isocyanide (D) A = Benzyl chloride, B = Benzyl isocyanide
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JEE-MAIN-2019 (9th Jan-First Shift)-PCM-13
PART–C (MATHEMATICS)
61.
The value of
cos x
3
dx is:
0
4 3 4 (D) 3
(A) 0 (C)
(B)
2 3
62.
The maximum volume (in cu.m) of the right circular cone having slant height 3 m is (A) 6 (B) 3 3 4 (C) (D) 2 3 3
63.
For x 2 n 1, n N (the set of natural numbers), the integral
x.
2 sin x
(A) loge (C)
64.
1 sin 2 x
2
2
1 sec 2 x 2 1 c 2
x 2 1 1 loge sec 2 c 2 2
(B)
1 loge sec x 2 1 c 2
x2 1 (D) loge sec c 2
If y y x is solution of the differential equation x 1 y is equal to 2 7 (A) 64 49 (C) 16
65.
dx is 1
2 sin x 2 1 sin 2 x 2 1
dy 2y x 2 satisfying y 1 1, then dx
1 4 13 (D) 16 (B)
Axis of a parabola lies along x – axis. If its vertex and focus are at distance 2 and 4 respectively from the origin, on the positive x – axis then which of the following points does not lie on it? (A) 5, 2 6 (B) (8, 6)
(C) 6, 4 2
(D) (4, –4)
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JEE-MAIN-2019 (9th Jan-First Shift)-PCM-14 x2 y2 . If the eccentricity of the hyperbola 1 is greater than 2, 2 cos2 sin2 then the length of its latus rectum lies in the interval: 3 (A) 3, (B) , 2 2 3 (C) 2, 3 (D) 1, 2
66.
Let 0
67.
For x R 0, 1 , let f1 x
1 1 , f2 x 1 x and f3 x be three given functions. If x 1 x a function, J (x) satisfies f2oJof1 x f3 x then J x is equal to: 1 f3 x x (D) f1 x
(A) f3 x
(B)
(C) f2 x 68.
2 Let a ˆi ˆj, b ˆi ˆj kˆ and c be a vector such that a c b 0 and a.c 4, then c is
equal to: 19 (A) 2
(B) 9
(C) 8 69.
70.
71.
(D)
If a, b and c be three distinct numbers in G.P. and a b c x b then x can not be (A) –2 (B) –3 (C) 4 (D) 2
3 2 3 If cos1 cos1 , x then x is equal to: 4 3x 4x 2 145 145 (A) (B) 12 10 146 145 (C) (D) 12 11 Equation of a common tangent to the circle, x 2 y 2 6x 0 and the parabola, y 2 4x , is: (A) 2 3y 12x 1 (B) 3y x 3 (C) 2 3y x 12
72.
17 2
(D)
3y 3x 1
The system of linear equation x y z 2, 2x 3y 2z 5, 2x 3y a 2 1 z a 1 then (A) is inconsistent when a = 4
(B) has a unique solution for a 3
(C) has infinitely many solutions for a = 4
(D) inconsistent when a 3
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JEE-MAIN-2019 (9th Jan-First Shift)-PCM-15
73.
If the fractional part of the number (A) 6 (C) 4
2403 k is , then k is equal to: 15 15 (B) 8 (D) 14
74
The equation of line passing through (–4, 1, 3), parallel to the plane x 2y z 5 0 and x 1 y 3 z 2 intersecting the line is: 3 2 1 x 4 y 3 z 1 x 4 y 3 z 1 (A) (B) 2 1 4 1 1 3 x 4 y 3 z 1 x 4 y 3 z 1 (C) (D) 3 1 1 1 1 1
75.
Consider the set of all lines px qy r 0 such that 3p 2q 4r 0 . Which one of the following statements is true? 3 1 (A) The lines are concurrent at the point , . 4 2 (B) Each line passes through the origin. (C) The lines are all parallel (D) The lines are not concurrent
76.
lim
1 1 y4 2
y 0
y4
(A) exists and equals
(C) exists and equals
1
1
(B) exists and equals
4 2
2 2
1
2 1
(D) does not exist
2 2
77.
The plane through the intersection of the plane x y z 1 and 2x 3y z 4 0 and parallel to y – axis also pass through the point: (A) (–3, 0, –1) (B) (–3, 1, 1) (C) (3, 3, –1) (D) (3, 2, 1)
78.
If denotes the acute angle between the curves, y 10 x 2 and y 2 x 2 at a point of their intersection, then tan is equal to 4 9 7 (C) 17
(A)
8 15 8 (D) 17
(B)
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JEE-MAIN-2019 (9th Jan-First Shift)-PCM-16
79.
80.
cos sin If A , then the matrix A 50 when , is equal to 12 sin cos 3 1 1 3 2 2 2 2 (A) (B) 1 3 1 3 2 2 2 2 3 1 1 3 2 2 2 2 (C) (D) 1 3 3 1 2 2 2 2
If the Boolean expression p q ~ pq is equivalent to p q , where , , , then the ordered pair , is :
81.
82.
(A) ,
(B) ,
(C) ,
(D) ,
5 students of a class have an average height 150 cm and variance 18 cm2. A new student, whose height is 156 cm, joined them. The variance (in cm2) of the height of these six students is: (A) 16 (B) 22 (C) 20 (D) 18 4 2 For any , , the expression 3 sin cos 6 sin cos 4 sin6 equals: 4 2 (A) 13 4 cos2 6 sin2 cos2 (B) 13 4 cos6
(C) 13 4 cos2 6cos4
(D) 13 4 cos4 2 sin2 cos2
83.
The area (in sq. units) bounded by the parabola y x 2 1, the tangent at the point (2, 3) to it and the y – axis is: 8 32 (A) (B) 3 3 53 14 (C) (D) 3 3
84.
Let a1,a2 ,.....,a30 be an A.P., S ai and T a 2i1 . If a5 27 a and S 2T 75 ,
30
15
i 1
then a10 is equal to (A) 52 (C) 47
i 1
(B) 57 (D) 42
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JEE-MAIN-2019 (9th Jan-First Shift)-PCM-17
85.
86.
if x 1 5, a bx, if 1 x 3 Let f :R R be a function defined as f x . Then f is: b 5x, if 3 x 5 30, if x5 (A) continuous if a 5 and b 5 (B) continuous if a 5 and b 10 (C) continuous if a 0 and b 5 (D) not continuous for any values of a and b
3 2i sin Let A 0 , : purely imaginary . Then the sum of the elements in A 2 1 2i sin is: 5 (A) (B) 6 2 3 (C) (D) 4 3
87.
Consider a class of 5 girls and 7 boys. The number of different teams consisting of 2 girls and 3 boys that can be formed from this class, if there are two specific boys A and B, who refuse to be the members of the same team, is: (A) 500 (B) 200 (C) 300 (D) 350
88.
Let and be two roots of the equation x 2 2x 2 0 , then 15 15 is equal to: (A) –256 (B) 512 (C) –512 (D) 256
89.
Three circles of radii a, b, c ( a < b < c ) touch each other externally. If they have x – axis as a common tangent, then: 1 1 1 1 1 1 (A) (B) a b c b a c (C) a, b, c are in A.P.
90.
(D)
a, b, c are in A.P.
Two cards are drawn successively with replacement from a well shuffled deck of 52 cards. Let X denote the random variable of number of aces obtained in the two drawn cards. Then P X 1 P X 2 equals: 49 169 24 (C) 169
(A)
52 169 25 (D) 169
(B)
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JEE-MAIN-2019 (9th Jan-First Shift)-PCM-18
JEE (Main) – 2019 ANSWERS PART A – PHYSICS 1.
D
2.
B
3.
A
4.
B
5.
C
6.
A
7.
D
8.
D
9.
B
10.
B
11.
A
12.
A
13.
A
14.
B
15.
B
16.
B
17.
D
18.
C
19.
B
20.
C
21.
BONUS
22.
D
23.
A
24.
A
25.
A
26.
B
27.
C
28.
B
29.
C
30.
C
PART B – CHEMISTRY 31.
A
32.
D
33.
C
34.
A
35.
D
36.
C
37.
D
38.
A
39.
C
40.
C
41.
C
42.
B
43.
C
44.
A
45.
B
46.
D
47.
D
48.
B
49.
C
50.
C
51.
A
52.
D
53.
C
54.
B
55.
A
56.
A
57.
C
58.
A
59.
B
60.
D
PART C – MATHEMATICS 61.
A
62.
D
63.
D
64.
C
65.
B
66.
A
67.
A
68.
A
69.
D
70.
A
71.
B
72.
D
73.
B
74.
C
75.
A
76.
A
77.
D
78.
B
79.
C
80.
A
81.
C
82.
B
83.
A
84.
A
85.
D
86.
D
87.
C
88.
A
89.
A
90.
D
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JEE-MAIN-2019 (9th Jan-First Shift)-PCM-19
HINTS AND SOLUTIONS PART A – PHYSICS 1.
If u = –10 cm v = +10 cm f = 5 cm 1 2 Glass plate shift = t 1 1.5 1 = 0.5 cm 3 So, new u = 10 – 0.5 = 9.5 cm 1 1 1 v u f 1 1 1 v 9.5 5 After solving we get, 47.5 47.5 2.5 v= . Hence, shift 10 0.55 cm 4.5 4.5 4.5
2.
Fact based.
3.
i = ne AVd 1.5 = 9 × 1028 × 1.6 × 10–19 × 5 × Vd Vd = 0.02
4.
Lets considered mass of each rod is m for stable equilibrium the torque about point O should be zero. Torque balance about O a a mg sin mg cos a sin 2 2 1 tan = 3 1 tan1 3
O a
mg
a mg
5.
6.
dx dy y ; x dt dt dx y dy x y2 = x 2 + c (VRMS )He MAr (VRMS )Ar MHe
40 10 = 3.16 4
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JEE-MAIN-2019 (9th Jan-First Shift)-PCM-20
7.
Magnetic field at centre of an area subtending angle at the centre
0I . 4r
1 1 B 0 10 2 2 5 10 4 3 10 4 = B 104 105 10 5 30 3 8.
K
When Vmax acceleration = 0 F x K
m
F
Apply work energy theorem W sp + W F = K.E.
1 1 F2 F 2 1 2 Kx 2 F.x K.E. ; K 2 mumax 2 2 K K 2 F2 1 2 mumax 2K 2 9.
;
F Vmax mK
Electric field kQx E 2 (x R2 )3/2 dE For maxima 0 dx R After solving we get, x 2
10.
I1 I2 16 ; I I 1 1 2 3 I1 5 I2
11.
I1 I2 I1 I2
4 1
I1 25 I2 9
1240 (KE)I 4x …(1) 350 1240 (KE)II x …(2) 540 (1) – (2) 1240 1240 3.542 2.296 = 3x 350 540 1.246 = 3x ; x = 0.41 = 2.296 – 0.41 = 1.886
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JEE-MAIN-2019 (9th Jan-First Shift)-PCM-21
12.
TA TB
T1 T2 3R 3 120 = 45 8R 5 8 5
T1
R/2 A
B R/2
R
T2
R/4
R/4 R
13.
As temperature at point A and C is same. Internal energy change will be same. Q – W = Q – W 60 – 30 = Q – 10 Q = 40 J
14.
Sin 90° = sin 1 Sin sin = 1.5 sin r tan = 1.5 1.5 tan 3 1 Sin 2 9 4
16.
17.
Use I = neAvd and
r
vd E
E C B E 6.3 10 27 B = 2.1 × 1019 C 3 108 kq d 2
18.
92 = 9 + 42 3 5
15.
Completely polarized
2
k 2 2
3d 2 4Q 4q 9 Q q 9
A 2 2 R A V R
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JEE-MAIN-2019 (9th Jan-First Shift)-PCM-22
dR 2d R dR Hence, 1%. R 19.
Activity = (number of atoms) 10 = A NA …(1) 20 = B NB …(2) NA = 2NB …(3) A 1 Solving we get, B 4
20.
v T/ M
g
g2 a2 60.5 g 60
2
1 a2 1 using by binomial approximation. 1 2 2 g 60 g a 30 1
21.
B(t) = 0.4 Sin 50 10 2
50 = 0.4 Sin = 0.4 100 Q Q q = R R 0.4 3.5 10 3 = = 140 mC 10 No option matching. 22.
Force on one pole m oI m = pole F 2 d2 x 2 strength Total force = 2F sin 2 o Im x oI mx = 2 2 2 2 [d2 a2 ] 2 d a d a = m2x = M = I a2 oIa 2 Total force = 2(d2 a2 )
oIa 2 2d2
F m
x
x
d x Current wire
–m F
d a
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JEE-MAIN-2019 (9th Jan-First Shift)-PCM-23 23.
For equilibrium of the block net force should be zero. Hence we can write. mg sin + 3 = P + friction mg sin + 3 = P + mg cos . After solving, we get, P = 32 N.
24.
Let’s consider a strip of thickness dx at a distance of x from the left end as shown in the figure. y d x a d y x a
C1 d
K y C2 x
dx a
0 adx k adx ; C2 0 (d y) y C1C2 k0 adx Ceq C1 C2 kd (1 k)y Now integrating it from 0 to a C1
a
0
25.
k0 adx d kd (1 k) x a
k0a2 nk d(k 1)
L = L T L Strain = T L
Y
;
F A T
26.
B = oH ; oni = oH 100 5.2 H 0.2 H = 2600 A/m
27.
Apply LMC (Linear Momentum Conservation) mv = (2m + M)v mv v 2m M Initial energy 1 mv 2 2 Final energy
1 mv (2m M) 2 2m M
2
Initial kinetic energy – Final kinetic energy = After solving, we get,
5 of initial kinetic energy. 6
M 4. m
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JEE-MAIN-2019 (9th Jan-First Shift)-PCM-24 28.
i3 + i2 = i1 20 v 10 v v 2 4 2 v = 10 V 10 i1 2 = 5 amp.
29.
Based on Kepler’s law. dA L dt 2m
30.
v
20V 20
10V 4
2
k ; = 0 × I T m2 3 k 2 T m 2 0 where I m 3 I 3 2 k = 0
m
L/3
2L/3
m/2
PART B – CHEMISTRY 31.
0 is calculated from the energies of absorbed radiation not from emitted radiation (complementary colour).
32.
Acid strength in this case varies directly with the electron withdrawing power of the groups attached to the -carbon of CH3COOH. The order of electron withdrawing tendency is NO2 > CN > F > Cl
33. H2 O R C N Al H R C N Al H RC N Al RCHO H2N Al
34.
Maximum number of unpaired electron of metal or metal ion in complexes = n = 5
s n n 2 35 5.916 5.92 35.
PV = nRT 200 10 = (0.5 + x)R 1000 4R On solving x 2R
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JEE-MAIN-2019 (9th Jan-First Shift)-PCM-25 36.
Materials those produce electric current when they are put under mechanical stress are called piezoelectric materials.
37.
Silicones are polymers and hydrophobic due to presence of alkyl groups. They are used as greases as some of them are cyclic. Br
38.
Br Br2
EtOH OEt Br
39.
40.
On moving down a group, electronegativity decreases and atomic radius increases for representative elements.
w 1000 Molality of Na 2 ( Na2SO4 contains two Na+ ions) W M 92 1000 2 8 23 1000
41
42.
The permissible level in ppm unit is Fe = 0.2 Mn = 0.05 Cu = 3 Zn = 5 Mn is higher
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JEE-MAIN-2019 (9th Jan-First Shift)-PCM-26
43.
Pb s SO 24 PbSO4 2 e
For 2F current passed, PbSO4 deposited = 303 g/mol 0.05 303 7.6 g 2 For 0.05 F: PbSO4 deposited = 44.
Gases having higher KH value are less soluble.
45.
R = k[A]x [B]y 6.93 10–3 = k (0.1)x (0.2)y (i) 6.93 10–3 = k (0.1)x (0.25)y (ii) 1.386 10–2 = k (0.2)x (0.3)y (iii) y = 0 (from (i) & (ii)), zero order w.r.t. B x = 1 (from (i) & (iii)) First order wrt A 6.93 10–3 = k (0.1) k = 6.93 10–3 min-1 Cl
46.
Cl Cl
O
NH2
H2N
NH NH2
O
O
HCl
O
Et3N HCl Et3 NHCl Cl
O Free radical polymerization
NH
n
NH2 O
47.
Due to larger size of Ba2+ ion, Ba(NO3)2 can not hold water molecules during crystallization.
48.
H2SO4 2NH4OH NH4 2 SO 4 2H2O 2 m.m
6 m.m 2 m.m
pOH 4.7 log
2 m.m
4 5 2
pH = 14 – 5 = 9
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JEE-MAIN-2019 (9th Jan-First Shift)-PCM-27
49.
x KP1/n m x 1 or log logK logP m n 1 2 1 Slope n 4 2 x P1/2 m
50.
CH CN 3 C CN 3 H
Negative charge of the conjugate base C CN 3 in extensively delocalized through the C N group. 51.
Copper pyrites is CuFeS2 Malachite: CuCO3.Cu(OH)2 Azurite : 2Cu(CO)3.Cu(OH)2 Dolomite: CaCO3.MgCO3
52.
For emission line nf < ni 1 1 1 1 RZ 2 2 2 R 2 2 n 8 ni nf 1 1 or, RH 2 64 n R R H 2H 64 n 1 R RH 2 H n 64 y = mx + c Slope = - RH
53.
The isotopes are: 1 2 3 1H, 1 H and 1 H P,D,T
54.
Both Li2 and Li2 have same bond order. But the number of antibonding electrons is less in Li2 than in Li2
55.
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JEE-MAIN-2019 (9th Jan-First Shift)-PCM-28 56.
The outermost electron configuration of Tl is 6s26p1. The 6s electrons are strongly attracted towards the nucleus due to its more penetrating power and deshielding of d and f-electrons. Hence 6s electrons do not participate in bonding.
57.
58.
Let the gas is expanded from V1 to V at T1 and from V2 to V at T2 At T1 V |W 1| = nRT1 n nRT nV nV1 V1 Similarly at T2 |W 2| = nRT2(nV - nV2) W 1 = nRT1 nV – nRT1 nV1 W 2 = nRT2nV – nRT2nV2 Slope of W 2 > Slope of W 1 As nRT2 > nRT1(T2 > T1) The intercept of W 2 is more negative than that of W 1 because V2 > V1.
59.
In(III), nitrogen atom undergoes sp3, in(I) sp2 hybridization. In(II), the lone pair participate in resonance. CH2Cl
60. HCHO HCl
CH2NC
AgCN
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JEE-MAIN-2019 (9th Jan-First Shift)-PCM-29
PART C – MATHEMATICS 2
cos x
61.
3
cos x
3
0
2
dx
3
2 cos x dx 0
2
3
2 cos x dx 0
2 4 2 (By wallis formula) 3 3 62.
3 r 2 h2 9
Volume of cone is
1 2 r h 3
1 h 9 h2 3 dv 1 9 3h2 0 dh 3 9 3h2 0 h2 3, h 3 1 V 6 3 2 3 3 V
2 sin x
1 1 cos x
h
r
1
2 sin x 2 1 1 cos x 2 1 63.
x
2
2
x 2 1 sin 2 dx x x2 1 cos 2
x 2 1 x tan dx 2 x2 1 Let t 2xdx 2dt 2 tan t dt n sec t c x2 1 n sec c 2
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JEE-MAIN-2019 (9th Jan-First Shift)-PCM-30
64.
dy 2y x 2 dx dy 2 yx dx x
x
This is linear differential equation in
dy dx
2
dx Integrating factor e x x2 Solution of differential equation is yx 2 x 3 dx x4 c 4 Curve passes through (1, 1) 3 then c 4 4 x 3 yx 2 4 1 Put x 2 yx 2
4
1 3 2 1 y 4 4 49 y 16
65.
Vertex is (2, 0) a=2 Any general point on given parabola can be taken as 2 2t 2 , 4t t R .
(8, 6) does not lie on this. 66.
x2 y2 1 cos2 sin2 e 2 (given) sin2 4 cos2 1 tan2 4 tan2 3 , 3 2 sin2 2 tan sin Latus rectum 2 cos e2 4 1
for , , 2 tan sin is increasing function 3 2 Hence latus rectum 3, FIITJEE Ltd., FIITJEE House, 29-A, Kalu Sarai, Sarvapriya Vihar, New Delhi -110016, Ph 46106000, 26569493, Fax 26513942 website: www.fiitjee.com.
JEE-MAIN-2019 (9th Jan-First Shift)-PCM-31 67.
x R 0,1
1 1 , f2 x 1 x, f3 x x 1 x Given f2 J f1 x f3 x f1 x
1 J f1 x f3 x
J f1 x 1 f3 x 1
1 1 x
x x 1 1 x 1 J x x 1 1 1 x 1 J x f3 x 1 x J f1 x
68.
ˆ c xiˆ yjˆ zkˆ a ˆi ˆj, b ˆi ˆj k, ac b 0 ˆi ˆj kˆ
1 1 0 ˆi ˆj kˆ 0 x
y
z
ˆi z ˆj z kˆ y x
1 z 0 z 1,
Also x y 1, and a.c 4 x y 4
x
3 5 , y 2 2
2 9 25 38 19 c x 2 y 2 z2 1 4 4 4 2 69.
70.
a ar ar 2 xar 1 r2 r 1 since a 0 so x; 1 r x r r 1 r , 2 2, x , 1 3, r 3 2 3 cos1 cos1 x 4 3x 4x 2 2 3 4 9 cos 1 1 2 1 3x 4x 9x 16x 2 2
1 9x 2 4 16x 2 9 2x 2 12x 2
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JEE-MAIN-2019 (9th Jan-First Shift)-PCM-32
6 9x 2 4 16x 2 9 Square both side 36 144x 4 81x 2 64x 2 36 144x 4 145x 2 145x 2 145 x4 x ,0 144 12 3 145 x hence x 4 12 ty x t 2
71.
3 t2
3
1 t2
t 3 3y x 3 By applying Crammer’s Rule 1 1 1
72.
D 2 3
2 2
2 3 a 1
3 a2 1 6 2 a 2 1 4 2
2
a 1 2 a 3 If a 3 system has unique solution If a 3 2x 3y 2z 3 1 Hence system is inconsistent for a 3 x y z 1 2x 3y 2z 1
100
73.
2403 23.2400 8. 1 15 15 15 15
8
100
2
C0 100C1 15 100C2 15 ....... 15
8 2 8 100 C1 15 100C2 15 ...... 15 Remainder is 8.
74.
x 4 y 3 z 1 a b c L x 2y z 5 0 x 1 y 3 z 2 L intersects 3 2 1 Let the line L be
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JEE-MAIN-2019 (9th Jan-First Shift)-PCM-33 3
0
1
a
b
c 0
3 2 1 2a 0b 6c 0 Also a 2b c 0 a b c 3 1 1 x 4 y 3 z 1 L is 3 1 1
px qy r 0
75.
3p 2q px qy 0 4 3 2 p x q y 0 4 4 3 1 x and y . 4 2 76.
n
1 x So,
1 nx when x 0
1 y4 1 2
lim y 0
y4 2
y4 2 2 y4
y4 2 1 1 8 2 1 4 y 8 4 2
77.
Equation of required plane is x y z 1 2x 3y z 4 0
1 2 x 1 3 y 1 0 since given plane is parallel to y – axis 3 1 0
1 3
Hence equation of plane is x 4z 7 0 78.
y x 2 2 and y 10 x 2 Meet at 2, 6 m1 4 and m2 4 8 tan 15
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JEE-MAIN-2019 (9th Jan-First Shift)-PCM-34
79.
cos sin A sin cos cos sin cos sin A2 sin cos sin cos cos 2 sin 2 A2 sin 2 cos 2 By using symmetry cos 50 sin 50 A 50 sin 50 cos 50
At
A 50
80.
12
25 25 3 sin cos sin cos 6 6 6 6 2 25 25 1 sin cos sin cos 6 6 6 6 2
1 2 3 2
Check all option repeatedly (i) A B ~ A B A B ~ A B A B A B i is correct
(ii) A B ~ A B A ~ A B f B f
(iii) A B ~ A B B (iv) A B ~ A B B A ~ A B f f only (1) is correct
81.
Let 5 students are x1, x 2 , x 3 , x 4 , x 5 Given x 2 i
x 5
x x 5
2
5 i
750
150
………..(1)
i 1
x 18 5
2 i
2
150 18 5
xi2 22500 18 5
x
2 i
112590
…………(2)
i 1
Height of new student = 156 (Let x6) Then x1 x 2 x 3 x 4 x5 x 6 750 156
x new
x1 x 2 x 3 x 4 x 5 x 6 906 151 6 6
……….(3)
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JEE-MAIN-2019 (9th Jan-First Shift)-PCM-35
x Variance (new)
2
i
x new
2
6 from equation (2) and (3) 2 112590 156 2 variance (new) 151 22821 22801 20 6 82.
2
3 1 sin2 6 1 sin 2 4 sin6
3 1 2 sin 2 sin2 2 6 6 sin 2 4 sin6 2
6
9 3 sin 2 4 sin
9 12 sin2 cos2 4 1 cos2
3
9 12 1 cos2 cos2 4 1 3 cos2 3 cos4 cos6 2
4
2
4
6
13 12cos 12cos 12 cos 12cos 4cos 13 4 cos6 3
83.
Area
3
xdy xdy
5
1
3
3
y5 y 1dy 4 1 5
y2 5y 2 4
3
3
2 3/2 y 1 3
(2, 3)
3
1
–1
5
9 25 2 15 2 25 16 8 4 3 3
30
84.
–5
15
S ai , T a2i1 , a5 27, S 2T 75 i 1
i 1
Let ai a i 1 D
S a1 a2 a3 .................. a30 T a1 a3 a5 ............... a29 2T 2a1 2a3 2a5 .................. 2a29 S 2T a2 a1 a 4 a3 a 6 a5 .............. a30 a29 75 = 15D But S 2T 75 15D 75 D 5 Now a5 27 a 4D 27 a 27 20 a 7 now a10 a 9D 7 45 52
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JEE-MAIN-2019 (9th Jan-First Shift)-PCM-36 85.
For x = 1 R.H.L = a + b L.H.L = 5 So to be continuous at x = 1 a+b=5 …………(i) for x = 3 R.H.L. = b + 15 L.H.L = a + 3b b + 15 = a + 3b a + 2b = 15 ………….(ii) for x = 5 R.H.L = 30 L.H.L = b + 25 b + 25 = 30 b = 5. From equation (ii) a = 10 but a 10 and b 5 does not satisfied equation (i) So f (x) is discontinuous for a R and b R
86.
z
3 2i sin 1 2i sin 1 2i sin 1 2i sin
3 4 sin 8i sin 2
z
1 4 sin2 For purely imaginary real part should be zero. i.e. 3 4 sin2 0 . 3 2 2 2 2 , , , Sum of all values is 3 3 3 3 3 3 3
i.e. sin
87.
Number of ways = Total number of ways without restriction – When two specific boys are in 7 C3 5C2 350 If team without any restriction, total number of ways of forming team is two specific boys B1, B2 are in same team then total number of ways of forming team 5 5 equals to C1 C2 50 ways total ways = 350 – 50 = 300 ways
88.
x 2 2x 2 0 x 1 1
2
3 i 4
x 1 i 2 e
15 , 15
15
2
3 2cos 15. 4
1 28 2 256 2
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JEE-MAIN-2019 (9th Jan-First Shift)-PCM-37 C1
89.
C3
Length of direct common tangent for circle C1 and C2 is AB
a b
2
a b
2
For C2 and C3 Length of direct common tangent is BC
a c
2
a c
b
2
For C1 and C3 Length of direct common tangent is AC
b c
2
b c
A
C2 a B
c C
2
AB BC AC
a b
2
2
a b
b c
2
b c
a c
2
2
a c
2
ab ac bc 1 1 1 c b a 90.
4 48 24 2 52 52 169 4 4 1 P x 2 52 52 169 25 P x 1 P x 2 169 P x 1
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FIITJEE Solutions to JEE(Main)-2019 Test Date: 9th January 2019 (Second Shift)
PHYSICS, CHEMISTRY & MATHEMATICS Paper - 1 Time Allotted: 3 Hours
Maximum Marks: 360
Please read the instructions carefully. You are allotted 5 minutes specifically for this purpose.
Important Instructions: 1.
The test is of 3 hours duration.
2.
This Test Paper consists of 90 questions. The maximum marks are 360.
3.
There are three parts in the question paper A, B, C consisting of Physics, Chemistry and Mathematics having 30 questions in each part of equal weightage. Each question is allotted 4 (four) marks for correct response.
4.
Out of the four options given for each question, only one option is the correct answer.
5.
For each incorrect response 1 mark i.e. ¼ (one-fourth) marks of the total marks allotted to the question will be deducted from the total score. No deduction from the total score, however, will be made if no response is indicated for an item in the Answer Box.
6.
Candidates will be awarded marks as stated above in instruction No.3 for correct response of each question. One mark will be deducted for indicating incorrect response of each question. No deduction from the total score will be made if no response is indicated for an item in the answer box.
7.
There is only one correct response for each question. Marked up more than one response in any question will be treated as wrong response and marked up for wrong response will be deducted accordingly as per instruction 6 above.
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JEE-MAIN-2019 (9th Jan-Second Shift)-PCM-2
PART –A (PHYSICS) 1.
At a given instant, say t = 0, two radioactive substances A and B have equal activates. R R The ratio B of their activities. The ratio B of their activates after time t itself decays RA RA with time t as e–3t. If the half-life of A is n2, the half-life of B is: (A) 4n2 (C)
n2 4
(B)
n2 2
(D) 2n2
2.
A power transmission line feeds input power at 2300 V to a step down transformer with its primary windings having 4000 turns. The output power is delivered at 230 V by the transformer. If the current in the primary of the transformer is 5A and its efficiency is 90%, the output current would be: (A) 50 A (B) 45 A (C) 35 A (D) 25 A
3.
The energy associated with electric field is (UE) and with magnetic field is (UB) for an electromagnetic wave in free space. Then: U (A) UE B (B) UE > UB 2 (C) UE < UB (D) UE = UB
4.
A force acts on a 2 kg object so that its position is given as a function of time as x = 3t2 + 5. What is the work done by this force in first 5 seconds? (A) 850 J (B) 950 J (C) 875 J (D) 900 J
5.
A particle having the same charge as of electron moves in a circular path of radius 0.5 cm under the influence of a magnetic field of 0.5 T. If an electric field of 100 V/m makes it to move in a straight path, then the mass of the particle is (given charge of electron = 1.6 × 10–19 C) (A) 9.1 × 10–31 kg (B) 1.6 × 10–27 kg –19 (C) 1.6 × 10 kg (D) 2.0 × 10–24 kg
6.
Two point charges q1( 10 C) and q2 ( 25 C) are placed on the x-axis at x = 1 m and x = 4 m respectively. The electric field (in V/m) at a point y = 3 m on y-axis is, 1 9 10 9 Nm2C2 take 4 0 2 ˆ 10 (A) (63iˆ 27i) (B) ( 63iˆ 27iˆ ) 10 2 ˆ 10 2 (C) (81iˆ 81i)
(D) ( 81iˆ 81iˆ ) 10 2
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JEE-MAIN-2019 (9th Jan-Second Shift)-PCM-3 7.
Expression for time in terms of G (universal gravitational constant), h (Planck constant) and c (speed of light) is proportional to: (A) (C)
hc 5 G Gh c5
(B) (D)
c3 Gh Gh c3
8.
Ge and Si diodes start conducting at 0.3 V and 0.7 V respectively. In the following figure if Ge diode connection are reversed, the value of Vo changes by: (assume that the Ge diode has large breakdown voltage) (A) 0.8 V (B) 0.6 V (C) 0.2 V (D) 0.4 V
9.
The top of a water tank is open to air and its water level is maintained. It is giving out 0.74 m3 water per minute through a circular opening of 2 cm radius in its wall. The depth of the centre of the opening from the level of water in the tank is close to: (A) 6.0 m (B) 4.8 m (C) 9.6 m (D) 2.9 m
10.
The energy required to take a satellite to a height ‘h’ above Earth surface (radius of Earth = 6.4 × 103 km) is E1 and kinetic energy required for the satellite to be in a circular orbit at this height is E2. The value of h for which E1 and E2 are equal is (A) 1.6 × 103 km (B) 3.2 × 103 km 3 (C) 6.4 × 10 km (D) 1.28 × 104 km
11.
Two Carnot engines A and B are operated in series. The first one, A, receives heat at T1(= 600 K) and rejects to a reservoir at temperature T2. The second engine B receives heat rejected by the first engine and, in turns, rejects to a heat reservoir at T3 (=400 K). Calculate the temperature T2 if the work outputs of the two engines are equal: (A) 600 K (B) 400 K (C) 300 K (D) 500 K
12.
A series AC circuit containing an inductor (20 mH), a capacitor (120 F) and a resistor (60 ) is driven by an AC source of 24 V/50 Hz. The energy dissipated in the circuit in 60 s is: (A) 5.65 × 102 J (B) 2.26 × 103 J (C) 5.17 × 102 J (D) 3.39 × 103 J
13.
A particle is executing simple harmonic motion (SHM) of amplitude A, along the x-axis, about x = 0. When its potential energy (PE) equals kinetic energy (KE), the position of the particle will be A A (A) (B) 2 2 2 A (C) (D) A 2 FIITJEE Ltd., FIITJEE House, 29-A, Kalu Sarai, Sarvapriya Vihar, New Delhi -110016, Ph 46106000, 26569493, Fax 26513942 website: www.fiitjee.com.
JEE-MAIN-2019 (9th Jan-Second Shift)-PCM-4 14.
A mass of 10 kg is suspended vertically by a rope from the roof. When a horizontal force is applied on the rope at some point, the rope deviated at an angle of 45° at the roof point. If the suspended mass is at equilibrium, the magnitude of the force applied is (g = 10 ms–2) (A) 200 N (B) 140 N (C) 70 N (D) 100 N
15.
A 15 g mass of nitrogen gas is enclosed in a vessel at a temperature 27°C. Amount of heat transferred to the gas, so that rms velocity of molecules is doubled, is about: [Take R = 8.3 J/K mole] (A) 0.9 kJ (B) 6 kJ (C) 10 kJ (D) 14 kJ
16.
In a Young’s double slit experiment, the slits are placed 0.320 mm apart. Light of wavelength = 500 nm is incident on the slits. The total number of bright fringes that are observed in the angular range –30° 30° is: (A) 640 (B) 320 (C) 321 (D) 641
17.
Two plane mirrors are inclined to each other such that a ray of light incident to the first mirror (M1) and parallel to the second mirror (M2) is finally reflected from the second mirror (M2) parallel to the first mirror (M1). The angle between the two mirrors will be: (A) 45° (B) 60° (C) 75° (D) 90°
18.
A rod of length 50 cm is pivoted at one end. It is raised such that if makes an angle of 30° fro the horizontal as shown and released from rest. Its angular speed when it passes through the horizontal (in rad s–1) will be (g = 10 ms–2). 30 (A) (B) 30 2 20 30 (C) (D) 2 2
19.
A carbon resistance has a following colour code. What is the value of the resistance? (A) 530 k 5% (B) 5.3 M 5% (C) 6.4 M 5% (D) 64 k 10%
20.
One of the two identical conducing wires of length L is bent in the form of a circular loop and the other one into a circular coil of N identical turns. If the same current is passed in both, the ratio of the magnetic field at the central of the loop (BL) to that at the centre of B the coil (BC), i.e. L will be BC 1 (A) N (B) N 1 (C) N2 (D) 2 N FIITJEE Ltd., FIITJEE House, 29-A, Kalu Sarai, Sarvapriya Vihar, New Delhi -110016, Ph 46106000, 26569493, Fax 26513942 website: www.fiitjee.com.
JEE-MAIN-2019 (9th Jan-Second Shift)-PCM-5 21.
A rod of mass ‘M’ and length ‘2L’ is suspended at its middle by a wire. It exhibits torsional oscillations; If two masses each of ‘m’ are attached at distance ‘L/2’ from its centre on both sides, it reduces the oscillation frequency by 20%. The value of ratio m/M is close to: (A) 0.77 (B) 0.57 (C) 0.37 (D) 0.17
22.
Charge is distributed within a sphere of radius R with a volume charge density A (r) 2 e 2r /a , where A and a are constants. If Q is the total charge of this charge r distribution, the radius R is: a 1 Q (A) a log 1 (B) log 2aA 2 1 Q 2aA a 1 a 1 (C) log (D) log 1 2 2 2aA 1 Q 2aA
23.
A parallel palate capacitor with square plates is filled with four dielectrics of dielectric constants K1, K2, K3, K4 arranged as shown in the figure. The effective dielectric constant K will be: (K 1 K 3 ) (K 2 K 4 ) K1 K 2 K 3 K 4 (K K 2 ) (K 3 K 4 ) (C) K 1 K1 K 2 K 3 K 4
(A) K
(K1 K 2 ) (K 3 K 4 ) 2(K 1 K 2 K 3 K 4 ) (K1 K 4 ) (K 2 K 3 ) (D) K 2(K 1 K 2 K 3 K 4 )
(B) K
24.
The pitch and the number of divisions, on the circular scale, for a given screw gauge are 0.5 mm and 100 respectively. When the screw gauge is fully tightened without any object, the zero of its circular scale lies 3 divisions below the mean line. The readings of the main scale and the circular scale for a thin sheet, are 5.5 mm and 48 respectively, the thickness of this sheet is (A) 5.755 mm (B) 5.950 mm (C) 5.725 mm (D) 5.740 mm
25.
A musician using an open flute of length 50 cm produces second harmonic sound waves. A person runs towards the musician from another end of a hall at a speed of 10 km/h. If the wave speed is 330 m/s, the frequency heard by the running person shall be close to: (A) 666 Hz (B) 753 Hz (C) 500 Hz (D) 333 Hz
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JEE-MAIN-2019 (9th Jan-Second Shift)-PCM-6 26.
In a car race on straight road, car A takes a time ‘t’ less than car B at the finish and passes finishing point with a speed ‘v’ more than that of car B. Both the cars start from rest and travel with constant acceleration a1 and a2 respectively. Then ‘v’ is equal to 2a1a1 (A) t (B) 2a1a 2 t a1 a 2 a a2 (C) a1a2 t (D) 1 t 2
27.
The magnetic field associated with a light wave is given, at the origin, by B = B0 [sin(3.14 × 107)ct + sin(6.28 × 107)ct] If this light falls on a silver plate having a work function of 4.7 eV, what will be the maximum kinetic energy of the photo electrons? (A) 6.82 eV (B) 12.5 eV (C) 8.52 eV (D) 7.72 eV
28.
In the given circuit the internal resistance of the 18 V cell is negligible. If R1 = 400 , R3 = 100 and R4 = 500 and the reading of an ideal voltmeter across R4 is 5 V, then the value of R2 will be (A) 300 (B) 450 (C) 550 (D) 230
29.
In a communication system operating at wavelength 800 nm, only one percent of source frequency is available as signal bandwidth. The number of channels accommodated for transmitting TV signals of band width 6 MHz are (Take velocity of light c = 3 × 108 m/s, h = 6.6 × 10–34 J-s) (A) 3.75 × 106 (B) 3.86 × 106 5 (C) 6.25 × 10 (D) 4.87 × 105
30.
The position co-ordinates of a particle moving in a 3-D coordinates system is given by x = a cost y = a sint and z = at The speed of the particle is: (A)
2 a
(B) a
(C)
3 a
(D) 2a
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JEE-MAIN-2019 (9th Jan-Second Shift)-PCM-7
PART –B (CHEMISTRY) 31.
The entropy change associated with the conversion of 1 kg of ice at 273 K to water vapours at 383 K is: (Specific heat of water liquid and water vapours are 4.2 kJ K–1 kg–1 and 2.0 kJ K–1 kg–1, heat of liquid fusion and vapourisation of water are 334 kJ kg–1 and 2491 kJ kg–1, respectively) (log 273 = 2.436, log 373 = 2.572, log 383 = 2.583) (A) 7.90 kJ K–1 kg–1 (B) 2.64 kJ K–1 kg–1 –1 –1 (C) 8.49 kJ K kg (D) 9.26 kJ K–1 kg–1
32.
For the following reaction the mass of water produced from 445 g of C57H110O6 is 2 C57H110O 6 s 163 O2 g 114 CO2 g 110H2O I (A) 490 g (B) 445 g (C) 495 g (D) 890 g
33.
The major product formed in the following reaction is:
(A)
(B)
(C)
(D)
34.
Which of the following conditions in drinking water causes methemoglobinemia? (A) > 50 ppm of lead (B) > 50 ppm of chloride (C) > 50 ppm of nitrate (D) > 100 ppm of sulphate
35.
The major product of the following reaction is:
(A)
(B)
(C)
(D)
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JEE-MAIN-2019 (9th Jan-Second Shift)-PCM-8 36.
37.
The major product obtained in the following reaction is:
(A)
(B)
(C)
(D)
The major product of the following reaction is:
(A)
(B)
(C)
(D)
38.
The correct match between item I and item II is Item I Item II (a) Benzaldehyde (p) Mobile phase (b) Alumina (q) Adsorbent (c) Acetonitrile (r) Adsorbate (A) a q, b p, c r (B) a r, b q, c p (C) a q, b r, c p (D) a p, b r, c q
39.
The metal that forms nitride by reacting directly with N2 of air is (A) K (B) Li (C) Rb (D) Cs
40.
For coagulation of arsenious sulphide sol, which one of the following salt solution will be most effective? (A) BaCl2 (B) AlCl3 (C) NaCl (D) Na3PO4 FIITJEE Ltd., FIITJEE House, 29-A, Kalu Sarai, Sarvapriya Vihar, New Delhi -110016, Ph 46106000, 26569493, Fax 26513942 website: www.fiitjee.com.
JEE-MAIN-2019 (9th Jan-Second Shift)-PCM-9 41.
The complex that has highest crystal field splitting energy() is (A) [Co(NH3)5(H2O)]Cl3 (B) K2[CoCl4] (C) [Co(NH3)5Cl]Cl2 (D) K3[Co(CN)6]
42.
The pH of rain water is approximately (A) 5.6 (C) 7.0
43.
(B) 7.5 (D) 6.5
Consider the following reversible chemical reactions: K1 A 2 g B 2 g ....... 1 2 AB g K2 3 A 2 g 3B2 g ....... 2 6 AB g
The relation between K1 and K2 is 1 (A) K1K2 = 3 3 (C) K 2 K1 44.
(B) K 2 K 13 (D) K1K2 = 3
The correct sequence of amino acids present in the tripeptide given below is
(A) Val – Ser – Thr (C) Leu – Ser – Thr
(B) Thr – Ser – Val (D) Thr – Ser – Leu
45.
For the reaction, 2A B products , when the concentrations of A and B both were doubled, the rate of the reaction increased from 0.3 mol L–1 s–1 to 2.4 mol L–1 s–1. When the concentration of A alone is doubled, the rate increased from 0.3 mol L–1 s–1 to 0.6 mol L–1 s–1. Which one of the following statements is correct? (A) Total order of the reaction is 4 (B) Order of the reaction with respect to B is 2 (C) Order of the reaction with respect to B is 1 (D) Order of the reaction with respect to A is 2
46.
The products formed in the reaction of cumene with O2 followed by treatment with dil. HCl are:
(A)
(B)
(C)
(D)
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JEE-MAIN-2019 (9th Jan-Second Shift)-PCM-10 47.
The tests performed on compound X and their inferences are: Test Interference (a) 2, 4-DNP test Colorued precipitate (b) Iodoform test Yellow precipitate (c) Azo-dye test No dye formation Compound ‘X’ is
(A)
(B)
(C)
(D)
48.
If the standard electrode potential for a cell is 2 V at 300 K, the equilibrium constant (K) for the reaction 2 Zn s Cu2 aq Zn aq Cu s At 300 K is approximately (R = 8 JK–1 mol–1, F = 96000 C mol–1) (A) e-80 (B) e–160 320 (C) e (D) e160
49.
The temporary hardness of water is due to (A) Na2SO4 (B) NaCl (C) Ca(HCO3)2 (D) CaCl2
50.
In which of the following processes, the bond order has increased and paramagnetic character has changed to diamagnetic? (A) NO NO (B) N2 N2 (C) O 2 O2
51.
(D) O 2 O 22
Which of the following combination of statements is true regarding the interpretation of the atomic orbitals? (1) An electron in an orbital of high angular momentum stays away from the nucleus than an electron in the orbital of lower angular momentum. (2) For a given value of the principal quantum number, the size of the orbit is inversely proportional to the azimuthal quantum number h (3) According to wave mechanics, the ground state angular momentum is equal to 2 (4) The plot of Vs r for various azimuthal quantum numbers, shows peak shifting towards higher r value (A) (1), (4) (B) (1), (2) (C) (1), (3) (D) (2), (3)
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JEE-MAIN-2019 (9th Jan-Second Shift)-PCM-11 52.
Which of the following compounds is not aromatic?
(A)
(B)
(C)
(D)
53.
Good reducing nature of H3PO2 is attributed to the presence of (A) Two P – OH bonds (B) One P – H bond (C) Two P – H bonds (D) One P – OH bond
54.
The correct statement regarding the given Ellingham diagram is
(A) At 1400oC, Al can be used for the extraction of Zn from ZnO (B) At 500oC, coke can be used for the extraction of Zn from ZnO (C) Coke cannot be used for the extraction of Cu from Cu2O (D) At 800oC Cu can be used for the extraction of Zn from ZnO 55.
The transition element that has lowest enthalpy of atomisation is (A) Fe (B) Cu (C) V (D) Zn
56.
The increasing basicity order of the following compounds is (1)
(2)
(3)
(4)
(A) (4) < (3) < (2) < (1) (C) (1) < (2) < (3) < (4)
(B) (4) < (3) < (1) < (2) (D) (1) < (2) < (4) < (3)
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JEE-MAIN-2019 (9th Jan-Second Shift)-PCM-12 57.
When the first electron gain enthalpy (egH) of oxygen is -141 kJ/mol, its second electron gain enthalpy is (A) a more negative value than the first (B) almost the same as that of the first (C) negative, but less negative than the first (D) a positive value
58.
At 100oC, copper(Cu) has FCC unit cell structure with cell edge length x A . What is the approximate density of Cu(in g cm–3) at this temperature? [Atomic mass of Cu = 63.55 u] 205 105 (A) 3 (B) 3 x x 211 422 (C) 3 (D) 3 x x
59.
A solution containing 62 g ethylene glycol in 250 g water is cooled to -10oC. If Kf for water is 1.86 K kg mol–1, the amount of water(in g) separated as ice is (A) 48 (B) 32 (C) 64 (D) 16
60.
Homoleptic octahedral complexes of a metal ion ‘M3+’ with three monodentate ligands L1, L2 and L3 absorb wavelengths in the region of green, blue and red respectively. The increasing order of the ligand strength is (A) L3 < L1 < L2 (B) L3 < L2 < L1 (C) L1 < L2 < L3 (D) L2 < L1 < L3
o
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JEE-MAIN-2019 (9th Jan-Second Shift)-PCM-13
PART–C (MATHEMATICS) 61.
The sum of the following series 9 12 22 3 2 12 12 22 32 42 15 12 22 .... 52 1 6 ... up to 15 terms, is: 7 9 11 (A) 7820 (B) 7830 (C) 7520 (D) 7510
62.
For each x R, let lim
x x x sin x x
x 0
x
be the greatest integer less than or equal to x. Then
is equal to
(A) sin1 (C) 1 63.
(B) 0 (D) sin 1
Let f : 0,1 R be such that f xy f x f y for all x, y 0,1 , and f 0 0 . If y y x satisfies the differential equation,
equal to (A) 4 (C) 5
dy 1 3 f x with y 0 1, then y y is dx 4 4 (B) 3 (D) 2
64.
If x sin1 sin10 and y cos1 cos10 , then y – x is equal to: (A) (B) 7 (C) 0 (D) 10
65.
If 0 x (A) 2 (C) 3
, then the number of values of x for which sin x – sin 2x + sin 3x = 0, is 2 (B) 1 (D) 4
66.
93 Let z0 be a root of the quadratic equation, x 2 x 1 0 . If z 3 6iz81 0 3iz 0 , then arg z is equal to (A) (B) 4 3 (C) 0 (D) 6
67.
The area of the region A x, y :0 y x x 1 and 1 x 1 in sq. units, is: (A)
2 3
(C) 2
1 3 4 (D) 3 (B)
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JEE-MAIN-2019 (9th Jan-Second Shift)-PCM-14 68.
x 4y 7z g, 3y 5z h, 2x 5y 9z k
If the system of linear equation consistent, then: (A) g h k 0 (C) g h 2k 0
is
(B) 2g h k 0 (D) g 2h k 0 3
69.
70.
1 t6 The coefficient of t 4 in the expansion of is 1 t (A) 12 (B) 15 (C) 10 (D) 14 If both the roots of the quadratic equation x 2 5x 4 0 are real and distinct and they lie in the interval 1, 5 , then m lies in the interval. (A) (4, 5) (C) (5, 6)
(B) (3, 4) (D) (–5, –4)
71.
Let S be the set of all triangle in the xy – plane, each having one vertex at the origin and the other two vertices lie on coordinate axes with integral coordinates. If each triangle in S has area 50 sq. units, then the number of elements in the set S is: (A) 9 (B) 18 (C) 32 (D) 36
72.
Let a, b and c be the 7th, 11th and 13th terms respectively of a non – constant A.P. If a these are also the three consecutive terms of a G.P. then is equal to: c 1 (A) (B) 4 2 7 (C) 2 (D) 13
73.
The logical statement ~ ~ p q p r ~ q r is equivalent to: (A) p r ~ q (B) ~ p ~ q r (D) p ~ q r
(C) ~ p r
x y z and perpendicular to the 2 3 4 x y z x y z plane containing the straight lines and is: 3 4 2 4 2 3 (A) x 2y 2z 0 (B) x 2y z 0 (C) 5x 2y 4z 0 (D) 3x 2y 3z 0
74.
The equation of the plane containing the straight line
75.
A data consists of n observations: n
x1, x 2 ,......, x n . If
x i 1
2
i
n
1 9n and
x
2
i
1 5n , then the standard deviation of this
i 1
data is: (A) 5
(B)
(C)
(D) 2
7
5
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JEE-MAIN-2019 (9th Jan-Second Shift)-PCM-15
76.
et If A e t et
77.
If f x
e t cos t
e t sin t
e cos t e sin t e sin t e cos t Then A is 2e t sin t 2e t cos t (A) Invertible only if t (B) not invertible for any t R 2 (C) invertible for all t R (D) invertible only if t t
t
5x 8 7x 6
x
2
1 2x7
2
t
t
dx, x 0 and f 0 0, then the value of f 1 is:
1 2 1 (C) 4
1 2 1 (D) 4
(A)
(B)
3
78.
Let f be a differentiable function R to R such that f x f y 2 x y 2 , for all x, y R . 1
If f 0 1 then
2
f x dx
is equal to
0
1 2 (D) 1
(A) 0
(B)
(C) 2 79.
d2 y at t , is: 2 dx 4 1 (B) 3 2 1 (D) 6 2
If x 3 tan t and y 3 sec t, then the value of (A)
3
2 2 1 (C) 6
80.
The number of natural numbers less than 7,000 which can be formed by using the digits 0, 1, 3, 7, 9 (repetition of digits allowed) is equal to: (A) 250 (B) 374 (C) 372 (D) 375
81.
If the circles x 2 y 2 16x 20y 164 r 2 and x 4 y 7 36 intersect at two distinct points, then: (A) 0 r 1 (B) 1 r 11 (C) r 11 (D) r 11
82.
A hyperbola has its centre at the origin, passes through the point (4, 2) and has transverse axis of length 4 along the x – axis. Then the eccentricity of the hyperbola is: 3 2 (A) (B) 2 3
2
(C)
3
2
(D) 2
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JEE-MAIN-2019 (9th Jan-Second Shift)-PCM-16
83.
Let A 4, 4 and B 9,6 be points on the parabola y 2 4x . Let C be chosen on the arc AOB of the parabola, where O is the origin, such that the area of ACB is maximum. Then, the area (in sq. units) of ACB , is: 3 (A) 31 (B) 32 4 1 1 (C) 30 (D) 31 2 4
84.
Let the equation of two sides of a triangle be 3x 2y 6 0 and 4x 5y 20 0 . If the orthocentre of this triangle is at (1, 1), then the equation of its third side is: (A) 122y 26x 1675 0 (B) 26x 61y 1675 0 (C) 122y 26x 1675 0 (D) 26x 122y 1675 0
85.
An urn contains 5 red and 2 green balls. A ball is drawn at random from the urn. If the drawn ball is green, then a red ball is added to the urn and if the drawn ball is red, then a green ball is added to the urn; the original ball is not returned to the urn. Now, a second ball is drawn at random from it. The probability that the second ball is red, is: 26 32 (A) (B) 49 49 27 21 (C) (D) 49 49
86.
If the lines x ay b, z cy d and x a ' z b ', y c ' z d' are perpendicular, then: (A) cc ' a a ' 0 (B) aa ' c c ' 0 (C) ab ' bc ' 1 0 (D) bb ' cc ' 1 0 ˆ b b ˆi b ˆj 2kˆ and c 5iˆ ˆj 2kˆ be three vectors such that the Let a ˆi ˆj 2k, 1 2 projection vector of b on a is a . If a b is perpendicular to c , then b is equal to:
87.
(A) (C) 88.
89.
The number of all possible positive integral values of for which the roots of the quadratic equation, 6x 2 11x 0 are rational numbers is: (A) 2 (B) 5 (C) 3 (D) 4 2x Let A x R : x is not a positive int eger Define a function f : A R as f x x 1 then f is (A) injective but nor surjective (B) not injective (C) surjective but not injective (D) neither injective nor surjective 3
90.
(B) 4 (D) 6
22 32
If
0
(A) 2 (C) 4
tan 2k sec
d 1
1 2
, k 0 , then the value of k is:
1 2 (D) 1 (B)
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JEE-MAIN-2019 (9th Jan-Second Shift)-PCM-17
JEE (Main) – 2019 ANSWERS PART A – PHYSICS 1.
C
2.
B
3.
D
4.
D
5.
D
6.
A
7.
C
8.
D
9.
B
10.
B
11.
D
12.
C
13.
C
14.
D
15.
C
16.
D
17.
B
18.
B
19.
A
20.
D
21.
C
22.
B
23.
A
24.
C
25.
A
26.
C
27.
D
28.
A
29.
C
30.
A
PART B – CHEMISTRY 31.
D
32.
C
33.
C
34.
C
35.
C
36.
D
37.
C
38.
B
39.
B
40.
B
41.
D
42.
A
43.
C
44.
A
45.
B
46.
C
47.
B
48.
D
49.
C
50.
A
51.
A
52.
A
53.
C
54.
A
55.
D
56.
B
57.
D
58.
D
59.
C
60.
A
PART C – MATHEMATICS 61.
A
62.
A
63.
B
64.
A
65.
A
66.
A
67.
C
68.
B
69.
B
70.
(Bonus)
71.
D
72.
B
73.
A
74.
B
75.
B
76.
C
77.
D
78.
D
79.
D
80.
B
81.
B
82.
A
83.
D
84.
D
85.
B
86.
B
87.
D
88.
C
89.
A
90.
A
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JEE-MAIN-2019 (9th Jan-Second Shift)-PCM-18
HINTS AND SOLUTIONS PART A – PHYSICS 1.
R = Ro e–t RB Ro e B t e ( B A )t e3t R A Ro e B t B A 3 n2 n2 3. TB n2 n2 TB 4
2.
Ps PP Es is = Eiip (0.9) (2300) (5) is = 45 A. (230)
3.
B
E C
1 oE 2 2 B2 E2 E2 UB ( o o ) UE 2o 2oC2 2 o
UE
4.
x = 3t2 + 5 v = 6t W = k 1 1 = (2) (30)2 2(0)2 2 2 = 900 J
5.
eE = evB
6.
eBr E B m eB 2r m E (1.6 1019 ) (0.5)2 (0.5 10 2 ) m 2 10 24 kg. 100
kq kq 10 ( 25) E 31 r1 32 r2 = k 10 6 ( ˆi 3ˆj) ( 4iˆ 3ˆj) r1 r2 125 10 10
1 1 = (9 103 ) ( ˆi 3 ˆj) ( 4iˆ 3 ˆj) 5 10
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JEE-MAIN-2019 (9th Jan-Second Shift)-PCM-19
1 4 ˆ 3 3 ˆ 7 ˆ 3 ˆ = (9 10 3 ) i i 9000 i j 10 10 10 5 10 5 = (63iˆ 27ˆj) (100)
7.
t = G a h b cc Mo Lo T = (M–1 L3 T–2)a (ML2T–1)b (LT–1)c –a + b = 0 a = b 3a + 2b + c = 0 c = –5a –2a – b – c = 1 1 1 5 a ; b ; c 2 2 2
8.
VOi 12 0.3 11.7 V VOf 12 0.7 11.3 V
9.
10.
Vo = –0.4 V
dV dV A A 2gh dt dt 2 0.74 2 (3.14) 2(9.8)h 60 100 h = 4.92 m E1
GMm GMm R h R 2
1 GM GMm E2 m 2 Rh 2(R h)
E1 = E2 ; h
R 2
11.
W1 = W2 600 – T2 = T2 – 400 T2 = 500 K
12.
E Pt
13.
PE = KE 1 1 m2 (A 2 x 2 ) = m2 x 2 2 2 A x 2
E2 (24)2 Rt (60) (60) 518 J. Z2 602 (8.33 2)2
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JEE-MAIN-2019 (9th Jan-Second Shift)-PCM-20 14.
T cos 45° = mg T sin 45° = F F = mg = 100 N.
15.
Q
16.
Xmax = d sin = 0.32 sin 30 = 0.16 mm Xmax 0.16 10 3 n 500 10 9 0.16 10 6 1600 = 320 500 5 Number of BFs = (2n + 1) = 641
f nRT 2 5 15 = (8.3) (1200 300) = 10000 J. 2 28
17.
= 60°
18.
mg
1 1 m 2 2 2 2 2 3
19. 20.
21.
3g 30 2
R = 530 k 5%
oi 2R oN i BC 2(R / N) BL 1 2 BC N BL
f
1 C 1 & 0.8 f 2 2 ML 2 3
C 2
ML mL2 2 3
ML2 mL2 25 2 3 2 16 ML 3
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JEE-MAIN-2019 (9th Jan-Second Shift)-PCM-21
22.
25 3m 1 16 2M 9 3m 16 2 M m 3 0.37 M 8
R A 2r Q 4 r 2 dr 2 e a (4r 2 ) dr 0r 2R a = 4A 1 e a 2 a Q R log 1 2 2Aa
L2 L2 0K 3 2 2 2 0L (K K ) C1 1 3 d d d 2 2 2 L L2 oK 2 oK 4 2 2 2 oL (K K ) C2 2 4 d d d 2 2 1 1 1 c c1 c 2 d d d 2 2 2 oKL oL (K1 K 3 ) oL (K 2 K 4 ) 0K 1
23.
0.5 mm 0.015 mm 100 0.5 MSR = 5.5 48 100 = 5.74 mm. Thickness = 5.74 – 0.015 = 5.725 mm
24.
Zero error = 0 3
25.
f
2 330 vs 660 Hz 2 0.5 50 330 18 s 50 f f 660 1 (660) 330 18 330 s = 666 Hz.
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JEE-MAIN-2019 (9th Jan-Second Shift)-PCM-22
2 2 t a2 a1
a1a 2 2 t a1 a2
2a1 2a 2
2 v
26.
27.
a1a2 t
1 a1 a2
( a1a 2 ) t
KEmax = hmax – =
(6.6 10 34 ) (6.28 107 ) (3 108 ) 4.7 1.6 10 19 2 3.14
= 12.37 – 4.7 = 7.67 eV 28.
5V
100
400
500
R2
12 6 6 400 600 R 2 1 1 1 200 600 R2 R2 = 300
18 V
29.
30.
c 3 108 3 1015 Hz 8 107 8 3 1013 (0.01) f 8 n 6 106 6 106 1 = 107 6.25 105 16 f
dx a sin t dt dy vy a cos t dt dz vz a dt vx
v v 2x v 2y v 2z a 2
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JEE-MAIN-2019 (9th Jan-Second Shift)-PCM-23
PART B – CHEMISTRY 31.
H2O s H2O H2O H2O g H2O g 1 kg 1 kg 1 kg at 273 K at 273 K at 373 K at 373 K at 383 K S = S1 + S2 + S3 + S4 334 373 2491 383 4.2n 2n = 9.267 kJ Kg–1 K-1 273 273 373 373
32.
2 C57H110O6 s 163 O 2 g 114 CO 2 g 110H2O I
Moles of C57H110O6 Moles of H2O 2 110 445 890
Mass of H2O 18
2 110 Mass of H2O = 495 g 33.
O
O
O
NaOH Ph C CH3 Ph C CH2 Ph C CH2
enolate ion
O
O
O
O
OH
O
H2O Ph C CH2 CH3 C H CH3 CH CH2 C Ph CH3 CH CH2 C Ph
34.
Fact based
35.
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JEE-MAIN-2019 (9th Jan-Second Shift)-PCM-24 36.
Nucleophilicity of NH2> OH OH
NH2
37.
OH
CH3CO 2 O/Py R.T
OH
O
NH2
NHCOCH3
NH2
Br2 /h
Br KOH
O
NH
CH3
38.
Acetonitrile is used as mobile phase for most of the reverse chromatography. Benzaldehyde is adsorbed on alumina.
39.
The only alkali metal which forms nitride by reacting directly with N2 is ‘Li’.
40.
As2S3 is a negatively charged sol. so AlCl3 will be most effective.
41.
As CN– is a strong field ligand. K3[Co(CN)6] will have maximum ‘’.
42.
Fact based.
43.
K1 A 2 g B2 g 2 AB g
....... 1
K2 3 A 2 g 3B 2 g ....... 2 6 AB g
Reaction(2) = -3 reaction(1) 3
1 K 2 K 2 K 13 K1
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JEE-MAIN-2019 (9th Jan-Second Shift)-PCM-25 44.
H 2O
Val 45.
46.
Ser
2A B products Rate = K[A]x[B]y r = K[A]x[B]y - - - - (i) 0.3 = K[A]x[B]y - - - - (1) 2.4 = K[2A]x[2B]y - - - - (2) 0.6 = K[2A]x[B]y - - - - (3) From (1), (2) & (3) x = 1, y = 2 Overall order = 2 + 1 = 3 Order w.r.t A = 1 Order w.r.t B = 2 H3C
CH3
O O
O2
47.
48.
Thr
OH
H
O dil.HCl
-COCH3 is present it will show both 2, 4-DNP & iodoform test. Due to steric inhibition of resonance. I.P of ‘N’ is not involved in delocalization so coupling reaction will not take place. 2 Zn s Cu2 aq Zn aq Cu s
-nFEcell = -RTnK
nK =
2 96500 2 = 160.83 8 300
K = e160 49.
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JEE-MAIN-2019 (9th Jan-Second Shift)-PCM-26
50.
NO NO B.O 0.5 3 Para Dia O2 O2 B.O 2 2.5 Para Para
51.
Refer Theory
N2 N2 B.O 3 2.5 Dia Para O2 O22 B.O 2 1 Para Dia
52. is anti aromatic
53.
Refer theory
54.
4 Al 6 ZnO 2 Al2O3 6 Zn
H for the above reaction is –ve. 55.
Due to weak metallic bonding.
56.
Correct order of basic strength is NH2(Et)2 > EtNH2 > NMC3 > Ph – NH – CH3
57.
2nd electron gain enthalpy of oxygen is positive.
58.
d
59.
ZM Na a3 4 63.55
6.023 1023 x 10
8 3
422 gm / cm3 x3
Let moles of H2O separated as ice = x gm Tf = iKfm 10 = 1 1.86
62 62 250 x 1000
x = 64 gm 60.
L1 L2 L3 Green Blue Red absorbed wave length Order of Red > Green > Blue L3 > L1 > L2 Strength of ligand 1/ Strength of ligand L2 > L1 > L3
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JEE-MAIN-2019 (9th Jan-Second Shift)-PCM-27
PART C – MATHEMATICS
3 n 1 3 1
2
61.
Tn
3.
n
2
2n 1 n 1 2n 1 6 2n 1
Tn S15
22 .... n2
1 15 3 n n2 2 n 1
n2 n 1
2 2 1 15 15 1 15 16 31 2 2 6
= 7820 62.
lim
x x x sin x x
x 0
x0 x 1 lim x x 1 sin 1 sin1 x 0 x x x 63.
f xy f x .f y f 0 1 as f 0 0 f x 1
dy f x 1 dx y xc At, x 0, y 1 c 1 y x 1 3 1 3 1 y y 1 1 3 4 4 4 4 64.
2
3
10
x sin 1 sin10 3 10
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JEE-MAIN-2019 (9th Jan-Second Shift)-PCM-28
2
3 10
y cos1 cos10 4 10 yx
65.
sin x sin 2x sin 3x 0 sin x sin 3x sin 2x 0 2 sin x.cos x sin 2x 0 sin 2x 2 cos x 1 0
sin 2x 0 or cos x 66.
z0 or 2 (where is a non – real cube root of unity) 81
z 3 6i 3i z 3 3i arg z 4 67.
1 x 0, 2 3
93
y
The graph is a follows 0
1
x
2
1
1 dx x 2 1 dx 2
y x2 1
0
y x2 1
68.
P1 x 4y 7z g 0 P2 3x 5y h 0 P3 2x 5y 9z k 0 Here 0 2P1 P2 P3 0 when 2g h k 0
69.
1 t 1 t 1 t 3t 3t 1 t
6 3
18
–1
1
3
6
12
3
coefficient of t 4 in 1 t
3
is
3 4 1
C4 6C2 15
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JEE-MAIN-2019 (9th Jan-Second Shift)-PCM-29 70.
x 2 mx 4 0 , 1,5 1
(1)
D 0 m2 16 0 m , 4 4,
(2)
f 1 0 5 m 0 m , 5
5
29 f 5 0 29 5m 0 m , 5 b m (4) 1 5 1 5 m 2,10 2a 2 m 4, 5 No option correct : Bonus * If we consider , 1, 5 then option (1) is correct. (3)
71.
Let A ,0 and B 0, be the vectors of the given triangle AOB 100 Number of triangles 4 (number of divisors of 100) 4 9 36
72.
a A 6d b A 10d c A 12d a, b, c are in G.P. 2
A 10d A 6d a 12d
A 14 d
A a A 6d d 6 14 4 A 12 14 c A 12d 12 d 6
73.
~ ~ p q p r ~ q r p ~ q p r ~ q r p ~ q r ~ q r p ~ q r p r ~ q
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JEE-MAIN-2019 (9th Jan-Second Shift)-PCM-30
74.
Vector along the normal to the plane containing the lines
x y z x y z and is 3 4 2 4 2 3
8iˆ ˆj 10kˆ . Vector perpendicular to the vectors 2iˆ 3 ˆj 4kˆ and 8iˆ ˆj 10kˆ is 26iˆ 52ˆj 26kˆ So, required plane is 26x 52y 26z 0 x 2y z 0 75.
x x
2
1 9n
i
……….(1)
2
1 5n
i
……….(2)
1 2 xi2 1 7n
x
2 i
6 n (1) . (2) 4 x i 4n
xi n
x
i
1 n variance = 6 – 1 = 5 standard diviation 5
1
76.
cos t
sin t
t
A e 1 cos t sin t sin t cos t 1
2 sin t
2cos t
e t 5 cos2 t 5 sin2 t t R
5e t 0 t R 77.
5x 8 7x 6
x
2
1 2x 7
2
dx
5x 6 7x 8 1 1 7 5 2 x x
2
As f 0 0, f x
f 1 78.
dx
1 C 1 1 2 5 7 x x
x7 2x7 x 2 1
1 4
f x f y 2 x y
3/2
divide both side by x y f x f y xy
2. x y
1/ 2
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JEE-MAIN-2019 (9th Jan-Second Shift)-PCM-31 Apply limit x y
f ' y 0 f ' y 0 f y c f x 1 1
1.dx 1 0
79.
dx 3 sec 2 t dt dy 3 sec t tan t dt dy tan t sin t dx sec t d2 y dt cos t 2 dx dx cos t cos3 t 1 1 2 3 sec t 3 3.2 2 6 2
80.
a1 a2 a3 Number of numbers 5 3 1 a 4 a1 a2 a3 2 ways for a 4 Numbers of numbers 2 53 Required number0020 53 2 53 1 = 374
81.
x 2 y 2 16x 20y 164 r 2 A 8,10 ,R1 r
x 4
2
2
y 7 36
B (4, 7), R2 6 R1 R 2 AB R1 R 2 1 r 11
82.
Given hyperbola is x2 y2 1 4 b2 Satisfying the point 4 b2 3 2 e 3
(4, 2)
(4,
2)
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JEE-MAIN-2019 (9th Jan-Second Shift)-PCM-32 83.
For maximum area, tangent at the point c must be parallel to chord BC. 1 t 2
B (9, 6) 2
c (t , 2t)
A (4, –4)
84.
A
Equation of AB is 3x 2y 6 0 Equation of AC is 4x 5y 20 0 . Equation of BE is 2x 3y 5 0 Equation of CF is 5x 4y 1 0 Equation of BC is 26x 122y 1675
(1,1)
B
85.
E
F
D
C
E1 : Event of drawing a Red ball and placing a green ball in the bag E 2 : Event of drawing a green ball and placing a red ball in the bag
E E E : Event of drawing a red ball in second draw P E P E1 P P E 2 P E1 E2 5 4 2 6 32 7 7 7 7 49 86.
87.
Line x ay b, z cy d x b y z d a 1 c Line x a ' z b ', y c ' z d' x b ' y d' z a' c' 1 Given both the lines are perpendicular aa ' c ' c 0 a.b Projection of b on a a a b1 b2 2 ………….(1) and a b c a b .c 0
5b1 b2 10 …………(2) from (1) and (2) b1 3 and b2 5 then b b12 b22 2 6
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JEE-MAIN-2019 (9th Jan-Second Shift)-PCM-33 88.
89.
D must be perfect square 121 24 2 maximum value of is 5 1 I 2 I 3 integral values 3 I 4 I 5 I
1 f x 2 1 x 1 2 f ' x 2 x 1
f is one – one but not onto 90.
1 2k
/3
0
1
2k k2
tan sec
d
2 cos
1 2k
/3 0
/3
0
sin cos
d
2 1 1 k 2
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FIITJEE Solutions to JEE(Main)-2019 Test Date: 10th January 2019 (First Shift)
PHYSICS, CHEMISTRY & MATHEMATICS Paper - 1 Time Allotted: 3 Hours
Maximum Marks: 360
Please read the instructions carefully. You are allotted 5 minutes specifically for this purpose.
Important Instructions: 1.
The test is of 3 hours duration.
2.
This Test Paper consists of 90 questions. The maximum marks are 360.
3.
There are three parts in the question paper A, B, C consisting of Physics, Chemistry and Mathematics having 30 questions in each part of equal weightage. Each question is allotted 4 (four) marks for correct response.
4.
Out of the four options given for each question, only one option is the correct answer.
5.
For each incorrect response 1 mark i.e. ¼ (one-fourth) marks of the total marks allotted to the question will be deducted from the total score. No deduction from the total score, however, will be made if no response is indicated for an item in the Answer Box.
6.
Candidates will be awarded marks as stated above in instruction No.3 for correct response of each question. One mark will be deducted for indicating incorrect response of each question. No deduction from the total score will be made if no response is indicated for an item in the answer box.
7.
There is only one correct response for each question. Marked up more than one response in any question will be treated as wrong response and marked up for wrong response will be deducted accordingly as per instruction 6 above.
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JEE-MAIN-2019 (10th Jan-First Shift)-PCM-2
PART –A (PHYSICS) 1.
In an electron microscope, the resolution that can be achieved is of the order of the wavelength of electrons used. To resolve a width of 7.5 × 1012 m, the minimum electron energy required is close to: (A) 500 keV (B) 100 keV (C) 1 keV (D) 25 keV
2.
To mop-clean a floor, a cleaning machine presses a circular mop of radius R vertically down with a total force F and rotates it with a constant angular speed about its axis. If the force F is distributed uniformly over the mop and the floor is , the torque, applied by the machine on the mop is: (A) FR/3 (B) FR/6 2 (C) FR/2 (D) FR 3
3.
A potentiometer wire AB having length L and resistance 12 r is joined to a cell D of emf and internal resistance r. A cell C having emf /2 and internal resistance 3r is connected. The length AJ at which the galvanometer as shown in figure shows no deflection is: 11 11 (A) L (B) L 12 24 13 5 (C) L (D) L 24 12
4.
A TV transmission tower has a height of 140 m and the height of the receiving antenna is 40 m. What is the maximum distance upto which signals can be broadcasted from this tower in LOS (Line of Sight) mode? (Given : radius of earth = 6.4 × 106 m). (A) 65 km (B) 48 km (C) 80 km (D) 40 km
5.
In the cube of side ‘a’ shown in the figure, the vector from the central point of the face ABOD to the central point of the face BEFO will be: 1 1 (A) a kˆ ˆi (B) a ˆi kˆ 2 2 1 1 (C) a ˆj ˆi (D) a ˆj kˆ 2 2
6.
A uniform metallic wire has a resistance of 18 and is bent into an equilateral triangle. Then, the resistance between any two vertices of the triangle is: (A) 4 (B) 8 (C) 12 (D) 2
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JEE-MAIN-2019 (10th Jan-First Shift)-PCM-3 7.
A block of mass m is kept on a platform which starts from rest with constant acceleration g/2 upward, as shown in figure. Work done by normal reaction on block in time t is: m g2 t 2 m g2 t 2 (A) (B) 8 8 3m g2 t 2 (C) 0 (D) 8
8.
In a Young’s double slit experiment with slit separation 0.1 mm, one observes a bright 1 fringe at angle rad by using light of wavelength 1. When the light of wavelength 2 is 40 used a bright fringe is seen at the same angle in the same set up. Given that 1 and 2 are in visible range (380 nm to 740 nm), their values are: (A) 625 nm, 500 nm (B) 380 nm, 525 nm (C) 380 nm, 500 nm (D) 400 nm, 500 nm
9.
Two guns A and B can fire bullets at speed 1 km/s and 2 km/s respectively. From a point on a horizontal ground, they are fired in all possible directions. The ratio of maximum areas covered by the bullets fired by the two guns, on the ground is: (A) 1 : 16 (B) 1 : 2 (C) 1 : 4 (D) 1 : 8
10.
The density of a material in SI units is 128 kg m3. In certain units in which the unit of length is 25 cm and the unit of mass 50 g, the numerical value of density of the material is: (A) 40 (B) 16 (C) 640 (D) 410
11.
A magnet of total magnetic moment 10 2 ˆi A-m2 is placed in a time varying magnetic field, B ˆi cos t where B = 1 Tesla and = 0.125 rad/s. The work done for reversing the direction of the magnetic moment at t = 1 second, is: (A) 0.01 J (B) 0.007 J (C) 0.028 J (D) 0.014 J
12.
A heat source at T = 103 K is connected to another heat reservoir at T = 102 K by a copper slab which is 1 m thick. Given that the thermal conductivity of copper is 0.1 WK1 m1, the energy flux through it in the steady state is: (A) 90 Wm2 (B) 120 Wm2 2 (C) 65 Wm (D) 200 Wm2
13.
A parallel plate capacitor is of area 6 cm2 and a separation 3 mm. The gap is filled with three dielectric materials of equal thickness (see figure) with dielectric constants K1 = 10, K2 = 12 and K3 = 14. The dielectric constant of a material which when fully inserted in above capacitor, gives same capacitance would be: (A) 4 (B) 14 (C) 12 (D) 36
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JEE-MAIN-2019 (10th Jan-First Shift)-PCM-4 14.
A charge Q is distributed over three concentric spherical shell of radii a, b, c (a < b < c) such that their surface charge densities are equal to one another. The total potential at a point at distance r from their common centre, where r < a, would be: Q a2 b2 c 2 Q ab bc ca (A) (B) 12 0 abc 4 0 a3 b3 c 3 (C)
15.
Q 4 0 a b c
(D)
Q a b C
4 0 a 2 b 2 c 2
Three Carnot engines operate in series between a heat source at a temperature T1 and a heat sink at temperature T4 (see figure). There are two other reservoirs at temperature T2 and T3, as shown, with T1 > T2 > T3 > T4. The three engines are equally efficient if: 1
(A) T2 T1T4 2 ;T3 T12T4
1 3
1
1 3
1
1 3
1
1 4
(B) T2 T12 T4 3 ;T3 T1T24 (C) T2 T1T42 3 ;T3 T12T4 (D) T2 T13 T4 4 ;T3 T1T43 16.
A satellite is moving with a constant speed in circular orbit around the earth. An object of mass ‘m’ is ejected from the satellite such that it just escapes from the gravitational pull of the earth. At the time of ejection, the kinetic energy of the object is: (A) 2 m2 (B) m2 1 3 (C) m2 (D) m2 2 2
17.
Water flows into a large tank with flat bottom at the rate of 104 m3s1. Water is also leaking out of a hole of area 1 cm2 at its bottom. If the height of the water in the tank remains steady, then this height is: (A) 5.1 cm (B) 1.7 cm (C) 4 cm (D) 2.9 cm
18.
A string of length 1 m and mass 5 g is fixed at both ends. The tension in the string is 8.0 N. The string is set into vibration using an external vibrator of frequency 100 Hz. The separation between successive nodes on the string is close to: (A) 10.0 cm (B) 33.3 cm (C) 16.6 cm (D) 20.0 cm
19.
A train moves towards a stationary observer with speed 34 m/s. The train sounds a whistle and its frequency registered by the observer is f 1. If the speed of the train is reduced to 17 m/s, the frequency registered is f 2. If speed of sound is 340 m/s, then the ratio f1/f2 is: (A) 18/17 (B) 19/18 (C) 20/19 (D) 21/20
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JEE-MAIN-2019 (10th Jan-First Shift)-PCM-5 20.
A plano convex lens of refractive index 1 and focal length f 1 is kept in contact with another plano concave lens of refractive index 2 and focal length f 2. If the radius of curvature of their spherical faces is R each and f1 = 2f2, then 1 and 2 are related as: (A) 1 + 2 = 3 (B) 21 + 2 = 1 (C) 32 + 1 = 1 (D) 22 + 1 = 1
21.
To get output ‘1’ at R, for the given logic gate circuit the input values must be: (A) X = 0, Y = 1 (B) X = 1, Y = 1 (C) X = 1, Y = 0 (D) X = 0, Y = 0
22.
If the magnetic field of a plane electromagnetic wave is given by (The speed of light = x 3 × 108 m/s) B 100 10 6 sin 2 2 1015 t then the maximum electric field c associated with it is: (A) 6 × 104 N/C (B) 3 × 104 N/C 4 (C) 4 × 10 N/C (D) 4.5 × 104 N/C
23.
Using a nuclear counter the count rate of emitted particles from a radioactive source is measured. At t = 0 it was 1600 counts per second and t = 8 seconds it was 100 counts per second. The count rate observed, as counts per second, at t = 6 seconds is close to: (A) 200 (B) 150 (C) 400 (D) 360
24.
A solid metal cube of edge length 2 cm is moving in a positive y-direction at a constant speed of 6 m/s. There is a uniform magnetic field of 0.1 T in the positive z-direction. The potential difference between the two faces of the cube perpendicular to the x-axis, is: (A) 12 mV (B) 6 mV (C) 1 mV (D) 2 mV
25.
Two electric dipoles A, B with respective dipole moments dA 4 qa ˆi and dB 2 qa ˆi are placed on the x-axis with a separation R, as shown in the figure. The distance from A at which both of them produce the same potential is: 2R R (A) (B) 2 1 2 1 (C)
R 2 1
(D)
2R 2 1
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JEE-MAIN-2019 (10th Jan-First Shift)-PCM-6 26.
In the given circuit the cells have zero internal resistance. The currents (in Amperes) passing through resistance R1 and R2 respectively, are: (A) 1, 2 (B) 2, 2 (C) 0.5, 0 (D) 0, 1
27.
A 2 W carbon resistor is color coded with green, black, red and brown respectively. The maximum current which can be passed through this resistor is: (A) 20 mA (B) 100 mA (C) 0.4 mA (D) 63 mA
28.
A piece of wood of mass 0.03 kg is dropped from the top of a 100 m height building. At the same time, a bullet of mass 0.02 kg is fired vertically upward, with a velocity 100 ms 1 , from the ground. The bullet gets embedded in the wood. Then the maximum height to which the combined system reaches above the top of the building before falling below is: (g = 10 ms2) (A) 20 m (B) 30 m (C) 40 m (D) 10 m
29.
An insulating thin rod of length has a linear charge density (x) 0
30.
A homogeneous solid cylindrical roller of radius R and mass M is pulled on a cricket pitch by a horizontal force. Assuming rolling without slipping, angular acceleration of the cylinder is: 3F F (A) (B) 2mR 3mR
x on it. The rod is rotated about an axis passing through the origin (x = 0) and perpendicular to the rod. If the rod makes n rotations per second, then the time averaged magnetic moment of the rod is: (A) n3 (B) n3 3 (C) n3 (D) n3 4
(C)
F 2mR
(D)
2F 3mR
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JEE-MAIN-2019 (10th Jan-First Shift)-PCM-7
PART –B (CHEMISTRY) 31.
The total number of isomers for a square planar complex [M(F)(Cl)(SCN)(NO2)] is (A) 16 (B) 8 (C) 4 (D) 12
32.
A process that has H = 200 J mol–1 and S = 40 JK–1 mol–1. Out of the values given below, choose the minimum temperature above which the process will be spontaneous: (A) 20 K (B) 12 K (C) 5 K (D) 4 K
33.
The values of Kp/Kc for the following reactions at 300 K are respectively: (At 300 K, Rt = 24.62 dm3 atm mol–1) N2 g O2 g 2NO g N2O 4 g 2NO 2 g N2 g 3H2 g 2NH3 g (A) 1, 24.62 dm3 atm mol–1, 606.0 dm6 atm2 mol–2 (B) 1, 24.62 dm3 atm mol–1, 1.65 10–3 dm–6 atm–2 mol2 (C) 1, 4.1 10–2 dm–3 atm–1 mol, 606.0 dm6 atm2 mol–2 (D) 24.62 dm3 atm mol–1, 606.0 dm6 atm2 atm2 mol–2, 1.65 10–3 dm–6 atm–2 mol2
34.
The total number of isotopes of hydrogen and number of radioactive isotopes among them, respectively are (A) 3 and 1 (B) 3 and 2 (C) 2 and 1 (D) 2 and 0
35.
Water filled in two glasses A and B have BOD values of 10 and 20 respectively. The correct statement regarding them is (A) B is more polluted than A (B) A is suitable for drinking, whereas B is not (C) both A and B are suitable for drinking (D) A is more polluted than B
36.
Which primitive unit cell has unequal edge length (a b c) and all axial angles different from 90o? (A) Triclinic (B) Hexagonal (C) Monoclinic (D) Tetragonal
37.
The major product of the following reaction is:
(A)
(B)
(C)
(D)
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JEE-MAIN-2019 (10th Jan-First Shift)-PCM-8 38.
Consider the given plots for a reaction obeying Arrhenius equation (0oC < T < 300oC) : (k and Ea are rate constant and activation energy respectively)
Choose the correct option: (A) I is right but II is wrong (C) I is wrong but II is right 39.
40.
(B) Both I and Ii are correct (D) Both I and II are wrong
The major product formed in the reaction given below will be:
(A)
(B)
(C)
(D)
Wilkinson catalyst is (A) [(Ph3P)3IrCl] (C) [(Ph 3P)3RhCl]
(B) [Et3P)3RhCl] (D) [(Et3P)3IrCl]
41.
If dichloromethane(DCM) and water(H2O) are used for differential extraction, which one of the following statements is correct? (A) DCM and H2O would stay as lower and upper layer respectively in the S.F (B) DCM and H2O will make turbid/colloidal mixture (C) DCM and H2O would stay as upper and lower layer respectively in the separating funnel(S.F) (D) DCM and H2O will be miscible clearly
42.
Which diccarboxylic acid in presence of a dehydrating agent is least reactive to give anhydride?
(A)
(B)
(C)
(D)
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JEE-MAIN-2019 (10th Jan-First Shift)-PCM-9 43.
The decreasing order of ease of alkaline hydrolysis for the following ester is
(A) III > II > IV > I (C) IV > II > III > I 44.
(B) III > II > I > IV (D) II > III > I > IV
Which of the graphs shown below does not represent the relationship between incident light and the electron ejected from metal surface?
(A)
(B)
(C)
(D)
45.
Which of the following is not an example of heterogeneous catalytic reaction? (A) Ostwald’s process (B) Combustion of coal (C) Hydrogenation of vegetable oils (D) Haber’s process
46.
The effect of lanthanoid contraction in the lanthanoid series of elements by the large means (A) increase in both atomic and ionic radii (B) decrease in atomic radii and increase in ionic radii (C) decrease in both atomic and ionic radii (D) increase in atomic radii and decrease in ionic radii
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JEE-MAIN-2019 (10th Jan-First Shift)-PCM-10 47.
Which hydrogen in compound(E) is easily replaceable during bromination reaction in presence of light?
(A) - hydrogen (C) - hydrogen 48.
49.
(B) - hydrogen (D) - hydrogen
The major product of the following reaction is
(A)
(B)
(C)
(D)
The correct structure of product ‘P’ in the following reaction is
(A)
(B)
(C)
(D)
50.
The type of hybridisation and number of lone pair(s) of electrons of Xe in XeOF4 respectively are (A) sp3d2 and 1 (B) sp3d and 2 3 2 (C) sp d and 2 (D) sp3d and 1
51.
The electronegativity of aluminium is similar to (A) carbon (B) beryllium (C) boron (D) lithium
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JEE-MAIN-2019 (10th Jan-First Shift)-PCM-11 52.
Consider the following Zn2 2e Zn s ;Eo 0.76 V Ca 2 2e Ca s ;Eo 2.87 V Mg2 2e Mg s ;Eo 2.36 V Ni2 2e Ni s ;Eo 0.25 V
The reducing power of the metals increases in the order (A) Ca < Zn < Mg < Ni (B) Ni < Zn < Mg < Ca (C) Zn < Mg < Ni < Ca (D) Ca < Mg < Zn < Ni 53.
The chemical nature of hydrogen peroxide is (A) oxidising agent in acidic medium, but not in basic medium (B) reducing agent in basic medium but not in acidic medium (C) oxidising and reducing agent in acidic medium but not in basic medium (D) oxidising and reducing agent in both acidic and basic medium
54.
A mixture of 100 m mol of Ca(OH)2 and 2 g of sodium sulphate was dissolved in water and the volume was made upto 100 mL. The mass of calcium sulphate formed and the concentration of OH– in resulting solution, respectively are (Molar mass of Ca(OH)2, Na2SO4 and CaSO4 are 74, 143 and 136 g mol–1 respectively; Ksp of Ca(OH)2 is 5.5 10–6) (A) 1.9 g, 0.28 mol L–1 (B) 13.6 g, 0.28 mol L–1 –1 (C) 1.9g, 0.14 mol L (D) 13.6 g, 0.14 mol L–1
55.
Liquids A and B form and ideal solution in the entire composition range. At 350 K, the vapour pressures of pure A and pure B are 7 103 Pa and 12 103 Pa, respectively. The composition of the vapour in equilibrium with a solution containing 40 mole percent of A at this temperature is (A) xA = 0.37; xB = 0.63 (B) xA = 0.28; xB = 0.72 (C) xA = 0.4; xB = 0.6 (D) xA = 0.76; xB = 0.24
56.
The major product ‘X’ formed in the following reaction is:
(A)
(B)
(C)
(D)
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JEE-MAIN-2019 (10th Jan-First Shift)-PCM-12
57.
The metal used for making X-ray tube window is (A) Mg (B) Na (C) Be (D) Ca
58.
The increasing order of pKa values of the following compounds is
(A) C < B < A < D (C) D < A < C < B 59.
(B) B < C < D < A (D) B < C < A< D
Hall-Herolut’s process is given by (A) Cu2 aq H2 g Cu s 2H aq (B) Cr2O3 2 Al Al2O3 2 Cr (C) 2Al2O3 3C 4 Al 3CO2 Coke, 1673 K (D) ZnO C Zn CO
60.
Two pi and half sigma bonds are present in (A) O 2 (B) N2 (C) O2
(D) N2
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JEE-MAIN-2019 (10th Jan-First Shift)-PCM-13
PART–C (MATHEMATICS) 61.
62.
An unbiased coin is tossed. If the outcome is a head then a pair of unbiased dice is rolled and the sum of the numbers obtained on the is noted. If the toss of the coin results in tail then a card from a well-shuffled pack of nine cards numbered 1, 2, 3,….., 9 is randomly picked and the number on the card is noted. The probability that the noted number is either 7 or 8 is: 13 15 (A) (B) 36 72 19 19 (C) (D) 72 36 3 The shortest distance between the point ,0 and the curve y x, x 0 , is: 2 5 3 (A) (B) 2 2 3 5 (C) (D) 2 4
63.
The plane passing through the point (4, 1, 2) and parallel to the lines x 2 y 2 z 1 x2 y3 z4 and also passes through the point: 3 1 2 1 2 3 (A) (1, 1< 1) (B) (1, 1, 1) (C) (1, 1, 1) (D) (1, 1, 1)
64.
The mean of five observations is 5 and their variance is 9.20. If three of the given five observations are 1, 3 and 8, then a ratio of other two observations is: (A) 10 : 3 (B) 4 : 9 (C) 5 : 8 (D) 6 : 7
65.
If 5, 5r, 5r2 are the lengths of the sides of a triangle, then r cannot be equal to: 3 5 (A) (B) 4 4 7 3 (C) (D) 4 2
66.
If
67.
3
20 Ci1 k 20 , then k equals: 20 Ci Ci1 21 i 1 (A) 400 (C) 200 20
(B) 50 (D) 100
3 The sum of all values of 0, satisfying sin2 2 cos4 2 is: 4 2 5 (A) (B) 4 3 (C) (D) 2 8 FIITJEE Ltd., FIITJEE House, 29-A, Kalu Sarai, Sarvapriya Vihar, New Delhi -110016, Ph 46106000, 26569493, Fax 26513942 website: www.fiitjee.com.
JEE-MAIN-2019 (10th Jan-First Shift)-PCM-14 68.
Consider the quadratic equation c 5 x 2 2cs c 4 0,c 5. Let S be the set of all integral values of c for which one root of the equation lies in the interval (0, 2) and its other root lies in the interval (2, 3). Then the number of elements in S is: (A) 18 (B) 12 (C) 10 (D) 11
69.
If
70.
In a class of 140 students numbered 1 to 140, all even numbered students opted Mathematics course, those whose number is divisible by 3 opted Physics course and those whose number is divisible by 5 opted Chemistry course. Then the number of students who did not opt for any of the three courses is: (A) 102 (B) 42 (C) 1 (D) 38
71.
If the third term in the binomial expansion of 1 xlog2 x
dy 3 1 4 y , x , and y , then y equals: 2 2 dx cos x cos x 3 3 4 3 4 1 1 (A) e 6 (B) 3 3 4 1 (C) (D) e3 3 3
value of x is: 1 (A) 4 1 (C) 8
5
equals 2560, then a possible
(B) 4 2 (D) 2 2
72.
If the parabolas y2 = 4b(x – c) and y2 = 8ax have a common normal, then which one of the following is a valid choice for the ordered triad (a, b, c)? 1 (A) ,2,3 (B) (1, 1, 3) 2 1 (C) ,2,0 (D) (1, 1, 0) 2
73.
If the system of equations xyz5 x 2y 3z 9 x 3y z has infinitely many solutions, then equals: (A) 21 (B) 8 (C) 18 (D) 5
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JEE-MAIN-2019 (10th Jan-First Shift)-PCM-15 74.
For each t R, let [t] be the greatest integer less than or equal to t. Then, 1 x sin 1 x sin 2 1 x lim x 1 1 x 1 x (A) equals 1 (C) equals 1
75.
2 4d sin 2 Let d R, and A 1 sin 2 d , [0, 2]. If the minimum 5 2 sin d sin 2 2d value of det(A) is 8, then a value of d is: (A) 5 (B) 7 (C) 2 2 1 (D) 2 2 2
76.
(B) equals 0 (D) does not exist
Let z1 and z2 be any two non-zero complex numbers such that 3 z1 4 z2 . If z
3z1 2z2 then: 2z2 3z1
(A) Re(z) = 0
b
a
(C)
79.
5 2
(D) Im(z) = 0
Let I x 4 2x 2 dx. If I is minimum then the ordered pair (a, b) is: (A) 0, 2
78.
(B) z
1 17 2 2
(C) z
77.
2, 2
(D)
(B) 2,0
2, 2
A point P moves on the line 2x – 3y + 4 = 0. If Q(1, 4) and R(3, 2) are fixed points, then the locus of the centroid of PQR is a line: 3 (A) with slope (B) parallel to x-axis 2 2 (C) with slope (D) parallel to y-axis 3
max | x |, x 2 , x 2 Let f(x) . Let S be the set of points in the interval (4, 4) at 2 x 4 8 2 x , which f is not differentiable. Then S: (A) is an empty set (B) equals {2, 1, 0, 1, 2} (C) equals {2, 1, 1, 2} (D) equals {2, 2}
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JEE-MAIN-2019 (10th Jan-First Shift)-PCM-16 80.
If a circle C passing through the point (4, 0) touches the circle x2 + y2 + 4x – 6y = 12 externally at the point (1, 1), then the radius of C is: (A) 2 5 (B) 4 (C) 5
(D)
57
81.
Let f : R R be a function such that f(x) = x3 + x2f(1) + xf(2) + f(3), x R. Then f(2) equals: (A) 4 (B) 30 (C) 2 (D) 8
82.
Let
83.
ˆ 4iˆ 3 ˆj 6kˆ and c 3iˆ 6 ˆj 1 kˆ be three vectors a 2iˆ 1ˆj 3k,b 2 3 such that b 2a and a is perpendicular to c. Then a possible value of (1, 2, 3) is:
(A) (1, 3, 1)
1 (B) ,4,0 2
1 (C) ,4, 2 2
(D) (1, 5, 1)
Let A be a point on the line r 1 3 ˆi 1 ˆj 2 5 kˆ and B(3, 2, 6) be a point in the space. Then the value of for which the vector AB is parallel to the plane x – 4y + 3z = 1 is: 1 1 (A) (B) 4 8 1 1 (C) (D) 2 4
sin
n
84.
Let n 2 be a natural number and 0 < < /2. Then
sin n 1
sin
1 n
cos
d is equal to:
(where C is a constant of integration) n 1 (A) 2 1 n 1 sinn1
n 1 n
n 1 (C) 2 1 n 1 sinn1
n 1 n
C
n 1 (B) 2 1 n 1 sinn1
C
n 1 (D) 2 1 n 1 sinn1
n 1 n
n 1 n
C
C
85.
Consider a triangular plot ABC with sides AB = 7 m, BC = 5 m and CA = 6 m. A vertical lamp-post at the mid point D of AC subtends an angle 30° at B. The height (in m) of the lamp-post is: 3 2 (A) 21 (B) 21 2 3 (C) 2 21 (D) 7 21
86.
The equation of a tangent to the hyperbola 4x2 – 5y2 = 20 parallel to the line x – y = 2 is: (A) x – y + 1 = 0 (B) x – y + 7 = 0 (C) x – y + 9 = 0 (D) x – y – 3 = 0 FIITJEE Ltd., FIITJEE House, 29-A, Kalu Sarai, Sarvapriya Vihar, New Delhi -110016, Ph 46106000, 26569493, Fax 26513942 website: www.fiitjee.com.
JEE-MAIN-2019 (10th Jan-First Shift)-PCM-17 87.
If the line 3x + 3y – 24 = 0 intersects the x-axis at the point A and the y-axis at the point B, then the incentre of the triangle OAB, where O is the origin, is: (A) (3, 4) (B) (2, 2) (C) (4, 3) (d) (4, 4)
88.
The sum of all two digit positive numbers which when divided by 7 yield 2 or 5 as remainder is: (A) 1256 (B) 1465 (C) 1365 (D) 1356
89.
If the area enclosed between the curves y = kx2 and x = ky2, (k > 0), is 1 square unit. Then k is: 3 1 (A) (B) 2 3 2 (C) 3 (D) 3
90.
Consider the statement: “ P n : n2 n 41 is prime.” Then which one of the following is true? (A) Both P(3) and P(5) are true (B) P(3) is false but P(5) is true (C) Both P(3) and P(5) are false (D) P(5) is false but P(3) is true
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JEE-MAIN-2019 (10th Jan-First Shift)-PCM-18
JEE (Main) – 2019 ANSWERS PART A – PHYSICS 1.
D
2.
D
3.
C
4.
A
5.
C
6.
A
7.
D
8.
A
9.
A
10.
A
11.
BONUS
12.
A
13.
C
14.
D
15.
B
16.
B
17.
A
18.
D
19.
B
20.
A
21.
C
22.
B
23.
A
24.
A
25.
A
26.
C
27.
A
28.
C
29.
C
30.
D
PART B – CHEMISTRY 31.
D
32.
C
33.
B
34.
A
35.
A
36.
A
37.
A
38.
A
39.
BONUS
40.
C
41.
A
42.
A
43.
B
44.
C
45.
B
46.
C
47.
B
48.
D
49.
B
50.
A
51.
C
52.
B
53.
D
54.
A
55.
B
56.
D
57.
C
58.
D
59.
C
60.
D
PART C – MATHEMATICS 61.
C
62.
A
63.
B
64.
B
65.
C
66.
D
67.
C
68.
D
69.
A
70.
D
71.
A
72.
B
73.
B
74.
B
75.
A
76.
No option is correct
77.
D
78.
C
79.
B
80.
C
81.
C
82.
B
83.
A
84.
A
85.
B
86.
A
87.
B
88.
D
89.
B
90.
A
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JEE-MAIN-2019 (10th Jan-First Shift)-PCM-19
HINTS AND SOLUTIONS PART A – PHYSICS 2
1.
2
P h 2m 2m 2 (6.62 1034 )2 1 = eV 31 12 2 2 9.1 10 (7.5 10 ) 1.6 1019
k
Alternate method 150 V
7.5 × 10–12 =
150 V
80 kV 3 80 Energy = keV 25 keV 3 V
2.
x
dx
d = ( dN)x F = 2 2 dx x R R
d 0
3.
VAJ I R AJ E E x 12r 3 12r 3r L
4.
5.
D 2 hTR 2hRR
a a 1 0, , 2 2 a a 2 , ,0 2 2 aˆ a ˆ r2 r1 i k 2 2 ˆi kˆ Unit vector = 2
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JEE-MAIN-2019 (10th Jan-First Shift)-PCM-20 6. 6
6
Rnet = 4
6
7.
W = F s cos g 1 g = m g t2 2 2 2
8.
For maxima, dsin = n d ( Sin when is small) n 2500 = nm n Take n = 4, 5 for = 625 nm and 500 nm.
9.
Area (Range)2 v4 4 A1 1 A2 2
10.
11.
12.
13.
128 kg 64 (2kg) m3 10 cm3 64 (2 kg) 503 = 6 3 10 (50 cm) 64 50 50 50 2 kg (2 kg) = = 8 3 100 100 100 (50 cm) (50 cm3 ) Work done by external agent = Uf – Ui = 2 MB
Q kA (T2 T1 ) t 1 Q k (T2 T1 ) A t Cnet = C1 + C2 + C3
kAEo AE AE AE k1 o k 2 o k 3 o d 3 d 3 d 3 d
k
k1 k 2 k 3 12 3
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JEE-MAIN-2019 (10th Jan-First Shift)-PCM-21 14.
Q1 + q 2 + Q3 = Q Q3 Q1 Q2 =k 2 2 4a 4 b 4 c 2
…(1) …(2)
Subs. Q1, Q2, Q3 in (1)
15.
k
Q 4 (a b 2 c 2 )
v
kQ1 kQ2 kQ3 a b c
2
n1 = n2 = n3 T T T 1 2 1 3 1 4 T1 T2 T3
T2 T3 T4 T1 T2 T3
T2T3 = T1T4 and
T32 T4 T2
Solve for T2 and T3. 16.
Initially, kinetic energy =
1 1 GMe mv 2 m 2 2 r
By conservation of M.E., GMem KE 0 r GMem KE = mv 2 r 17.
18.
dV a 2gh 0 (for maximum height) dt 2 10 8 h = 5.1 cm 2ga2 2 9.8 10 8 n T 2 m On solving, n = 5, 5 loops are formed in 1 m. f
19.
Separation between successive nodes =
340 f1 fo ; f2 fo 340 17 f1 306 18 17 18 or or f2 323 19 17 19
1 m = 20 cm 5
340 340 34
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JEE-MAIN-2019 (10th Jan-First Shift)-PCM-22
20.
21.
1 1 1 1 1 1 (1 1) ; (2 1) f1 R f 2 R 1 ( 1) 1 1 ; 2 f1 R f2 R 1 2 f2 2f1 ; f1 f2 1 1 2( 2 1) R R 1 + 22 = 3 R = P Q (x y) (xy)
= x y xy = x y xy = xy 22.
E = CB
23.
In 8 seconds count rate becomes 4 half lives = 8s 1 half lie = 2s
In 6s or 3 half lives, count rate = 24.
25.
1600 = 200 23
E = vB v = Ed = dvB P1
P2 r
OV
v1 = v 2 k(2qa) k(4qa) x2 (R x)2 R–x= 2x R x= 2 1
x
26.
1 times. 16
OV
OV
20
20
O
0.5 A
OV
10 V
10 V
10 V
OV
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JEE-MAIN-2019 (10th Jan-First Shift)-PCM-23
27.
By colour coding, R = 50 102 I2R = P P I 20 mA R
28.
u=0
Ycm from ground =
40 m
0.02 100 = 40 m/s 0.05 V2 40 40 H = cm = 80 m 2g 20 Height above building = 80 – 40 = 40 m
COM
Vcm = 60 m
100 m/s
29.
0.03 100 = 60 m 0.05
dM = di A dq 2 = x 2 = dx x 2 2 L
M dM 0
30.
F – f = ma 1 fR = mR2 2 a = R
PART B – CHEMISTRY 31.
Cl
F
Cl
M
= 3 GI Total = 12 GI
NO 2
F
Cl
M
M SCN
NO 2
Cl
F NCS
ONO
= 3 GI
F M
SCN = 3 GI
ONO
NCS = 3 GI
H T
32.
S
33.
KP = KC(RT)ng
34.
Fact based
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JEE-MAIN-2019 (10th Jan-First Shift)-PCM-24 35.
More polluted water has high biological oxygen demand.
36.
Fact based.
37. Major product will be
Due to E2 elimination & extended
conjugation
38.
K Ae Ea /RT (i) is true but (ii) is false (ii) should be
K
T
39.
40.
Fact based
41.
Density of DCM is higher than H 2O.
42.
Adipic acid will give unstable 7 membered anhydride.
43.
Electron withdrawing group enhances the rate of hydrolysis.
44.
K.E = h - h0, So (B) is not correct.
45.
Fact based.
46.
Atomic & ionic radii decreases
47.
Allylic radical will form in presence of sunlight.
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JEE-MAIN-2019 (10th Jan-First Shift)-PCM-25 48.
49.
O
50. F
F Xe
F
F
51.
Diagonal relationship.
52.
Lesser the SRP value higher the reducing power.
53.
Fact based.
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JEE-MAIN-2019 (10th Jan-First Shift)-PCM-26 54.
55.
PA APA0 YAPT PT APA0 BPB0 0.4 7 103 + 0.6 12 103 = 104 0.4 7 103 = YA 104 YA = 0.28 YB = 1 -0.28 = 0.72
56.
NaBH4 don’t reduce ester.
57.
Fact based
58.
Higher the Ka value lower the pKa value.
59.
Hall-Herolut’s process is given by 2Al2O3 3C 4 Al 3CO2 3 2 2Al2O3 4Al 6O
At cathode: 4 Al3 12e 4Al At anode: 6O2 3 O 2 g 12e
2C 3 O2 3 CO2 60
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JEE-MAIN-2019 (10th Jan-First Shift)-PCM-27
PART C – MATHEMATICS 61.
P (7 or 8)
P H P 7 or 8 P T P 7 or 8
62.
1 11 1 2 11 1 19 2 36 2 9 72 9 72
3 Let P be the point nearest to , 0 , then normal 2 3 at P will pass through , 0 . 2
t2 t Let Co – ordinates of P be s , 4 2 Hence equation of normal is y tx
t t2 2 4
3 The line passes through , 0 2 3 3t t t t2 (–2, 0 are rejected) 2 2 4 hence nearest point is (1, 1) 2
5 2 3 2 1 1 0 2
distance
63.
ˆi ˆj kˆ n n1 n2 3 1 2 7iˆ 7ˆj 7kˆ 1
2
3
Equation of plane is 7 x 4 7 y 1 7 z 2 0 7x 7y 7z 7 0 x y z 1 64.
1 3 8 x y 5 25 12 x y x y 13
2
x
i
………..(1)
2
N 1 9 64 x 2 y 2 9.2 25 5 34.2 5 74 x 2 y 2
171 74 x 2 y 2 97 x 2 y 2
……………(2)
2
x 2 y 2 2xy 169 97 2xy xy 36
x y
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JEE-MAIN-2019 (10th Jan-First Shift)-PCM-28 T = 4, 9 So ratio is 65.
4 9 or 9 4
5,5r,5r 2 sides of triangle, 5 5r 5r 2 ………………(1) 2 5 5r 5r ………………(2) 2 5r 5r 5 ………………(3) 2 From r r 1 0 1 5 1 5 , ………..(4) r 2 2 From (2) r2 r 1 0 r R ……….(5) From (3) r2 r 1 0 1 5 1 5 r So, r 0 2 2 1 5 1 5 …………(6) r , , 2 2 1 5 1 5 From (4), (5), (6) r , 2 2 now check options 3
20 CI1 20 20 CI CI1 i 1 20 20 CI1 CI1 I Now 20 20 21 CI CI1 CI 21 Let given sum be S, so 3 2 20 i 1 20.21 100 S 3 3 I1 21 21 2 21 20
66.
Given S
67.
k k 100 21
sin2 2 cos4 2
3 4
Let cos2 2 t
1 cos2 2 cos4 2 t
3 4
1 1 cos2 2 2 2
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JEE-MAIN-2019 (10th Jan-First Shift)-PCM-29
2cos2 2 1 0 cos 4 0 4 2n 1 2 3 2n 1 , 0, 8 8 8 2 Sum of values of is 2 68.
Case – 1 c 50 f 0 0
………….(i) ……………(ii)
c40 f 2 0
4 c 5 4c c 4 0
c 24 f 2 0
………..(iii)
9 c 5 6c c 4 0
49 ………(iv) 4 Here i ii iii iv 4c 49 0 c
49 c , 24 4 Case – II …………….(i) c 50 f 0 0
……………..(ii) c4 f 2 0 c 24 ………..(iii) f 3 0 c 49 ………..(iv) 4 c
49 c , 24 4
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JEE-MAIN-2019 (10th Jan-First Shift)-PCM-30
69.
dy 3 sec 2 x y sec 2 x dx This is linear differential equation 3 sec 2 x dx Integrating factor e e3 tan x
Hence y.e3 tan x e
3 tan x
.sec 2 x dx
e3 tan x c 3 1 y Ce3 tan x 3 4 4 1 Given y Ce 3 3 3 4 3 3 Ce 1 1 Hence y e3 .e3 e6 3 3 4 y e3 tan x
70.
140 n p 46 3 140 n C 28 5 140 n M 70 2 n P C M n P n C n M n P C n C M n M P n P M C 140 140 140 140 46 28 70 15 10 6 30 144 9 14 23 4 102 So required number of student 140 102 38
71.
In the expansion of 1 xlog2 x
third term say T3 5C2 xlog2 x
xlo2 x
2
5
2
2560
256
taking lograthium to the base 2 on both sides 2
2 log2 x 8 log2 x 2
1 4 1 Here x 4 x 4,
72.
Normal to there 2 curves are y m x c 2bm bm3
y mx 4am 2am3 FIITJEE Ltd., FIITJEE House, 29-A, Kalu Sarai, Sarvapriya Vihar, New Delhi -110016, Ph 46106000, 26569493, Fax 26513942 website: www.fiitjee.com.
JEE-MAIN-2019 (10th Jan-First Shift)-PCM-31 If they how a common normal (c + 2b) – 4a = (2a – b) m2 (m = 0 corresponds to axis) c m2 20 2a b c 2 2a b According to the question answer is (1, 2, 3, 4) but my be in question they want common normal other than x – axis, hence answer is (B) 73.
xyz5 x 2y 3z 9, x 3y z 1 1 1 D 1 2 3 0 2 9 3 3 2 0 5 1 3 0 1 1 5
Now, D3 1 2 9 0 2 27 9 5 3 2 0 13 1 3
at 5, b 13 above 3 planes from common line
74.
Lim
1 x sin 1 x sin 1 x 2 1 x 1 x
x 1
Lim
1 x sin x 1 sin 2
x 1 1 x 1 sin x 1 Lim 1 1 0 x 1 x 1 x 1
2 75.
sin 2
4d
d sin 2 2 sin d sin 2 2d 4 d sin 2 d sin
A 1 5 2 1 1
0
(New R3 R3 2R2 R1 )
0 2
4 d d sin2 4 d 2 sin2 2
Because minimum value of A 8 d 2 9 d 1 or –5
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JEE-MAIN-2019 (10th Jan-First Shift)-PCM-32
76.
3z1 2 2z2 3z1 2cos 2 sin i 2z2 2z 1 1 2 cos sin 3z1 2 2 2z 3z 5 3 Given, z 1 2 cos sin i 3z2 2z1 2 2 Let
Which is neither purely real nor purely imaginary and z depends on .
b
77.
x
4
2x 2 dx
a
From figure min area is 2, 2
78.
Let point P is , and centroid of PQR is (h, k), then 3h 1 3 and 3k 4 2 3h 4 and 3k 2 Because , lies on 2x 3y 4 0 2 3h 4 3 k 2 4 0
locus is 6x 9y 2 0 whose slope is
2 3
79.
From the graph we can easily conclude that f x is non – derivable at x 2, 1,0,1,2
80.
Tangent at (1, –1) is x 1 y 1 2 x 1 3 y 1 12 0 3x 4y 7 2
2
Required circle is x 1 y 1 3x 4y 7 0 It pass through (4, 0) 9 1 12 7 0 2
required circle is x 2 y 2 8x 10y 16 0 Radius 16 25 16 5 81.
f x x 3 ax 2 bx c f ' x 3x 2 2ax b f " x 6x 2a
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JEE-MAIN-2019 (10th Jan-First Shift)-PCM-33 f "' x 6 a f ' 1 3 2a b
a b 3
b f " 12 2a 2a b 12 c f "' 3 c 6 and a 5, b 2 f x x 2 5x 2 2x 6 f 2 8 20 4 6 2
82.
Because b 2a , so 3 2 21
……….(i)
Because a is perpendicular to c so 6 61 3 3 1 0
………..(ii)
1, 2 , 3 1,3 21, 1 21 where 1 R
1 , 4, 0 satisfied above triplet. 2 83.
Let A is 1 3, 1, 2 5 AB 3 2 i 3 j 4 5 kˆ which is parallel to plane x 4y 3z 1 1 3 2 4 3 3 4 5 0
8 2 0
84.
sinn sin
1 n
cos
sinn1
t
n
1 4
d
n
t dt
(Put sin t )
t n 1 1
1
1 n 1 n t 1 n1 1 n1 t t n1 dt dt t tn n 1 1 Put 1 n 1 z dt dz, t tn 1
1
n 1 2 1 n 1 sinn1
h 1 n
n 1
1 n 1 t1n 1 zn n I z dz c n 1 n2 1 1 1 n 1 n
c
c
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JEE-MAIN-2019 (10th Jan-First Shift)-PCM-34
85.
Length of median BM A
1 1 2 2 BC2 BA 2 AC 2 25 49 36 2 2 T (Top of tower)
3 7
M
h
3 30o
B
5
C
1 112 112 28 2 7 2 4
Let h be height of tower, given tan30o
86.
M
2 7
Hyperbola is
h 2 h
h2
7 28 3 3
x2 y2 1 5 4
Equation of its tangent in slope from is y mx 5m2 4 Hence tangent with slope 1 is y x 1 87.
ax bx 2 cx3 ay1 by 2 cy 2 I 1 , abc abc 8 60 6 0 10 0 10 0 6 8 8 0 , 2, 2 24 24
B (0, 6)
10 6 8 C (0, 0)
88.
A (8, 0)
Two digit numbers of the form 7 2 are 16, 23 …………….,93 Two digit numbers of the form 7 5 are 12, 19,…………….,96 12 13 Sum of all the above numbers equals to 16 93 12 96 654 702 1356 2 2
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JEE-MAIN-2019 (10th Jan-First Shift)-PCM-35
89.
y kx 2 , x ky 2 3
1 1 x k k 2 x 4 x 0 or x 3 x , 0 k k 1 1 Point of intersection are , and 0,0 k k
1 1 , k k
1/k
1/k
x 1 x 3/2 kx 3 Area kx1 dx 1 k 3 k 3/2 0 2 1 1 2 1 k2 3k 3 1 k 3
90.
P n n2 41 P 3 9 3 41 47 P 5 25 5 41 61
Hence P 3 and P 5 are both prime
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FIITJEE Solutions to JEE(Main)-2019 Test Date: 10th January 2019 (Second Shift)
PHYSICS, CHEMISTRY & MATHEMATICS Paper - 1 Time Allotted: 3 Hours
Maximum Marks: 360
Please read the instructions carefully. You are allotted 5 minutes specifically for this purpose.
Important Instructions: 1.
The test is of 3 hours duration.
2.
This Test Paper consists of 90 questions. The maximum marks are 360.
3.
There are three parts in the question paper A, B, C consisting of Physics, Chemistry and Mathematics having 30 questions in each part of equal weightage. Each question is allotted 4 (four) marks for correct response.
4.
Out of the four options given for each question, only one option is the correct answer.
5.
For each incorrect response 1 mark i.e. ¼ (one-fourth) marks of the total marks allotted to the question will be deducted from the total score. No deduction from the total score, however, will be made if no response is indicated for an item in the Answer Box.
6.
Candidates will be awarded marks as stated above in instruction No.3 for correct response of each question. One mark will be deducted for indicating incorrect response of each question. No deduction from the total score will be made if no response is indicated for an item in the answer box.
7.
There is only one correct response for each question. Marked up more than one response in any question will be treated as wrong response and marked up for wrong response will be deducted accordingly as per instruction 6 above.
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JEE-MAIN-2019 (10th Jan-Second Shift)-PCM-2
PART –A (PHYSICS) 1.
The Wheatstone bridge shown in Fig. here, gets balanced when the carbon resistor used as R1 has the colour code (Orange, Red, Brown). The resistors R2 and R4 are 80 and 40 , respectively. Assuming that the colour code for the carbon resistors gives their accurate values, the colour code for the carbon resistor, used as R3 would be: (A) Brown, Blue, Brown (B) Brown, Blue, Black (C) Red, Green, Brown (D) Grey, Black, Brown
2.
Consider the nuclear fission Ne20 2He4 C12 .Given that the binding energy / nucleon of Ne20, He4 and C12 are, respectively, 8.03 MeV, 7.07 MeV, and 7.86 MeV, identify the correct statement: (A) energy of 12.4 MeV will be supplied (B) 8.3 MeV energy will be released (C) energy of 3.6 MeV will be released (D) energy of 11.9 MeV has to be supplied
3.
A hoop and a solid cylinder of same mass and radius are made of a permanent magnetic material with their magnetic moment parallel to their respective axes. But the magnetic moment of hoop is twice of solid cylinder. They are placed in a uniform magnetic field in such a manner that their magnetic moments make a small angle with the field. If the oscillation periods of hoop and cylinder are Th and TC respectively, then : (A) Th TC (B) Th 2TC (C) Th 1.5TC
(D) Th 0.5C
4.
An unknown metal of mass 192 g heated to a temperature of 100°C was immersed into a brass calorimeter of mass 128 g containing 240 g of water at a temperature of 8.4°C. Calculate the specific heat of the unknown metal if water temperature stabilizes at 21.5°C. (Specific heat of brass is 394 J kg–1 K–1) (A) 458 J kg–1 K–1 (B) 1232 J kg–1 K–1 –1 –1 (C) 916 J kg K (D) 654 J kg–1 K–1
5.
Two identical spherical balls of mass M and radius R each are stuck on two ends of a rod of length 2R and mass M (see figure). The moment of inertia of the system about the axis passing perpendicularly through the centre of the rod is : 137 17 (A) MR2 (B) MR 2 15 15 209 152 (C) MR2 (D) MR 2 15 15
6.
The self produced emf of a coil is 25 volts. When the current in it is changed at uniform rate from 10 A to 25 A in 1 s, the change in the energy of the inductance is : (A) 740 J (B) 437.5 J (C) 540 J (D) 637.5 J
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JEE-MAIN-2019 (10th Jan-Second Shift) PCM-3 7.
The actual value of resistance R, shown in the figure is 30 . This is measured in an experiment as shown using the standard V formula R , where V and I are the readings of the voltmeter I and ammeter, respectively. If the measured value of R is 5% less, then the internal resistance of the voltmeter is : (A) 600 (B) 570 (C) 35 (D) 350
8.
At some location on earth the horizontal component of earth’s magnetic field is 18 10–6 T. At this location, magnetic needle of length 0.12 m and pole strength 1.8 A m is suspended from its mid-point using a thread, it makes 45° angle with horizontal in equilibrium. To keep this needle horizontal, the vertical force that should be applied at one of its ends is : (A) 3.6 105 N (B) 1.8 105 N (C) 1.3 105 N
9.
Two vectors A and B have equal magnitudes. The magnitude of A B is ‘n’ times the magnitude of A B . The angle between A and B is:
2
n 1 (A) cos1 2 n 1 n2 1 (C) sin1 2 n 1 10.
11.
12.
(D) 6.5 10 5 N
n 1 (B) cos1 n 1 n 1 (D) sin1 n 1
A metal plate of area 1 10–4 m2 is illuminated by a radiation of intensity 16 mW/m2. The work function of the metal is 5eV. The energy of the incident photons is 10 eV and only 10% of it produces photo electrons. The number of emitted photo electrons per second and their maximum energy, respectively, will be : [1 eV = 1.6 10–19 J] (A) 1014 and 10 eV (B) 1012 and 5 eV 11 (C) 10 and 5 eV (D) 1010 and 5 eV A particle which is experiencing a force, given by F 3 i 12 j, undergoes a displacement of d 4 i . If the particle had a kinetic energy of 3 J at the beginning of the displacement, what is its kinetic energy at the end of the displacement? (A) 9 J (B) 12 K (C) 10 J (D) 15 J
Consider a Young’s double slit experiment as shown in figure. What should be the slit separation d in terms of wavelength such that the first minima occurs directly in front of the slit (S1)? (A) (B) 2( 5 2) ( 5 2) (C) (D) 2(5 2) (5 2) FIITJEE Ltd., FIITJEE House, 29-A, Kalu Sarai, Sarvapriya Vihar, New Delhi -110016, Ph 46106000, 26569493, Fax 26513942 website: www.fiitjee.com.
JEE-MAIN-2019 (10th Jan-Second Shift)-PCM-4 13.
The eye can be regarded as a single refracting surface. The radius of curvature of this surface is equal to that of cornea (7.8 mm). This surface separates two media of refractive indices 1 and 1.34. Calculate the distance from the refracting surface at which a parallel beam of light will come to focus. (A) 1 cm (B) 2 cm (C) 4.0 cm (D) 3.1 cm
14.
A current of 2 mA was passed through an unknown resistor which dissipated a power of 4.4 W. Dissipated power when an ideal power supply of 11 V is connected across it is : (A) 11 105 W (B) 11 103 W (C) 11 104 W
(D) 11 105 W
15.
The diameter and height of a cylinder are measured by a meter scale to be 12.60.1 cm and 34.2 0.1 cm, respectively. What will be the value of its volume in appropriate significant figures? (A) 426481 cm3 (B) 426481.0 cm3 (C) 426080 cm3 (D) 430080 cm3
16.
For equal point charges Q each are placed in the xy plane at (0, 2), (4, 2), (4, –2) and (0, –2). The work required to put a fifth change Q at the origin of the coordinate system will be : Q2 1 Q2 1 (A) 1 (B) 1 4 0 4 0 3 3 (C)
Q2 2 2 0
(D)
Q2 4 0
17.
The modulation frequency of an AM radio station is 250 kHz, which is 10% of the carrier wave. If another AM station approaches you for license what broadcast frequency will you allot? (A) 2750 kHz (B) 2900 kHz (C) 2250 kHz (D) 2000 kHz
18.
A closed organ pipe has a fundamental frequency of 1.5 kHz. The number of overtones that can be distinctly heard by a person with this organ pipe will be : (Assume that the highest frequency a person can hear is 20,000 Hz) (A) 6 (B) 4 (C) 7 (D) 5
19.
For the circuit shown below, the current through the Zener diode is :
(A) 9 mA (C) Zero
(B) 5 mA (D) 14 mA
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JEE-MAIN-2019 (10th Jan-Second Shift) PCM-5 20.
The electric field of a plane polarized electromagnetic wave in free space at time t = 0 is given by an expression E(x, y) 10 ˆj cos[(6x 8z)] The magnetic field B (x, z, t) is given by : (c is the velocity of light) 1 ˆ 1 ˆ (A) 6k 8iˆ cos[(6x 8z 10 ct)] (B) 6k 8iˆ cos[(6x 8z 10ct)] c c 1 ˆ 1 ˆ (C) 6k 8iˆ cos[(6x 8z 10 ct)] (D) 6k 8iˆ cos[(6x 8z 10 ct)] c c
21.
Charges – q and + q located at A and B, respectively, constitute an electric dipole. Distance AB = 2a, O is the mid point of the dipole and OP is perpendicular to AB. A charge Q is placed at P where OP = y and y >> 2a. The charge Q experiences an electrostatic force F. If Q is now moved along the equatorial line to P such that y y OP ' , the force on Q will be close to: 2a 3 3
(A) 3F (C) 9F
F 3 (D) 27F (B)
22.
Two stars of masses 3 1031 kg each, and at distance 2 1011 m rotate in a plane about their common centre of mass O. A meteorite passes through O moving perpendicular to the star’s rotation plane. In order to escape from the gravitational field of this double star, the minimum speed that meteorite should have at O is : (Take Gravitational constant G = 6.67 10–11 Nm2 kg–2) (A) 2.4 104 m/s (B) 1.4 105 m/s (C) 3.8 104 m/s (D) 2.8 105 m/s
23.
Half mole of an ideal monoatomic gas is heated at constant pressure of 1 atm from 20°C to 90°C. Work done by has is close to: (Gas constant R = 8.31 J/molK) (A) 581 J (B) 291 J (C) 146 J (D) 73 J
24.
A parallel plate capacitor having capacitance 12 pF is charged by a battery to a potential difference of 10 V between its plates. The charging battery is now disconnected and a porcelain slab of dielectric constant 6.5 is slipped between the plates. The work done by the capacitor on the slab is : (A) 692 pJ (B) 508 pJ (C) 560 pJ (D) 600 pJ FIITJEE Ltd., FIITJEE House, 29-A, Kalu Sarai, Sarvapriya Vihar, New Delhi -110016, Ph 46106000, 26569493, Fax 26513942 website: www.fiitjee.com.
JEE-MAIN-2019 (10th Jan-Second Shift)-PCM-6 25.
A particle starts from the origin at time t = 0 and moves along the positive x-axis. The graph of velocity with respect to time is shown in figure. What is the position of the particle at time t = 5s? (A) 10 m (B) 6 m (C) 3 m (D) 9 m
26.
A rigid massless road of length 3l has tow masses attached at each end as shown in the figure. The rod is pivoted at point P on the horizontal axis (see figure). When released from initial horizontal position, its instantaneous angular acceleration will be :
g 13l g (C) 2l (A)
g 3l 7g (D) 3l (B)
27.
Two forces P and Q, of magnitude 2F and 3F, respectively, are at an angle with each other. If the force Q is doubled, then their resultant also gets doubled. Then, the angle is : (A) 120° (B) 60° (C) 90° (D) 30°
28.
A cylindrical plastic bottle of negligible mass of filled with 310 ml of water and left floating in a pond with still water. If pressed downward slightly and released, it starts performing simple harmonic motion at angular frequency . If the radius of the bottle is 2.5 cm then is close to: (density of water = 103 kg/m3) (A) 3.75 rad s–1 (B) 1.25 rad s–1 –1 (C) 2.50 rad s (D) 5.00 rad s–1
29.
A particle executes simple harmonic motion with an amplitude of 5 cm. When the particle is at 4 cm from the mean position, the magnitude of its velocity is SI units is equal to that of its acceleration. Then, its periodic time in seconds is : 4 3 (A) (B) 3 8 8 7 (C) (D) 3 3
30.
Two kg of a monoatomic gas is at a pressure of 4 104 N/m2. The density of the gas is 8 kg/m3. What is the order of energy of the gas due to its thermal motion? (A) 103 J (B) 105 J 4 (C) 10 J (D) 106 J
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JEE-MAIN-2019 (10th Jan-Second Shift) PCM-7
PART –B (CHEMISTRY) 31.
The ground state energy of hydrogen atom is –13.6 eV. The energy of second excited state of He+ ion in eV is: (A) –54.4 (B) –3.4 (C) –6.04 (D) –27.2
32.
Haemoglobin and gold sold are examples of: (A) positively and negatively charged sols, respectively (B) positively charged sols (C) negatively charged sols (D) negatively and positively charged sols, respectively
33.
The major product of the following reaction is:
(A)
(B)
(C)
(D)
34.
The amount of sugar (C12H22O11) required to prepare 2 L of its 0.1 M aqueous solution is: (A) 136.8 g (B) 17.1 g (C) 68.4 g (D) 34.2 g
35.
Among the following reactions of hydrogen with halogens, the one that requires a catalyst is: (A) H2 I2 2HI (B) H2 Cl2 2HCl (C) H2 Br2 2HBr (D) H2 F2 2HF
36.
5.1 g NH4SH is introduced in 3.0 L evacuated flask at 327°C. 30% of the solid NH4SH decomposed to NH3 and H2S as gases. The Kp of the reaction at 327°C is (R = 0.082 L atm mol–1 K–1, Molar mass of S = 32 g mol–1, molar mass of N = 14 g mol–1) (A) 0.242 10–4 atm2 (B) 1 10–4 atm2 –3 2 (C) 4.9 10 atm (D) 0.242 atm2
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JEE-MAIN-2019 (10th Jan-Second Shift)-PCM-8 37.
The reaction that is NOT involved in the ozone layer depletion mechanism in the stratosphere is :
uv (A) CF2Cl2 (g) Cl(g) CF2Cl(g)
(B) Cl O(g) O(g) Cl(g) O2 (g)
(C) CH4 2 O3 3CH2 O 3H2O
hv (D) HOCl(g) OH(g) C l(g)
38.
In the cell Pt(s) H2 (g,1bar) HCl(aq) AgCl(s) Ag(s) Pt(s) the cell potential is 0.92 V when a 10–6 molal HCl solution is used. The standard electrode potential of (AgCl / Ag,Cl ) electrode is : 2.303RT 0.06V at 298K Given, F (A) 0.94 V (B) 0.76 V (C) 0.40 V (D) 0.20 V
39.
The 71st electron of an element X with an atomic number of 71 enters into the orbital: (A) 6p (B) 4f (C) 5d (D) 6s
40.
The correct match between item ‘I’ and item ‘II’ is: Item ‘I’ Item ‘II’ (compound) (reagent) (a) Lysine (p) 1-naphtol (b) Furfural (q) Ninhydrin (c) Benzyl alcohol (r) KMnO4 (d) Styrene (s) Ceric ammonium nitrate (A) (a)(q); (b) (p); (c) (s); (d) (r) (B) (a)(q); (b) (p); (c) (r); (d) (s) (C) (a)(r); (b) (p); (c) (q); (d) (s) (D) (a)(q); (b) (r); (c) (s); (d) (p)
41.
An aromatic compound ‘A‘ having molecular formula C7H6O2 on treating with aqueous ammonia and heating forms compound ‘B’. The compound ‘B’ on reaction with molecular bromine and potassium hydroxide provides compound ‘C’ having molecular formula C6H7N. The structure of ‘A’ is : (A) (B)
(C)
42.
(D)
The process with negative entropy change is: (A) Dissociation of CaSO4(s) to CaO(s) and SO3(g) (B) Sublimation of dry ice (C) Dissolution of iodine in water (D) Synthesis of ammonia from N2 and H2
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JEE-MAIN-2019 (10th Jan-Second Shift) PCM-9 43.
An ideal gas undergoes isothermal compression from 5 m3 to 1 m3 against a constant external pressure of 4 NM–2. Heat released in this process is used to increase the temperature of 1 mole of Al. If molar heat capacity of Al is 24 J mol–1K–1, the temperature of Al is increases by: 3 (A) K (B) 2 K 2 2 (C) K (D) 1 K 3
44.
Elevation in the boiling point for 1 molal solution of glucose is 2 K. The depression in the freezing point for 2 molal solution of glucose in the same solvent is 2 K. The relation between Kb and Kf is : (A) K b 1.5K f (B) K b K f (C) K b 0.5K f
45.
46.
47.
(D) K b 2K f
The major product of the following reaction is :
(A)
(B)
(C)
(D)
Sodium metal on dissolution in liquid ammonia gives a deep blue solution due to the formation of: (A) sodium-ammonia complex (B) sodamide (C) sodium ion-ammonia complex (D) ammoniated electrons k1 For an elementary chemical reaction, A 2 2A, the expression for k 1
(A) k1[A 2 ] k 1[A]2
(B) 2k1[A 2 ] k 1[A]2
(C) k1[A 2 ] k 1[A]2
(D) 2k1[A 2 ] 2k 1[A]2
d[A] is dt
48.
Which of the following tests cannot be used for identifying amino acids? (A) Biuret test (B) Barfoed test (C) Ninhydrin test (D) Xanthoproteic test
49.
The difference in the number of unpaired electrons of a metal ion in its high-spin and low-spin octahedral complexes is two. The metal ion is : (A) Ni2 (B) Fe2 (C) Co2 (D) Mn2
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JEE-MAIN-2019 (10th Jan-Second Shift)-PCM-10 50.
51.
The major product obtained in the following reaction is:
(A)
(B)
(C)
(D)
The pair that contains two P – H bonds in each of the oxoacids is : (A) H4P2O5 and H4P2O6 (B) H3PO2 and H4P2O5 (C) H3PO3 and H3PO2
52.
Which is the most suitable reagent for the following transformation?
(A) Tollen’s reagent (C) CrO 2 Cl2 / Cs2 53.
(D) H4P2O5 and H3PO3
(B) I2 / NaOH (D) alkaline KMnO4
What is the IUPAC name of the following compound?
(A) 3-Bromo-1, 2-dimethylbut-1-ene (B) 3-Bromo-3-methyl-1, 2-dimethylprop-1-ene (C) 2-Bromo-3-methylpent-3-ene (D) 4-Bromo-3-methylpent-2-ene 54.
The number of 2-centre-2-electron and 3-centre-2-electron bonds in B2H6, respectively, are : (A) 2 and 1 (B) 4 and 2 (C) 2 and 2 (D) 2 and 4
55.
In the reaction of oxalate with permanganate in acidic medium, the number of electrons involved in producing one molecule of CO2 is : (A) 1 (B) 10 (C) 2 (D) 5
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JEE-MAIN-2019 (10th Jan-Second Shift) PCM-11 56.
A reaction of cobalt(III) chloride and ethylenediamine in a 1 : 2 mole ratio generates two isomeric products A (violet coloured) and B (green coloured). A can show optical activity, but, B is optically inactive. What type of isomers does A and B represent? (A) Geometrical isomers (B) Coordination isomers (C) Linkage isomers (D) Ionisation isomers
57.
The electrolytes usually used in the electroplating of gold and silver, respectively, are : (A) [Au(CN2 ] and [Ag(CN)2 ] (B) [Au(Cn)2 ] and [AgCl2 ] (C) [Au(OH)4 ] and [Ag(OH)2 ]
(D) [Au(NH3 )2 ] and [Ag(CN)2 ]
58.
A compound of formula A2B3 has the hcp lattice. Which atom forms the hcp lattice and what fraction of tetrahedral voids is occupied by the other atoms : 2 1 (A) hcp lattice - A, Tetrahedral voids - B (B) hcp lattice - A, Tetrahedral voids - B 3 3 2 1 (C) hcp lattice - B, Tetrahedral voids - A (D) hcp lattice - B, Tetrahedral voids – A 3 3
59.
The major product of the following reaction is :
60.
(A)
(B)
(C)
(D)
What will be the major product in the following mononitration reaction?
(A)
(B)
(C)
(D)
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JEE-MAIN-2019 (10th Jan-Second Shift)-PCM-12
PART–C (MATHEMATICS) 61.
The value of such that sum of the squares of the roots of quadratic equation, x 2 (3 )x 2 has the lest value is : 15 (A) (B) 1 8 4 (C) (D) 2 9
62.
The value of cos
cos 3 cos 10 sin 10 is : 2 2 2 2 2
1 512 1 (C) 256
1 1024 1 (D) 2
(A)
(B)
63.
The curve amongst the family of curves represented by the differential equation, (x 2 y 2 )dx 2xy dy 0 which passes through (1, 1), is : (A) a circle with centre on the x-axis (B) an ellipse with major axis along the y-axis (C) a circle with centre on the y-axis (D) a hyperbola with transverse axis along the x-axis
64.
Let f : (–1, 1)R be a function defined by f(x) max x , 1 x 2 . if K be the set of all
points at which f is not differentiable, then K has exactly : (A) five elements (B) one element (C) three elements (D) two elements 10
65.
66.
The positive value of for which the co-efficient of x2 in the expression x 2 x 2 is x 720, is : (A) 4 (B) 2 2 (C) 5 (D) 3 2
The tangent to the curve, y xe x passing, through the point (1, e) also passes through the point : 4 (A) (2, 3e) (B) ,2e 3 5 (C) ,2e (D) (3, 6e) 3
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JEE-MAIN-2019 (10th Jan-Second Shift) PCM-13 67.
Let N be the set of natural numbers and two functions f and g be defined as f, g : NN such that n 1 if n is odd f(n) 2 n if n is even 2 and g(n) n ( 1)n . Then fog is : (A) onto but not one-one. (B) one-one but not onto. (C) both one-one and onto. (D) neither one-one nor onto.
68.
The number of values of (0, ) for which the system of linear equations x 3y 7z 0 x 4y 7z 0 (sin 3)x (cos 2)y 2z 0 has a non-trivial solution, is : (A) three (B) two (C) four (D) one Let ( 2) a b and (4 2) a 3 b be two given vectors where a and b are non collinear. The value of for which vectors and are collinear, is : (A) –4 (B) –3 (C) 4 (D) 3
69.
70.
Two sides of a parallelogram are along the lines, x + y = 3 and x – y + 3 = 0. If its diagonals intersect at (2, 4), then one of its vertex is : (A) (3, 5) (B) (2, 1) (C) (2, 6) (D) (3, 6)
71.
If
x
1
2 2 ' f(t) dt x t f(t) dt, then f 12 is : 0
x
24 25 4 (C) 5
18 25 6 (D) 25
(A)
(B)
5
72.
73.
5
3 i 3 i Let z . If R(z) and I(z) respectively denote the real and 2 2 2 2 imaginary parts of z, then : (A) I(z) = 0 (B) R(z) > 0 and I(z) > 0 (C) R(z) < 0 an I(z) > 0 (D) R(z) = –3 1 , then the minimum 3 number of independent shots at the target required by him so that the probability of 5 hitting the target at least once is greater than , is : 6 (A) 3 (B) 6 (C) 5 (D) 4
If the probability of hitting a target by a shooter, in any shot, is
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JEE-MAIN-2019 (10th Jan-Second Shift)-PCM-14
74.
If
5
x e
4x3
dx
1 4 x3 e f(x) C, where C is a constant of integration, then f(x) is equal 48
to : (A) 2x 3 1 (C) 2x 3 1
(B) 4x 3 1 (D) 4x 3 1
75.
If the area of an equilateral triangle inscribed 2 2 x y 10x 12y c 0 is 27 3 sq. units then c is equal to : (A) 13 (B) 20 (C) –25 (D) 25
76.
Consider the following three statements : P : 5 is a prime number. Q : 7 is a factor of 192. R : L.C.M. of 5 and 7 is 35. Then the truth value of which one of the following statements is true? (A) (~ P) (Q R) (B) (P Q) (~ R) (C) (~ P) (~ Q R) (D) P (~ Q R)
77.
The length of the chord of the parabola x2 = 4y having equation x 2y 4 2 0 is : (A) 3 2 (C) 8 2
78.
circle,
b 1 2 det(A) 2 Let A b b 1 b where b > 0. Then the minimum value of is : b 1 b 2 (A) 2 3 (B) 2 3
(D)
3
y2 x2 Let S (x, y) R2 : 1 , where r 1. Then S represents : 1 r 1 r 2 (A) a hyperbola whose eccentricity is , when 0 r 1. 1 r 2 (B) an ellipse whose eccentricity is , when r 1. r 1 2 (C) a hyperbola whose eccentricity is , when 0 r 1. r 1 1 (D) an ellipse whose eccentricity is , when r 1. r 1 25
80.
the
(B) 2 11 (D) 6 3
(C) 3 79.
in
if
50
Cr 50 r C25 r K
50
C25 , then K is equal to :
r 0
(A) (25)2 (C) 224
(B) 225 – 1 (D) 225
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JEE-MAIN-2019 (10th Jan-Second Shift) PCM-15 81.
82.
83.
The plane which bisects the line segment joining the points (–3, –3, 4) and (3, 7, 6) at right angles, passes through which one of the following points? (A) (–2, 3, 5) (B) (4, –1, 7) (C) (2, 1, 3) (D) (4, 1, –2) n 19 The value of cot cot 1 1 2p is : n1 p 1 21 (A) 19 22 (C) 23
19 21 23 (D) 22
(B)
If mean and standard deviation of 5 observations x1, x 2 , x3 , x 4 , x 5 are 10 and 3, respectively, then the variance of 6 observations x1, x 2 ,, x 3 and –50 is equal to : (A) 509.5 (B) 586.5 (C) 582.5 (D) 507.5
84.
Let f be a differential function such that 3 f(x) 1 f '(x) 7 ,(x 0) and f(1) 4. Then lim x f : x 0 4 x x 4 (A) exists and equals . (B) exists and equals 4. 7 (C) does not exist. (D) exists and equals 0.
85.
Two vertices of a triangle are (0, 2) and (4, 3). If its orthocenter is at the origin, then its third vertex lies in which quadrant? (A) third (B) second (C) first (D) fourth
86.
The value of
2
to t, is : 1 (A) (7 5) 12 3 (C) (4 3) 20 87.
dx , where [t] denotes the greatest integer less than or equal [x] [sin x] 4
2
1 (7 5) 12 3 (D) (4 3) 10
(B)
On which of the following lines lies the point of intersection of the line, x4 y5 z3 and the plane, x + y + z = 2? 2 2 1 x 3 4 y z 1 x4 y5 z5 (A) (B) 3 3 2 1 1 1 x 1 y 3 z 4 x2 y3 z3 (C) (D) 1 2 5 2 2 3
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JEE-MAIN-2019 (10th Jan-Second Shift)-PCM-16 88.
Let a1,a 2 ,a3 , ,a10 be in G.P. with ai 0 for i 1, 2, ,10 and S be the set of pairs (r,k),r,k N (the set of natural numbers) for which loge a1r a2k
loge a2r a3k
loge a3r a4k
loge a4r a5k loge a7r a8k
loge a5r a 6r loge a8r a9k
loge a 6r a7k 0 loge a9r a10k
Then the number of elements in S, is : (A) 4 (C) 2
(B) infinitely many (D) 10
89.
With the usual notation, in ABC, if A + B = 120°, a 3 1, then the ratio A : B, is : (A) 7 : 1 (B) 5 : 3 (C) 9 : 7 (D) 3 : 1
90.
A helicopter is flying along the curve given by y x
3
2
7, (x 0). A solider positioned at
1 the point ,7 wants to shoot down the helicopter when it is nearest to him. Then this 2 nearest distance is : 5 1 7 (A) (B) 6 3 3 1 7 1 (C) (D) 6 3 2
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JEE-MAIN-2019 (10th Jan-Second Shift) PCM-17
JEE (Main) – 2019 ANSWERS PART A – PHYSICS 1.
A
2.
No option is correct
3.
A
4.
B
5.
A
6.
A
7.
B
8.
D
9.
A
10.
C
11.
D
12.
A
13.
D
14.
A
15.
B
16.
B
17.
D
18.
A
19.
A
20.
B
21.
D
22.
D
23.
B
24.
B
25.
D
26.
A
27.
A
28.
B
29.
C
30.
C
PART B – CHEMISTRY 31.
C
32.
A
33.
C
34.
C
35.
A
36.
A
37.
C
38.
D
39.
C
40.
A
41.
A
42.
D
43.
C
44.
D
45.
D
46.
D
47.
D
48.
B
49.
C
50.
C
51.
B
52.
B
53.
D
54.
B
55.
B
56.
A
57.
A
58.
A
59.
B
60.
D
PART C – MATHEMATICS 61.
D
62.
A
63.
A
64.
C
65.
A
66.
B
67.
A
68.
B
69.
A
70.
D
71.
A
72.
D
73.
C
74.
B
75.
D
76.
D
77.
D
78.
A
79.
B
80.
B
81.
D
82.
A
83.
D
84.
B
85.
B
86.
C
87.
C
88.
B
89.
A
90.
C
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JEE-MAIN-2019 (10th Jan-Second Shift)-PCM-18
HINTS AND SOLUTIONS PART A – PHYSICS 1.
R1 = 32 × 10 = 320 Ry 40 320 R1 R3 = = 160 R2 80 Colour code of R3 be Brown, Blue, Brown.
2.
Q = (B.E.)R – (B.E.)P = 20 × 8.03 – (8 × 7.07 + 12 × 7.86) = 160.6 – (56.56 + 94.32) Q = +9.72 meV 9.72 MeV released.
3.
T 2
I B
Th I R c Tc Ic h
2
=
1 1 2
Th = T c 4.
5.
6.
Heat loss = Heat gain 192 × S(100 – 21.5) = (128 × 0.394 + 240 × 4.2) (21.5 – 8.4) 192 × 78.5 × S = 1058.432 × 13.1 S = 0.91995 J/g K–1 S = 919.95 J/kg K–1
4R2 2 I 2 MR 2 M4R2 M 12 5 137 1 4 = MR2 8 = MR2 3 5 15
LT 25 t 25 1 5 15 3 1 2 2 1 5 U L If I1 (252 10 2 ) 2 2 3 5 = 525 4.7.5 J 6
L
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JEE-MAIN-2019 (10th Jan-Second Shift) PCM-19
7.
8.
9.
30 R 30 00.95 R 30 R = 570 = F × 0.06 = 1.8 × 0.012 × 18 × 10–6 F = 6.48 × 10–5 A B n A B
A2 + B2 + 2AB cos = n2(A2 + B2 – 2AB cos ) 2a2 (n2 1) cos (1 n2 ) 2a 2 n2 1 cos 2 n 1 10.
11.
12.
13.
14.
A B a
16 10 3 10 4 1.6 1011 10 10 10 19 (K.E.)max = (10 – 5) eV = 5eV n
K.E. – 3 = F d K.E. = 3 + (3iˆ 12ˆj) (4iˆ ) K.E. = 3 + 12 = 15 J
(2d)2 (d)2 2d
( 5 2)d
d
2
2
2( 5 2)
2 1 2 1 v u R 1.34 1 0.34 v 7.8 1.34 7.8 v nm = 3.074 cm 0.34 P 4.4 2 I 4 10 6 V 2 11 11 4 106 P W R 4.4 = 11 × 10–5 W R
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JEE-MAIN-2019 (10th Jan-Second Shift)-PCM-20
15.
2 Dh 4 = 4260 cm2 v D h 2 v D h v R2h
0.1 0.1 = 2 v 12.6 34.2
2x426 426 12.6 34.2 = 67.61 + 12.459 = 80.075 v = 4260 80 cm3 =
16.
Y
(0, 2)
Q
Q (4, 2)
0 Q Q (0, –2)
17.
18.
19.
20.
1 1 1 2 Q2 4 0 2 2 2 5 Q2 1 1 4 0 5
W VQ
x
0 Q (4, –2)
The interval between two carrier frequencies should be at least two times of AM frequency.
250 kHz 13.33 1.5 kHz Possible hormones 1, 3, 5, 7, 9, 11, 13 i.e 6.
50 5 mA 10k 120 50 i5 k = 14 mA 5k i2 (14 5) mA = 9 mA i10 k
E 10ˆj cos(6x 8z 10ct) E 10 Bo o C C W = 10 C ˆ Eˆ Bˆ C
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JEE-MAIN-2019 (10th Jan-Second Shift) PCM-21 ˆi
ˆj
kˆ
0
1
0
Bx
By
3 4 Bz ˆi 0 ˆj Bxkˆ ˆi ˆj 5 5 3 4 Bz , By 0, Bz 5 5 1 ˆ cos (6x 8z 10 ct) B ( 8iˆ 6k) C
21.
22.
Bz
6iˆ 8 ˆj 10
1 r3 Required force = 27 F.
F
1 2( GMm) mv 2 0 2 r 4GM 4 6.67 10 11 3 1031 V2 r 2 1011 V 20 2 10 4 m / s = 2.828 × 105 m/s
23.
W = nRT 1 = 8.31 70 2 = 290.85 J
24.
W
25.
rt = 5 = area 1 = 2 2 2 2 3 1 m 2 = (2 + 4 + 3) m = 9 m.
Q2 Q2 2c 2ck Q2 1 = 1 2c k 1 1 = 12 100 pJ 1 2 6.5 12 100 11 = pJ = 507.69 pJ 2 13
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JEE-MAIN-2019 (10th Jan-Second Shift)-PCM-22
I 5Mog 4Mo g = 5Mo 2 2Mo 42 Mo g = 13 Mo 2 g = 13
26.
27.
2 P Q P 2Q 13 + 12 cos = 10 + 6 cos 1 cos = 2 = 120°.
28.
Agx = Frestoring r2gx = n2x r 2g v
g 3.14 10 2.5 10 2 v 310 10 6 = 2.5 × 10–2 × 102 10 =r
= 2.5 10 29.
f
2.5 10 = 1.25 2
v 4 a4
w
A 2 x2
4
(w 2 x)4
w 25 16 w 2 4 3 w 4 2 4 8 T 2 w 3 3
30.
1 M Vm2 2 1 3P = 2 2
E
=
3 4 10 4 = 1.5 × 104 J 8
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JEE-MAIN-2019 (10th Jan-Second Shift) PCM-23
PART B – CHEMISTRY 2
31.
EHe
32.
n eV z2 4 13.6 9
E 13.6
Fact based O O
33.
HO - C - C - OH HO
OH
OH O O
O
C
O
n
HO
34.
0.1
nC12H22O11
2 nC12H22O11 0.2
Wt C12H22O11 0.2 342 68.4
35.
First reaction will be requiring a catalyst among halogens oxidizing power decrease down the group.
36.
NH4HS s NH3 g H2S g 5.1 g 0.1 g mol – 0.03
0.03 mol
0.03 mol
0.98 0.98 V = 3L, T = 327oC 2 2 K P PNH3 PH2S PV = nRT KP
0.98 0.98 2 2
P 3 = 0.06 0.0821 600
0.06 0.0821 200 3 P = 0.98
P KP = 0.243 37.
CH4 is not present in stratosphere.
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JEE-MAIN-2019 (10th Jan-Second Shift)-PCM-24
38.
H Cl E Eo 0.06 log 1/2 H2
106.10 6 11/2 o –12 0.92 = E - 0.06 log10 Eo + 0.06 12 Eo = 0.92 - 0.06 12 Eo = 0.92 – 0.72 EoAg / AgCl 0.20 0.92 Eo 0.06 log
39.
The electron configuration is [Xe]4f 145d16s2
40.
Lysine is an amino acid nindydrin test is used for amino acids. Furfural reacts with 1-napthol to give violet colouration. Benzyl alcohol undergoes reaction with ceric ammonium nitrate to give red colouration. CH = CH 2
Styrene
41.
, discharges color of KMnO4
CH3
CONH2
NH3 ,
[A]
42.
NH2
Br2 , KOH
[B]
CaSO 4 s CaO s SO3 g CO 2 s CO2 g I2 I2 aq N2 g 3 H2 g 2NH3 g N2 g 3 H2 2NH3 g S 2SNH3 SN2 3 SH2 There is decrease in number of moles of NH3 entropy is decreasing.
43.
q = PV q = 16 CP = 24 qP CP T 16 2 T K K 24 3 FIITJEE Ltd., FIITJEE House, 29-A, Kalu Sarai, Sarvapriya Vihar, New Delhi -110016, Ph 46106000, 26569493, Fax 26513942 website: www.fiitjee.com.
JEE-MAIN-2019 (10th Jan-Second Shift) PCM-25 44.
Tb = Kbm Tb = Kb 1 2 = Kb Tf = Kfm 2 = 2Kf = Kf Kf 1 Kb 2
45.
NaBH4 reduces both carbonyl group and imine.
46.
Na s x y NH3 Na NH3 x
e NH3 y Blue colour ammoniated electrons
47.
1 dA 2 K 1 A 2 K 1 A 2 dt d A 2 2K1 A 2 2K 1 A dt
48.
Barfoed test is used to detect monosaccirides.
49.
Co2+ high spin t 52g eg2 ‘3’ unpaired electrons Co2+ low spin t 62g e1g ‘1’ unpaired electron Difference is 3 – 1 = 2
50.
O
O
O
O 1. NaOEt 2.
O
O
OC2H5
O
H2O
HO COOEt
HO
O
O
P
P
P
H H3PO 2
52.
O COOEt
O
H
O
O NaOEt HO
51.
OC2H5
HO
OH
O
OH
COOEt
H
H4P 2O 5
Iodoform reaction can be used for this transformation. FIITJEE Ltd., FIITJEE House, 29-A, Kalu Sarai, Sarvapriya Vihar, New Delhi -110016, Ph 46106000, 26569493, Fax 26513942 website: www.fiitjee.com.
JEE-MAIN-2019 (10th Jan-Second Shift)-PCM-26 53.
4-Bromo-3-methyl pentane 2c - 2e
54.
H
H B H
H B
H 3c - 2e
55.
bond
H bond
5 C2O24 2KMnO 4 H 10 CO2 2Mn2 H2O After balancing 10 gain or loss of electron.
56.
57.
[Au(CN2 ] and [Ag(CN)2 ] both are soluble complexes.
58.
Formula of the compounds No. of octahedral voids are equal to the number of atoms forming lattice A occupy octahedral void i.e. 2/3 of them B forms crystal lattice A 2/3B A 2B3
59.
60.
‘EAS’ first ring is activated and second is deactivated NO2 attack at para position of activated ring.
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JEE-MAIN-2019 (10th Jan-Second Shift) PCM-27
PART C – MATHEMATICS 61.
2
2
2
2 2
3 2 2 2 9 6 4 2 2 4 5 For least value 2 sin 2n A cos A cos 2 A cos 22 A...........cos 2n1 A n 2 sin A
62.
Using formula
63.
x 2 dx 2xydy y 2 dx 0 x 2 dx y 2 dx 2xy dy
x
2
y 2 dx 2xy dy 0
x.2y dy y 2 dx dx x2 Integrals y2 x c x x 2 y 2 cx y
64.
A, B, C are sharp edges 1
–1
A
x
C
10
65.
x2 x 2 x Consider constant term r 10 r 10 Cr x 2 x 10 r 2r 0 2 10 5r 0 r2 10C2 2 720 4
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JEE-MAIN-2019 (10th Jan-Second Shift)-PCM-28
66.
2
y xe x (1, e) lies on this 2 2 dy Now xe x .2x e x .1 dx Put x = 1 m 2e e 3e Equation of tangent at (1, e) y e 3e x 1 y e 3ex 3e y 3ex 2e
4 3 , 2e satisfies it Answer is B
67.
f g 1 1 Many one f g 2 1
f g 2k k f g 2k 1 k 1 Onto
68.
sin3
1 1
cos 2
4
3 0
2 7 7 7 sin3 14 cos 2 14 0
sin3 2cos 2 2 0, sin 69.
1 2
2 a b 4 2 a 3b and are collinear
2
1
0 4 2 3 3 6 4 2 0 4 0 4
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JEE-MAIN-2019 (10th Jan-Second Shift) PCM-29 D (3, 6)
70.
C
Intersection point is A (0, 3) M 4, 6 B 1, 2 , D 3, 6
A (0, 3)
71.
x+y=3
B (1, 2)
Differentiability we get f x 2x x 2 f x
1 x2 2x f x f " x 2 2 1 x2 1 x2
1 24 f ' 2 25 5
72.
5
3 i i b e ei5 /6 2 2 5 3 z 2 cos 2 3 2 6 n
73.
5 2 1 6 3 n
1 2 6 3 n5
74.
4 x 3
5
x .e dx x .x e 2
3
4x 3
dx
3
4x t 12x 2dx dt 1 t et dt 12 4 1 t e t dt 48 1 t te 1.et c 48 3 1 4x3 e . 4x 3 e 4 x c 48
75.
r 25 36 c 36 c 25 FIITJEE Ltd., FIITJEE House, 29-A, Kalu Sarai, Sarvapriya Vihar, New Delhi -110016, Ph 46106000, 26569493, Fax 26513942 website: www.fiitjee.com.
JEE-MAIN-2019 (10th Jan-Second Shift)-PCM-30 76.
P is True Q is False R is True ~ Q is True ~ Q R is True P ~ Q R is True.
77.
x 2y 4 2
x 2 4y Solving we get point of intersection A 2 2, 2 , B 4 2,8
6 2 6 6 2
AB 78.
79.
2
3
Det A b2 3 det A 3 b b b Least value 2 3 y2 x2 1 1 r 1 r r 1 ellipse
2 r 1 e 1 r 1 r 1 25
80.
50
r 50 r r 1
25
r 1
81.
50 r 25 r 25
50 r 25 r 25
50 25 1 25 r 1 r 25 r
50 25 25 25 r 1
25
Cr 50 C25 225 1
A (–3, 3, 4), B (3, 7, 6) Mid point 0,2,5 n AB 6iˆ 10ˆj 2kˆ Equation of plane r .n a.n r . 6iˆ 10ˆj 2kˆ 0iˆ 2ˆj 5kˆ . 6iˆ 10ˆj 2kˆ
3x 5y z 15 (4, 1, –2) satisfies it Answer is D FIITJEE Ltd., FIITJEE House, 29-A, Kalu Sarai, Sarvapriya Vihar, New Delhi -110016, Ph 46106000, 26569493, Fax 26513942 website: www.fiitjee.com.
JEE-MAIN-2019 (10th Jan-Second Shift) PCM-31
82.
n 19 cot cot 1 1 2p p 1 n 1 19 cot cot 1 1 n2 n n1 19 1 cot tan1 2 1 n n n1
19 cot tan 1 n 1 tan1 1 n1 1 1 cot tan 20 tan 1
19 cot tan1 21 21 19 83.
x 50 2
ex 1 3 ex 2 5 5 1 2500 9 x2 5 5 x 2 545 2
New variable
84.
1 0 3045 507.5 6 6
3 f x . , x0 4 x 3 f 'x f x 7 4x f ' x 7
3
f x .e
4 x dx
(Linear)
3
7.e
4x dx
c
f x . x3/4 7.x 3/4 c x 7/4 c 7 4 f x 4x cx 3/4 7
1 4 f cx3/4 x x 1 Lt xf Lt 4 c x 7/4 4 x 0 x x 0
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JEE-MAIN-2019 (10th Jan-Second Shift)-PCM-32 (h, k)
85.
k 3 0 h4 k 4 0 h 32
k3
4h k
h
(0, 0)
3 4 (0, 2)
/2
86.
(4, 3)
dx
x sin x 4
/2
0
2 dx dx / 2 x 1 4 0 x 4
1
0
1
dx dx dx / 2 2 1 4 1 1 1 4 0 4
/2
dx
1 4 1
1 1 1 2 1 2 4 52 9 3 5 20 87.
Put 2 4, 2 5, 3 in x y z 2 2 4 2 5 3 2 5 10 2 P 0, 1, 1 Now put in options Answer is C
88.
For any value of r determinant is zero.
89.
a 3 1 b 3 1 sin A 3 1 3 1 2 3 2 3 sinB 2 3 1 sin A 32 sin 120 A
sin A 32 sin12cos A cos12 sin A 1 3 2 3 1 cot A 2 2 FIITJEE Ltd., FIITJEE House, 29-A, Kalu Sarai, Sarvapriya Vihar, New Delhi -110016, Ph 46106000, 26569493, Fax 26513942 website: www.fiitjee.com.
JEE-MAIN-2019 (10th Jan-Second Shift) PCM-33
3 cot A 1 1 2 32 3 2 1 3 cot A 1 3 2 2 3 cot A 4 2 3 1 3 cot A 3 2 3 cot A 3 2 cot A 2 3 tan15
A 105o B 15o 90.
dy 3 x dx 2 2 Slope of normal 3 x 3/2 Let point is x1, x1 2
y x 3/2 2
Normal y x13/2
2 2 x x 3 x 1
1
Now put (1, 7) and solve it. 1 x1 3 1 1 P , 7 , A 1, 7 3 3 3
AD
1 7 6 3
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FIITJEE Solutions to JEE(Main)-2019 Test Date: 11th January 2019 (First Shift)
PHYSICS, CHEMISTRY & MATHEMATICS Paper - 1 Time Allotted: 3 Hours
Maximum Marks: 360
Please read the instructions carefully. You are allotted 5 minutes specifically for this purpose.
Important Instructions: 1.
The test is of 3 hours duration.
2.
This Test Paper consists of 90 questions. The maximum marks are 360.
3.
There are three parts in the question paper A, B, C consisting of Physics, Chemistry and Mathematics having 30 questions in each part of equal weightage. Each question is allotted 4 (four) marks for correct response.
4.
Out of the four options given for each question, only one option is the correct answer.
5.
For each incorrect response 1 mark i.e. ¼ (one-fourth) marks of the total marks allotted to the question will be deducted from the total score. No deduction from the total score, however, will be made if no response is indicated for an item in the Answer Box.
6.
Candidates will be awarded marks as stated above in instruction No.3 for correct response of each question. One mark will be deducted for indicating incorrect response of each question. No deduction from the total score will be made if no response is indicated for an item in the answer box.
7.
There is only one correct response for each question. Marked up more than one response in any question will be treated as wrong response and marked up for wrong response will be deducted accordingly as per instruction 6 above.
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JEE-MAIN-2019 (11th Jan-First Shift)-PCM-2
PART –A (PHYSICS) 1.
A body is projected at t = 0 with a velocity 10 ms–1 at an angle of 600 with the horizontal. The radius of curvature of its trajectory at t = 1s is R. Neglecting air resistance and taking acceleration due to gravity g = 10 ms–2, the radius of R is: (A) 10.3 m (B) 2.8 m (C) 2.5 m (D) 5.1 m
2.
A hydrogen atom, initially in the ground state is excited by absorbing a photon of wavelength 980 Å. The radius of the atom in the excited state, in terms of Bohr radius a0, will be: (hc = 12500 eV- Å) (A) 25a0 (B) 9a0 (C) 16a0 (D) 4a0
3.
In the given circuit the current through Zener Diode is close to: (A) 0.0 mA (B) 6.7 mA (C) 4.0 mA (D) 6.0 mA
4.
There are two long co – axial solenoids of same length I. The inner and outer coils have radii r1 and r2 and number of turns per unit length n1 and n2 respectively. The ratio of mutual inductance to the self – inductance of the inner – coil is:
5.
(A)
n1 n2
(B)
n2 r1 . n1 r2
(C)
n2 r2 2 . n1 r12
(D)
n2 n1
A particle undergoing simple harmonic motion has time dependent displacement given by x t A sin be (A) 1/9 (C) 2
6.
t . The ratio of kinetic to potential energy o the particle at t = 210s will 90 (B) 1 (D) 3
A satellite is revolving in a circular orbit at a height h from the earth surface, such that h < < R where R is the radius of the earth. Assuming that the effect of earth’s atmosphere can be neglected the minimum increase in the speed required so that the satellite could escape from the gravitational field of earth is: (A)
2gR
(B)
gR
(C)
gR 2
(D)
gR
2 1
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JEE-MAIN-2019 (11th Jan-First Shift)-PCM-3
7.
x2 ; where x is kt
The force of interaction between two atoms is given by F = exp
the distance, k is the Boltzmann constant and T is temperature and and are two constants. The dimension of is: 0 2 4 2 4 (A) M L T (B) M LT 2 2 2 2 (C) MLT (D) M L T 8.
In a Young’s double slit experiment, the path difference, at a certain point on the screen, between two interfering waves is 1/8th of wavelength. The ratio of the intensity at this point to that at the centre of a bright fringe is close to: (A) 0.74 (B) 0.85 (C) 0.94 (D) 0.80
9.
In the circuit shown, The switch S1 is close at time t = 0 and the switch S2 is kept open. At some later time (t0), the switch S1 is opened and S2 is closed. The behaviour of the current I as a function of time ‘t’ is given by:
10.
(A)
(B)
(C)
(D)
Three charges Q, +q and +q are placed at the vertices of a right – angle isosceles triangle as shown below. The net electrostatic energy of the configuration is zero, if the value of Q is:
(A) +q (C)
q 1 2
(B)
2q 2 1
(D) –2q
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JEE-MAIN-2019 (11th Jan-First Shift)-PCM-4
11.
12.
The variation of refractive index of a crown glass thin prism with wavelength of the incident light is shown. Which of the following, graph is the correct one, if Dm is the angle of minimum deviation? (A)
(B)
(C)
(D)
The equilateral triangle ABC is cut from a thin solid sheet of wood. (See figure) D, E and F are the mid points of its sides as shown and G is the centre of the triangle. The moment of inertia of the triangle about an axis passing through G and perpendicular to the plane of the triangle is I0. If the smaller triangle DEF is removed from ABC, the moment of inertia of the remaining figure about the same axis is I. Then:
15 I0 16 9 (C) I I0 16
3 I0 4 I (D) I 0 4
(A) I
(B) I
13.
A slab is subjected to the two forces F1 and F2 of same magnitude F as shown in the figure. Force
F2 is in XY – plane while force F1 acts along z – axis at the point 2 i 3 j . The moment of these
forces about point O will be:
(C) 3ˆi 2 ˆj 3kˆ F (A) 3ˆi 2 ˆj 3kˆ F
(D) 3ˆi 2 ˆj 3kˆ F (B) 3ˆi 2 ˆj 3kˆ F
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JEE-MAIN-2019 (11th Jan-First Shift)-PCM-5 14.
Ice at –200 C is added to 50 g of water at 400 C. When the temperature of the mixture reaches 0o C, it is found that 20 g of ice is still unmelted. The amount of ice added to the water was close to (Specific heat of water = 4.2 J/g/0C) Heat of fusion of water at 0oC = 334 J/g) (A) 50 g (B) 100 g (C) 60 g (D) 40 g
15.
A particle is moving along a circular path with a constant speed of 10 ms–1. What is the magnitude of the change in velocity of the particle, when it moves through an angle of 60o+ around the centre of the circle? (A) 10 3 m / s (B) 0 (C) 10 2 m / s
16.
(D) 10 m/s
An amplitude modulated signal is given by
V t 10 1 0.6 cos 2.2 104 t sin 5.5 105 t . Here t is in seconds. The seconds. The sideband frequencies (in kHz) are. [Given = 22/7] (A) 1785 and 1715 (B) 1785 and 1715 (C) 89.25 and 85.75 (D) 892.5 and 857.5 17.
An object is at a distance of 20 m from a convex lens of focal length 0.3 m. The lens forms an image of the object. If the object moves away from the lens at a speed of 5 m/s, the speed and direction of the image will be: (A) 2.26 x 10–3 m/s away from the lens (B) 0.92 x 10–3 m/s away from the lens -3 (C) 3.22 x 10 m/s towards the lens (D) 1.16 x 10-3 m/s towards the lens
18.
A gas mixture consists of 3 moles of oxygen and 5 moles or argon at temperature T. Considering only translational and rotational modes, the total internal energy of the system is: (A) 15 RT (B) 12 RT (C) 4 RT (D) 20 RT
19.
Two equal resistances when connected in series to a battery, consume electric power of 60 W. If these resistances are now connected in parallel combination to the same battery, the electric power consumed will be. (A) 60 W (B) 240 W (C) 120 W (D) 30 W
20.
The resistance of the meter bridge AB in given figure is 4 . With a cell of emf = 005 V and rheostat resistance Rh = 2 the null point is obtained at some point J. When the cell is replaced by another one of emf = 2 the same null point J is found for Rh = 6 . The emf 2 is:
(A) 0.4V (C) 0.6V
(B) 0.3V (D) 0.5V
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JEE-MAIN-2019 (11th Jan-First Shift)-PCM-6 21.
An electromagnetic wave of intensity 50 Wm–2 enters in a medium of refractive index’ n’ without any loss . The ratio of the magnitudes of electric fields, and the ratio of the magnitudes of magnetic fields of the wave before and after entering into the medium are respectively. Given by:
1 1 , n n 1 (C) n, n (A)
22.
(B)
n, n
1 , n n
(D)
In the figure shown below, the charge on the left plate of the F capacitor is –30C. The charge on the right place of the 6 F capacitor is:
(A) –12C (C) –18C
(B) +12C (D) +18C
23.
A rigid diatomic ideal gas undergoes an adiabatic process at room temperature. The rational between temperature and volume for the process is TVx = constant, then x is: (A) 3/5 (B) 2/5 (C) 2/3 (D) 5/3
24.
A liquid of density is coming out of a hose pipe of radius a with horizontal speed and hits a mesh. 50% of the liquid passes through the mesh unaffected. 25% looses all of its momentum and 25% comes back with the same speed. The resultant pressure on the mesh will be:
1 2 4 1 (C) 2 2 (A)
25.
3 2 4
(D) 2
If the deBroglie wavelength of an electron is equal to 10-3 times the wavelength of a photon of frequency 6 x1014 Hz, then the speed of electron is equal to: (Speed of light = 3 x 10 = –34 -31 8 = m/s ; Planck’s constant = 6.63 x 10 J.s ; Mass of electron = 9.1 x 10 kg) (A) 1.1 x 106 m/s (C) 1.8 x 106 m/s
26.
(B)
(B) 1.7 x 106 m/s (D) 1.45 x 106 m/s
The give graph shown variation (with distance r from centre) of: (A) Electric field of a uniformly charged sphere (B) Potential of a uniformly charged spherical shell (C) Potential of a uniformly charged sphere (D) Electric field of a uniformly charged spherical shell
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JEE-MAIN-2019 (11th Jan-First Shift)-PCM-7 27.
Equation of travelling wave on a stretched string of linear density 5 g/m is y = 0.03 sin (450 t – 9x) where distance and time are measured in SI united. The tension in the string is: (A) 10 N (B) 7.5 N (C) 12.5 N (D) 5 N
28.
In an experiment, electrons are accelerated, from rest, by applying, a voltage of 500 V. Calculate the radius of the path if a magnetic field 100 mT is then applied. [Charge of the electron = 1.6 x 10-19 C Mass of the electron = 9.1 x 10-31 kg] (A) 7.5 x 10-3 m (B) 7.5 x 10-2 m (C) 7.5 m (D) 7.5 x 10-4 m
29.
A body of mass 1 kg falls freely from a height of 100 m, on a platform of mass 3 kg which is mounted on a spring having spring constant k = 1.25 x 106 N/m. The body sticks to the platform and the spring’s maximum compression is found to be x. Given that g = 10 ms-2, the value of x will be close to: (A) 40 cm (B) 4 cm (C) 80 cm (D) 8 cm
30.
In a Wheatstone bridge (see figure) Resistance P and Q are approximately equal. When R = 400 , the bridge is balanced. On interchanging P and Q, the value of R, for balance is 405 . The value of X is close to: (A) 401.5 ohm (B) 404.5 ohm (C) 403.5 ohm (D) 402.5 ohm
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JEE-MAIN-2019 (11th Jan-First Shift)-PCM-8
PART –B (CHEMISTRY) 31.
Among the following compounds, which one is found in RNA?
(A)
(B)
(C)
(D)
32.
The correct match between items I and II is: Item – I Item – II (Mixture) (Separation method) (a) H2O : Sugar p. Sublimation (b) H2O : Aniline q. Recrystallization (c) H2O : Toluene r. Stem distillation s. Differential extraction (A) a - d, b – r, c – p (B) a – q, b – r, c – s (C) a – r, b – p, c – s (D) a – q, b – r, c – p
33.
Which compounds out of the following is/are not aromatic? (a)
(b)
(c)
(d)
(A) b, c and d (C) b 34.
(B) c and d (D) a and c
A solid having density of 9 x 103 kg m
-3
forms face centred cubic crystale of edge length
200 2 pm. What is the molar mass of the solid? [Avogadro constant 6 x 1023 and mol–1, 3] (A) 0.0432 kg mol–1 (C) 0.0305 kg mol–1
(B) 0.0216 kg mol–1 (D) 0.4320 kg mol-1
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JEE-MAIN-2019 (11th Jan-First Shift)-PCM-9
35.
For the cell Zn s Zn 2 aq Mx aq M s , different half cells and their standard electrode potentials are given below:
If E
0 zn2 / zn
0.76 V, which cathode will given a maximum value of E0cell per electron
transferred? (A) Ag / Ag
(B) Fe 3 /Fe 2
(C) Au3 /Au
(D) Fe 2 / Fe
36.
The correct match between item I and item II is: Item – I Item – II a. Norethindrone p. Anti – biotic b. Ofloxacin q. Anti – fertility c. Equanil r. Hypertention s. Analgesics (A) a – q, b – r, c – s (B) a – q, b – p, c – r (C) a – r, b – p, c – s (D) a – r, b – p, c – r
37.
The polymer obtained from the following reactions is:
38.
(A)
(B)
(C)
(D)
NaH is an example of: (A) electron rich hydride (C) saline hydride
(B) metallic hydride (D) molecular hydride
39.
The element the usually does NOT show variable oxidation states is: (A) Cu (B) Ti (C) Sc (D) V
40.
An example of solid sol is: (A) Paint (C) Butter
41.
(B) Gem stones (D) Hair cream
The concentration of dissolved oxygen (DO) is cold water can go upto: (A) 14 ppm (B) 8 ppm (C) 10 ppm (D) 16 ppm
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JEE-MAIN-2019 (11th Jan-First Shift)-PCM-10
42.
The correct statements among (a) to (d) regarding H2 as a fuel are: (a) it produces less pollutants than petrol (b) A cylinder of compressed dihydrogen weighs 30 times more than a petrol tank producing the same amount of energy. (c) Dihydrogen is stored in tanks of metal alloys like NaNi5 (d) On combustion, values of energy released per gram of liquid dihydrogen and LPG are 50 and 142 kJ, respectively. (A) (b) and (d) only (B) (a) and (d) only (C) (b), (c) and (d) only (D) (a), (b) and (c) only
43.
The major product of the following reaction is:
(A)
(B)
(C)
(D)
44.
The freezing point of diluted milk sample is found to be – 0.20C, while it should have been 0.50C for pure milk. How much water been added to pure milk to make the diluted sample? (A) 1 cup of water to 2 cups of pure milk (B) 3 cups of water to 2 cups of pure milk (C) 1 cup of water to 3 cups of pure milk (D) 2 cups of water to 3 cups of pure milk
45.
If a reaction follows the Arrhenius equation, the plot Ink v 1/(RT) gives straight line with a gradient (–y) unit. The energy required to activate the reactant is: (A) y/R unit (B) y unit (C) yR unit (D) –y unit
46.
A 10 g effervescent tablet containing sodium bicarbonate and oxalic acid releases 0.25 ml of CO2 at T = 298.15 K and p = 1 bar. If molar volume of CO2 is 25.0 L under such condition, what is the percentage of sodium bicarbonate in each tablet? [Molar mass of NaHCO3 = 84 g mol–1] (A) 0.84 (B) 33.6 (C) 16.8 (D) 8.4
47.
The amphoteric hydroxide is: (A) be (OH)2 (C) Mg(OH)2
48.
(B) Ca(OH)2 (D) Sr(OH)2
Peroxyacetyl nitrate (PAN), an eye irritant is produced by: (A) classical smog (B) acid rain (C) organic waste (D) photochemical smog FIITJEE Ltd., FIITJEE House, 29-A, Kalu Sarai, Sarvapriya Vihar, New Delhi -110016, Ph 46106000, 26569493, Fax 26513942 website: www.fiitjee.com.
JEE-MAIN-2019 (11th Jan-First Shift)-PCM-11 49.
The major product of the following reactions is:
(A)
(B)
(C)
(D)
50.
An organic compound is estimated through Dumus method and was found to evolve 6 mole of CO2 4 moles of H2O and 1 mole of nitrogen gas. The formula of the compound is: (A) C12H8N (B) C12H8N2 (C) C6H8N2 (D) C6H8N
51.
Consider the reaction
N2 g 3H2 g 2NH3 g The equilibrium constant of the above reaction is K3. If pure ammonia is left to dissociate, the partial pressure of ammonia at equilibrium is given by (Assume that
PNH3 Ptotal at equilibrium) (A) (C)
33 / 2 K1p/ 2 P2 16 K P2 1/ 2 p
4
(B)
K1p/ 2 P 2 16 3 K1p/ 2 P2 3/2
(D)
4
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JEE-MAIN-2019 (11th Jan-First Shift)-PCM-12 52.
53.
The major product of the following reaction is:
(A)
(B)
(C)
(D)
Two blocks of the same metal having same mass and at temperature T1 and T2 respectively, are brought in contact with each other and allowed to attain thermal equilibrium at constant pressure. The change in entropy, S, for this process is:
T1 T2 2 (A) Cp In 4T1T2 T1 T2 4T1T2
(C) 2Cp In
1 2 T T 1 2 (B) 2Cp In T1T2 T1 T2 (D) 2Cp In 2T1T2
54.
Match the metal column I with the coordination compounds/enzymes column II Column I Column II Metals Coordination compounds/enzymes a. Co i. Wilkinson catalyst b. Zn ii. Chlorophyll c. Rh iii. Vitamin B12 d. Mg iv. Carbonic anhydrase (A) a – iii, b – iv, c – i, d - ii (B) a – i, b – ii, c – iii, d – iv (C) a – ii, b – i, c – iv, d – iii (D) a – iv, b – iii, c – i, d – ii
55.
For the chemical reaction X Y , the standard reaction Gibbs energy depends on temperature T (in K) as
3 r G0 in kJ mol1 120 T 8
The major component of the reaction mixture at T is: (A) Y if T = 300 K (B) Y if T = 280 K (C) X if T = 350 K (D) X if T = 315 K
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JEE-MAIN-2019 (11th Jan-First Shift)-PCM-13
56.
Match the ores (column A) with the metals (column B) Column A Column B Ores Metals I. Siderite a. Zinc II. Kaolinite b. Copper III. Malachite c. Iron IV. Calamine d. Aluminium (A) I – a, II – b, III – c, IV – d (B) I – c, II – d, III – b, IV – a (C) I – c, II – d, III – a, IV – b (D) I – b, II – c, III – d, IV – a
57.
Heat treatment of muscular pain involves radiation of wavelength of about 900 nm. Which spectral line of H – atom is suitable for this purpose?
RH 1 105 cm 1,h 6.6 10 34 Js c 3 108 ms 1 (A) Paschen, 3 (C) Balmer, 2
(B) Paschen, 5 3 (D) Lyman, 1
58.
The chloride that CANNOT get hydrolysed is (A) PbCI4 (B) CCI4 (C) SnCI4 (D) SiCI4
59.
The correct order of the atomic radii of C, Cs, AI and S is: (A) C < S < AI < Cs (B) S < C < Cs < AI (C) S < C < AI, Cs (D) C < S < Cs < AI
60.
The major product of the following reaction is COCH3 i KMnO 4 /KOH, ii H2SO4 dil H3C COCOOH (A)
COOH (B)
HOOC
OHC COOH
(C)
COCH3 (D)
HOOC
HOOC
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JEE-MAIN-2019 (11th Jan-First Shift)-PCM-14
PART–C (MATHEMATICS) 61.
If the system of linear equations
2x 2 y 3z a 3x y 5z b x 3y 2z c Where a, b, c are non zero real numbers, has more than one solution, then: (A) b – c + a = 0 (B) b – c – a = 0 (C) a + b + c = 0 (D) b + c –a = 0 62.
Let fk x
1 sink x cosk x for k = 1, 2, 3, … Then for all x R, the value of k
f4 x f6 (x) is equal to: 1 12 1 (C) 12
1 4 5 (D) 12
(A)
63.
A square is inscribed in the circle x 2 y 2 6 x 8y 103 0 with its sides parallel to the coordinate axes. Then the distance of the vertex of the square which is nearest to the origin is: (A) 6 (B) 137 (C)
64.
(B)
41
(D) 13
If q is false and p q r is true, then which one of the following statements is a tautology? (A) p r p r (B) p r p r (C) p r
(D) p r
65.
The area (in sq. units) of the region bounded by the curve x 2 4 y and the straight line x 4 y 2 is: (A) 5/4 (B) 9/8 (C) 7/8 (D) 3/4
66.
The value of the integral
2
or equal to x) is: (A) 0 (C) 4
sin2 x x 1 dx (where [x] denotes the greatest integer less than 2 2 (B) sin 4 (D) 4 – sin 4
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JEE-MAIN-2019 (11th Jan-First Shift)-PCM-15
67.
68.
1 d each, 2 1 1 10 items gave outcome each and the remaining 10 items gave outcome d each. If 2 2 4 the variance of this outcome data is then d equals: 3 2 (A) (B) 2 3 5 (C) (D) 2 2 x 3 y 2 z 1 The plane containing the line and also containing its projection on 2 1 3 The outcome of each of 30 items was observed; 10 items gave an outcome
the plane 2x + 3y – z = 5, contains which one of the following points? (A) (2, 2, 0) (B) (–2, 2, 2) (C) (0, –2, 2) (D) (2, 0, –2) 69.
Two integers are selected at random from the set {1, 2, …, 11}. Given that the sum of selected numbers is even, the conditional probability that both the numbers are even is: 7 1 (A) (B) 10 2 2 3 (C) (D) 5 5
70.
Two circles with equal radii intersecting at the points (0, 1) and (0, –1). The tangent at the point (0, 1) to one of the circles passes through the centre of the other circle. Then the distance between the centres of these circles is: (A) 1 (B) 2 (C) 2 2 (D) 2
71.
The value of r for which (A) 15 (C) 11
72.
Cr 20 C0 20 Cr 120C1 20 Cr 2 20C2 ... 20 C0 20Cr is maximum is: (B) 20 (D) 10
If tangents are drawn to the ellipse x 2 2 y 2 2 at all points on the ellipse other than its four vertices than the mid points of the tangents intercepted between the coordinate axes lie on the curve: 1 1 x2 y2 (A) 1 (B) 1 4x2 2y 2 4 2 (C)
73.
20
1 1 2 1 2 2x 4y
(D)
x2 y2 1 2 4
1, 2 x 0 Let f x 2 and g x f x f x , Then, in the interval 2, 2 ,g is: x 1, 0, x 2 (A) differentiable at all points (B) not continuous (C) not differentiable at two points (D) not differentiable at one point
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JEE-MAIN-2019 (11th Jan-First Shift)-PCM-16
74.
If x loge loge x x 2 y 2 4 y 0 , then (A) (C)
1 2e
dy at x = e is equal to: dx (B)
2 4 e2 1 2e
2e 1 2 4 e2
(D)
4 e2
e 4 e2
m 1 x2 2 dx A x 1 x C , for a suitable chosen integer m and a function A(x), x4 where C is a constant of integration, then (A(x))m equals: 1 1 (A) (B) 9 27x 3x 3 1 1 (C) (D) 6 27x 9x 4
75.
If
76.
Let [x] denote the greatest integer less than or equal to x. Then: lim
tan sin2 x x sin x x
2
x2
x 0
(A) does not exist (C) equal + 1 77.
(B) equals (D) equals 0
The maximum value of the function f x 3x 3 18x 2 27x 40 on the set S =
x R : x
2
30 11x is
(A) – 122 (C) 122 78.
Let f : R R be defined by f x 1 1 (A) , 2 2 1 1 (C) R , 2 2
79.
(B) –222 (D) 222 x , x R. Then the range of f is: 1 x2
(B) R 1,1 (D) 1,1 0
The sum of an infinite geometric series with positive terms is 3 and the sum of the cubes 27 of its terms is . Then the common ratio of this series is: 19 1 2 (A) (B) 3 3 2 4 (C) (D) 9 9
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JEE-MAIN-2019 (11th Jan-First Shift)-PCM-17 80.
The straight line x + 2y = 1 meets the coordinate axes at A and B. A circle is drawn through A, B and the origin. Then the sum of perpendicular distances from A and B on the tangent to the circle at the origin is: 5 (A) (B) 2 5 2 5 (C) (D) 4 5 4
81.
In a triangle, the sum of lengths of two sides is x and the product of the lengths of the same two sides is y. If x 2 c 2 y , where c is the length of the third side of the triangle, then the circumradius of the triangle is: 3 c (A) y (B) 2 3 c y (C) (D) 3 3
82.
Equation of a common tangent to the parabola y 2 4x and the hyperbola xy = 2 is: (A) x y 1 0 (B) x 2y 4 0 (C) x 2y 4 0 (D) 4x 2y 1 0
83.
The sum of the real values of x for which the middle term in the binomial expansion of 8
x3 3 equals 5670 is: 3 x (A) 0 (C) 4
84.
(B) 6 (D) 8 a3 a Let a1, a2, …, a10 be a G.P. If 25 , then 9 equals: a1 a5 (B) 4 52
3
(D) 2(52)
(C) 5
85.
(A) 54
0 2q r Let A p q r . If AAT = I3 , P then p is: p q r 1 1 (A) (B) 5 3 1 1 (C) (D) 2 6
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JEE-MAIN-2019 (11th Jan-First Shift)-PCM-18
86.
If y (x) is the solution of the differential equation
dy 2x 1 2x y e , x 0 where y(1)= dx x
1 2 e , then: 2
87.
(B) y loge 2
1 (C) y(x) is decreasing in ,1 2
(D) y(x) is decreasing in (0, 1)
The direction ratios of normal to the plane through the points (0, –1, 0) and (0, 0, 1) and making an angle with the plane y – z + 5 = 0 4 (A) 2, –1, 1 (B) 2, 2, 2 (C)
88.
89.
loge 2 4
(A) y loge 2 loge 4
(D) 2 3,1, 1
2,1, 1
ˆ ˆi ˆj 4kˆ and c 2iˆ 4 ˆj 2 1 kˆ be coplanar vectors. Then the Let a ˆi 2ˆj 4k,b non – zero vector a c is: (A) 10iˆ 5ˆj (B) 14iˆ 5ˆj (C) 14iˆ 5 ˆj (D) 10iˆ 5 ˆj
If one real root of the quadratic equation 81x 2 kx 256 0 is cube of the other root, then a value of k is: (A) – 81 (B) 100 (C) 144 (D) – 300 3
90.
1 x iy Let 2 i 3 27 (A) 91 (C) 85
i
1 , where x and y are real numbers, then y – x equals:
(B) – 85 (D) – 91
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JEE-MAIN-2019 (11th Jan-First Shift)-PCM-19
JEE (Main) – 2019 ANSWERS PART A – PHYSICS 1.
B
2.
C
3.
A
4.
D
5.
No match (Correct answer is
6.
D
7.
B
8.
B
9.
No match
10.
B
11.
A
12.
A
13.
B
14.
D
15.
D
16.
C
17.
D
18.
A
19.
B
20.
B
21.
D
22.
D
23.
B
24.
B
25.
D
26.
B
27.
C
28.
D
29.
No match (Correct answer is 2 cm)
30.
D
1 ) 3
PART B – CHEMISTRY 31.
B
32.
B
33.
B
34.
C
35.
C
36.
B
37.
C
38.
C
39.
C
40.
B
41.
C
42.
D
43.
D
44.
B
45.
B
46.
A
47.
A
48.
D
49.
A
50.
C
51.
A
52.
B
53.
A
54.
A
55.
D
56.
B
57.
A
58.
B
59.
A
60.
C
PART C – MATHEMATICS 61.
B
62.
A
63.
C
64.
B
65.
B
66.
A
67.
D
68.
D
69.
C
70.
B
71.
B
72.
C
73.
D
74.
B
75.
A
76.
A
77.
C
78.
A
79.
B
80.
A
81.
B
82.
C
83.
A
84.
A
85.
C
86.
C
87.
BC
88.
D
89.
D
90.
A
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JEE-MAIN-2019 (11th Jan-First Shift)-PCM-20
HINTS AND SOLUTIONS PART A – PHYSICS 1.
5 3
at t = 1 ux = 5, uy = 5 3
10 o
60
vy = 5 3 10 ;
30o
5
tan (2 3 ) = –30°
a
a g 10
2.
v2 102 a (10 cos 30o ) 10 20 = 2 = m 3 3 52 (10 5 3 )2 200 100 3 = 2.8 m 10 cos 10 0.965
R
12 – 500(2i1 + i2) – 10 = 0 2 1 2i1 i2 500 250 < i1 when zenor has break down. So, i2 = 0.
2i1 – i2
R1 = 500
i1 i1
12V R2
i2
R2 = 500
M 0nn1n2 n12 L 0 n12r12
=
5.
t
Energy supplied 12400 E = 12.65 eV 900 En – E1 = 12.65 1 (13.6) 1 2 = 12.65 n 2 n 14.3 n4 r n2
3.
4.
vx 5
K
n2 n1
1 1 1 mv 2 ; U kx 2 m2 x2 2 2 2 cos(wt) k v2 2 2 U x sin (wt) = cot2 210 90
2
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JEE-MAIN-2019 (11th Jan-First Shift)-PCM-21
= cot 2 2 3 2
1 1 = . 3 3
6.
v = vf – vi 2gMe gMe = Re Re = ( 2 1) gR e
7.
Power of e should be dimensionless. So, [] = (Tk) L2 = [] (ML2 T–2) () = (M–1 T2) 1 E KT 2 2 2 (ML T ) ; (E) = [KT) () = (F) (M–1 T2) () = (MLT–2)
8.
x =
8
2 8 4 R Ires = I + I + 2I cos 4 1 = 2I 1 2I 1.7 2 I 2I 1.7 0.85 res Iman 4I Phase P
I
9.
t
Above is the correct graph for growth and decay of current. 10.
kQq kq2 kQq =0 a a a 2 q Q 1 1 2 UTotal
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JEE-MAIN-2019 (11th Jan-First Shift)-PCM-22 11.
Prism formula Dm = Sm = (n – 1) A (for thin prism) So, answer is 1.
12.
I m2
(let = mass present area)
I1 4
and I2 2 I So, I2 16
…(1) 4
…(2)
Moment of inertia of remaining sheet = I = 13.
0 1 2
I 16
15 I 16
F F 3 ˆ ˆ = 6iˆ ˆi j + (2iˆ 3ˆj) (Fk) 2 2 ˆ ˆ ˆ = 3F k( 2Fj 3Fi )
ˆ = F(3iˆ 2ˆj 3k)
14.
15.
16.
m × 0.5 × 20 + (m – 20) × 80 = 50 × 1 × 40 90 m – 1600 = 2000 90 m = 3600 m = 40 gm v = 2v sin 2 = 2 × 10 × sin(30o) = 10 m/s
fso = fc fm
c m 2 (5.5 0.22) 105 = 22 2 7 = 89.25, 85.75 =
17.
1 1 1 V 20 30 1 1 1 200 3 197 V 0.30 20 60 60
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JEE-MAIN-2019 (11th Jan-First Shift)-PCM-23 dV
0
dV V 2 du 3 2 = ( 5) 0.00116 m / s dt u dt 197
dt V
2
du
dt u2
2
18.
Utotal = UO2 UAr
3 5 RT 5 3 RT 2 2 = 15 RT =
19.
Assuming constant voltage supply
P 60
V2 R1 R 2
And P
20.
(1)
V 2 V 2 2V 2 = 4P = 4 × 60 R1 R2 R1 = 240 W
0.5 =
6 x (2 L)
…(1)
E2
6 x (6 L)
…(2)
So dividing equation (1) and (2) E2 2 4 3 0.5 6 4 5 21.
E2 = 0.3 volt.
Ei C …(1) Bi Ef c …(2) Bf n EiB f 1 Ef Bi n Ei 1 Bi Ef n B f 1 n e 0 r 1 : n n
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JEE-MAIN-2019 (11th Jan-First Shift)-PCM-24 22.
–18
4V + 6V = 30 V=3
+18
6F –30 +30
–12
+12
–30 +30
4F
23.
24.
Equation of adiabatic process TV 2/f = constant 2 2 =x f 5 Area = A
1 1 Av 2 2av 2 4F 3 4 Av 2 A 4 F
100% 50%
25% 25%
25.
–3
e = photon × 10 h c 103 s mv h v mC 10 3 =
6.62 10 34 6 1014
9.1 10 31 3 108 10 3 = 1.45 × 106
26.
27.
The potential inside a uniformly charged shell is constant, while it decrease hyperbolically outside. 9x y = 0.03 sin 450 t 450 450 So, v = 50 m/s 9 T Also, v
50
T
5 103 T = 2500 × 5 × 10–3 = 12.5 N 28.
ke
p2 500 e 2Me
& R
…(i)
p 1000 me e 10 10 9.1 10 31 eB eB 1.6 10 19
= 100 × 7.541 × 10–6
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JEE-MAIN-2019 (11th Jan-First Shift)-PCM-25 29.
v 2 10 100 = 20 5
1 kg
1 20 5 4 =5 5
100 m
COLM v p
3 kg
COTME 2 1 1 2 6 30 4 (5 5 ) 1.25 10 2 2 k 30 1 2 = 4 10 n kn k 2 900 1200 1 2 250 40x kx 6 2 1.25 10 k 2 x 2cm 30.
P 400 …(1) Q S Q 405 and …(2) P 5 Solving S2 = 400 × 405 S = 402.5
PART B – CHEMISTRY 31.
Fact based
32.
Fact based
33.
b antiaromatic d Non aromatic (no cyclic conjugation)
34.
Density
ZM NAV a3
Here Z = 4, a = 200 2
3 Kg NAV = 6.02 1023 , d 9 10 m3
35.
o EoCell E Cathode EoZn2 / Zn
Since Au3+ Au has maximum value. 36. 37
38.
Fact based. NaNO HCl
2 NH2
H O
2 N2 (unstable)
OH
ester polymer
NaH is an example of saline hydride. FIITJEE Ltd., FIITJEE House, 29-A, Kalu Sarai, Sarvapriya Vihar, New Delhi -110016, Ph 46106000, 26569493, Fax 26513942 website: www.fiitjee.com.
JEE-MAIN-2019 (11th Jan-First Shift)-PCM-26 39.
Sc3+ has noble gas configuration hence only +3 exists.
40.
An example of solid sol is gem stones.
41.
In cold water, dissolved oxygen can reach a concentration upto 10 ppm
42.
Fact based.
43.
O
O
O OEt
CH2NH2
CN
DIBAL - H
44.
45.
46.
NH
OEt
H2 /Ni
N
1 1 0.5 ,0.2 2 x 0.5 x Here ; x = 5, hence 3 cup 0.2 2 Ea nA RT Slope = - Ea = -y nK
Let NaHCO3 = x gm Then, H2C2O4 = (10 – x) gm x 84 2 NaHCO3 Na2CO3 H2O CO2
nNaHCO3
nCO2
x 168
Total CO2 =
10 x nH2C2 O4 90 H2C2O4 H2O CO2 CO 10 x nCO2 90
x 10 x 0.25 168 90 25
On solving ‘x’ %
x 100 = 10 x 10
47.
Be – O – H Both bond has same dissociation energy.
48.
Fact based. FIITJEE Ltd., FIITJEE House, 29-A, Kalu Sarai, Sarvapriya Vihar, New Delhi -110016, Ph 46106000, 26569493, Fax 26513942 website: www.fiitjee.com.
JEE-MAIN-2019 (11th Jan-First Shift)-PCM-27 49.
–SO3H will be replaced by Br this is called ipso effect.
50.
CO2 = 6 mole, N1 = 1 mole Catom= 6, Natom = 2 Hence C6H8N2
51.
N2 3H2 2NH3 m equ x 3x P1 2 P1 PT = 4x K p x 27 3 P x 4 P1 27x 4K p 2
1/ 2
27 K p
52.
33/ 2 K1/p 2p 2 PT 4 16
Cl
Cl
Br Alc KOH
HBr
O
O
O
T1 T2 C
53.
STotal CP n
54.
Co Vitamin B12 Zn Carbonic anhydrase Rh Wilkinson catalyst Mg Chlorophyll
55.
2 T1
Tautomerise
P
n
HO
T1 T2 2 T2
3 Go 120 T 0 8 Then T = 320 K Hence T > 320 K Y formed T < 320 K X formed
56.
Siderite Iron Kaolinite Aluminium Malachite Copper Calamine Zinc
57.
[i] 900 nm = 9000 A It is in far infra red region hence paschen.
58.
Central atom has no vacant orbital.
59.
On moving down size increases.
60.
It is the case of side chain oxidation.
o
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JEE-MAIN-2019 (11th Jan-First Shift)-PCM-28
PART C – MATHEMATICS 61.
2x 2y 3z a 3x y 5z b x 3y 2z c
(1) (2) (3)
2x 2y 3z x 3y 2z 3x y 5z 0 ac b 0
62.
sin4 x cos4 x 1 2 sin2 x .cos 2 x 1 1 2 sin x.cos2 x 4 4 4 2 2 2 2 2 6 6 sin x cos x 1 3 sin x.cos x sin x cos x F6 x 6 6 1 1 sin2 x.cos2 x 6 2 1 1 64 2 1 F4 x f6 x 4 6 24 24 12 F4 x
63.
Centre (3, –4) Radius 9 16 103 128 8 2 5, 4 will be nearer to the (0, 0) Ans.
(3, 4)
(–5, 4) 8
5 2 42 25 16 41
8 (–5, –12)
64.
As p q r is true If p q is true and r is true. As q is false p q can not be true This case is not possible
Or
(11, 4)
8 (3, –4) 8
(3, –12)
(11, –12)
p q is F and r is F.
Their will be two cases for p, q, r. T, F, F, or F, F, F respectively Now check the options.
65.
x 2 4y ……..(1) x 2 4y ……..(2) Solve (1) and (2) x 2 x2 x2 x 2 0 x 2 x 1 0 x 2, 1
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JEE-MAIN-2019 (11th Jan-First Shift)-PCM-29 2
x 2 x2 Area dx 4 4 1 2
1 x2 x3 2x 4 2 3 1 1 8 1 1 2 4 2 4 3 2 3 1 10 3 12 2 1 10 7 4 3 6 4 3 6 1 27 9 4 6 8 sin2 x 2 sin x 0 1 x 1 x dx 2 2 2
66.
2 sin2 x 2 sin x dx dx 0 x 1 x 0 1 0 1 2 2
67.
Variance remains some if same number is subtracted from each observation. (subtract 10 from each observation) 2
2
10 d 10 0 10 d 30
2
2
10 d 10 0 10 d 4 30 3
20d2 4 30 3 d2 2 d 2
68.
Equation of required plane is a x 3 b y 2 c z 1 0 [as it contains the point (3, –2, 1)] 2a b 3c 0 2a 3b c 0 a b c 8 8 8 a b c 1 1 1 equation of plane is x 3 y 2 z 1 0 xyz4
(3, –2, 1)
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JEE-MAIN-2019 (11th Jan-First Shift)-PCM-30
69.
P
both number is even number of ways selecting two numbers such that their sum is even. 5
70.
5
C2 10 10 2 6 C2 C2 10 15 25 5
y
The two circle will be orthogonal OD = 1 OA OB OD 1 AB 2
D (0, 1) r
r 45o
A
O
x
B
(0, –1)
71.
20
Cr .
20
C0 20Cr 1 . 20C1 ...... 20 C0 . 20Cr
Selecting r student from 20 boys and 20 girls 40Cr 40
72.
Cr will be maximum if r = 20.
Equation of tangent is x y cos sin 1 a b a A is , 0 cos
B
b B is 0, sin Let P (h, k) is mid point a 2h cos
2k
a cos ,b sin P A
b sin
cos2 sin2 1 a2 b2 2 2 1 4h 4k 2 1 2 2 1 4x 4y 1 1 2 2 1 2x 4y
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JEE-MAIN-2019 (11th Jan-First Shift)-PCM-31
73.
2 x 0 1 f x 2 x 1 0 x 2 1 2 x 0 fx 2 x 1 0 x 2
f x x 2 1 2 x 2
1 x 2 1 2 x 0 g x f x f x : 2 2 x 1 x 1 0 x 2 x2 g x 0 2 2 x 1
2 x 0 0 x 1 1 x 2
y
O
1
x 2
g x is not differentiable at x 1
74.
1 1 dy . loge loge x 2x 2y. 0 n x x dx Put x e dy 1 2e 2y 0 dx dy 2e 1 (1) dx 2y Put x = e in original equation O e2 y 2 4 x.
y 2 4 e2 Now put in (i) dy 2e 1 dx 2 4 e 2 2
75.
1 x dx x4 1 Let 2 1 t x
x
1 1 1 x2 dx 3 4 x x
1 1dx x2
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JEE-MAIN-2019 (11th Jan-First Shift)-PCM-32
2 dx dt x3 3/2
Integration becomes
1 1 t t dt 2 2 3 2
3/ 2
1 1 2 1 3x 3/2 1 3 1 x2 3x 3 1 3 1 x2 3x 3 3 1 1 A x 3 27x 9 3x
76.
Lt
tan sin2 x sin2 x
x 0
. sin
2
x2
x
x sin x x x
2
2
sin x x x x 1 . 1 Lt 1 . …………(i) x 0 x x x x x x Lt 1 1 x 0 x x 2
Lt
x 0
x x
x
x 0 x
0
Put in equation (i) Limit does not exist.
77.
x 2 11x 30 0 x 6 x 5 0 3
6
5 5x6
2
f x 3x 18x 27x 40 f ' x 9x 2 36x 27
9 x 2 4x 3 9 x 3 x 1 Nature of f(x)
f x will be maximum when x 6
+ 1 3
+
– 3
2
f 6 3 6 18 6 27 6 40 36 18 18 162 40 122
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JEE-MAIN-2019 (11th Jan-First Shift)-PCM-33
78.
79.
x x 1 2 yx x y 0 D0 1 4y 2 0 1 y2 4 1 1 y . 2 2 y
2
a 3 1 r Cube both sides a3 27 ……….(1) 3 1 r and
a3 27 3 1 r 19
(1) / (2) gives
r 80.
81.
……….(2) 1 r3 3
1 r
19
2 3
1 Equation of circle x x 1 y y 0 2 y x2 y2 x 0 2 Equation of tangent at (0, 0) x0 y0 x .0 y .0 0 2 22 2x y 0 ……..(1) Sum of distance of A and B from Line (i) is 1 2 5 5 2 2 5 5 2 5
B 1 0, 2
(1, 0) (0, 0)
A x + 2y = 1
x2 c 2 y
a b
2
c 2 ab
a 2 b 2 c 2 ab a2 b 2 c 2 1 2ab 2 1 cos c 2
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JEE-MAIN-2019 (11th Jan-First Shift)-PCM-34
c
2 3
sin C
3 2
c c c 2R R sinc 2 sinc 3 82.
Equation of tangent to parabola y 2 4x is y mx
1 m
………(1)
Now solve it with xy 2
1 mx m x 2 1 mx 2 x 2 0 m For common tangent D = 0 1 2 8m 0 m 1 m 2 1 Put it in equation (i) y x 2 2 2x y 4 0 83.
5th term will be the middle term. 4
4
x3 3 t 41 8C4 5670 3 x 8C 4 . x 8 5670 8765 8 x 5670 43 2 567 x8 81 7 x 8 81 0 Real value of x 3 84.
a3 a1r 2 r2 a1 a1
r 2 25 a a r8 2 Now 9 1 4 r 4 25 5 4 a5 a1r 85.
A is orthogonal matrix 4q2 r 2 p2 q2 r 2 1
……..(1)
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JEE-MAIN-2019 (11th Jan-First Shift)-PCM-35
p2 q2 r 2 0 2
……..(2)
2
and 2q r 0 Solving (1), (2) and (3) 1 p2 2 1 p 2 1
86.
………(3)
2 dx I.F. e x e 2x .x Solution will be y xe 2x e 2x . xe 2x .dx c
xy e 2x
x2 c 2
1 1 1 1 e 2 .e 2 c (Put x = 1 and y e 2 ) 2 2 2 c 0 x y e 2x 2 dy 2xe 2x e 2x e2x 1 2x dx 2 2 dy 1 0 x dx 2 dy 1 0 if x , 1 dx 2 dr = (a, b, c)
87.
d. r. of AB = (0, 1, 1) a.0 b.1 c.1 0 bc 0
B (0, –1, 0) (0, 0, 1) A
……….(i) bc 0 a.0 b 1 c 1 cos 4 a 2 b2 c 2 . 2 1
bc
a b2 c 2 b2 c 2 2bc a2 2bc a2 2b b 2
a 2 2b2
2
. 2
……….(ii) a 2b
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JEE-MAIN-2019 (11th Jan-First Shift)-PCM-36 a b 2 1
From (i) a
OR
b 1
2 a b c 2 1 1 a b c 2 1 1
dr. will be 88.
a 1 1 2
a b c 2 1 1
from (i)
2, 1, 1 or
b c 0 2 4 4 2 4 1R
2, 1, 1
0
3 R3 2R1
1 2
4
1
4
0
2
0 0 9
2
9 2 0
2 OR 2 5 ˆi ˆj kˆ a c 1 2 4
2 4 2 1
ˆi 2 2 2 16 ˆj 2 1 8
2 9 2iˆ ˆj 5 2iˆ ˆj
89.
3
K 81
256 81 4 3 From (1) and (2) 4 64 K 3 27 81 K 300
(put 2 )
…………(1)
4
………..(2)
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JEE-MAIN-2019 (11th Jan-First Shift)-PCM-37 3
90.
i x iy 2 3 27 i3 i2 x iy 3 2 i 1 23 3 2 3 2 . 27 9 3 27 i 2 x iy 8 4i 27 3 27 x 2 8 27 3 y 1 and 4 27 27 yx 1 2 48 27 27 3 1 27 4 18 27 109 18 27 91 27 y x 91
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FIITJEE Solutions to JEE(Main)-2019 Test Date: 11th January 2019 (Second Shift)
PHYSICS, CHEMISTRY & MATHEMATICS Paper - 1 Time Allotted: 3 Hours
Maximum Marks: 360
Please read the instructions carefully. You are allotted 5 minutes specifically for this purpose.
Important Instructions: 1.
The test is of 3 hours duration.
2.
This Test Paper consists of 90 questions. The maximum marks are 360.
3.
There are three parts in the question paper A, B, C consisting of Physics, Chemistry and Mathematics having 30 questions in each part of equal weightage. Each question is allotted 4 (four) marks for correct response.
4.
Out of the four options given for each question, only one option is the correct answer.
5.
For each incorrect response 1 mark i.e. ¼ (one-fourth) marks of the total marks allotted to the question will be deducted from the total score. No deduction from the total score, however, will be made if no response is indicated for an item in the Answer Box.
6.
Candidates will be awarded marks as stated above in instruction No.3 for correct response of each question. One mark will be deducted for indicating incorrect response of each question. No deduction from the total score will be made if no response is indicated for an item in the answer box.
7.
There is only one correct response for each question. Marked up more than one response in any question will be treated as wrong response and marked up for wrong response will be deducted accordingly as per instruction 6 above.
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JEE-MAIN-2019 (11th Jan-Second Shift)-PCM-2
PART –A (PHYSICS) 1.
A particle moves from the point
5.0iˆ 4.0ˆj
2.0iˆ 4.0ˆj m,
at t = 0 with an initial velocity
ms–1+. It is acted upon by a constant force which produces a constant
acceleration 4.0iˆ 4.0 ˆj ms–2. What is the distance of the particle from the origin at time 2s? (A) 15m (C) 5m
(B) 20 2 m (D) 10 2 m
2.
A thermometer graduated according to a linear scale reads a value x0 when in contact with boiling water, and x0/3 when in contact with ice. What is the temperature of an object in 0C, if this thermometer in the contact with the object reads x0/2? (A) 25 (B) 60 (C) 40 (D) 45
3.
A galvanometer having a resistance of 20 and 30 divisions on both sides has figure of merit 0.005 ampere /division. The resistance that should be connected in series such that it can be used as a voltmeter upto 15 volt is: (A) 100 (B) 120 (C) 80 (D) 125
4.
In the experimental setup of metre bridge shown in the figure, the null point is obtained at a distance of 40cm from A. If a 10 resistor is connected in series with R1, the null point shifts by 10cm. The resistance that should be connected in parallel with (R1 + 10) such that the null point shifts back to its initial position is:
(A) 20 (C) 60 5.
(B) 40 (D) 30
A circular disc D1 of mass M and radius R has two identical discs D2 and D3 of the same mass M and radius R attached rigidly as its opposite ends (see figure). The moment of inertia of the system about the axis OO’, passing through the centre of D1 as shown in the figure, will :
(A) MR2 4 (C) MR2 5
(B) 3MR2 2 (D) MR 2 3
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JEE-MAIN-2019 (11th Jan-Second Shift)-PCM-3
6.
The magnitude of torque on a particle of mass 1kg is 2.5 Nm about the origin. If the force acting on it is IN, and the distance of the particle from the origin is 5m, the angle between the force and the position vector is (in radians): (A) (B) 6 3 (C) (D) 8 4
7.
A copper wire is wound on a wooden frame, whose shape is that of an equilateral triangle. If the linear dimension of each side of the frame is increased by a factor of 3, keeping the number of turns of the coil per unit length of the frame the same, then self inductance of the coil: (A) decreases by a factor of 9 (B) increases by a factor of 27 (C) increases by a factor of 3 (D) decreases by a factor of 9 3
8.
A particle of mass m is moving in a straight line with momentum p. Starting at time t = 0, a force F = kt acts in the same direction on the moving particle during time interval T so that its momentum changes from p to 3p. Here k is a constant. The value of T is: k p (A) 2 (B) 2 p k (C)
2k p
(D)
2p k
9.
A paramagnetic substance in the form of a cube with sides 11 cm has a magnetic dipole moment of 20 x 10–6 J/T when a magnetic intensity of 60 x 103 A/m is applied. Its magnetic susceptibility is: (A) 3.3 x 10–2 (B) 40.3 x 10-2 (C) 2.3 x 10–2 (D) 3.3 x 10-4
10.
A simple pendulum of length 1 m is oscillating with an angular frequency 10 rad/s. The support of the pendulum starts oscillating up and down with a small angular frequency of 1 rad/s and an amplitude of 10–2 m. The relative change in the angular frequency of the pendulum is best given by: (A) 10-3 rad/s (B) 1 rad/s -1 (C) 10 rad/s (D) 10-5 rad/s
11.
The circuit shown below contains two ideal diodes, each with a forward resistance of 50. If the battery voltage is 6V, the current through the 100 resistance (in Amperes) is:
(A) 0.036 (C) 0.027
(B) 0.020 (D) 0.030
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JEE-MAIN-2019 (11th Jan-Second Shift)-PCM-4 12.
An electric field of 1000V/m is applied to an electric dipole at angle of 45°. The value of electric dipole moment is 10-29 C.m. What is the potential energy of the electric dipole? (A) 20 1018 J (B) 7 10 27 J 29 (C) 10 10 J (D) 9 10 20 J
13.
A metal bal of mass 0.1 kg is heated upto 5000C and dropped into a vessel of heat capacity 800 JK–1 and containing 0.5 kg water. The initial temperature of water and vessel is 300C. What is the approximate percentage increment in the temperature of the water? [Specific heat Capacities of water and metal are, respectively 4200 Jkg1K 1 and 400 jkg1K 1 ] (A) 15% (B) 30% (C) 25% (D) 20%
14.
The region between y = 0 and y = d contains a magnetic field B Bzˆ . A particle of mass mv m and charge q enters the region with a velocity v v ˆi . If d = , the acceleration of 2qB the charged particle at the point of its emergence at the other side is: qvB 1 ˆ 3 ˆ qvB 3 ˆ 1 ˆ (A) j (B) i j i m 2 2 m 2 2 qvB ˆj ˆi qvB ˆj ˆi (C) (D) m 2 m 2
15.
A pendulum is executing simple harmonic motion and its maximum kinetic energy is K1. If the length of the pendulum is doubled and it performs simple harmonic motion with the same amplitude as in the first case, its maximum kinetic energy is K2 then: K (A) K 2 2K1 (B) K 2 1 2 K (C) K 2 1 (D) K 2 K1 4
16.
Two rods A and B of identical dimensions are at temperature 30oC. If a heated upto 180oC and B upto T0C, then the new lengths are the same. If the ratio of the coefficients of linear expansion of A and B is 4:3, then the value of T is: (A) 2300C (B) 270oC (C) 200oC (D) 250oC
17.
If speed (V), acceleration (A) and force (F) are considered as fundamental units, the dimension of Young’s modulus will be: (A) V 2 A 2F2 (B) V 2 A 2F2 (C) V 4 A 2F (D) V 4 A 2F
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JEE-MAIN-2019 (11th Jan-Second Shift)-PCM-5
18.
A string is wound around a hollow cylinder of mass 5 kg and radius 0.5m. If the string is now pulled with a horizontal force of 40 N, and the cylinder is rolling without slipping on a horizontal surface (see figure), then the angular acceleration of the cylinder will be (Neglect the mass and thickness of the string)
(A) 20 rad/s2 (C) 12 rad/s2
(B) 16 rad/s2 (D) 10 rad/s2
19.
A 27 mW lager beam has a cross – sectional area of 10 mm2. The magnitude of the maximum electric field in this electromagnetic wave is given by: [Given permittivity of space 0 9 10 12 SI units, speed of light c = 3 x 108 m/s] (A) 2 kV/m (B) 0.7 kV/m (C) 1 kV/m (D) 1.4 kV/m
20.
In the circuit shown, the potential difference between A and B is:
(A) 1V (C) 3V
(B) 2V (D) 6V
21.
The mass and the diameter of a planet are three times the respective values for the Earth. The period of oscillation of a simple pendulum on the Earth is 2s. The period of oscillation of the same pendulum on the planet would be: 3 2 (A) s (B) s 2 3 3 (C) s (D) 2 3 s 2
22.
An amplitude modulated signal is plotted below:
Which one of the following best described the above signal? (A) 9 sin 2.5 105 t sin 2 10 4 t V (B) 1 9 sin 2 10 4 t sin 2.5 105 t V
(C) 9 sin 2 10 t sin 2.5 10 t V 4
5
(D) 9 sin 4 10 t sin 5 10 t V 4
5
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JEE-MAIN-2019 (11th Jan-Second Shift)-PCM-6 23.
In a process, temperature and volume of one mole of an ideal monoatomic gas are varied according to the relation VT = K, where I is a constant. In this process the temperature of the gas is increased by T. The amount of heat absorbed by gas is (R is gas constant) 1 1 (A) RT (B) KRT 2 2 3 2K (C) RT (D) T 2 3
24.
When 100g of a liquid A at 1000C is added to 50g of a liquid B at temperature 75oC, the temperature of the mixture becomes 90oC. The temperature of the mixture, if 100g of liquid A at 100oC is added to 50g of liquid B at 50oC, will be: (A) 85oC (B) 60oC o (C) 80 C (D) 70oC
25.
In a hydrogen like atom, when an electron jumps from the M – shell to the L-shell the wavelength of emitted radiation is . If an electron jumps from N-shell to the L-shell the wavelength of emitted radiation will be: 27 16 (A) (B) 20 25 25 20 (C) (D) 16 27
26.
A monochromatic light is incident at a certain angle on an equilateral triangular prism and suffers minimum deviation. If the refractive index of the material of the prism is 3 , then the angle of incidence (A) 90o (B) 30o o (C) 60 (D) 45o
27.
In a double – slit experiment, green light (5303 A ) falls on a double slit having a separation of 19.44 m and a width of 4.05m. The number of bright fringes between the first and the second diffraction minima is: (A) 10 (B) 05 (C) 04 (D) 09
28.
Seven capacitors, each of the capacitance 2F, are to be connected in a configuration to 6 obtain an effective capacitance of F. Which of the combinations, shown in figures 13 below, will achieve the desired value?
0
(A)
(B)
(C)
(D)
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JEE-MAIN-2019 (11th Jan-Second Shift)-PCM-7
29.
A particle of mass m and charge q is in an electric and magnetic field given by: E 2iˆ 3 ˆj;B 4 ˆj 6kˆ The charged particle is shifted from the origin to the point P(x = 1; y = 1) along a straight path. The magnitude of the total work done is: (A) (0.35)q (B) 5q (C) (2.5)q (D) (0.15)q
30.
In a photoelectric experiment, the wavelength of the light incident on a metal is changed from 300 nm to 400 nm. The decrease in the stopping potential is close to hc 1240 nm V e (A) 0.5 V (B) 1.5 V (C) 1.0 V (D) 2.0 V
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JEE-MAIN-2019 (11th Jan-Second Shift)-PCM-8
PART –B (CHEMISTRY) 31.
The reaction:
MgO s C s Mg s CO g , for which r H0 491.1 kJ mol1 and r S0 198.0 JK 1 mol1 is not feasible at 298 K. Temperature above which reaction will be feasible is: (A) 2040.5 K (C) 2480. K 32.
(B) 1890.0K (D) 2380.K
The correct match between Item I and Item II is Item I Item II A. Allosteric effect P. Molecule binding to the active site of enzyme B. Competitive inhibitor Q. Molecule crucial for communication in the body C. Receptor R. Molecule binding to a site other than the active site of enzyme D. Poison S. Molecule binding to the enzyme covalently (A) A R, B P, C Q, D S (C) A R, B P, C S, D Q
(B) A P, B R, C Q, D S (D) A P, B R, C S, D Q
33.
The coordination number of Th in K 4 Th C2O 4 4 OH2 2 is: C2 O 42 Oxalato (A) 14 (B) 6 (C) 8 (D) 10
34.
The major product obtained in the following reaction is:
(A)
(B)
(C)
(D)
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JEE-MAIN-2019 (11th Jan-Second Shift)-PCM-9
35.
The standard reaction Gibbs energy for a chemical reaction at an absolute temperature T is given by
r G0 A BT Where A and B are non – zero constant. Which of the following is TRUE about this reaction? (A) Endothermic if A > 0 (B) Exothermic if A > 0 and B < 0 (C) Endothermic if A < 0 and B > 0 (D) Exothermic if B < 0 36.
The radius of the largest sphere which fits properly at the centre of the edge of a body centered cubic unit cell is: (Edge length is represented by ‘a’) (A) 0.0027a (B) 0.047 a (C) 00137a (D) 0.07a
37.
The hydride that is NOT electron deficient is: (A) SiH4 (B) B 2H6 (C) GaH3
38.
(D) AIH3
Given the equilibrium constant: KC of the reaction:
0
Cu s 2 Ag aq Cu2 aq 2 Ag s is 10 x 1015=, calculate the Ecell of the reaction
of 298 K 2.303
RT at 298K 0.059 V F
(A) 0.04736 mV (C) 0.4736 V
(B) 0.4736 mV (D) 0.04736 V
39.
The correct option with respect to the Pauling electronegativity values of the elements is: (A) Te > Xe (B) Ga > Ge (C) Si > AI (D) P > S
40.
Which of the following compounds will produce a precipitate with AgNO3? (A) (B)
(C)
41.
(D)
The de Broglie wavelength () associated with a photoelectron varies with the frequency (v) of the incident radiation as, [v0 is threshold frequency]: (A)
(C)
1 v v0
(B)
1
v v0 4
1
v v0
1
3 2
(D)
1 1
v v0 2
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JEE-MAIN-2019 (11th Jan-Second Shift)-PCM-10 42.
Which of the following compounds reacts with ethylmagnesium bromide and also decolourizes bromine water solution: (A) (B)
(C)
43.
(D)
In the following compound
The favourable site/s for protonation is/are (A) a and e (B) b, c and d (C) a and d (D) a 44.
Taj Mahal is being slowly disfigured and discoloured. This is primarily due to: (A) global warming (B) acid rain (C) water pollution (D) soil pollution
45.
The relative stability of +1 oxidation state of group 13 elements follows the order: (A) AI Ga TI In (B) TI In Ga AI (C) Ga AI In TI (D) AI Ga In TI
46.
For the equilibrium
2H2O H3O OH , the value of G0 at 298 K is approximately: (A) 100 kJ mol–1 (C) 80 kJ mol–1 47.
(B) – 80 kJ mol-1 (D) –100 kJ mol-1
The reaction that does NOT define calcinations is: (A) Fe 2 O3 .XH2O Fe 2 O3 XH2O (B) 2Cu2 S 3O 2 2Cu2 O 2SO 2 (C) ZnCO3 ZnO CO 2 (D) CaCO3 .MgCO3 CaO MgO 2CO 2
48.
A compound ‘X’ on treatment with Br2/NaOH, provided C3H9N, which gives positive carbylamine test. Compound ‘X’ is: (A) CH3COCH2NHCH3 (B) CH3CH2COCH2NH2 (C) CH3CH2CH2CONH 2
(D) CH3CON CH3 2
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JEE-MAIN-2019 (11th Jan-Second Shift)-PCM-11
49.
Among the colloids cheese (C), milk (M) and smoke (S), the correct combination of the dispersed phase and dispersion medium, respectively is: (A) C: liquid in solid; M: liquid in solid; S: solid in gas (B) C : liquid in solid; M: liquid in liquid; S: solid in gas (C) C : solid in liquid ; M : liquid in liquid ; S : gas in solid (D) C : solid in liquid ; M : solid in liquid; S : solid in gas
50.
The homopolymer formed from 4 – hydroxybutanoic acid is: (A)
(B)
(C)
(D)
51.
K2HgI4 is 40% ionized in aqueous solution. The value of its van’t Hoff factor (i) is: (A) 1.6 (B) 1.8 (C) 2.0 (D) 2.2
52.
25 mL of the given HCI solution requires 20 mL of 0.1 M sodium carbonate solution. What is the volume of this HCI solution required to titrate 30 mL of 0.0 M aqueous NaOH solution? (A) 25 mL (B) 75 mL (C) 50 mL (D) 12.5 mL
53.
The reaction 2X B is a zeroth order reaction. If the initial concentration of X is 0.2M, the half life is 6 h. When the initial concentration of X is 0.5 M, the time required to reach its final concentration of 0.2 M will be: (A) 9.0 h (B) 12.0 h (C) 18.0 h (D) 7.2 h
54.
Match the following items in column I with the corresponding items in column II. Column I Column II I. Na2CO3 . 10 H2O A. Portland cement ingredient II. Mg (HCO3)2 B. Castner – Kellner process III. NaOH C. Solvay process IV. Ca3AI2O6 D. Temporary hardness (A) I – B, II – C, III – A, IV – D (B) I – C, II – B, III – D, IV – A (C) I – D, II – A, III – B, IV – C (D) I – C, II – D, III – B, IV – A
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JEE-MAIN-2019 (11th Jan-Second Shift)-PCM-12 55.
56.
The major product obtained in the following conversion is:
(A)
(B)
(C)
(D)
The major product of the following reaction is:
(A)
(B)
(C)
(D)
57.
The higher concentration of which gas in air can cause stiffness of flower buds? (A) NO2 (B) CO2 (C) SO2 (D) CO
58.
The correct match between Item I and Item II is Item I Item II A. Ester test P. Tyr B. Carbylamine test Q. Asp C. Phthalein dye test R. Ser S. Lys (A) A – Q, B – S, C – P (B) A – R, B – Q, C – P (C) A – R, B – S, C – Q (D) A – Q, B – S, C – R
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JEE-MAIN-2019 (11th Jan-Second Shift)-PCM-13
59.
The number of bridging CO ligand(s) and Co-Co bond(s) in Co2 (CO)8 respectively are: (A) 2 and 1 (B) 2 and 0 (C) 0 and 2 (D) 4 and 0
60.
4KOH,O2 A 2B 2 H2O
Green
4HCI 3B 2C MnO 2 2H2O
Purple
H2O,KI 2C 2 A 2KOH D
In the above sequence of reactions, A and D, respectively are: (A) Ki and KMnO4 (B) MnO2 and KIO3 (C) KIO3 and MnO2 (D) KI and K2MnO4
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JEE-MAIN-2019 (11th Jan-Second Shift)-PCM-14
PART–C (MATHEMATICS) 61.
lim x 0
x cot 4 x sin2 x cot 2 2x
is equal to:
(A) 0 (C) 4 62.
63.
(B) 2 (D) 1
2
7 cot 1 x 10 0 , lie in the interval:
(A) ,cot 5 cot 4,cot 2
(B) cot 2,
(C) ,cot 5 cot 2,
(D) cot 5,cot 4
If a hyperbola has length of its conjugate axis equal to 5 and the distance between its foci is 13, then the eccentricity of the hyperbola is:
13 12 13 (C) 6 (A)
64.
1
All x satisfying the inequality cot x
(B) 2 (D)
13 8
If the area of the triangle whose one vertex is at the vertex of the parabola, y 2 4 x a2 0 and the other two vertices are the points of intersection of the
parabola and y – axis, is 250 sq. units, then a value of ‘a’ is:
1/ 3
(B) 5 2
(A) 5 5 (C) 10 65.
2/3
Two lines
(D) 5
x 3 y 1 z 6 x5 y2 z3 and intersect at the point R. The 1 3 1 7 6 4
reflection f R in the xy – plane has coordinates: (A) (2, –4, –7) (B) (2, 4, 7) (C) (2, –4, 7) (D) (–2, 4, 7) 66.
Contrapositive of the statement “If two numbers are not equal, then their squares are not equals” is: (A) If the squares of two numbers are not equal, then the numbers are equal (B) If the squares of two numbers are equal, then the numbers are not equal (C) If the squares of two numbers are equal, then the numbers are equal (D) If the squares of two numbers are not equal, then the numbers are not equal
67.
If in a parallelogram ABDC, the coordinates of A, B and C are respectively (1, 2), (3, 4) and (2, 5), then the equation of the diagonal AD is: (A) 5x 3y 1 0 (B) 5x 3y 11 0 (C) 3x 5y 7 0 (D) 3x 5y 13 0
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JEE-MAIN-2019 (11th Jan-Second Shift)-PCM-15
68.
69.
The integral
/4
/ 6
dx equals: sin 2x tan5 x cot 5 x
(A)
1 1 tan1 20 9 3
(C)
40
1 1 tan1 9 l 3 10 4 1 1 (D) tan1 5 4 3 3 (B)
Let x, y be positive real numbers and m, n positive integers. The maximum value of the expression
xm yn is: 1 x 2m 1 y 2n
1 2 mn (D) 6mn
(A) 1 (C)
(B)
1 4
2
70.
n
q 1 q 1 q 1 Let Sn 1 q q ... q and Tn 1 ... where q is a 2 2 2 real number and q 1. If 101 C1 101 C2 .S1 ... 101 C101.S100 T100 then is equal to: 2
n
99
(A) 2 (C) 200 71.
(B) 202 (D) 2100
Let and be the roots of the quadratic equation n 1 n x sin x sin cos 1 cos 0 0 45 , and < . Then n is n0
2
0
equal to:
1 1 1 cos 1 sin 1 1 (C) 1 cos 1 sin
1 1 1 cos 1 sin 1 1 (D) 1 cos 1 sin
(A)
72.
(B)
A bag contains 30 white balls and 10 red balls. 16 balls are drawn one by one randomly from the bag with replacement. If X be the number of white balls drawn, then
mean of X is equal to: s tandard deviation of X (A) 4
(B) 4 3
(C) 3 2
(D)
4 3 3
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JEE-MAIN-2019 (11th Jan-Second Shift)-PCM-16
73.
Let z be a complex number such that z z 3 i where i
1 . Then z is equal
to: (A) (C)
74.
34 3 41 4
5 3 5 (D) 4 (B)
a bc
2a
2a
2b
bca
2b
2c
2c
c ab
If
2
= a b c x a b c , x = and a + b+ +c 0, then x is equal to: (A) abc (C) 2(a + b + c) 75.
Let
(B) – (a + b + c) (D) –2 (a + b+ c)
3ˆi j, ˆi 3ˆj and ˆi 1 ˆj respectively be the position vectors of the points A,
B and C with respect to the origin O. If the distance of C from the bisector of the acute angle between OA and OB is
3 , then the sum of all possible values of is 2
(A) 4 (C) 2
(B) 3 (D) 1
76.
If 19th terms of non – zero A.P. is zero, then its (49th term) : (29th term) is: (A) 4:1 (B) 1:3 (C) 3:1 (D) 2:1
77.
If
x 1 dx f x 2x 1 C , where C is a constant of integration of integration, 2x 1
then f(x) is equal to:
1 x 1 3 2 (C) x 4 3
2 x 2 3 1 (D) x 4 3
(A)
78.
(B)
Let a function f : 0, 0, be defined by f x 1 (A) not injective but it is surjective (C) neither injective nor surjective
79.
1 . Then f is: x
(B) injective only (D) both injective as well as surjective
Let K be the set of all real values of x where the function f x sin x x 2 x cos x is not differentiable. Then the set K is equal to: (A) (en empty set) (C) {0}
(B) {} (D) {0, }
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JEE-MAIN-2019 (11th Jan-Second Shift)-PCM-17
80.
The area (in sq. units) in the first quadrant bounded by the parabola, y x 2 1 , the tangent to it at the point (2, 5) and the coordinate axes is:
8 3 187 (C) 24
37 24 14 (D) 3
(A)
81.
Given
(B)
cos A cos cosC bc c a ab for a ABC with usual nation. If , 11 12 13
then the ordered tried (, , ) has a value: (A) (7, 19, 25) (B) (3, 4, 5) (C) (5, 12, 13) (D) (19, 7, 25) 82.
The solution of the differential equation (A) loge
2x xy 2y
(C) loge 83.
1 x y xy2 1 x y
(D) loge
2y 2 y 1 2x
Let the length of the latus rectum of an ellipse with its major axis long x – axis and center at the origin, be 8. If the distance between the foci of this ellipse is equal to the length of the length of its minor axis, then which one of the following points lies on it?
(C) 4,
3
(D) 4
(A) 4, 2, 2 2
84.
dy 2 x y when y(1) = 1, is: dx 1 x y (B) loge 1 x y 2 x 1
3, 2
3
(B) 4 3, 2 2
2, 2
let S = {1, 2, … 20}. A subset B of S is said to be “nice”, if the sum of the elements of B is 203. Then the probability that a randomly chosen subset of S is ‘nice’ is:
7 2 20 4 (C) 20 2
5 2 20 6 (D) 20 2
(A)
(B)
85.
If the point (2, , ) lies on the plane which passes through the points (3, 4, 2) and (7, 0, 6) and is perpendicular to the plane 2x 5y 15 , then 2 3 is equal to: (A) 12 (B) 7 (C) 5 (D) 17
86.
Let x 10 x 10
50
(A) 12.50 (C) 12.25
50
a0 a1x a 2 x 2 ... a50 x50 , for xR; then
a2 is equal to: a0
(B) 12.00 (D) 12.75
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JEE-MAIN-2019 (11th Jan-Second Shift)-PCM-18 87.
The number of functions f from {1, 2, 3, …, 20} only {1, 2, 3, …, 20} such that f(k) is a multiple of 3, whenever k is a multiple of 4, is: (A) 65 15 ! (B) 5! 6 ! (C) 15 ! 6 !
(D) 56 15
88.
A circle cuts a chord of length 4a on the x – axis and passes through a point on the y – axis, distant 2b from the origin. Then the locus of the center of this circle, is: (A) a hyperbola (B) an ellipse (C) a straight line (D) a parabola
89.
Let f x
x 2
a x
2
dx b2 d x
2
,r R f, where a, b and d are non – zero real
constant. Then: (A) f is an increasing function of x (B) f is a decreasing function of x (C) f is not a continuous function of x (D) f is neither increasing nor decreasing function of x 90.
Let A and B be two invertible matrices of order 3 x 3. If det (ABAT) = 8 and det(AB–1) = 8, then det (BA–1 BT) is equal to:
1 4 1 (C) 16 (A)
(B) 1 (D) 16
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JEE-MAIN-2019 (11th Jan-Second Shift)-PCM-19
JEE (Main) – 2019 ANSWERS PART A – PHYSICS 1.
B
2.
A
3.
C
4.
C
5.
B
6.
A
7.
C
8.
B
9.
D
10.
A
11.
B
12.
B
13.
D
14.
BONUS
15.
A
16.
A
17.
D
18.
B
19.
D
20.
B
21.
D
22.
B
23.
A
24.
C
25.
D
26.
C
27.
B
28.
B
29.
B
30.
C
PART B – CHEMISTRY 31.
C
32.
A
33.
D
34.
C
35.
A
36.
D
37.
A
38.
C
39.
B
40.
B
41.
D
42.
B&C
43.
B
44.
B
45.
D
46.
C
47.
B
48.
C
49.
B
50.
D
51.
B
52.
A
53.
C
54.
D
55.
A
56.
D
57.
C
58.
A
59.
A
60.
B
PART C – MATHEMATICS 61.
D
62.
B
63.
A
64.
D
65.
A
66.
C
67.
A
68.
B
69.
C
70.
D
71.
C
72.
B
73.
B
74.
B
75.
D
76.
C
77.
D
78.
C
79.
A
80.
B
81.
A
82.
B
83.
B
84.
B
85.
B
86.
C
87.
C
88.
D
89.
A
90.
C
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JEE-MAIN-2019 (11th Jan-Second Shift)-PCM-20
HINTS AND SOLUTIONS PART A – PHYSICS 1.
2.
3.
1 S (5iˆ 4 ˆj)2 (4ˆj 4ˆj)4 2 = 10iˆ 8ˆj 8iˆ 8ˆj r2 r1 18iˆ 16ˆj r2 20iˆ 20ˆj rf 20 2 x t x0 100o C x100 x 0 x0 x0 2 3 100o C x0 x 3 = 25°C
t
Rg = 20 NL = Ng = N = 30 1 FOM = = 0.005 A/Div.
1 Current sentivity = CS = 0.005 1 Igmax = 0.005 × 30 = 15 × 10–2 = 0.15 15 = 0.15[20 + R] 100 = 20 + R R = 80. 4.
R1 2 …(i) R2 3 R1 10 1 R2 R1 + 10 = R2 …(ii) 2R 2 R 10 = R2 ; 10 2 3 3 R2 = 30 & R1 = 20 30 R 30 R 2 30 3 R = 60
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JEE-MAIN-2019 (11th Jan-Second Shift)-PCM-21
MR 2 MR2 2 MR2 2 4 2 2 MR MR = 2MR2 = 3 MR2 2 2
5.
I
6.
2.5 = 1 × 5 sin 1 Sin = 0.5 = 2 6
7.
Total length L will remain constant L = (3a) N (N = total turns) And length of winding = (d) N = (d = diameter of wire) Self inductance = 0n2A
a
a a
3 a2 = 0n 2 dN 4
a2 N a So self inductance will become 3 times.
8.
dp F kt dt
3P
P
2p
9.
10.
T
dP kt dt 0
KT 2 p ; T2 2 k
I H Magnetic moment I Volume 20 10 6 I = 20 N/m2 6 10 20 1 103 3 60 10 3 = 0.33 × 10–3 = 3.3 × 10–4
Angular frequency of pendulum geff 1 geff 2 geff
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JEE-MAIN-2019 (11th Jan-Second Shift)-PCM-22
1 g 2 g [s = angular frequency of support] 1 g 2 g
1 2(As5 ) 2 10 1 10 2 = 10–3 10
11. 12.
I
6 0.002 (D2 is in reverse bias) 300
U = P E = –PE cos = –(10–29) (103) cos 45° = –0.707 × 10–26 J = –7 × 10–27 J
13.
0.1 × 400 × (500 – T) = 0.5 × 4200 × (T – 30) + 800 (T – 30) 40(500 – T) = (T – 30) (2100 + 800) 20000 – 40T = 2900 T – 30 × 2900 20000 + 30 × 2900 = T(2940) T = 30.4°C T 6.4 100 100 20% T 30
14.
BONUS
15.
Maximum kinetic energy at lowest point B is given by K = mg (1 – cos )
m
where = angular amp. K1 = mg (1 – cos )
A
K2 = mg(2) (1 – cos ) K2 = 2K1 16.
B
1 = 2
1T1 = 2T2
1 T1 4 T 30 ; 2 T2 3 180 30 T = 230°C
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JEE-MAIN-2019 (11th Jan-Second Shift)-PCM-23
17.
F F y ; [Y] A A Now from dimension ML F F 2 ; L T2 T M 4 2 V F V L2 2 T M A A 2 4 F v L2 2 2 2 F = MA MA A V4 L2 2 A [F] [Y] F1V 4 A 2 [A]
18.
40 + f = m(R) …(i) 40 × R – f × R = mR=2 40 – f = mR …(ii) From (i) and (ii) 40 = 16 mR
19.
Intensity of EM wave is given by Power 1 I 0E02C Area 2 27 10 3 1 = 9 10 2 E 2 3 108 10 106 2 E = 2 103 kV / m = 1.4 kV/m
20.
Potential difference across AB will be equal to battery equivalent across CD. E1 E 2 E3 1 2 3 r1 r2 r3 VAB VCD 1 1 1 1 1 1 1 1 1 r1 r2 r3 1 1 1 6 = = 2V 3
21.
g
GM R2 2
2 Mp R e 1 1 3 ge Me Rp 3 3 1 Also, T g
gp
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JEE-MAIN-2019 (11th Jan-Second Shift)-PCM-24
22.
Tp Te
ge 3 gp
Tp 2 3s
Analysis of graph says (1) Amplitude varies as 8 – 10 V or 9 1 (2) Two time period 100 s (signal wave) and 8 s (carrier wave) 2t 2t Hence signal is 9 1 sin sin T1 T2 = 9 1 sin (2 × 104t) sin 2.5 × 105 t
23.
VT = K
PV V k nR PV2 = K R C CV (For polytropic process) 1 x R 3R R C 1 2 2 2 Q = nC T
24.
100 × SA × [100 – 90] = 50 × SB × (90 – 75) 2SA = 1.5 SB 3 S A SB 4 Now, 100 × SA × [100 –T] = 50 × SB (T – 50) 3 2 (100 T) (T 50) 4 300 – 3T = 2T – 100 400 = 5T T = 80
25.
For M L steel 1 1 K5 1 K 2 2 3 36 2 for N L 1 1 K 3 1 K 2 2 4 16 2 20 27
26.
i=e r1 = r2 =
A 300 2
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JEE-MAIN-2019 (11th Jan-Second Shift)-PCM-25
by Snell’s law 1 sini 3
1 3 2 2
i = 60 27.
28.
For diffraction Location of 1st minima D y1 0.2469D a Location of 2nd minima 2D y2 0.4938D a Now for interference Path for interference Path difference at P. dy 4.8 D Path difference at P. dy 9.6 D So orders of maxima in between P and Q is 5, 6, 7, 8, 9 So 5 bright fringes all present between P & Q.
y2
P y1
6 F 13 Therefore three capacitors most be in parallel to get 6 in 1 1 1 1 1 1 Ceq 3C C C C C Ceq
Ceq 29.
Q
3C 6 F 13 13
Fnet dE q(v B) = (2q ˆi 3q ˆi) q(v B) W = Fnet S
= 2q + 3q = 5q 30.
hc eV1 1 hc eV2 2 (i) – (ii)
…(i) …(ii)
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JEE-MAIN-2019 (11th Jan-Second Shift)-PCM-26
1 1 hc e(V1 V2 ) 1 2 hc 2 1 V1 V2 e 1 2 = (1240 nm V) =
100 nm 300 nm 400 nm
12.4 1 V. 12
PART B – CHEMISTRY 31.
In order to be spontaneous GO should be –ve GO = HO –TSO 0 = 491.1 103 – T 198 491100 T 2480 198 If temp is above 2480 K, the reaction will be spontaneous.
32.
Fact based, go through definition.
33.
Th is a metal having large size and oxalate is a bidentate ligand hence its co-ordination number in given complex is 10. O
34.
Since LiAlH4 is a strong reducing agent, it will reduce –NO2, C O H and ketonic group but cannot reduce normal alkene >C=C<
35.
GO = HO – TSO GO = A – BT In endothermic reaction H = +ve. Hence, A = +ve
ll
36.
3a 4R 3a 4 2(R + r) = a 3a 2 r a 4 R
3a 2r a 2 3a 2a 3a 2r a 2 2 2a 1.7329 .268a r = .067a 4 4
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JEE-MAIN-2019 (11th Jan-Second Shift)-PCM-27
37.
SiH4 has complete octet hence it is not an electron deficient hydride.
38.
Go nFEocell 2.303RT log K eq 2F Eocell 2.303RT log K eq
2.303RT log 10 1015 F .059 16 8 .059 .472 2
2 E ocell Eocell
39.
Electronegativity increases from left to right in a period and decreases down the group. Br
40. On ionization
41.
, this compound will produce Aromatic cation, which is stable.
h mv According to Einstein’s theory of photoelectric effect: h ho + KE
1 h = h o + mv 2 2 2h o mv 2 2h o m
v2 1
v o 2
h 1
m o 2
1 1
o 2 42.
Option B and option D both will react with Grignard reagent and decolourizes Br2/H2O. (IIT has given option D only)
43.
After protonation at b or c or d the conjugate acid is stabilized by resonance.
44.
Acid rain reacts with marble. Hence, The Taj Mahal which made up of marble is discoloured.
45.
Inert pair effect gradually increases down the group. Hence, stability of lower oxidation state increases down the group.
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JEE-MAIN-2019 (11th Jan-Second Shift)-PCM-28 46.
Go 2.303RT log K eq
= –2.303 8.314 298 log 10–14 = –2.303 8.314 298 –14 = 79,881.87 80 KJ mol–1 47.
Calcination takes place in absence of air. Hence step 2 is not defining it.
48.
Br2/NaOH converts amide into primary amine having one carbon atom less, which gives carbylamine test.
49.
Go through different types of colloid and their examples.
50.
4-hydroxy butanoic acid undergoes intermolecular esterification to give polymer.
51.
K 2 Hg 4 2K Hg 4
2
1
2
Total number of particle = 1 + 2 1 2 Hence, Van’t Hoff factor = 1 1 2 0.4 = 1 0.8 1.8 1 52.
Apply law of equivalence: 25 N = 30 0.1 2 30 0.2 6 1.2 NHCl 0.2 25 5 5 nd For the 2 titration 1.2 VHCl 30 0.2 5 6 5 30 VHCl = 25 ml 1.2 1.2
53.
For zero order reaction Co – Ct = Kt 0.5M – 0.2M = Kt 0.3 = Kt ‘K’ can be calculated by C t1/2 o 2K 0.2 6 2K
----- (1)
0.2 2 10 1 1 12 12 60 Putting the value of K in eq (1) 0.3 0.3 t 60 0.3 18 Hr 1 K 60 K
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JEE-MAIN-2019 (11th Jan-Second Shift)-PCM-29
54.
Fact based
55.
Attack on C
C
is more preferred than benzene ring.
56.
First Markonikov’s addition one alkene followed by intramolecular Friedel-Craft alkylation takes place.
57.
SO2 gas causes stiffness of flower buds?
58.
Go through structure of amino acids.
59.
Go through structure of [Co2(CO)8] O OC OC
C
CO
Co Co CO
OC
C
CO
O
60.
4KOH, O2 MnO2 2K 2MnO4 2H2O
Green
4 HCl
3 K 2MnO 4 2KMnO 4 MnO 2 2H2O Purple
H2O, K
2KMnO 4 2MnO 2 2KOH KO3
PART C – MATHEMATICS 61.
62.
x cos 4x sin2 2x sin2 x.cos2 2x.sin 4x 4x cos 4x . cos2 x sin 4x cos2 2x 1 as x 0
cot x 5 cot x 2 0 1
1
cot 1 x ,2 5,
Put 0 cot 1 x cot 1 x 0,2 x cot 2,
63.
2b 5 and 2ae 13 2
ae
a 2 b2 2
a2 ae b2
169 25 4 4
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JEE-MAIN-2019 (11th Jan-Second Shift)-PCM-30 a6 ae 13 e a 12
64.
y 2 4 x a2
Vertices of triangle are a2 ,0 and (0, 2a) and (0, –2a)
1 2 a 2 a3 125
4a 250
Area
65.
Points on the given lines are 3, 3 1, 6 and 7 5, 6 2, 4 3 3 7 5 3 1 6 2 1, 1 Point R is (2, –4, 7) Image of R under xy – plane is (2, –4, –7)
66.
Contra positive of p q is ~ q ~ p Answer is C C (2, 5)
67.
5 9 E is , 2 2
E
5 Slope of AD 3 Equation of AD is y 2
5 x 1 3
5x 3y 1 0 4
68.
I 6
D
A (1, 2)
B (3, 4)
sec 2 x dx
2 tan x tan5 x cot 5 x
Put tan x t 1
1
3
1
dt 1 2t t 5 5 t 4 t dt
2t 1
10
1
3
Put t5 y 1 1 I tan1 y 52 3 10 1 1 1 tan 5/ 2 10 4 3 FIITJEE Ltd., FIITJEE House, 29-A, Kalu Sarai, Sarvapriya Vihar, New Delhi -110016, Ph 46106000, 26569493, Fax 26513942 website: www.fiitjee.com.
JEE-MAIN-2019 (11th Jan-Second Shift)-PCM-31
69.
xm y n
1 x 1 y 2m
2n
1 1 n 1 m x m y n x y 1 Put x m m 2 x 1 1 1 2 m x xm
Maximum Value
1 4
101
70.
101
Cr sr 1
r 1
101
101Cr r 1
qr 1 q 1
101 1 101 101 r 101 C q Cr r q 1 r 1 r 1 1 101 1 q 1 2101 1 q 1
101 101 1 q 2 100 2 q1 100 2
71.
Using quadratic formula, 2
x
cos sin 1 cos sin 1
4 sin cos
2 sin
cos sin 1
2
cos sin 1
2 sin cos , cos ec cos , cos ec
n0
n
1
n
n
n
n
cos ec sin n0
n 0
1 1 1 cos 1 sin (C) is the correct answer.
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JEE-MAIN-2019 (11th Jan-Second Shift)-PCM-32 72.
There are 30 white balls and 10 red balls 30 3 P (white ball) p 40 4 1 q 4 mean x np s tan darddeviation x npq np q
73.
3 16 4 4 3 1 4
z x iy x 2 y 2 x iy 3 i y 1 x 2 1 x 3 x 2 1 9 6x x 2 4 x 3 5 z x2 y2 3
74.
R1 R1 R2 R3 1
a b c 2b 2c
1
1
bac
2b
2c
c ab
c 3 c 3 c1, c 2 c 2 c1 1
0
0
a b c 2b a b c 0 2c 0 a b c
a b c
3 2
a b c x a b c
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JEE-MAIN-2019 (11th Jan-Second Shift)-PCM-33
75.
y
Equation of angle bisector of DA and OB is y=x 1 3 Given that, 2 2 2 1 3 2, 1 Sum of values of 1
A
B 1, 3
a 18d 0 a 18d t 49 a 48d 18d 48d t 29 a 28d 18d 28d
77.
x
O
76.
3, 1
30d 3 10d
Put 2x 1 t 2 x 1 dx 2x 1 t2 3 t 3 3t dt C 6 2 2
t2 3 t C 6 2 x4 2x 1 C 3 78.
1 x Neither one - one nor Onto y 1
y=1
(1, 0)
79.
At x or x f x sin x x 2 x cos x f x is differentiable For x 0 f x sin x x 2 x cos x
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JEE-MAIN-2019 (11th Jan-Second Shift)-PCM-34 f ' x cos x 1 2 cos x 2 x sin x f ' 0 2 For x 0 f x sin x x 2 x cos x f ' x cos x 1 2 cos x 2 x sin x f ' 0 2 LHD = RHD f is differentiable at x = 0.
80.
Equation of tangent at (2, 5) is
y5 x 2 1 2
(2, 5)
or y 4x 3 2
Required Area x 2 1 dx 0
1 3 2 . 5 2 4
x3 25 2 x 0 3 8 8 9 37 3 8 24
(3/4,0) (0, –3)
81.
bc c a ab abc 11 12 13 18 a 7k, b 6k, c 5k b2 c 2 a2 1 2bc 5 19 5 cosB , cos C 25 7 1 19 5 5 35 7 7 19 25 35 35 35 : : 7 : 19 :25 cos A
82.
uxy du dy 1 dx dx du 1 u2 dx du 1 u2 dx
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JEE-MAIN-2019 (11th Jan-Second Shift)-PCM-35
du dx 1 u2 1 1 u log xc 2 1 u 1 1 x y log xc 2 1 x y Put x 1 c 1 1 x y log 2 x 1 1 x y
83.
x2 y2 1 a2 b2 Given that 2b 2ae 2b2 8 b ae and a a 1 e 2 4, a2 e2 a2 1 e2
Consider
1 2 a 8, b 4 2 e2
Hence equation of ellipse is
4 84.
x2 y2 1 64 32
3, 2 2 lies on this
Sum of all elements of S = 210 So X be a nice set if x S 7 , S 1, 6 , S 2, 5 ,S 3, 4 , S 1,2,4
5 220 (2) is the answer. P x
85.
A 7, 0, 6 and B (3, 4, 2) AB 4iˆ 4 ˆj 4kˆ Also 2iˆ 5ˆj is parallel to the plane
ˆi
ˆj
kˆ
Normal perpendicular to the required plane is 1 1 1 5iˆ 2jˆ 3kˆ 2 5 0 Equation of the plane is 5 x 7 2 y 0 3 z 6 0 5x 2y 3z 17
2, ,
lies on this 10 2 3 17 2 3 7
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JEE-MAIN-2019 (11th Jan-Second Shift)-PCM-36
86.
50
10 x
a0 1050
50
10 x
2
a 2 50C2 10
a2 a0 87.
48
2 48 50 C2 10 2 1052 (2)
12.25
k 4,8,12,16,20 f k can takes the values 3,6,9,12,15,18
Number of ways 6C5 .5! Total number of onto functions 6C5 .5! 15! 6!15!
88.
k 2 4a2 r 2 and 2
h 0
2
k 2b r 2 2
h2 k 2b k 2 4a2 h2 4bk 4b 2 4a2 Locus is x 2 4 by b2 a 2
r
(0, 2b)
h (k)
r
Parabola.
k 2a
89.
f x x x3 a
f ' x
1 2 2
a
x
2
a2
d x b2 d x
1 2 2
2
x 2 a2
b2
b
2
d x
2
2
b2 d x
ve 90.
2
ABA T A .B . A T A B
AB 1 8 A 8 B 1
BA B
B
2
A
B
2
8B
B 8
1 16
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FIITJEE Solutions to JEE(Main)-2019 Test Date: 12th January 2019 (First Shift)
PHYSICS, CHEMISTRY & MATHEMATICS Paper - 1 Time Allotted: 3 Hours
Maximum Marks: 360
Please read the instructions carefully. You are allotted 5 minutes specifically for this purpose.
Important Instructions: 1.
The test is of 3 hours duration.
2.
This Test Paper consists of 90 questions. The maximum marks are 360.
3.
There are three parts in the question paper A, B, C consisting of Physics, Chemistry and Mathematics having 30 questions in each part of equal weightage. Each question is allotted 4 (four) marks for correct response.
4.
Out of the four options given for each question, only one option is the correct answer.
5.
For each incorrect response 1 mark i.e. ¼ (one-fourth) marks of the total marks allotted to the question will be deducted from the total score. No deduction from the total score, however, will be made if no response is indicated for an item in the Answer Box.
6.
Candidates will be awarded marks as stated above in instruction No.3 for correct response of each question. One mark will be deducted for indicating incorrect response of each question. No deduction from the total score will be made if no response is indicated for an item in the answer box.
7.
There is only one correct response for each question. Marked up more than one response in any question will be treated as wrong response and marked up for wrong response will be deducted accordingly as per instruction 6 above.
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JEE-MAIN-2019 (12th Jan-First Shift)-PCM-2
PART –A (PHYSICS) 1.
An ideal gas occupies a volume of 2 m3 at a pressure of 3 × 106 Pa. The energy of the gas is: (A) 9 × 106J (B) 6 × 104J 8 (C) 10 J (D) 3 × 102J
2.
A travelling harmonic wave is represented by the equation y(x, t) = 103sin(50t + 2x), where x and y are in meter and t is in seconds. Which of the following is a correct statement about the wave? (A) The wave is propagating along the negative x-axis with speed 25 ms1. (B) The wave is propagating along the positive x-axis with speed 100 ms1. (C) The wave is propagating along the positive x-axis with speed 25 ms1. (D) The wave is propagating along the negative x-axis with speed 100 ms1.
3.
An ideal battery of 4 V and resistance R are connected in series in the primary circuit of a potentiometer of length 1 m and resistance 5 . The value of R, to give a difference of 5 mV across 10 cm of potentiometer wire, is: (A) 490 (B) 480 (C) 395 (D) 495
4.
In a meter bridge, the wire of length 1 m has a nondR uniform cross-section such that, the variation of its d dR 1 resistance R with length is . Two equal d resistances are connected as shown in the figure. The galvanometer has zero deflection when the jockey is at point P. What is the length AP? (A) 0.2 m (B) 0.3 m (C) 0.25 m (D) 0.35 m
5.
A passenger train of length 60 m travels at a speed of 80 km/hr. Another freight train of length 120 m travels at a speed of 30 km/hr. The ratio of times taken by the passenger train to completely cross the freight train when: (i) they are moving in the same direction, and (ii) in the opposite directions is: 11 5 (A) (B) 5 2 3 25 (C) (D) 2 11
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JEE-MAIN-2019 (12th Jan-First Shift)-PCM-3 6.
Two electric bulbs, rated at (25 W, 220 V) and (100 W, 220 V), are connected in series across a 220 V voltage source. If the 25 W and 100 W bulbs draw powers P1 and P2 respectively, then: (A) P1 = 16 W, P2 = 4 W (B) P1 = 16 W, P2 = 9 W (C) P1 = 9 W, P2 = 16 W (D) P1 = 4 W, P2 = 16 W
7.
A straight rod of length L extends from x = a to x = L + a. The gravitational force it exerts on a point mass ‘m’ at x = 0, if the mass per unit length of the rod is A + Bx2, is given by: 1 1 1 1 (A) Gm A BL (B) Gm A BL aL a a a L 1 1 (C) Gm A BL aL a
1 1 (D) Gm A BL a a L
8.
What is the position and nature of image formed by lens combination shown in figure? (f1, f2 are focal lengths) (A) 70 cm from point B at left; virtual (B) 40 cm from point B at right; real 20 (C) cm from point B at right; real 3 (D) 70 cm from point B at right ; real
9.
A point source of light, S is placed at a distance L in front of the centre of plane mirror of width d which is hanging vertically on a wall. A man walks in front of the mirror along a line parallel to the mirror, at a distance 2 L as shown below. The distance over which the man can see the image of the light source in the mirror: (A) d (B) 2d d (C) 3d (D) 2
10.
A light wave is incident normally on a glass slab of refractive index 1.5. If 4% of light gets reflected and the amplitude of the electric field of the incident light is 30 V/m, then the amplitude of the electric field for the wave propagating in the glass medium will be: (A) 30 V/m (B) 10 V/m (C) 24 V/m (D) 6 V/m
11.
For the given cyclic process CAB as shown for a gas, the work done is: (A) 30 J (B) 10 J (C) 1 J (D) 5 J
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JEE-MAIN-2019 (12th Jan-First Shift)-PCM-4 12.
13.
14.
15.
Determine the electric dipole moment of the system of three charges, placed on the vertices of an equilateral triangle, as shown in the figure: ˆj ˆi ˆi ˆj (A) 3 q (B) q 2 2 (C) 2 q ˆj (D) 3 q ˆj The position vector of the centre of mass r cm of an asymmetric uniform bar of negligible area of cross-section as shown in figure is: 13 5 5 13 (A) r cm L xˆ L yˆ (B) r cm L xˆ L yˆ 8 8 8 8 3 11 11 3 (C) r cm L xˆ L yˆ (D) r cm L xˆ L yˆ 8 8 8 8 A person standing on an open ground hears the sound of a jet aeroplane, coming from north at an angle 60° with ground level. But he finds the aeroplane right vertically above his position. If is the speed of sound, speed of the plane is: 3 2 (A) (B) 2 3 (C) (D) 2 Two light identical springs of spring constant k are attached horizontally at the two ends of a uniform horizontal rod AB of length l and mass m. the rod is pivoted at its centre ‘O’ and can rotate freely in horizontal plane. The other ends of the two springs are fixed to rigid supports as shown in figure. The rod is gently pushed through a small angle and released. The frequency of resulting oscillation is: 1 3k 1 2k (A) (B) 2 m 2 m (C)
16.
1 6k 2 m
1 k 2 m
A simple pendulum, made of a string of length l and a bob of mass m, is released from a small angle 0. It strikes a block of mass M, kept on a horizontal surface at its lowest point of oscillations, elastically. It bounces back and goes up to an angle 1. Then M is given by: 1 m 1 (A) 0 (B) m 0 2 0 1 0 1 1 (C) m 0 0 1
17.
(D)
(D)
m 0 1 2 0 1
A 100 V carrier wave is made to vary between 160 V and 40 V by a modulating signal. What is the modulation index? (A) 0.3 (B) 0.5 (C) 0.6 (D) 0.4
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JEE-MAIN-2019 (12th Jan-First Shift)-PCM-5 18.
A cylinder of radius R is surrounded by a cylindrical shell of inner radius R and outer radius 2R. The thermal conductivity of the material of the inner cylinder is K1 and that of the outer cylinder is K2. Assuming no loss of heat, the effective thermal conductivity of the system for heat flowing along the length of the cylinder is: K K2 (A) 1 (B) K1 K 2 2 2K1 3K 2 K 3K 2 (C) (D) 1 5 4
19.
The least count of the main scale of a screw gauge is 1 mm. The minimum number of divisions on its circular scale required to measure 5 m diameter of a wire is: (A) 50 (B) 200 (C) 100 (D) 500
20.
The output of the given logic circuit is: (A) AB AB (B) AB AB (C) AB (D) AB
21.
As shown in the figure, two infinitely long, identical wires are bent by 90° and placed in such a way that the segments LP and QM are along the x-axis, while segments PS and QN are parallel to the y-axis. If OP = OQ = 4 cm, and the magnitude of the magnetic field at O is 104 T, and the two wires carry equal current (see figure), the magnitude of the current in each wire and the direction of the magnetic field at O will be (0 = 4 × 107 NA2): (A) 20 A, perpendicular out of the page (B) 40 A, perpendicular out of the page (C) 20 A, perpendicular into the page (D) 40 A, perpendicular into the page
22.
A particle of mass m moves in a circular orbit in a central potential field U(r)
23.
A proton and an -particle (with their masses in the ratio of 1 : 4 and charges in the ratio of 1:2 are accelerated from rest through a potential difference V. If a uniform magnetic field (B) is set up perpendicular to their velocities, the ratio of the radii rp : r of the circular paths described by them will be: (A) 1: 2 (B) 1 : 2 (C) 1 : 3 (D) 1: 3
1 2 kr . If 2 Bohr’s quantization conditions are applied, radii of possible orbits and energy levels vary with quantum number n as: 1 (A) rn n, En n (B) rn n, En n 1 (C) rn n, En n (D) rn n2 , En 2 n
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JEE-MAIN-2019 (12th Jan-First Shift)-PCM-6 24.
Let the moment of inertia of a hollow cylinder of length 30 cm (inner radius 10 cm and outer radius 20 cm), about its axis be I. The radius of a thin cylinder of the same mass such that its moment of inertia about its axis is also I, is: (A) 12 cm (B) 16 cm (C) 14 cm (D) 18 cm
25.
In the figure shown, a circuit contains two identical resistors with resistance R = 5 and an inductance with L = 2 mH. An ideal battery of 15 V is connected in the circuit. What will be the current through the battery long after the switch is closed? (A) 5.5 A (B) 7.5 A (C) 3 A (D) 6 A
26.
In the figure shown, after the switch ‘S’ is turned from position ‘A’ to position ‘B’, the energy dissipated in the circuit in terms of capacitance ‘C’ and total charge ‘Q’ is: 1 Q2 3 Q2 (A) (B) 8 C 8 C 2 5Q 3 Q2 (C) (D) 8 C 4 C
27.
A satellite of mass M is in a circular orbit of radius R about the centre of the earth. A meteorite of the same mass, falling towards the earth collides with the satellite completely inelastically. The speeds of the satellite and the meteorite are the same, just before the collision. The subsequent motion of the combined body will be: (A) such that it escapes to infinity (B) in an elliptical orbit (C) in the same circular orbit of radius R (D) in a circular orbit of a different radius
28.
The galvanometer deflection, when key K1 is closed but K2 is open, equals 0 (see figure). On closing K2 also and adjusting R2 to 5 , the deflection in galvanometer becomes 0 . The resistance of the galvanometer is, then, given by 5 [Neglect the internal resistance of battery]: (A) 5 (B) 22 (C) 25 (D) 12
29.
A particle A of mass ‘m’ and charge ‘q’ is accelerated by a potential difference of 50 V. Another particle B of mass ‘4 m’ and charge ‘q’ is accelerated by a potential difference of 2500 V. The ratio of de-Broglie wavelength A is close to: B (A) 10.00 (B) 0.07 (C) 14.14 (D) 4.47
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JEE-MAIN-2019 (12th Jan-First Shift)-PCM-7 30.
There is a uniform spherically symmetric surface charge density at a distance Ro from the origin. The charge distribution is initially at rest and starts expanding because of mutual repulsion. The figure that represents best the speed V(R(t)) of the distribution as a function of its instantaneous radius R(t) is: (A) (B)
(C)
(D)
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JEE-MAIN-2019 (12th Jan-First Shift)-PCM-8
PART –B (CHEMISTRY) 31.
In the Hall-heroult process, aluminium is formed at the cathode. The cathode is made out of (A) Pure Aluminium (B) Carbon (C) Copper (D) Platinum
32.
The correct order for acid strength of compounds: CH CH, CH3 – C CH, CH2 = CH2 is as follows: (A) CH CH > CH2 = CH2 > CH3 – C CH (B) CH3 – C CH > CH CH > CH2 = CH2 (C) CH3 – C CH > CH2 = CH2 > HC CH (D) HC CH > CH3 – C CH > CH2 = CH2
33.
K In a chemical reaction A 2B 2C D , the initial concentration of B was 1.5 times of A but the equilibrium concentrations of A and B were found to be equal. The equilibrium, constant(K) for the aforesaid chemical reaction is (A) 4 (B) 16 (C) 1/4 (D) 1
34.
Given Gas: H2 CH4 CO2 SO2 Critical: 33 190 304 630 Temp/K On the basis of data given, predict which of the following gases shows least adsorption on a definite amount of charcoal (A) SO2 (B) CH4 (C) CO2 (D) H2
35.
Mn2(CO)10 is an organometalic compound due to the presence of (A) Mn – C bond (B) Mn – Mn bond (C) Mn – O bond (D) C – O bond
36.
The major product of the following reaction is:
(A)
(B)
(C)
(D)
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JEE-MAIN-2019 (12th Jan-First Shift)-PCM-9 37.
A metal on combustion in excess air forms X, X upon hydrolysis with water yields H2O2 and O2 along with another product. The metal is: (A) Na (B) Rb (C) Mg (D) Li
38.
The molecules that has minimum/no role in the formation of photochemical smog is: (A) N2 (B) CH2 = O (C) O3 (D) NO
39.
The pair of metal ions that can give a spin-only magnetic moment of 3.9 BM for the complex [M(H2O)6]Cl2 is (A) V2+ and Co2+ (B) V2+ and Fe2+ 2+ 2+ (C) Co and Fe (D) Cr2+ and Mn2+
40.
For a disatomic ideal gas in a closed system, which of the following plots does not correctly describe the relation between various thermodynamic quantities?
(A)
(B)
(C)
(D)
41.
Among the following compounds most basic amino acid is (A) Asparagine (B) Lysine (C) Serine (D) Histidine
42.
The increasing order of reactivity of the following compounds towards reaction with alkyl halide directly is:
(a) (A) (b) < (a) < (c) < (d) (C) (b) < (a) < (d) < (c)
(c)
(b)
(d)
(B) (a) < (b) < (c) < (d) (D) (a) < (c) < (d) < (b)
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JEE-MAIN-2019 (12th Jan-First Shift)-PCM-10 43.
Which of the following has lowest freezing point?
(A)
(B)
(C)
(D)
44.
50 mL of 0.5 M oxalic acid is needed to neutralize 25 ml of sodium hydroxide solution. The amount of NaOH in 50 mL of the given sodium hydroxide solution is (A) 40 g (B) 10 g (C) 20 g (D) 80 g
45.
The volume of gas A has is twice than that of gas B. The compressibility factor of gas A is thrice that that of gas B at same temperature. The pressure of gases for equal per moles are (B) 2PA = 3PB (A) 3PA = 2PB (C) PA = 3 PB (D) PA = 2PB
46.
The hardness of water sample (in terms of equivalents of CaCO3) containing 10–3 M CaSO4 i s (molar mass of CaSO4 = 136 g mol–1) (A) 10 ppm (B) 50 ppm (C) 90 ppm (D) 100 ppm
47.
(A) CH3CH2COCH3 + PhMgX (C) PhCOCH3 + CH3CH2MgX
(B) PhCOCH2CH3 + CH3MgX (D) HCHO + PhCH(CH3)CH2MgX
48.
Freezing point of a 4% aqueous solution of X is equal to freezing point of 12% aqueous solution of Y. If molecular weight of X is A then molecular weight of Y is (A) 3A (B) 2A (C) A (D) 4A
49.
The metal d-orbital that can directly facing the ligands in K3[Co(CN)6] are (A) d xy and dx2 y2 (B) dx2 y2 and dz2 (C) dxz ,dyz and dz2
(D) dxy, dxz and dyz
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JEE-MAIN-2019 (12th Jan-First Shift)-PCM-11
50.
51.
Decomposition of X exhibits a rate constant for 0.05 g/year. How many years are required for the decomposition of 5 g of X into 2.5 g? (A) 50
(B) 25
(C) 20
(D) 40
dE dT The standard electrode potential E– and its temperature coefficient for a cell are 2 V and -5 10–4 VK–1 at 300 K respectively. The cell reaction is : 2 Zn s Cu2 aq Zn aq Cu s Standard reaction enthalpy (rH–)…….. (A) -412.8 (C) 1920
52.
(B) -384.0 (D) 206.4 kJ
The major product of the following reaction is
(A)
(B)
(C)
(D)
53.
The element with Z = 120(not yet discovered) will be an/a (A) Inner transition metal (B) Alkaline earth metal (C) Alkali metal (D) Transition metal
54.
In the following reaction HCl Aldehyde Alcohol Acetal Aldehyde Alcohol t HCHO BuOH CH3CHO MeOH The best combination is (A) CH3CHO and tBuOH (C) CH3CHO and MeOH
(B) HCHO and MeOH (D) HCHO and tBuOH
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JEE-MAIN-2019 (12th Jan-First Shift)-PCM-12 55.
Two solids dissociate as follows 2 A s B g C g ;Kp1 x atm 2 D s C g E g ;K p2 y atm
The total pressure when both the solids dissociate simultaneously is (A) x y atm (B) 2 x y atm
(C) (x + y) atm 56.
(D) x2 + y2 atm
In the following reactions, products A and B are:
(A)
(B)
(C)
(D)
o
57.
What is the work function of the metal if the light of wavelength 4000 A generates photoelectrons of velocity 6 105 ms–1 from it? (Mass of electron = 9 10–31 kg Velocity of light = 3 108 ms–1 Planck’s constant = 6.626 10–34 Js Charge of electron = 1.6 10–19 JeV–1) (A) 0.9 eV (B) 3.1 eV (C) 2.1 eV (D) 4.0 eV
58.
Iodine reacts with concentrated HNO3 to yield along with other products. The oxidation state of iodine in Y is (A) 5 (B) 7 (C) 3 (D) 1
59.
Poly - -hydroxybutyrate-co- -hydroxyvalerate(PHBV) is a copolymer of ________ (A) 3-hydroxybutanoic acid and 4-hydroxypentanoic acid (B) 2-hydroxybutanoic acid and 3-hydroxypentanoic acid (C) 3-hydroxybutanoic acid and 2-hydroxypentanoic acid (D) 3-hydroxy butanoic acid and 3-hydroxypentanoic acid
60.
Water sample with BOD vales of 4 ppm and 18 ppm respectively are (A) Clean and clean (B) Highly polluted and clean (C) Clean and highly polluted (D) Highly polluted and highly polluted. FIITJEE Ltd., FIITJEE House, 29-A, Kalu Sarai, Sarvapriya Vihar, New Delhi -110016, Ph 46106000, 26569493, Fax 26513942 website: www.fiitjee.com.
JEE-MAIN-2019 (12th Jan-First Shift)-PCM-13
PART–C (MATHEMATICS) 61.
An ordered pair (, ) for which the system of linear equations (1 + )x + y + z = 2 x + (1 + )y + z = 3 x + y + 2z = 2 has a unique solution, is: (A) (2, 4) (B) (3, 1) (C) (4, 2) (D) (1, 3)
62.
The product of three consecutive terms of a G.P. is 512. If 4 is added to each of the first and the second of these terms, the three terms now form an A.P. Then the sum of the original three terms of the given G.P. is: (A) 36 (B) 32 (C) 24 (D) 28
63.
The Boolean expression ((p q) (p q)) ( p q) is equivalent to: (A) p q (B) p (q) (C) (p) (q) (D) p (q)
64.
Consider three boxes, each containing 10 balls labelled 1, 2, ….., 10. Suppose one ball is randomly drawn from each of the boxes. Denote by ni, the label of the ball drawn from the ith box, (i = 1, 2, 3). Then, the number of ways in which the balls can be chosen such that n1 < n2 < n3 is: (A) 120 (B) 82 (C) 240 (D) 164
65.
A tetrahedron has vertices P(1, 2, 1), Q(2, 1, 3), R(1, 1, 2) and O(0, 0, 0). The angle between the faces OPQ and PQR is: 17 19 (A) cos1 (B) cos1 31 35 9 7 (C) cos1 (D) cos1 35 31
66.
The maximum area (in sq. units) of a rectangle having its base on the x-axis and its other two vertices on the parabola, y = 12 – x2 such that the rectangle lies inside the parabola, is: (A) 36 (B) 20 2 (C) 32 (D) 18 3
67.
If the straight line, 2x – 3y + 17 = 0 is perpendicular to the line passing through the points (7, 17) and (15, ), then equals: 35 (A) (B) 5 3 35 (C) (D) 5 3
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JEE-MAIN-2019 (12th Jan-First Shift)-PCM-14 68.
The sum of the distinct real values of ˆ ˆi ˆj k, ˆ ˆi ˆj kˆ are co-planar, is: ˆi ˆj k, (A) 1 (B) 0 (C) 1 (D) 2
69.
Let P(4, 4) and Q(9, 6) be two points on the parabola, y2 = 4x and let X be any point on the arc POQ of this parabola, where O is the vertex of this parabola, such that the area of PXQ is maximum. Then this maximum Area (in sq. units) is: 75 125 (A) (B) 2 4 625 125 (C) (D) 4 2 1 0 0 Let P 3 1 0 and Q = [qij] be two 3 × 3 matrices such that Q – P5 = I3. Then 9 3 1
70.
q21 q31 is equal to: q32 (A) 10 (C) 15
71.
,
for
which
the
vectors,
(B) 135 (D) 9
Let y = y(x) be the solution of the differential equation, x = loge 4 – 1, then y(e) is equal to: e (A) 2 e (C) 4
(B) (D)
dy y x loge x,(x 1). If 2y(2) dx
e2 2
e2 4
72.
The area (in sq. units) of the region bounded by the parabola, y = x2 + 2 and the lines, y = x + 1, x = 0 and x = 3, is: 15 21 (A) (B) 4 2 17 15 (C) (D) 4 2
73.
In a random experiment, a fair die is rolled until two fours are obtained in succession. The probability that the experiment will end in the fifth throw of the die is equal to: 200 150 (A) 5 (B) 5 6 6 225 175 (C) 5 (D) 5 6 6
74.
Let C1 and C2 be the centres of the circles x2 + y2 – 2x – 2y – 2 = 0 and x2 + y2–6x– 6y+14 =0 respectively. If P and Q are the points of intersection of these circles, then the area (in sq. units) of the quadrilateral PC1QC2 is: (A) 8 (B) 6 (C) 9 (D) 4 FIITJEE Ltd., FIITJEE House, 29-A, Kalu Sarai, Sarvapriya Vihar, New Delhi -110016, Ph 46106000, 26569493, Fax 26513942 website: www.fiitjee.com.
JEE-MAIN-2019 (12th Jan-First Shift)-PCM-15
75.
The maximum value of 3 cos 5 sin for any real value of is: 6 79 (A) 19 (B) 2 (C) 34 (D) 31
76.
Considering only the principal values of inverse functions, the set A x 0 : tan 1 2x tan1 3x 4 (A) contains two elements (B) contains more than two elements (C) is a singleton (D) is an empty set
77.
If be the ratio of the roots of the quadratic equation in x, 3m2x2 + m(m – 4)x + 2 = 0, 1 then the least value of m for which 1, is: (A) 2 3 (B) 4 3 2 (C) 2 2
(D) 4 2 3
78.
If a variable line, 3x + 4y – = 0 is such that the two circles x2 + y2 – 2x – 2y + 1 = 0 and x2 + y2 – 18x – 2y + 78 = 0 are on its opposite sides, then the set of all values of is the interval: (A) (2, 17) (B) [13, 23] (C) [12, 21] (D) (23, 31)
79.
For x > 1, if (2x)2y = 4e2x – 2y, then 1 loge 2x x loge 2x loge 2 x x loge 2x loge 2 (C) x
(A)
80.
dy is equal to: dx
(B) loge 2x (D) x loge 2x
The integral cos loge x dx is equal to: (where C is a constant of integration) x sin loge x cos loge x C 2 x (C) cos loge x sin loge x C 2
(A)
81.
2
(B) x cos loge x sin loge x C (D) x cos loge x sin loge x C
A ratio of the 5th term from the beginning to the 5th term from the end in the binomial 1 1 expansion of 2 3 is: 1 3 2 3 1
1
(A) 1: 2 6 3
(B) 1: 4 16 3
1
(C) 4 36 3 : 1
1
(D) 2 36 3 : 1
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JEE-MAIN-2019 (12th Jan-First Shift)-PCM-16 1 2 3 ..... k 5 2 . If S12 S 22 ..... S10 A, then A equal to: k 12 (A) 283 (B) 301 (C) 303 (D) 156
82.
Let Sk
83.
The perpendicular distance from the origin to the plane containing the two lines, x2 y2 z5 x 1 y 4 z 4 and , is: 3 5 7 1 4 7 (A) 11 6 (B) 11 6 (D) 6 11
(C) 11 84.
If the sum of the deviations of 50 observations from 30 is 50, then the mean of these observations is: (A) 30 (B) 51 (C) 50 (D) 31
85.
Let S be the set of all points in (, ) at which the function, f(x) = min {sinx, cosx} is non-differentiable. Then S is a subset of which of the following? 3 3 (A) ,0, (B) , , , 4 4 4 4 4 4 3 3 (C) , , , (D) , , , 2 2 4 2 4 4 2 4
86.
Let f and g be continuous functions on [0, a] such that f(x) = f(a – x) and g(x) + g(a – x) = a
4, then
f(x)g(x)dx is equal to: 0
a
a
(A) 4 f(x)dx
(B)
(C) 2 f(x)dx
(D) 3 f(x)dx
0 a
0
a
0
87.
0
cot3 x tan x is: x 4 cos x 4 lim
(A) 4 (C) 8 2 88.
f(x)dx
(B) 4 2 (D) 8
z ( R) is a purely imaginary number and |z| = 2, then a value of is: z (A) 2 (B) 1 1 (C) (D) 2 2
If
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JEE-MAIN-2019 (12th Jan-First Shift)-PCM-17 89.
Let S = {1, 2, 3, ….., 100}. The number of non-empty subsets A of S such that the product of elements in A is even is: (A) 2100 – 1 (B) 250 (250 – 1) (C) 250 – 1 (D) 250 + 1
90.
If the vertices of a hyperbola be at (2, 0) and (2, 0) and one of its foci be at (3, 0), then which one of the following points does not lie on this hyperbola? (A) 6,2 10 (B) 2 6,5
(C) 4,
15
(D) 6,5 2
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JEE-MAIN-2019 (12th Jan-First Shift)-PCM-18
JEE (Main) – 2019 ANSWERS PART A – PHYSICS 1.
Insufficient information
2.
C
3.
C
4.
C
5.
A
6.
A
7.
D
8.
D
9.
C
10.
C
11.
B
12.
D
13.
A
14.
D
15.
C
16.
C
17.
C
18.
D
19.
B
20.
C
21.
C
22.
A
23.
A
24.
B
25.
D
26.
B
27.
B
28.
B
29.
C
30.
C
PART B – CHEMISTRY 31.
B
32.
D
33.
A
34.
A
35.
A
36.
C
37.
B
38.
D
39.
A
40.
A
41.
B
42.
A
43.
D
44.
Bonus
45.
B
46.
D
47.
A
48.
A
49.
B
50.
A
51.
A
52.
A
53.
B
54.
B
55.
C
56.
A
57.
C
58.
A
59.
D
60.
C
PART C – MATHEMATICS 61.
A
62.
D
63.
C
64.
A
65.
B
66.
C
67.
D
68.
A
69.
B
70.
A
71.
C
72.
D
73.
D
74.
D
75.
A
76.
C
77.
B
78.
C
79.
A
80.
C
81.
C
82.
C
83.
B
84.
D
85.
B
86.
C
87.
D
88.
A
89.
B
90.
D
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JEE-MAIN-2019 (12th Jan-First Shift)-PCM-19
HINTS AND SOLUTIONS PART A – PHYSICS 1.
Cannot determine, degree of freedom must be given.
2.
y = 10–3 sin (50t + 2x) Wave is travelling along negative x-axis 50 Wave speed = = 25 m/s. k 2
3. 4V
i
R
4 5R
VAB = i(5) = i
VAB 20 0.1 2 (0.1) L 5 R 1 5R 2 Now, = 5 × 10–3 5R R = 395 VAP
P
0.1 V
A
4.
20 5R
5 1m
dR k d
B
k = constant
d R = 2k resistance of wire AB. R/ 2 L d Again, dR k L Length AP 0 0 R k2 L1/2 ; k = k2 L1/2 2 1 L m = 0.25 m 4
R
0
5.
dR k
1
0
0.06 km
PASSENGER TRAIN 30 km/hr
80 km/hr 0.12 km
FREIGHT TRAIN
Time taken if both moving in same direction t1
distance 0.12 0.06 0.18 speed 80 30 50
Time taken if moving in opposite direction. distance 0.12 0.06 0.18 t2 = speed 80 30 110 t1 11 t2 5 FIITJEE Ltd., FIITJEE House, 29-A, Kalu Sarai, Sarvapriya Vihar, New Delhi -110016, Ph 46106000, 26569493, Fax 26513942 website: www.fiitjee.com.
JEE-MAIN-2019 (12th Jan-First Shift)-PCM-20
6.
V2 P (220)2 R1 1936 25 (220)2 R2 484 100 220 1 i A R1 R 2 11 Power dissipated through R1 = P1 = i2R1 = 16 W Power dissipated through R2 = P2 = i2R2 = 4 W
Resistance, R
7.
x=a x=0
x
R1
R2
i 220 V
L
dx
Mass of element = dm = (A + Bx2)dx Field due to element at x = 0 G(dm) GA dE 2 GB dx 2 x x Total field a L 1 aL E GA dx GB dx 2 a a x 1 1 = G A BL a aL So, force = mE 1 1 = Gm A BL a aL 8.
Image by convex lens:
1 1 1 v u f
;
1 1 1 v 20 5
20 cm 3 Image by concave lens: v
20 14 u 2 cm 3 3
1 1 1 ; v u f v = 70 cm
1 3 1 v 14 5
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JEE-MAIN-2019 (12th Jan-First Shift)-PCM-21 9.
h = d
d
d
L
S
d/2 L
d d d = 3d. 2 2
h
d/2 d
10.
1 0E02C 2 1 0.96 I = E02 V 2 I
…(i) …(ii) 2
E v 0.96 0 E0 0 C 2
E 1 0.96 0 E0 0 1.5 2
E 1 0.96 0 r …(iii) 1.5 E0 1 C and v ; v o o r r r r
C ; r 1.5 v From equation (iii) r r
2
E 0.96 o (1.5)2 Eo Eo 24 V /m 11.
; r 1 for transparent medium.
1 1.5
Work done = Area of loop 1 = (4) (5) = 10 J 2
12.
P qL Pnet 2P cos 30o ( ˆj) = P 3 ( ˆj)
Y
X
P
30° 30°
=
3 qL ( ˆj)
P
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JEE-MAIN-2019 (12th Jan-First Shift)-PCM-22 13.
Three parts of rod can be considered as point masses. Y
2m (L. L)
m
L 2L, 2
m 5L
2m r1 m r2 m r3 rcm 4m rcm
X
2 , 0
ˆ m 2L ˆi L ˆj m 5L ˆi 2m(L ˆi Lj) 2 2 4m
13 ˆ 5 ˆ L i Lj 2 = 2 4 13 ˆ 5 ˆ = Li Lj 8 8
14.
B
A 60°
60°
15.
kx
kx
VP Speed of plane V Speed of sound V cos 60° = VP V VP 2
C
Torque on rod at displacement from mean position is very small. L x 2 L L2 kL2 2kx 2k 2 4 2 Now, I
6k kL2 mL2 ; 2 12 m 1 6k 2 2 m
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JEE-MAIN-2019 (12th Jan-First Shift)-PCM-23 16.
Just before collision speed of m v 2gL(1 cos o )
Just after collision speed of M v1 2gL(1 cos 1 )
Mm And v1 v ; M m
v1 M m v Mm
1 cos 1 M m 1 cos o M m
Sin ( 1 / 2) M m Sin (0 / 2) M m 1 M m o M m M1 + m1 = Mo – mo o M m 1 o 1 17.
AC = 100 AC + Am = 160 AC – Am = 40 AC = 100, Am = 60 A m 0.6 AC
18.
Equivalent thermal resistance,
1 cos 2 2 sin2
1 1 1 R R1 R2 2 2 k(2R)2 k1R2 k 2 (2R) R L L L
19.
4k = k1 + 3k2 k 3k 2 k 1 4
Least count = 5 10 6
Pitch No. of divisions on circular scale
103 N
N = 200
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JEE-MAIN-2019 (12th Jan-First Shift)-PCM-24
20.
Y A.AB AB.B
A. A B AB .B AB 0
21.
i B 2 o (cos 1 cos 2 ) 4nd B = 10–4
1 = 90°
o = 4 × 10 d = 4 × 10–2
22.
U
–7
2 = 180°
i = 20 A (into the page)
1 2 kr 2
Force, F
dU kr dr
For circular motion
mv 2 kr r
…(i)
nh 2 nh r2 2 km mvr
And
…(ii)
r n Total energy, E = k + U 1 1 = mv 2 kr 2 2 2
1 2 1 2 kr kr 2 2 2 E = kr En =
23.
mP = m qp = q m = 4 m q = 2q Radius of circular path, r
[From equation (i)]
kp = qV = k k = 2qV = 2k
mv 2km qB qB
rP 1 r 2
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JEE-MAIN-2019 (12th Jan-First Shift)-PCM-25 24.
r = 10 cm R = 20 cm
R
mass per unit area m (R 2 r 2 )
Mass = m r
Consider an element of radius x and thickness dx
x
r R
Mass of element dm = 2 x(dx) Moment of inertia of element, dI = (dm)x2
R
I = 2 x3 dx r
2 4 4 (R r ) 4 m = (R4 – r4) 2 2 (R r ) 2 m 2 2 I = (R r ) …(i) 2 Moment of inertia of thin cylinder of same mass, I = mro2 …(ii) =
25.
m 2 2 (R r ) 2 ro2 = 250 ; ro 16 cm. mro2
After long time, inductor will behave like resistance less wire, 15 15 i Req (5 / 2)
i 5 15 V
=6A
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5
JEE-MAIN-2019 (12th Jan-First Shift)-PCM-26 26.
Before switch is shifted: Energy stored, 1 Ui CE2 2 and charge stored, Q = CE
E
C
After switch is shifted: N C
4C
3C
M
Q CE E 4C 4C 4 2 1 1 E Energy stored, Uf (4C) CE 2 2 8 4 3 Energy dissipated = Ui – Uf = CE2 8 VM VN
27.
v
vy v
m
m
After Collision
vx
2m
2mvx = mv
v 2 v vy 2 vx
v 2 So, path will be elliptical. v net v x 2 v y 2
28.
When K1 closed and K2 is open
i
V 220 R
220
R
i
G
…(i)
When k1 and k2 are closed. KVL i iR i 25 5 220 5 R V i iR i 25 5 220 5 R i(220 R) R = 22
V
5 220
iR 25
iR i 25 5
G i/5
From eq. (i)
R
V
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JEE-MAIN-2019 (12th Jan-First Shift)-PCM-27
29.
P
30.
h 2mqV 4 1 2500 10 2 14.14 1 1 50
Ui + Ki = Uf + Kf kQ 2 kQ 2 1 O mv 2 2R0 2R 2 V
kQ2 m
1 1 R0 R
PART B – CHEMISTRY 31.
Cathode is made up of carbon.
32.
Acidic strength Stability of conjugate base E.N. sp carbon > sp2 carbon > sp3 carbon
33.
A 2 B 2C D Initially conc. a 1.5a at eq. a–x 1.5(a–2x) 2x x at equilibrium a–x = 1.5a – 2x 0.5a = x a = 2x 2 2x x 4x 2 .x kC 4 2 2 a x 1.5a 2x x x
34.
Amount of gas adsorbed TC
35.
Organometalic compound contains at least one chemical bond between carbon and a metal.
36.
CN
CHO O i DBAL H E Cb
OH
1 O ii H3 O
CHO
Nitriles are selectively reduced by DIBAL–H to imines followed by hydrolysis to aldehydes similarly, esters are also reduced to aldehyde with DIBAL–H. 37.
Rb form super oxides on reaction with excess air Rb + O2 RbO2 2RbO2 + H2O 2RbOH + H2O2 + O2
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JEE-MAIN-2019 (12th Jan-First Shift)-PCM-28 38.
Photochemical smog is produced when ultraviolet light from sun reacts with oxides of nitrogen in atmosphere.
39.
H2O is weak field ligand and magnetic moment = Moment n n 2 BM Solving for ‘n’ we get it as 3; i.e., no. of unpaired electron = 3 Fe2+ = t2g4eg2 Co2+ = t2g5eg2 V2+ = t2g3eg0 M is V, Co
40.
CP does not changes with change in pressure
41.
Compound Histidine Serine Lysine Asparagine
42.
Lone pair over nitrogen will act as the reacting centre. So, more nucleophilic nitrogen will be more reactive towards alkyl halide.
43.
Stronger intermolecular forces result in higher melting point.
44.
Eq. of (COOH)2 = Eq. of NaOH 50 0.5 2 = 25 M 1 50 2 Mass of NaOH in 50 mL = 40 4 g 1000
45.
Z = PV/nRT ZnRT P V
Pl value 7.6 5.7 9.8 5.4
Z V PA Z A VB 3 1 3 PB ZB VA 1 2 2 2PA = 3PB at constant T and mol P
46.
As 1L solution have 10–3 mol CaSO4 Eq. of CaSO4 = eq. of CaCO3 In 1L solution nCaSO4 v.f. nCaCO3 v.f. 10 3 2 nCaCO3 2 nCaCO3 10 3 mol in 1 L
w CaCO3 100 10 3 g in 1 L solution hardness in terms of CaCO3 w CaCO3 100 10 3 g = 106 106 = 100 ppm w Total 1000g FIITJEE Ltd., FIITJEE House, 29-A, Kalu Sarai, Sarvapriya Vihar, New Delhi -110016, Ph 46106000, 26569493, Fax 26513942 website: www.fiitjee.com.
JEE-MAIN-2019 (12th Jan-First Shift)-PCM-29 47.
The reaction in option (A) will yield 1o-alcohol.
48.
Tf X Tf Y k f m X k tm Y 4 1000 12 1000 A 96 M 88 M = 3.27 A = 3A
49.
Given K3[Co(CN)6] is inner orbital complex with hybridization d2sp3 and octahedral geometry. Ligands are approaching metal along the axes. Hence, dx2 y2 , dz2 orbitals are directly in front of the ligands.
50.
According to unit of rate constant it is a zero order reaction then half life of reaction will be C 5g t1/2 0 = 50 years 2k 2 0.05g / year
51.
G nFEcell 2 96500 2 386 kJ dE 4 o S nF 2 96500 5 10 J / C 96.5 kJ dT at 298 K TS = 298 (–96.5 J) = – 28.8 kJ at constant T (=248 K) and pressure G = H – TS H = G + TS = –386 – 28.8 = -412.8 kJ
Cl
52. H3 CO
H3 CO Cl / CCl
2 4
Cl Cl Anhyd. AlCl
3
H3 CO
H3 CO
Cl
53.
[Og118] 8s2 is configuration for Z = 120, where Og is oganesson As per the configuration it is in IInd group.
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JEE-MAIN-2019 (12th Jan-First Shift)-PCM-30 1 Steric Crowding
54.
Rate
55.
A s B g C g P1
P1 P2
D s C g E g P1 P2
P2
k P1 x atm2 k P2 y atm2
k P1 P1 P1 P2 k P2 P2 P1 P2
kP1 kP2 P1 P2
2
x + y (P1 + P2)2 P1 P2 x y 2 P1 P2 x y
PTotal PB PC PE = 2 P1 P2 2 x y 56.
57.
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JEE-MAIN-2019 (12th Jan-First Shift)-PCM-31 58.
2 + 10HNO3 2 HO3 + 10NO2 + 4H2O Iodine in HIO3 has +5 oxidation state.
59.
PHBV is obtained by copolymerization of 3-Hydroxybutanioic acid & 3-Hydroxypentanoic acid.
60.
Clean water has BOD value of less than 5 ppm and highly polluted water has a BOD value of 17 ppm or more.
PART C – MATHEMATICS 61.
1 x y z 0 x 1 y z 0 x y 2z 0 1 1 D 1 1 2
C1 C1 C2 C3 1
1
D 2 1 1 1 1
2
R2 R 2 R1, R 3 R3 R1 1 1 D 2 0 1 0 2 0 0 1
62.
63.
For unique solution 2 0 2 a Let the numbers be ,a,ar r 3 Given a 512 a 8 8 Now given 4, 12,8r are in A.P. r 2 2r 5r 2 0 1 r or 2 2 Numbers are 4, 8, 16 or 16, 8, 4 Sum of numbers = 4 + 8 + 16 = 28
p q p p q ~ q ~ p q p p ~ q q ~ q ~ p q p p ~ q ~ p q p ~ q ~ p q ~ p ~ q
64.
10
C3 120
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JEE-MAIN-2019 (12th Jan-First Shift)-PCM-32 65.
Vector perpendicular to face ˆi ˆj kˆ
O
OAB 1 2 1 5iˆ ˆj 3kˆ 2 1 3
Vector perpendicular to face ˆi ˆj kˆ ABC 2
1
1 1
1 ˆi 5ˆj 3kˆ 2
Angle between two faces 5 5 9 19 cos 35 35 35
A (1, 2, 1)
B (2, 1, 3)
19 cos1 35 66.
A 2 12 2
C (–1, 1, 2)
(0, 12)
dA 0 2 12 3 2 0 2 d A 4 12 4 32
,12
,12
2
2
,0 67.
,0
Line perpendicular to 2x 3y 5 0 is 3x 2y c 0 Which is satisfied by point (7, 17) 3 7 2 17 c 0 c 55 equation of line is 3x 2y 55 0 3 15 2 55 0 2 55 45 5
1 1
68.
1 1 1
D 1 1 , R1 R2 R3 2 1 1 , 1 1
1 1 1
0
0 2
C3 C3 C1 and C2 C2 C1, 2 1 1 0 2 1 0 1 0 1 Hence sum of distinct value of 2 1 1
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JEE-MAIN-2019 (12th Jan-First Shift)-PCM-33
69.
m
PQ
64 2 94
X
1 1 , X , 1 4 4 (Coordinates of points X for maximum area) 125 area sq. units. 4
Q (9, 6)
2yy ' 4 2y 2 4 y 1, x
70.
71.
1 0 0 1 0 0 3 P 6 1 0 P 9 1 0 . P5 27 6 1 54 9 1 0 0 1 0 0 2 1 Q 15 1 0 0 1 0 15 135 15 1 0 0 1 135 q q31 15 135 10 So 21 q32 15 2
P (4, –4)
0 0 1 15 1 0 135 15 1 0 0 2 0 15 2
dy 1 y log x dx x dx
I.F. e x x yx xn x dx x2 x dx 2 2 2 2 x x xy n C 2 4 Putting x 2 n 4 1 1 c n2 2 2 2 c 0 x x y n x 2 4 e e e y e 2 4 4 xy nx
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JEE-MAIN-2019 (12th Jan-First Shift)-PCM-34
72.
Required area 3
x
2
0
3
2 x 1 dx x 2 x 1 dx 0
15 2
(0, 11) y x2 2
(0, 2) (0, 1) (0, 0)
73.
(3, 0)
P _ _ _ 44 P 4 _ _ 44 P not4 _ _ 44
74.
1 5 5 1 1 5 5 1 1 25 25 175 1 5 4 5 6 6 6 6 6 6 6 6 6 6 6 6
Since circles are orthogonal and have equal radii therefore the quadrilateral PC1QC 2 is a square. Hence area 2 2 4 sq.units
P
2
2 C2
C1
2
2 Q
75.
5 sin 3cos 6 5 sin cos cos sin 3 cos 6 6
5 3 1 sin cos 2 2 2
76.
5 3 1 2 76 Maximum value is 19 2 2 4 tan1 2x tan1 3x 4 2x 3x Taking tangent on both side, we get 1 1 6x 2 6x 2 5x 1 0 x 1 6x 1 0
1 {–1 is rejected as it does not satisfies the given equation} 6 Hence number of element in S is one. x
77.
Let roots are and now 1 1 1 2 2
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JEE-MAIN-2019 (12th Jan-First Shift)-PCM-35 2
2 2 m m 4 3.
2 2 2 3m 3m m2 8m 2 0 m 43 2 So least value of m 4 3 2
78.
3x 4y 0
7 31 0 (since centres lie opposite side) 7,31 ……………(1) 7 31 1 and 5 5 7 5 and 31 2 or 12 and 21 or 41 1 2 3
2 10 …………….(2) …………….(3)
12, 21
79.
2yn 2x n4 2x 2y 2y 1 n2x n 4 2x n 2 x . 1 n 2x 2 y 2 1 n 2x n 2x
n 2x
80.
n2 xn 2x n 2 x x
By parts
x I x cos log x sin log x dx x I x cos log x sin log x dx
I x cos log x x sin log x cos log x dx c I
x cos log x sin log x c 2
1 C4 2 31/3 5 th term frombegining 5 th term from end 4/3 10 C4 2 2
4
10
81.
26/3 6 1 31/3
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JEE-MAIN-2019 (12th Jan-First Shift)-PCM-36
82.
22 223 4/ 3 4
4/3 6
2 2
3
1/3
32/3 .28/3 4. 36
2
k 1 2 2 10 5 k 1 A 12 k 1 2
Sk
5A 3 11 12 23 5A 1 6 3 3 505 A 5 A = 303 22 32 .......112
x2 y2 z5
83.
Equation of plane is
3
5
7
1
4
7
0
x 2 7 y 2 14 z 5 7 0 x 2y z 11 0
The perpendicular distance from the origin to the plane is
0 0 0 11 1 4 1
11 6
50
84.
Given
x
i
30 50
i 1
x i 30 50 50
85.
x
i
50
31
Hence number of points where f(x) is non – differentiable are 2 3 which are and 4 4
3 4 4
a
86.
a
I f x g x dx f a x g a x dx 0
b
0
b
f x dx f a b x dx a a a
a
a
f x 4 g x dx I 4 f x dx I I 2 f x dx 0
0
0
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JEE-MAIN-2019 (12th Jan-First Shift)-PCM-37 87.
Using LH rule 3cot 2 x cos ec 2 x sec 2 x im 8 x 4 sin x 4 z z z z
88.
zz z z 2 zz z z 2
2 z 2
2
89.
Product is even when atleast one element of subset is even Hence required number of subsets = total subsets – number of subsets all whose elements are odd 2100 250
90.
Equation of hyperbola is
x2 y2 1 and ae = 3 4 b2 We know that a 2e 2 a2 b 2 9 4 b2 b 2 5 x2 y2 Hence equation of hyperbola is 1 4 5 x2 y2 Hence 6, 5 2 does not lie on 1 4 5
(-3, 0)
(-2, 0)
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FIITJEE Solutions to JEE(Main)-2019 Test Date: 12th January 2019 (Second Shift)
PHYSICS, CHEMISTRY & MATHEMATICS Paper - 1 Time Allotted: 3 Hours
Maximum Marks: 360
Please read the instructions carefully. You are allotted 5 minutes specifically for this purpose.
Important Instructions: 1.
The test is of 3 hours duration.
2.
This Test Paper consists of 90 questions. The maximum marks are 360.
3.
There are three parts in the question paper A, B, C consisting of Physics, Chemistry and Mathematics having 30 questions in each part of equal weightage. Each question is allotted 4 (four) marks for correct response.
4.
Out of the four options given for each question, only one option is the correct answer.
5.
For each incorrect response 1 mark i.e. ¼ (one-fourth) marks of the total marks allotted to the question will be deducted from the total score. No deduction from the total score, however, will be made if no response is indicated for an item in the Answer Box.
6.
Candidates will be awarded marks as stated above in instruction No.3 for correct response of each question. One mark will be deducted for indicating incorrect response of each question. No deduction from the total score will be made if no response is indicated for an item in the answer box.
7.
There is only one correct response for each question. Marked up more than one response in any question will be treated as wrong response and marked up for wrong response will be deducted accordingly as per instruction 6 above.
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JEE-MAIN-2019 (12th Jan-Second Shift)-PCM-2
PART –A (PHYSICS) 1.
2.
The moment of inertia of a solid sphere, about an axis parallel to its diameter and at a distance of x from it, is ‘I(x)’. Which one of the graphs represents the variation of I(x) with x correctly?
(A)
(B)
(C)
(D)
A load of mass M kg is suspended from a steel wire of length 2 m and radius 1.0 mm in Searle’s apparatus experiment. The increase in length produced in the wire is 4.0 mm. Now the load is fully immersed in a liquid of relative density 2. The relative density of the material of load is 8. The new value of increase in length of the steel wire is: (A) 3.0 mm (B) 4.0 mm (C) 5.0 mm (D) zero
3.
3 3 F,R 20 ,L H and R1 10 . Current in L-R1 path is 2 10 1 and C-R2 path is 2. The voltage of A.C source is given by, V 200 2 sin(100 t) volts. The phase difference between 1 and 2 is : (A) 60° (B) 30° (C) 90° (D) 0°
In the above circuit, C
4.
An ideal gas is enclosed in a cylinder at pressure of 2 atm and temperature, 300 K. The mean time between two successive collisions is 6 10–8 s. If the pressure is doubled and temperature is increased to 500 K, the mean time between two successive collisions will be close to : (A) 2 107 s (B) 4 10 8 s (C) 0.5 10 8 s (D) 3 106 s
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JEE-MAIN-2019 (12th Jan-Second Shift)-PCM-3 5.
In the figure, given that VBB supply can vary from 0 to 5.0 V, VCC = 5 V, dc = 200, RB = 100, k, RC = 1 k an VBE = 1.0 V, The minimum base current and the input voltage at which the transistor will go to saturation, will be respectively : (A) 25 A and 3.5 V (B) 20 A and 3.5 V (C) 25 A and 2.8 V (D) 20 A and 2.8 V
6.
In the circuit shown, find C if the effective capacitance of the whole circuit is to be 0.5 F. All values in the circuit are in F. 7 6 (A) F (B) F 11 5 7 (C) 4 F (D) F 10
7.
An alpha-particle of mass m suffers 1-deminsional elastic collision with a nucleus at rest of unknown mass. It is scattered directly backwards losing, 64% of its initial kinetic energy. The mass of the nucleus is : (A) 2 m (B) 3.5 m (C) 1.5 m (D) 4 m
8.
A 10 m long horizontal wire extends from North East to South West. It is falling with a speed of 5.0 ms–1, at right angles to the horizontal component of the earth’s magnetic field, of 0.3 10–4 Wb/m2. The value of the induced emf in wire is : (A) 1.5 10 3 V (B) 1.1 10 3 V (C) 2.5 10 3 V (D) 0.3 10 3 V
9.
To double the covering range of a TV transition tower, its height should be multiplied by : 1 (A) (B) 2 2 (C) 4 (D) 2
10.
A plano-convex lens (focal length f 2, refractive index 2, radius of curvature R) fits exactly into a plano-concave lens (focal length f 1, refractive index 1, radius of curvature R). Their plane surfaces are parallel to each other. Then, the focal length of the combination will be : R (A) f1 f2 (B) 2 1 2f1f2 (C) (D) f1 f2 f1 f2
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JEE-MAIN-2019 (12th Jan-Second Shift)-PCM-4 11.
A vertical closed cylinder is separated into two parts by a frictionless piston of mass m and of negligible thickness. The piston is free to move along the length of the cylinder .The length of the cylinder above the piston is l1, and that below the piston is l2, such that l1 > l2. Each part of the cylinder contains n moles of an ideal gas at equal temperature T. If the piston is stationary, its mass, m, will be given by: (R is universal gas constant and g is the acceleration due to gravity) RT l1 3l2 RT 2l1 l2 (A) (B) ng l1l2 g l1l2 nRT 1 1 nRT l1 l2 (C) (D) ng l2 l1 g l1l2
12.
Two satellites, A and B, have masses m and 2m respectively. A is in a circular orbit of radius R, and B is in a circular orbit of radius 2R around the earth. The ratio of their kinetic energies, TA/TB, is : 1 (A) (B) 1 2 1 (C) 2 (D) 2
13.
A long cylindrical vessel is half filled with a liquid. When the vessel is rotated about its own vertical axis, the liquid rises up near the wall. If the radius of vessel is 5 cm and its rotational speed is 2 rotations per second, then the difference in the heights between the centre and the sides, in cm, will be : (A) 2.0 (B) 0.1 (C) 0.4 (D) 1.2
14.
A block kept on a rough inclined plane, as shown in the figure, remains at rest upto a maximum force 2 N down the inclined plane. The maximum external force up the inclined plane that does not move the block is 10 N. The coefficient of static friction between the block and the plane is : [Take g = 10 m/s2] 3 3 (A) (B) 2 4 1 2 (C) (D) 2 3
15.
In a Frank-hertz experiment, an electron of energy 5.6 eV passes through mercury vapour and emerges with an energy 0.7 eV. The minimum wavelength of photons emitted by mercury atoms is close to : (A) 1700 nm (B) 2020 nm (C) 220 nm (D) 250 nm
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JEE-MAIN-2019 (12th Jan-Second Shift)-PCM-5 16.
A particle of mass 20 g is released with an initial velocity 5 m/s along the curve from the point A, as shown in the figure. The point A is at height h from point B. The particle slides along the frictionless surface. When the particle reaches point B, its angular momentum about O will be : [Take g = 10 m/s2] (A) 2 kg-m2/s (B) 8 kg-m2/s 2 (C) 6 kg-m /s (D) 3 kg-m2/s
17.
A galvanometer, whose resistance is 50 ohm, has 25 divisions in it. When a current of 4 10–4 A passes through it, its needle (pointer) deflects by one division. To use this galvanometer as a voltmeter of range 2.5 V, it should be connected to a resistance of : (A) 250 ohm (B) 200 ohm (C) 6200 ohm (D) 6250 ohm
18.
A soap bubble, blown by a mechanical pump at the mouth of a tube, increases in volume, with time, at a constant rate. The graph that correctly depicts the time dependence of pressure inside the bubble is given by :
19.
(A)
(B)
(C)
(D)
In the given circuit diagram, the currents, 1 = –0.3 A, 4 = 0.8 A and 5 = 0.4 A, are flowing as shown. The currents 2, 3 and 6, respectively are :
(A) 1.1 A, –0.4 A, 0.4 A (C) 0.4 A, 1.1 A, 0.4 A
(B) 1.1 A, 0.4 A, 0.4 A (D) –0.4 A, 0.4 A, 1.1 A
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JEE-MAIN-2019 (12th Jan-Second Shift)-PCM-6 20.
A resonance tube is old and has jagged end. It is still used in the laboratory to determine velocity of sound in air. A tuning fork of frequency 512 Hz produces first resonance when the tube is filled with water to a mark 11 cm below a reference mark, near the open end of the tube. The experiment is repeated with another fork of frequency 256 Hz which produces first resonance when water reaches a mark 27 cm below the reference mark. The velocity of sound in air, obtained in the experiment, is close to : (A) 322 ms–1 (B) 341 ms–1 –1 (C) 335 ms (D) 328 ms–1
21.
A paramagnetic material has 1028 atoms/m3. Its magnetic susceptibility at temperature 350 K is 2.8 10–4. Its susceptibility at 300 K is : (A) 3.267 104 (B) 3.672 10 4 (C) 3.726 104 (D) 2.672 104
22.
In a radioactive decay chain, the initial nucleus is
232 90
Th . At the end there are 6 -
particles and 4 -particles with are emitted. If the end nucleus is by : (A) A = 208; Z = 80 (B) A = 202; Z = 80 (C) A = 208; Z = 82 (D) A = 200; Z = 81 23.
X, A and Z are given
Let l, r, c and v represent inductance, resistance, capacitance and voltage, respectively. 1 The dimension of in S units will be : rcv (A) [LA 2 ] (B) [A 1 ] (C) [LTA]
24.
A Z
(D) [LT 2 ]
Formation of real image using a biconvex lens is shown below :
If the whole set up is immersed in water without disturbing the object and the screen positions, what will one observe on the screen? (A) Image disappears (B) Magnified image (C) Erect real image (D) No change 25.
When a certain photosensitive surface is illuminated with monochromatic light of frequency v, the stopping potential for the photo current is –V0/2. When the surface is illuminated by monochromatic light of frequency v/2, the stopping potential is –V0. The threshold frequency for photoelectric emission is : 5v 4 (A) (B) v 3 3 3v (C) 2 v (D) 2
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JEE-MAIN-2019 (12th Jan-Second Shift)-PCM-7 26.
A simple harmonic motion is represented by : y 5(sin 3 t 3 cos 3 t) cm The amplitude and time period of the motion are : 2 3 (A) 10 cm, s (B) 10 cm, s 3 2 3 2 (C) 5 cm, s (D) 5 cm, s 2 3
27.
Two particles A, B are moving on two concentric circles of radii R1 and R2 with equal angular speed . At t = 0, their positions and direction of motion are shown in the figure :
The relative velocity v A vB at t is given by : 2 (A) (R1 R2 ) ˆi (B) (R1 R2 ) ˆi (C) (R R ) ˆi (D) (R R ) ˆi 2
1
1
2
28.
The mean intensity of radiation on the surface of the Sun is about 108 W/m2. The rms value of the corresponding magnetic field is closet to : (A) 1 T (B) 102 T –2 (C) 10 T (D) 10–4 T
29.
A parallel plate capacitor with plates of area 1 m2 each, are at a separation of 0.1 m. If the electric field between the plates is 100 N/C, the magnitude of charge on each plate is: (A) 7.85 10 10 C (B) 6.85 10 10 C (C) 8.85 10 10 C (D) 9.85 10 10 C
30.
The charge on a capacitor plate in a circuit, as a function of time, is shown in the figure :
What is the value of current at t = 4 s? (A) zero (C) 2 A
(B) 3 A (D) 1.5 A
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JEE-MAIN-2019 (12th Jan-Second Shift)-PCM-8
PART –B (CHEMISTRY) 31.
8 g of NaOH is dissolved in 18 g of H2O. Mole fraction of NaOH in solution and molality (in mol kg–1) of the solution respectively are : (A) 0.2, 22.20 (B) 0.2, 11.11 (C) 0.167, 11.11 (D) 0.167, 22.20
32.
The magnetic moment of an octahedral homoleptic Mn() complex is 5.9 BM. The suitable ligand for this complex is : (A) ethylenediamine (B) CN (C) NCS (D) CO
33.
The element that does NOT show catenation is : (A) Ge (B) Si (C) Sn (D) Pb
34.
Among the following, the false statement is : (A) It is possible to cause artificial rain by throwing electrified sand carrying charge opposite to the one on clouds from an aeroplane. (B) Tyndall effect can be used to distinguish between a colloidal solution and a true solution. (C) Lyophilic sol can be coagulated by adding an electrolyte. (D) Latex is a colloidal solution of rubber particles which are positively charged.
35.
The correct structure of histidine in a strongly acidic solution (pH = 2) is :
36.
(A)
(B)
(C)
(D)
The major product of the following reaction is:
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JEE-MAIN-2019 (12th Jan-Second Shift)-PCM-9
37.
(A)
(B)
(C)
(D)
m for NaCl, HCl and NaA are 126.4, 425.9 and 100.5 S cm2mol–1, respectively. If the conductivity of 0.001 M HA is 5 105 Scm1, degree of dissociation of HA is : (A) 0.50 (B) 0.25 (C) 0.125 (D) 0.75
38.
Molecules of benzoic acid (C6H5COOH) dimerise in benzene. ‘w’ g of the acid dissolved in 30 g of benzene shows a depression in freezing point equal to 2 K. If the percentage association of the acid to form dimer in the solution is 80, then w is : (Given that Kf = 5 K kg mol–1, Molar mass of benzoic acid = 122 g mol–1) (A) 2.4 g (B) 1.0 g (C) 1.5 g (D) 1.8 g
39.
Chlorine on reaction with hot and concentrated sodium hydroxide gives : (A) Cl and ClO3 (B) Cl and ClO (C) ClO 3 and ClO 2
40.
(D) Cl and ClO2
The major product of the following reaction is :
(A)
(B)
(C)
(D)
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JEE-MAIN-2019 (12th Jan-Second Shift)-PCM-10
41.
The combination of plots which do not represent isothermal expansion of an ideal gas is:
(a)
(c) (A) (b) and (d) (C) (b) and (c) 42.
43.
(b)
(d) (B) (a) and (c) (D) (a) and (d)
The major product in the following conversion is :
(A)
(B)
(C)
(D)
The major product of the following reaction is :
(A)
(B)
(C)
(D)
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JEE-MAIN-2019 (12th Jan-Second Shift)-PCM-11 44.
The correct order of atomic radii is : (A) Nd > Ce > Eu > Ho (C) Ce > Eu > Ho > Nd
(B) Ho > Nd > Eu > Ce (D) Eu > Ce > Nd > Ho
45.
The element that shows greater ability to form p p multiple bonds, is: (A) Sn (B) C (C) Ge (D) Si
46.
The two monomers for the synthesis of Nylon 6, 6 are : (A) HOOC(CH2)4COOH, H2N(CH2)6NH2 (B) HOOC(CH2)6COOH, H2N(CH2)6NH2 (C) HOOC(CH2)4COOH, H2N(CH2)4NH2 (D) HOOC(CH2)6COOH, H2N(CH2)4NH2
47.
The aldehydes which will not form Grignard product with one equivalent Grignard reagents are: (a)
(b)
(c)
(d)
(A) (b), (d) (C) (b), (c), (d)
(B) (b), (c) (D) (c), (d)
48.
Given : (i) C(graphite) O 2 (g) CO2 (g); rH x kJ mol1 1 (ii) C(graphite) O 2 (g) CO(g); rH y kJ mol1 2 1 (iii) CO(g) O 2 (g) CO2 (g); rH z kJ mol1 2 Based on the above thermochemical equations, find out which one of the following algebraic relationships is correct? (A) x = y + z (B) z = x + y (C) y = 2z – x (D) x = y – z
49.
The volume strength of 1M H2O2 is : (Molar mass of H2O2 = 34 g mol–1) (A) 5.6 (B) 16.8 (C) 11.35 (D) 22.4
50.
The correct statement(s) among I to III with respect to potassium ions that are abundant within the cell fluids is/are : I. They activate many enzymes II. They participate in the oxidation of glucose to produce ATP III. Along with sodium ions, they are responsible for the transmission of nerve signals (A) I and II only (B) I and III only (C) I, II and III (D) III only
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JEE-MAIN-2019 (12th Jan-Second Shift)-PCM-12 51.
The compound that is NOT a common component of photochemical smog is : (A) O3 (B) H3C C OONO 2 O
(C) 52.
CH2 CHCHO
(D)
CF2 Cl2
For a certain reaction consider the plot of nk versus 1/T given in the figure. If the rate constant of this reaction at 400 K is 10–5 s–1, then the rate constant at 500 K is:
(A) 10–6 s–1 (C) 10 4 s1
(B) 2 10 4 s1 (D) 4 10 4 s1
53.
An open vessel at 27°C is heated until two fifth of the air (assumed as an ideal gas) in it has escaped from the vessel. Assuming that the volume of the vessel remains constant, the temperature at which the vessel has been heated is : (A) 500°C (B) 500 K (C) 750°C (D) 750 K
54.
The major product of the following reaction is :
55.
(A)
(B)
(C)
(D)
The increasing order of the reactivity of the following with LiAIH4 is : (a)
(b)
(c)
(d)
(A) (b) < (a) < (c) < (d) (C) (a) < (b) < (d) < (c)
(B) (b) < (a) < (d) < (c) (D) (a) < (b) < (c) < (d)
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JEE-MAIN-2019 (12th Jan-Second Shift)-PCM-13 56.
The pair that does NOT require calcination is : (A) ZnO and MgO (B) ZnO and Fe2O3xH2O (C) ZnCO3 and CaO (D) Fe2O3 and CaCO3MgCO3
57.
The major product of the following reaction is :
58.
(A)
CH3CH C CH2
(B)
(C)
CH3CH CHCH2NH2
(D)
CH3CH2 CH
CH2
NH2 NH2
CH3 CH2C CH
If K sp of Ag2 CO3 is 8 1012 , the molar solubility of Ag2CO3 in 0.1 M AgNO3 is : (A) 8 1012 M (C) 8 1010 M
(B) 8 10 11M (D) 8 1013 M
59.
The upper stratosphere consisting of the ozone layer protects us from the sun’s radiation that falls in the wavelength region of : (A) 200 – 315 nm (B) 400 – 550 nm (C) 0.8 – 1.5 nm (D) 600 – 750 nm
60.
If the de Brogile wavelength of the electron in nth Bohr orbit in a hydrogenic atom is equal to 1.5a0 (a0 is Bohr radius), then the value of n/z is : (A) 0.40 (B) 1.50 (C) 1.0 (D) 0.75
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JEE-MAIN-2019 (12th Jan-Second Shift)-PCM-14
PART–C (MATHEMATICS) 61.
2 sin1 x
lim
1
(A)
2 2
(C) 62.
63.
64.
2
(D)
Let a, b, and c be three unit vectors, out of which vectors b and c are non-parallel. If and are the angles which vector a makes with vectors b and c respectively and 1 a b c b, then is equal to : 2 (A) 30° (B) 90° (C) 60° (D) 45°
The integral
(A)
(C)
66.
(B)
Let f be a differentiable function such that f(1) 2 and f '(x) f(x) for all xR. If h(x) f(f(x)), then h '(1) is equal to : (A) 2e2 (B) 4e (C) 2e (D) 4e2 2x x e x e The integral loge x dx is equal to : x 1 e 1 1 1 1 1 (A) e 2 (B) 2 2 e 2 e 2e 3 1 1 3 1 (C) 2 (D) e 2 2 e 2e 2 2e
65.
is equal to :
1 x
x 1
3x13 2x11
2x
4
3x 2 1
x4
6 2x 4 3x 2 1 x4
3
2x 4 3x 2 1
3
C
C
4
dx is equal to (where C is a constant of integration)
(B)
(D)
x12
6 2x 4 3x 2 1 x12
3
2x 4 3x 2 1
3
C
C
If sin4 4 cos4 2 4 2 sin cos ; , [0, ], then cos( ) is equal to : (A) 0 (B) –1 (C) 2 (D) 2
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JEE-MAIN-2019 (12th Jan-Second Shift)-PCM-15 67.
If a curve passes through the point (1, –2) and has slope of the tangent at any point (x, x 2 2y y) on it as , then the curve also passes through the point : x (A) (3, 0) (B) 3, 0
(D) 2, 1
(C) (–1, 2)
68.
If an angle between the line,
x 1 y 2 z 3 and the plane, x 2y kz 3 is 2 1 2
2 2 cos1 , then a value of k is : 3 5 3 3 (C) 5
(A)
69.
3 5 5 (D) 3
(B)
n n 1 n lim 2 2 2 2 is equal to 2 2 n 3 5n n 1 n 2 (A) (B) tan 1(3) 4 (C) (D) tan 1(2) 2 x
70.
In a game, a man wins Rs. 100 if he gets 5 or 6 on a throw of a fair die and loses Rs. 50 for getting any other number on the die. If he decides to throw the die either till he gets a five or a six or to a maximum of three throws, then his expected gain/loss (in rupees) is : 400 (A) loss (B) 0 9 400 400 (C) gain (D) loss 3 3
71.
If a straight line passing through the point P(–3, 4) is such that its intercepted portion between the coordinate axes is bisected at P, then its equation is : (A) 3x 4y 25 0 (B) 4x 3y 24 0 (C) x y 7 0 (D) 4x 3y 0
72.
There are m men and two women participating in a chess tournament. Each participant plays two games with every other participant. If the number of games played by the men between themselves exceeds the number of games played between the men and the women by 84, then the value of m is : (A) 12 (B) 11 (C) 9 (D) 7
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JEE-MAIN-2019 (12th Jan-Second Shift)-PCM-16 73.
The tangent to the curve y x 2 5x 5, parallel to the line 2y 4x 1, also passes through the point : 7 1 1 (A) , (B) , 7 2 4 8 1 1 7 (C) ,7 (D) , 8 4 2
74.
Let S be the set of all real values of such that a plane passing through the points ( 2 ,1,1), (1, 2 ,1) and (1,1, 2 ) also passes through the point (–1, –1, 1). Then S is equal to: (A) 3 (B) 3, 3
(C) 1, 1 75.
(D) 3, 3
Let z1 and z2 be two complex numbers satisfying z1 9 and z2 3 4i 4 . Then the minimum value of z1 z 2 is : (A) 0 (C) 1
76.
(B) 2 (D) 2
The total number or irrational terms in the binomial expansion of 7 (A) 55 (C) 48
1 5
3
60 1 10
is :
(B) 49 (D) 54
77.
The equation of a tangent to the parabola, x 2 8y, which makes an angle with the positive direction of x-axis, is : (A) y x tan 2 cot (B) y x tan 2cot (C) x y cot 2 tan (D) x y cot 2 tan
78.
In a class of 60 students, 40 opted for NCC, 30 opted for NSS and 20 opted for both NCC and NSS. If one of these students is selected at random, then the probability that the student selected has opted neither for NCC nor for NSS is : 1 1 (A) (B) 6 3 2 5 (C) (D) 3 6
79.
The mean and the variance of five observations are 4 and 5.20, respectively. If three of the observations are 3, 4 and 4; then the absolute value of the difference of the other two observations, is : (A) 7 (B) 5 (C) 1 (D) 3
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JEE-MAIN-2019 (12th Jan-Second Shift)-PCM-17
80.
81.
sin 1 1 3 5 If A sin 1 sin ; then for all , , det (A) lies in the interval : 4 4 1 sin 1 5 5 (A) 1, (B) , 4 2 2 3 3 (C) 0, (D) , 3 2 2
If a circle of radius R passes through the origin O and intersects the coordinate axes at A and B, then the locus of the foot of perpendicular from O on AB is : (A) (x 2 y 2 )2 4R 2 x 2 y 2 (B) (x 2 y 2 )3 4R 2 x 2 y 2 (C) (x 2 y 2 )2 4Rx 2 y 2
(D) (x 2 y 2 )(x y) R2 xy
82.
The set of all values of for which the system of linear equations x 2y 2z x x 2y z y x y z (A) is a singleton (B) contains exactly two elements (C) is an empty set (D) contains more than two elements
83.
If the function f given by f(x) x3 3(a 2)x 2 3ax 7, for some aR is increasing in (0, f(x) 14 1] and decreasing in [1, 5), then a root of the equation, 0 (x 1) is : (x 1)2 (A) –7 (B) 5 (C) 7 (D) 6
84.
Let Z be the set of integers. If 2 A {x Z : 2( x 2)( x 5 x 6) 1} and B {x Z : 3 2x 1 9}, then the number of subsets of the set A B, is : (A) 215 (B) 218 12 (C) 2 (D) 210
85.
If n C4 , nC5 and nC6 are in A.P., then n can be : (A) 9 (B) 14 (C) 11 (D) 12
86.
Let S and S be the foci of an ellipse and B be any one of the extremities of its minor axis. If SBS is a right angled triangle with right angle at B and area (SBS) = 8 sq. units, then the length of a latus rectum of the ellipse is : (A) 4 (B) 2 2 (C) 4 2 (D) 2
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JEE-MAIN-2019 (12th Jan-Second Shift)-PCM-18 87.
If the angle of elevation of a cloud from a point P which is 25 m above a lake be 30° and the angle of depression of reflection of the cloud in the lake from P be 60°, then the height of the cloud (in meters) from the surface of the lake is : (A) 60 (B) 50 (C) 45 (D) 42
88.
The expression ~(~ p q) is logically equivalent to : (A) ~ p ~ q (B) p ~ q (C) ~ p q (D) p q 3
89.
90.
3
3
3
3 1 1 3 If the sum of the first 15 terms of the series 1 2 33 3 ..... is 4 2 4 4 equal to 225 k, then k is equal to : (A) 108 (B) 27 (C) 54 (D) 9
The number of integral values of m for which the quadratic expression, (1 + 2m)x2 – 2(1 + 3m)x + 4(1 + m), xR, is always positive, is : (A) 3 (B) 8 (C) 7 (D) 6
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JEE-MAIN-2019 (12th Jan-Second Shift)-PCM-19
JEE (Main) – 2019 ANSWERS PART A – PHYSICS 1.
D
2.
A
3.
No option is correct
4.
B
5.
A
6.
A
7.
D
8.
B
9.
C
10.
B
11.
D
12.
B
13.
A
14.
A
15.
D
16.
C
17.
B
18.
No option is correct
19.
B
20.
D
21.
A
22.
C
23.
B
24.
A
25.
D
26.
A
27.
C
28.
D
29.
C
30.
A
PART B – CHEMISTRY 31.
C
32.
C
33.
D
34.
C
35.
C
36.
A
37.
C
38.
A
39.
A
40.
B
41.
A
42.
B
43.
B
44.
D
45.
B
46.
A
47.
A
48.
A
49.
C
50.
D
51.
A
52.
C
53.
C
54.
D
55.
C
56.
A
57.
D
58.
C
59.
A
60.
D
PART C – MATHEMATICS 61.
B
62.
B
63.
D
64.
A
65.
B
66.
D
67.
B
68.
A
69.
D
70.
B
71.
B
72.
A
73.
B
74.
B
75.
A
76.
D
77.
C
78.
A
79.
A
80.
D
81.
B
82.
A
83.
C
84.
A
85.
B
86.
A
87.
B
88.
A
89.
B
90.
C
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JEE-MAIN-2019 (12th Jan-Second Shift)-PCM-20
HINTS AND SOLUTIONS PART A – PHYSICS 1.
I = ICM + Mx2 2 ICM = MR 2 5 2 I MR2 Mx 2 5
2.
In first case: mg A Y 1 In second case:
1
mg B A Y 1
3 mg mg B 4 2 YA YA
3.
2
mg YA
3 1 = 3 mm 4
For the first branch: 1 3 10 6 100 2 XC 103 tan 1 R2 20
1 –90°. For the second branch:
tan 2
4.
XL 3 R
2 60°. Phase difference between current in branch 1 and 2 = 150°. No option is correct. P The mean time between two collision T
T t1 P1 2 t 2 P2 T1
6 10 8 3 2 5 t 2
t 2 6 10 8
1 5 4 10 8 sec . 2 3
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JEE-MAIN-2019 (12th Jan-Second Shift)-PCM-21 5.
6.
When switched on: VCE = 0 VCC – RCIC = 0 V iC CC 5 10 3 A RC
IC = BiB iB = 25 A VBB = iBRB – VBE = 0
VBB = VBE + iBRB = 3.5 V
4 1 3 C 7C 1 Ceq = 0.5 F 7 7 2 C C 3 3 14C 7 C 3 3 7 C 11
7.
vo
(I) (II)
M
m
mvo = –mv1 + mv2 vo = v 1 + v2
2mv o v2 mM
m v2
M
2
1 1 M m 2 36 1 KEf mv12 m mv 02 v0 2 2 Mm 100 2
v1
M = 4 m.
8.
Emf = Bv sin 45° = (0.3 10 4 ) (10) 5
1 2
= 1.1 × 10–3 V 9.
Covering range of transition power = 2hR To double the range make height 4 times.
10.
1 (1 1) (1 2 ) f R R 1
R
1 (1 2 ) f R
2
R
f
R 1 2
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JEE-MAIN-2019 (12th Jan-Second Shift)-PCM-22 11. 1
nT P1A
m P2A
2
12.
nT
P2A – P1A = mg 1 P A P A m 1 1 2 2 g 1 2 1 nRT nRT m g 1 2 nRT 1 1 m g 1 2
1 GM m 2 R 1 GM KEB = (2m) 2 2R KE A 1 KEB KE A
w 2 x 2 (2 2)2 (0.05)2 2g 2g –4 = 25 × 8 × 10 = 2 cm
13.
y
14.
mg sin + 2 = mg cos 10 – mg sin = mg cos On adding: (1) + (2) 3 12 12 = 2mg ; mg 2 3 On (1) – (2): 1 8 2 mg ; mg = 8 ; 2
…(1) …(2)
3 2
12410 eV A o 12410 eV A o 4.9 eV 250 nm
15.
5.6 eV 0.7 eV 4.9 eV
16.
v 52 2gh 52 2 10 10 225 = 15 m/s h rmv 20 (20 103 kg) (15) = 6 kg m2/sec
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JEE-MAIN-2019 (12th Jan-Second Shift)-PCM-23 17.
Vo = igo (RG R) igo = 4 × 10–4 × 25 = 10–2 A
Vo = 2.5 V V 2.5 Rg R o 2 = 250 igo 10 18.
R = 200 .
V = kt =
4 3 R 3 1/3
3k R t 4
4T R Bonus
Pin = Patm + 19.
I3 + I5 = I4 I1 + I2 = I4 I3 + I6 = I2 + I1
20.
1 4(11 e)
I3 = I4 – I5 = 0.4 A I2 = I4 – I1 = 1.1 A I6 = I2 + I1 – I3 = 0.4 A
v 512
2 4(27 e)
v 256
11 e 1 27 e 2 22 + 2e = 27 + e e=5 21.
For paramagnetic materials
1 R
1 T2 2 T1
T1 350 1 2.8 10 4 T2 300 = 3.267 × 10–4
2
22.
232 90
6 Th
208 78
4 Y
208 82
23.
L [A 1 ] RCV
24.
If the water is filled focal length will decrease and image will disappear.
X
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JEE-MAIN-2019 (12th Jan-Second Shift)-PCM-24 25.
eVs h eVo h 2 2 h eVo 2 h 0 = 2h 2 2 3h 3 th 2 2
26.
…(1) …(2)
y 5(sin 3 t 3 cos 3 t) cm y = 10 sin(3t + ) A = 10 cm 2 T = sec 3
27.
R1
A
A at t = 0
at t =
2
B
R2 B
v A v B R1ˆi R2 ˆi = (R R ) ˆi 2
28.
I
1
B2o C 2o
B2o I 2o C 103 2 4 10 7 3 108 Bo 10 4 T B2o
29.
30.
q A o q = EAo q = 100 × 1 × 8.85 × 10–12 q = 8.85 × 10–10 C E
Current = slope of q – t graph = 0. [at t = 4 sec]
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JEE-MAIN-2019 (12th Jan-Second Shift)-PCM-25
PART B – CHEMISTRY 8 0.2 40 18 Moles of H2O = 1 18 0.2 Mole fraction of NaOH = 0.167 1.2 8 1000 Molality = 11.11 40 18
31.
Moles of NaOH =
32.
Homoleptic complexes contain identical ligands, e.g., [Mn(NCS)6]4–.
33.
Due to the lowest bond energy of Pb – Pb bond.
34.
Lyophobic sols are coagulated by adding electrolysis.
35.
The COO– group absorbs H+ in acidic medium (pH = 2)
36.
37.
0m (HA) = 100.5 + 425.9 – 126.4 = 400 K 1000 5 10 5 103 50 M 103 50 0.125 400 m0
38.
2A A2 1 2 0.8 1 0.8 2 i = 1 – 0.8 +
0.8 0.6 2
Tf = Kf i m = 5 0.6
x 1000 2 Since Tf 2 122 30
x = 2.44 g
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JEE-MAIN-2019 (12th Jan-Second Shift)-PCM-26 39.
6NaOH Cl2 5 NaCl NaClO3 3H2O
40.
41.
The plot (b) and (d) are incorrect. The correct ones are given below:
P
U
0
Vm
(b)
0
Vm
(c)
42
43
44.
The atomic radii are Eu = 199 pm Ce = 183 pm Nd = 181 pm Ho = 176 pm
45.
It is due to the smallest atomic size of carbon in the given options.
46.
The monomers are hexamethylene diamine and adipic acid.
47.
Each of (b) and (d) will react two moles of Grignard reagent.
48.
(ii) + (iii) = (i) Y+z=x
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JEE-MAIN-2019 (12th Jan-Second Shift)-PCM-27 49.
Volume strength = 11.35 M = 11.35 (STP)
50.
Active transport proteins exchanges Na+ ions for K+ ions across the plasma membrane of animal cells.
51.
O3 is not common component of London and Los Angeles smog. It is present only in Los Angeles smog
52.
n nA
Ea 4606 nA RT T k Ea 500 400 n 5 10 R 500 400 1 k n 5 4606 2.303 n10 2000 10 k n 5 n10 10 k 10 4
53.
n1T1 = n2T2 2n n 300 = n T2 5 3 300 T2 T2 = 500 K 5
54.
55.
The order is identical to nucleophilic substitution order: Acid chloride > Acid anhydride > Acid > Amide
56.
In calcination the ore is converted to metal oxide. ZnO and MgO are already in oxide form.
57.
Br EtOH Br
NaNH
2 Br
H CH3 CH2 C CH Br
EtO
58.
8 10–12 = (2S’ + 0.1)2S’ or S’ = 8 10–10 M
59.
In ozone layer the wavelength of U.V radiation is 200 – 340 nm.
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JEE-MAIN-2019 (12th Jan-Second Shift)-PCM-28 60.
2r = n 2r 2n2 a0 n 2 a0 n n Z Z 1.5a0 n a0 1.5a0 Z n 1.5 0.75 Z 2 2
PART C – MATHEMATICS 61.
2 sin1 x
lim
1 x
x 1
2 sin1 x
2 sin1 x 2
lim
x 1
1 x
2sin1 x
1
lim
2 sin1 x
2cos x
x 1
.
1
1 x 2 Assuming x cos 2
lim
62.
.
1
2 sin 2 2
0
f ' x f x
2
1 x R
Integrate and use f 1 2 f x 2e x 1 f ' x 2e x 1
h x f f x h' x f ' f x f ' x h' 1 f ' f 1 f 1 f ' 2 f ' 1 2e.2 4e e
63.
2x
e
x e 1 e loge x.dx 1 x loge x.dx 2x
x
x e Let t, v e x 1 1 1 1 3 1 1 dt dv 1 2 1 e 2 e 2 1 2 2 e 2 2e e e
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JEE-MAIN-2019 (12th Jan-Second Shift)-PCM-29
64.
a.c b a.b .c 21 b
b and c are linearly independent 1 a.c and a.b 0 2 (All given vectors are unit vectors) a c 60o and a b 90o
65.
30o
3x13 2x11
2x
4
3x 2 1
4
dx
2 3 x 3 x5 dx 3 1 4 2 x2 x 4
3 1 Let 2 2 4 t x x 1 dt 1 x12 4 3 C C 3 2 t 6t 6 2x 4 3x 2 1
66.
A.M. G.M. 1 sin4 4 cos 4 1 1 sin4 . 4 cos 4 .1.1 4 4 4 sin 4 cos2 2 4 2 sin cos given that sin4 4 cos4 2 4 2 sin cos
A.M. G.M. sin4 1 4 cos4 1 sin 1,cos , As , 0, 2 1 sin 1, cos 2 1 sin as 0, 2 cos cos 2 sin sin 2
67.
dy x 2 2y (Given) dx x dy y 2 x dx x 2
dx I.F. e x x 2 y . x 2 x . x 2 dx C
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JEE-MAIN-2019 (12th Jan-Second Shift)-PCM-30
x4 C y
Hence b passes through 1, 2 C
9 4
x4 9 4 4 Checking options, only option (B) satisfies. yx 2
68.
Direction Ratio of line are 2, 1, –2 Normal vector of plane is ˆi 2ˆj kkˆ 2iˆ ˆj 2kˆ . ˆi 2ˆj kkˆ sin 3 1 4 k2 2k sin …………(1) 3 k2 5
cos
2 2 3
(Given)
…………(2)
Using sin2 cos2 1 k 2
5 3
2n
69.
n 2 r 1 n r 2n 1 lim Using D.I. as limit of sum, we get x r2 r 1 n 1 2 n lim
x
2
2
dx tan1 2 2 1 x 0
70.
2 1 Let w denotes probability that outcome 5 or 6 w 6 3 4 2 Let, L denotes probability that outcome 1, 2, 3, 4 L 6 3 Expected Gain/Loss w 100 Lw 50 100 L2 w 50 50 100 L3 150 2
3
1 2 1 2 1 2 100 . 50 0 150 0 3 3 3 3 3 3
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JEE-MAIN-2019 (12th Jan-Second Shift)-PCM-31
71.
72.
x y 1 a b a b 3, 4 , 2 2 a 6, b 8 equation of line is 4x 3y 24 0 Let the line be
B (0, b) (–2, 4)
A (a, 0)
Let m – men, 2 – women m C2 2 mC1 2C1 .2 84 m2 5m 84 0 m 12 m 7 0 m = 12
73.
y x 2 5x 5 dy 7 2x 5 2 x dx 2 7 1 at x , y 2 4 29 7 1 Equation of tangent at , is 2x y 0 4 2 4 Now check options 1 x , y7 8
74.
All four points are coplanar so 1 2 2 0 2
2 1
0
2
2
2
2
2
0
1
3 0
1
2
3 75.
z1 9, z2 3 4i 4 C1 0,0 radius r1 9 C2 3, 4 , radius r2 4 C1C2 r1 r2 5 Circle touches internally z1 z 2 min 0 60 r
76.
r
General term Tr 1 60 Cr ,7 5 3 10 for rational term, r = 0, 10, 20, 30, 40, 50, 60 number of rational terms = 7 number of irrational terms = 54 FIITJEE Ltd., FIITJEE House, 29-A, Kalu Sarai, Sarvapriya Vihar, New Delhi -110016, Ph 46106000, 26569493, Fax 26513942 website: www.fiitjee.com.
JEE-MAIN-2019 (12th Jan-Second Shift)-PCM-32
x 2 8y dy x tan dx 4 x1 4 tan
77.
y1 2 tan2 Equation of tangent :y 2 tan2 tan x 4 tan x y cot 2 tan 78.
A opted NCC B opted NSS
P (neither A nor B)
79.
B
A
10 1 z 60 6
20
20
20
Mean x 4, 2 5.2,n 5, x1 3 x 2 4 x 3
x
i
20
x 4 x5 9 2 i
x x x
……….(i) 2
2 xi2 106
x 24 x 52 65
……….(ii) 2
Using (i) and (ii) x 4 x5 49 x4 x5 7
1
80.
sin
1
1
sin
sin
1
A sin 1
2 1 sin2
1 1 3 5 , sin 2 2 4 4 1 0 sin2 2 A 2, 3
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JEE-MAIN-2019 (12th Jan-Second Shift)-PCM-33
h k Equation of AB is hx ky h2 k 2 Slope of AB
81.
P (h, k)
h2 k 2 h2 k 2 A , 0 ,B 0, k h As, AB 2R
x
2
1
82.
1
3
y
h2 k 2
2
A
4R2h2k 2
2 3
B
4R 2 x 2 y 2
2
2 1 0
1
1
1
3
1 0 1 83.
f ' x 3x 2 6 a 2 x 3a f ' x 0 x 0, 1 f ' x 0 x 1, 5
f ' x 0 at x 1 a 5 2
f x 14 x 1 x 7 f x 14
x 1
2
x7
Hence root of equation
84.
x 2 x2 5x 6
A x z:2 x 2 x 2 5x 6
2
f x 14
x 1
2
0 is 7.
1
20 x 2,2,3
A 2,2,3 B x Z : 3 2x 1 9 B 0,1,2,3,4 Hence, A B has is 15 elements. So number of subsets of A B is 215 .
85.
2. nC5 nC4 nC6
n n n 5n5 4 n4 6n6 2 1 1 1 . 5 n 5 n 4 n 5 30
2.
n = 14 satisfying equation.
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JEE-MAIN-2019 (12th Jan-Second Shift)-PCM-34 86.
mSB .mS'B 1 b 2 a2 e 2 …………(i) 1 S 'B.SB 8 2 a 2e 2 b 2 16 …………..(ii) 2 2 b a 1 e2
B
S’
…………..(iii) using (i), (ii), (iii) a = 4 b2 2 1 e 2 2b 2 L.R 4 a 87.
x y 3x y 25 x 25 tan 60o y tan 30o
S (a, e 0)
(–a, e 0)
Cloud
………..(i)
3y 50 x 3x 50 x x 25m Height of cloud from surface 25 25 50m
30o
x
60o 25 m
25 m
Surface
88. p
q
~p
~p q
~ ~ p q
~ p ~ q
T F T F
T T F F
F T F T
T T T F
F F F T
F F F T
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JEE-MAIN-2019 (12th Jan-Second Shift)-PCM-35 3
89.
3
3
3
3 6 9 12 S ..........15 term 4 4 4 4 27 15 3 r 64 r 1
27 15 15 1 . 64 2
2
= 225 K (Given in question) K 27 90.
Expression is always positive it 2m 1 0 m
1 2
and D 0 m2 6m 3 0 …….(iii) 3 12 m 3 12 Common interval is 3 12 m 3 12 Integral value of m 0,1,2,3, 4,5,6
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FIITJEE Solutions to JEE(Main)-2019 Test Date: 8th April 2019 (First Shift)
PHYSICS, CHEMISTRY & MATHEMATICS Paper - 1 Time Allotted: 3 Hours
Maximum Marks: 360
Please read the instructions carefully. You are allotted 5 minutes specifically for this purpose.
Important Instructions: 1.
The test is of 3 hours duration.
2.
This Test Paper consists of 90 questions. The maximum marks are 360.
3.
There are three parts in the question paper A, B, C consisting of Physics, Chemistry and Mathematics having 30 questions in each part of equal weightage. Each question is allotted 4 (four) marks for correct response.
4.
Out of the four options given for each question, only one option is the correct answer.
5.
For each incorrect response 1 mark i.e. ¼ (one-fourth) marks of the total marks allotted to the question will be deducted from the total score. No deduction from the total score, however, will be made if no response is indicated for an item in the Answer Box.
6.
Candidates will be awarded marks as stated above in instruction No.3 for correct response of each question. One mark will be deducted for indicating incorrect response of each question. No deduction from the total score will be made if no response is indicated for an item in the answer box.
7.
There is only one correct response for each question. Marked up more than one response in any question will be treated as wrong response and marked up for wrong response will be deducted accordingly as per instruction 6 above.
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JEE-MAIN-2019 (8th Apr-First Shift)-PCM-2
PART –A (PHYSICS) 1.
If 10 22 gas molecules each of mass 10 26 kg collide with a surface (perpendicular to it) elastically per second over an area 1 m2 with a speed 10 4 m/s, the pressure exerted by the gas molecules will be of the order of: N N (A) 108 2 (B) 103 2 m m 4 N 16 N (C) 10 2 (D) 10 m m2
2.
A particle moves in one dimension from rest under the influence of a force that varies with the distance traveled by that varies with the distance traveled by the particle as shown in the figure. The kinetic energy of the particle after it has traveled 3 m is: (A) 2.5 J (B) 4 J (C) 5 J (D) 6.5 J
3.
An upright object is placed at a distance of 40 cm in front of a convergent lens of focal length 20 cm. A convergent mirror of focal length 10 cm is placed at a distance of 60 cm on the other side of the lens. The position and size of the final image will be: (A) 40 cm from the convergent mirror, same size as the object (B) 20 cm from the convergent mirror, same size as the object (C) 40 cm from the convergent lens, twice the size of the object (D) 20 cm from the convergent mirror, twice the size of the object
4.
Four particles A, B, C and D with masses mA m, mB 2m, mc 3m and mD 4m are at the corners of a square. They have accelerations of equal magnitude with directions as shown. The acceleration of the centre of mass of the particles is: a ˆ ˆ (A) ij (B) Zero 5 a ˆ ˆ (C) ij (D) a ˆi ˆj 5
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JEE-MAIN-2019 (8th Apr-First Shift)-PCM-3 5.
Two identical beakers A and B contain equal volumes of two different liquids at 60o C each and left to cool down. Liquid in A has density of 8 10 2 kg / m3 and specific heat of 2000 Jkg1 K 1 while liquid in B has density of 103 kgm3 and specific heat of 4000JKg1K 1 . Which of the following best describes their temperature versus time graph schematically? (assume the emissivity of both the beakers to be the same)
(A)
(B)
(C)
(D)
6.
A thin strip 10 cm long is on a U shaped wire of negligible resistance and it is connected to a spring of spring constant 0.5Nm1 (see figure). The assembly is kept in a uniform magnetic field of 0.1 T. If the strip is pulled from its equilibrium position and released, the number of oscillations it performs before its amplitude decreases by a factor of e is N. If the mass of the strip is 50 grams, its resistance 10 and air drag negligible, N will be close to: (A) 50000 (B) 10000 (C) 1000 (D) 5000
7.
A 20 Henry inductor coil is connected to a 10 ohm resistance in series as shown in figure. The time at which rate of dissipation of energy (Joule’s heat) across resistance is equal to the rate at which magnetic energy is stored in the inductor, is 2 ln 2 1 (C) ln 2 2
(A)
(B) ln 2 (D) 2ln 2
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JEE-MAIN-2019 (8th Apr-First Shift)-PCM-4 8.
A wire of length 2L is made by joining two wires A and b of same lengths but different radii r and 2r and made of the same material. It is vibrating at a frequency such that the joint of the two wires forms a node. If the number of antinodes in wire A is p and that in B is q then the ratio p : q is:
(A) 1 : 4 (C) 3 : 5
(B) 1 : 2 (D) 4 : 9
9.
A steel wire having a radius of 2.0 mm, carrying a load of 4 kg, is hanging from a ceiling. Given that g 3.1 ms2 , what will be the tensile stress that would be developed in the wire? (A) 6.2 10 6 Nm2 (B) 4.8 106 Nm2 6 2 (C) 5.2 10 Nm (D) 3.1 106 Nm2
10.
Voltage rating of a parallel plate capacitor is 500V. Its dielectric can withstand a V maximum electric field of 106 . The plate area is 10 4 m2 . What is the dielectric m constant if the capacitance is 15 pF? (given 0 8.86 10 12 C2 / Nm2 ) (A) 3.8 (B) 6.2 (C) 4.5 (D) 8.5
11.
An alternating voltage v t 220 sin100l volt is applied to a purely resistive load of 50 . The time taken for the current to rise from half of the peak value of the peak value is: (A) 2.2 ms (B) 3.3 ms (C) 5 ms (D) 7.2 ms
12.
The wavelength of the carrier waves in a modern optical fiber communication network is close to: (A) 1500 nm (B) 600 nm (C) 2400 nm (D) 900 nm
13.
Water from a pipe is coming at a rate of 100 litres per minute. If the radius of the pipe is 5 cm, the Reynolds number for the flow is of the order of : (density of water = 1000 kg/m3, coefficient of viscosity of water = 1 mPa s) (A) 103 (B) 10 6 2 (C) 10 (D) 10 4
14.
A boy’s catapult is made of rubber cord which is 42 cm long, with 6 mm diameter of cross – section and of negligible mass. The boy keeps a stone weighing 0.02 kg on it and stretches the cord by 20 cm by applying a constant force. When released, the stone flies off with a velocity of 20 ms–1. Neglect the change in the area of cross section of the cord while stretched. The Young’s modulus of rubber is closest to: (A) 103 Nm2 (B) 10 6 Nm2 (C) 108 Nm2
(D) 10 4 Nm2
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JEE-MAIN-2019 (8th Apr-First Shift)-PCM-5 15.
Two particles move at right angle to each other. Their de Broglie wavelengths are 1 and 2 respectively. The particles suffere perfectly inelastic collision. The de Broglie wavelength , of the final particle, is given by: 2 (A) 1 2 (B) 1 2 2 1 1 1 1 1 (C) (D) 2 2 2 1 2 1 2
16.
Four identical particles of mass M are located at the corners of a square of side ‘a’. What should be their speed if each of them revolves under the influence of other’s gravitational field in a circular orbit circumscribing the square? GM GM (A) 1.35 (B) 1.16 a a GM GM (C) 1.41 (D) 1.21 a a
17.
In figure, the optical fiber is l 2m long and has a diameter of d = 20 m. If a ray of light is incident on one end of the fiber at angle 1 40o , the number of reflection it makes before emerging from the other end is close to:
(A) 57000 (C) 66000 18.
19.
(B) 45000 (D) 55000
A circular coil having N turns and radius r carries a current. It is held in the XZ plane in a magnetic field Biˆ . The torque on the coil due to the magnetic field is: Br 2 1 (A) (B) zero N Br 2 1 (C) (D) Br 2IN N Ship A is sailing towards north – east with velocity 30 ˆi 50ˆj km/hr where ˆi points east and ˆj , north. Ship B is at a distance of 80 km east and 150 km north of Ship A and is sailing towards west at 10 km/hr. A will be at minimum distance from B ins: (A) 2.2 hrs. (B) 4.2 hrs. (C) 2.6 hrs. (D) 3.2 hrs.
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JEE-MAIN-2019 (8th Apr-First Shift)-PCM-6 20.
A plane electromagnetic wave travels in free space along the x – direction. The electric field component of the wave at a particular point of space and time is E Vm1 along y – direction. Its corresponding magnetic filed component, B would be: (A) 2 10 8 T along z – direction (B) 6 10 8 T along x – direction (C) 6 10 8 T along z- direction (D) 2 10 8 T along y – direction
21.
A thermally insulted vessel contains 150 g of water at 0o C . Then the air from the vessel is pumped out a adiabatically. A fraction of water turns into ice and the rest evaporates at 0o C itself. The mass of evaporated water will be closes to: (Latent heat of vaporization of water = 2.10 10 6 Jkg1 and Laten heat of Fusion of water 3.36 105 Jkg1 ) (A) 35 g (C) 130 g
(B) 150 g (D) 20 g
22.
Radiation coming from transition n = 2 to n = 1 of hydrogen atoms fall of He ions in n = 1 and n = 2 states. The possible. Transition of helium ions as they absorb energy from the radiation is: (A) n 2 n 4 (B) n 2 n 5 (C) n 2 n 3 (D) n 1 n 4
23.
A 200 resistor has a certain color code. If one replaces the red color by green in the code, the new resistance will be: (A) 500 (B) 400 (C) 300 (D) 100
24.
The reverse breakdown voltage of a Zener diode is 5.6 V in the given circuit .
The current Iz through the Zener is: (A) 10 mA (C) 7 mA 25.
(B) 15 mA (D) 17 mA
A thin circular plate of mass M and radius R has its density varying as p r p0 r with P0 as constant and r is the distance from its center. The moment of Inertia of the circular plate about an axis perpendicular to the plate and passing through its edge is I aMR2 . The value of the coefficient a is: 8 1 (A) (B) 5 2 3 3 (C) (D) 5 2 FIITJEE Ltd., FIITJEE House, 29-A, Kalu Sarai, Sarvapriya Vihar, New Delhi -110016, Ph 46106000, 26569493, Fax 26513942 website: www.fiitjee.com.
JEE-MAIN-2019 (8th Apr-First Shift)-PCM-7
26.
In SI units, the dimensions of (A) AT 3ML3/2 (C) A 2 T 3M1L2
27.
0 is: 0
(B) A 1TML3 (D) AT 2M1L1
For the circuit shown, with R1 1.0 , R 2 2.0 ,E1 2 V and E2 E 3 4 V, the potential difference between the points ‘a’ and ‘b’ is approximately (in V):
(A) 3.3 (C) 3.7
(B) 2.3 (D) 2.7
28.
A solid conducting sphere, having a charge Q, is surrounded by an uncharged conducting hollow spherical shell. Let the potential difference between the surface of the solid sphere and that of the outer surface of the hollow shell be V. If the shell is now given a charge of –4 Q, the new potential difference between the same two surface is: (A) 2 V (B) –2V (C) 4 V (D) V
29.
In an interference experiment the ratio of amplitudes of coherent waves is
a1 1 . The a2 3
ratio of maximum and minimum intensities of fringes will be: (A) 9 (B) 4 (C) 18 (D) 2 30.
The bob of a simple pendulum has mass 2g and a charge of 5.0 C . It is at rest in a V uniform horizontal electric field of intensity 2000 . At equilibrium, the angle that the m m pendulum makes with the vertical is: (take g 10 2 ) s (A) tan 1 2.0 (B) tan 1 0.2 (C) tan 1 5.0
(D) tan 1 0.5
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JEE-MAIN-2019 (8th Apr-First Shift)-PCM-8
PART –B (CHEMISTRY) 31.
Element ‘B’ forms ccp structure and ‘A’ occupies half of the octahedral voids, while oxygen atoms occupy all the tetrahedral voids. The structure of bimetallic oxide is (A) A 2B 2 O (B) AB 2 O 4 (C) A 4B 2 O (D) A 2BO 4
32.
Coupling of benzene diazonium chloride with 1 – naphthol in alkaline medium will give:
(A)
(B)
(C)
(D)
33.
The size of the iso-electronic species CI , Ar and Ca2+ is affected by (A) Principal quantum number of valence shell (B) Azimuthal quantum number of valence shell (C) electron – electron interaction in the outer orbitals (D) nuclear charge
34.
In the following compounds, the decreasing order of basic strength will be: (A) C 2H5NH2 NH3 C2H5 2 NH (B) NH3 C 2H5NH2 C2H5 2 NH (C) C2H5 2 NH C 2H5NH2 NH3
35.
(D) C2H5 2 NH NH3 C2H5NH2
The correct order of the spin only magnetic moment of metal ions in the following low 4
4
3
2
spin complexes, V CN6 , Fe CN6 , Ru NH3 6 , and Cr NH3 6 , is: (A) Cr 2 Ru3 Fe2 V 2 (B) V 2 Cr 2 Ru3 Fe2 2 2 3 2 (C) Cr V Ru Fe (D) V 2 Ru3 Cr 2 Fe2
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JEE-MAIN-2019 (8th Apr-First Shift)-PCM-9 36.
An organic compound ‘X’ showing the following solubility profile is:
(A) Oleic acid (C) Benzamide 37.
(B) o – Toluidine (D) m – Cresol
Diborane B 2H6 reacts independently with O2 and H2 O to produce, respectively: (A) H3BO3 and B2O3
(B) B2O3 and H3BO3
(C) HBO2 and H3BO3
(D) B2O3 and BH4
38.
The lanthanoide that would show colour is: (A) Gd3 (B) La3 3 (C) Lu (D) Sm3
39.
The major product of the following reaction
40.
(A)
(B)
(C)
(D)
100 mL of a water sample contains 0.81 g of calcium bicarbonate and 0.73 g of magnesium bicarbonate. The hardness of this water sample expressed in terms of equivalents of CaCO3 is: (molar mass of calcium bicarbonate is 162 gmol1 and magnesium bicarbonate is 146 g
mol1 ) (A) 10, 000 ppm (C) 5,000 ppm
(B) 1, 000 ppm (D) 100 ppm
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JEE-MAIN-2019 (8th Apr-First Shift)-PCM-10 41.
42.
An organic compound neither reacts with neutral ferric chloride solution nor with Fehling solution. It however, reacts with Grignard reagent and gives positive iodoform test. The compound is:
(A)
(B)
(C)
(D)
For the reaction 2A B C, the values of initial rate at different reactant concentrations are given in the table below: The rate law for the reaction is: Initial Rate (mol L1s1 ) A molL1 B molL1
0.05
0.05
0.045
0.10
0.05
0.090
0.20
0.10
0.72
(A) Rate = k[A]2[B]2 (C) Rate = k[A][B] 43.
(B) Rate = k[A][B]2 (D) Rate = k[A]2[B]
The major product of the following reaction is:
(A)
(B)
(C)
(D)
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JEE-MAIN-2019 (8th Apr-First Shift)-PCM-11 44.
Which is wrong with respect to our responsibility as a human being to protect our environment? (A) Restricting the use of vehicles (B) Using plastic bags (C) Setting up compost tin in gardens (D) Avoiding the use of floodlighted facilities.
45.
The major product of the following reaction
46.
(A)
(B)
(C)
(D)
For silver CP JK 1 mol1 23 0.01T . If the temperature (T) of 3 moles of silver is raised from 300 K to 1000 K at 1 atm pressure, the value of H will be close to: (A) 13 kJ (B) 62 kJ (C) 16 kJ (D) 21 kJ
47.
The correct order of hydration enthalpies of alkali metal ions is: (A) Li Na K Cs Rb (B) Na Li K Rb Cs (C) Na Li K Cs Rb (D) Li Na K Rb Cs
48.
Adsorption of a gas follows Freundlich adsorption isotherm. x is the mass of the gas x adsorbed on mass m of the adsorbent. The plot of log versus log p is shown in the m x given graph, is proportional to: m
(A) p2/3
(B) p2
(C) p3
(D) p3/2
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JEE-MAIN-2019 (8th Apr-First Shift)-PCM-12 49.
The vapour pressures of pure liquids A and B are 400 and 600 mm Hg, respectively at 298 K. On mixing the two liquids, the sum of their initial volumes is equal to the volume of the final mixture. The mole fraction of liquid B is 0.5 in the mixture. The vapour pressure of the final solution, the mole fractions of components A and B in vapour phase, respectively are: (A) 500 mmHg, 0.4, 0.6 (B) 500 mmHg, 0.5, 0.5 (C) 450 mmHg, 0.5, 0.5 (D) 450 mmHg, 0.4, 0.6
50.
0 Given that E0O2 /H2O 1.23V; E0S O2 /SO2 2.05 V;EBr 2
51.
8
2 /Br
4
strongest oxidizing agent is: (A) O2
(B) S2O82
(C) Au3
(D) Br2
1.09 V;E0Au3 /Au 1.4V . The
The quantum number of four electrons are given below: I. n 4,l 2,ml 2, ms 1/ 2 II. n 3,l 2, ml 1,m8 1/ 2 III. n 4,l 1, ml 0, ms 1/ 2 IV. n 3,l 1, mI 1; ms 1/ 2 The correct order of their increasing energies will be: (A) I < III < II < IV (B) I < II < III < IV (C) IV < II < III < I (D) IV < III < II < I
52.
If solubility product of Zr3 PO 4 4 is denoted by K SP and its molar solubility is denoted by S, then which of the following relation between S and K SP is correct? 1/7
K (A) S SP 6912
1/9
K Sp (C) S 929
1/6
K (B) S SP 144
1/7
K (D) S SP 216
53.
Which of the following amines can be prepared by Gabriel phthalimide reaction? (A) t – butylamine (B) n – butylamine (C) neo – pentylamine (D) triethylamine
54.
Which one of the following equations does not correctly represent the first law of thermodynamics for the given processes involving an ideal gas? (Assume non – expansion work is zero) (A) Adiabatic process: U w (B) Isochoric process: U q (C) Cyclic process: q w (D) Isothermal process: q w
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JEE-MAIN-2019 (8th Apr-First Shift)-PCM-13 55.
The following ligand is:
(A) tridentate (C) tetradentate 56.
(B) bidentate (D) hexadentate
The IUPAC name of the following compound is
(A) 3-Hydroxy – 4 – methylpentaonic acid (B) 4 – Methyl – 3 – hydroxypentanoic acid (C) 2 – Methyl – 3 – hydroxylpentan-5-oic acid (D) 4, 4 – Dimethyl – 3 – hydroxybutanoic acid 57.
In order to oxidize a mixture of one mole of each of FeC 2 O4 , Fe2 C2 O 4 3 ,FeSO 4 and Fe2 SO 4 3 in acidic medium, the number of moles of KMnO4 required is:
(A) 1 (C) 2
(B) 1.5 (D) 3
58.
Assertion : Ozone is destroyed by CFCs in the upper stratosphere. Reason : Ozone holes increase the amount of UV radiation reaching the earth. (A) Assertion and reason are incorrect (B) Assertion is false, but the reason is correct (C) Assertion and reasons are both correct, and the reason is the correct explanation for the assertion. (D) Assertion and reason are correct, but the reason is not the explanation for the assertion.
59.
Maltose on treatment with dilute HCI gives: (A) D – Fructose (B) D – Galactose (C) D – Glucose and D – Fructose (D) D – Glucose
60.
With respect to an ore, Ellingham diagram helps to predict the feasibility of its (A) Electrolysis (B) Thermal reduction (C) Vapour phase refining (D) Zone refining
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JEE-MAIN-2019 (8th Apr-First Shift)-PCM-14
PART–C (MATHEMATICS) 61.
The magnitude of the projection of the vector the plane containing the vectors ˆi ˆj kˆ and
3 2 3 (D) 2
(A) 3 6 (C)
2iˆ 3 ˆj kˆ on the vector perpendicular to ˆi 2ˆj 3kˆ , is:
(B)
6
62.
The shortest distance between the line y x and the curve y 2 x 2 is: 11 (A) (B) 2 4 2 7 7 (C) (D) 8 4 2
63.
If and be the roots of the equation x 2 2x 2 0 , then the least value of n for n
which 1 is: (A) 4 (C) 5
64.
(B) 2 (D) 3
All possible numbers are formed using the digits 1, 1, 2, 2, 2, 2, 3, 4, 4 taken all at a time. The number of such numbers in which the odd digits occupy even places is: (A) 180 (B) 175 (C) 162 (D) 160 5x 2x dx is equal to: sin 2 (where c is a constant of integration). (A) x 2 sin x 2 sin 2x c (C) x 2 sin x sin 2x c sin
65.
66.
(B) 2x sin x 2 sin 2x c (D) 2x sin x sin 2x c
Let O (0, 0) and A (0, 1) be two fixed points. Then the locus of a point P such that the perimeter of AOP , is 4, is: (A) 9x 2 8y 2 8y 16 (B) 8x 2 9y 2 9y 18 (C) 9x 2 8y 2 8y 16
(D) 8x 2 9y 2 9y 18
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JEE-MAIN-2019 (8th Apr-First Shift)-PCM-15
67.
68.
69.
3 5 If cos ,sin and 0 , , then tan 2 is equal to: 5 13 4 63 33 (A) (B) 52 52 63 21 (C) (D) 16 16
cos sin 0 1 32 Let A , R such that A . Then a value of is: sin cos 1 0 (A) 0 (B) 16 (C) (D) 32 64 1 x 2x If f x loge is equal to: , x 1, then f 2 1 x 1 x
(B) f x
(A) 2f x
(C) 2f x 2
2
(D) 2f x 2
70.
71.
3 cos x sin x dy If 2y cot 1 , x 0, then is equal to cos x 3 sin x 2 dx (A) x (B) x 6 3 (C) x (D) 2x 6 3
The sum of the solutions of the equation (A) 9 (C) 10
72.
lim x 0
73.
x 4 2 0, x 0 is equal to:
(B) 4 (D) 12 sin2 x
2 1 cos x
(A) 2 (C) 4
x 2 x
equals (B) 4 2 (D) 2 2
2. 20C0 5. 20 C1 8. 20C2 11. 20C3 .... 62. 20 C20 is equal to
(A) 223 (C) 224
(B) 226 (D) 225
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JEE-MAIN-2019 (8th Apr-First Shift)-PCM-16
74.
such that y 0 0 . If
ay 1
1 2 1 (C) 16
If f x
2
dy 2x x 2 1 y 1 dx
, then the value of ‘a’ is 32
(A)
75.
Let y y x be the solutions of the differential equation, x 2 1
(B) 1 (D)
1 4
2 x cos x and g x loge x, x 0 then the value of the integral 2 x cos x
/4
g f x dx is:
/ 4
76.
(A) loge 1
(B) loge 2
(C) loge e
(D) loge 3
The area (in sq. units) of the region A 26 3 53 (C) 6
(A)
x, y R R 0 x 3, 0 y 4, y x (B)
2
3x is:
59 6
(D) 8
77.
The mean and variance of seven observations are 8 and 16, respectively. If 5 of the observations are 2, 4, 10, 12, 14, then the product of the remaining two observations is: (A) 40 (B) 45 (C) 49 (D) 48
78.
The length of the perpendicular from the point (2, –1, 4) on the straight line, x3 y2 z is: 10 7 1 (A) greater than 2 but less than 3 (B) less than 2 (C) greater than 4 (D) greater than 3 but less than 4
79.
The contrapositive of the statement “If you are born in India, then you are a citizen of India”, is: (A) If you are a citizen of India, then you are born in India (B) If your are not a citizen of India, then you are not born in India (C) If you are no born in India, then you are not a citizen of India (D) If you are born in India, then you are not a citizen of India
80.
The sum of all natural numbers ‘n’ such that 100 n 200 and H.C. F (91, n) > 1 is: (A) 3221 (B) 3303 (C) 3203 (D) 3121
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JEE-MAIN-2019 (8th Apr-First Shift)-PCM-17 81.
If S1 and S 2 are respectively the sets of local minimum and local maximum points of the function. f x 9x 4 12x 3 36x 2 25, x R, then (A) S1 2, 1 ;S 2 0 (B) S1 2,0 ;S2 1 (C) S1 2; S2 0,1 (D) S1 1 ;S2 0,2
82.
The sum of the co – efficient of all even degree terms in x in the expansion of
6
6
x x3 1 x x3 1 , x 1 is equal to:
(A) 26 (C) 32
(B) 24 (D) 29
83.
A point on the straight line, 3x 5y 15 which is equidistant from the coordinate, axes will lie only in: (A) 4th quadrant (B) 1st , 2nd and 4th quadrants (C) 1st quadrant (D) 1st and 2nd quadrants
84.
If the tangents on the ellipse 4x 2 y 2 8 at the points (1, 2) and (a, b) are perpendicular to each other, then a 2 is equal to: 2 4 (A) (B) 17 17 64 128 (C) (D) 17 17
85.
3 1 If cos1 , tan1 , where 0 , , then is equal to: 2 5 3 9 9 (A) sin1 (B) cos1 5 10 5 10 9 9 (C) tan 1 (D) tan 1 14 5 10
86.
The equation of a plane containing the line of intersection of the planes 2x y 4 0 and y 2z 4 0 and passing through the point (1, 1, 0) is: (A) x 3y z 4 (B) 2x z 2 (C) x 3y 2z 2 (D) x y z 0
87.
The sum of the squares of the lengths of the chords intercepted on the circle, x 2 y 2 16, by the lines, x y n,n N , where N is the set of all natural numbers is: (A) 320 (B) 160 (C) 105 (D) 210
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JEE-MAIN-2019 (8th Apr-First Shift)-PCM-18 88.
The greatest value of c R for which the system of linear equations x cy cz 0 cx y cz 0 cx cy z 0 has a non – trivial solution, is: 1 (A) –1 (C) 2 (B) 2 (D) 0
89.
Let f : 0,2 R be a twice differentiable function such that f " x 0, for all x 0,2 . If x f x f 2 x , then is: (A) increasing on (0, 2) (B) decreasing on (0, 2) (C) decreasing on (0, 1) and increasing on (1, 2) (D) increasing on (0, 1) and decreasing on (1, 2)
90.
Let A and b be two non – null events such that A B . Then, which of the following statements is always correct? (A) P A B 1 (B) P A B P A (C) P A B P B P A
(D) P A B P A
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JEE-MAIN-2019 (8th Apr-First Shift)-PCM-19
JEE (Main) – 2019 ANSWERS PART A – PHYSICS 1. 5. 9. 13. 17. 21. 25. 29.
Bonus A D D Bonus D A B
2. 6. 10. 14. 18. 22. 26. 30.
D D D B D B C D
3. 7. 11. 15. 19. 23. 27.
Bonus D B D C A A
4. 8. 12. 16. 20. 24. 28.
A B A B A A D
34. 38. 42. 46. 50. 54. 58.
C D B B B B D
64. 68. 72. 76. 80. 84. 88.
A D B B D A B
PART B – CHEMISTRY 31. 35. 39. 43. 47. 51. 55. 59.
B B D A D C C D
32. 36. 40. 44. 48. 52. 56. 60.
C D A B A A A B
33. 37. 41. 45. 49. 53. 57.
D B C D A B C
PART C – MATHEMATICS 61. 65. 69. 73. 77. 81. 85. 89.
D C A D D A A C
62. 66. 70. 74. 78. 82. 86. 90.
C C D C D B D D
63. 67. 71. 75. 79. 83. 87.
A C C A B D D
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JEE-MAIN-2019 (8th Apr-First Shift)-PCM-20
HINTS AND SOLUTIONS PART A – PHYSICS 1.
Pressure is defined as normal force per unit area. Force is calculated as change in momentum/ time. By this answer is 2N/m2 None of the option matches so this question must be Bonus. Detailed solution is as following: Magnitude of change in momentum per collision = 2 mv Force N(2mv) Pressure = Area 1 22 26 10 2 10 10 4 = 1 = 2 N/m2
2.
According to work energy theorem. Work done by force on the particle = Change in KE Work done = Area under F-x graph = F dx =
(2 3) 1 2 W = KEfinal – KEinitial = 6.5 KEinitial = 0 KEfinal = 6.5 J 22
3.
There will be 3 phenomenon (i) Refraction from lens (ii) Reflection from mirror (iii) Refraction from lens After these phenomena. Image will be on object and will have same size. None of the option depicts so this question is Bonus.
1st refraction u = –40 cm ; f = +20 cm v = +40 cm (image I1) and m1 = –1 for reflection u = –20 cm ; f = –10 cm
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JEE-MAIN-2019 (8th Apr-First Shift)-PCM-21 v = –20 cm (image I2) and m2 = –1 2nd refraction u = –40 cm ; f = +20 cm v = +40 cm (image I3) and m3 = –1 Total magnification = m1 × m2 × m3 = –1 and final image is formed at distance 40 cm from convergent lens and is of same size as the object. 4.
a A a ˆi ; aB ajˆ aC a ˆi ; aD ajˆ ma aa mb ab mc ac mda d acm ma mb mc md ma ˆi 2mjˆ 3ma ˆi 4majˆ acm 10m a a a ˆ ˆ 2ma ˆi 2majˆ = = ˆi ˆj = ij 10m 5 5 5
5.
6.
dT eA(T 4 T04 ) dt dT eA 4 dT 4eAT03 (T T04 ) ; ( T) dt ms dt ms T = T0 + (Ti – T0) e–kt 4eAT03 where k ms 4eAT03 dT k ; k vs dt dT 1 dt s ASA = 2000 × 8 × 102 = 16 × 105 BSB = 4000 × 103 = 4 × 106 ASA < BSB dT dT dt A dt B ms
T0 2
m 2 k 10
A = A0e–1/
A0 , t= e 2m 2m 2 2 = 104 s t== b B R t 10 4 No of oscillation 5000. T0 2 / 10 for A
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JEE-MAIN-2019 (8th Apr-First Shift)-PCM-22 7.
LIDI = I2R E 1 E L ( e t / 2 ) (1 e t / 2 ) 10 10 2 10 e 1/ 2 1 e1/ 2 ; t = 2n2
8.
Let mass per unit length of wires are 1 and 2 respectively. Materials are same, so density is same. r 2L 4r 2L and 2 4 L L Tension in both are same = T, let speed of wave in wires are V1 and V2 V V V V V1 1 & f02 2 2L 2L 2L 4L Frequency at which both resonate is L.C.M. of V both frequencies i.e. . 2L
1
Hence number of loops in wires are 1 and 2 respectively
So, ratio of number of antinodes is 1 : 2. 9.
Tensile stress in wire will be Tensile force = Cross section Area mg 4 3.1 = Nm2 = 3.1 × 106 Nm–2 2 R 4 10 6
10.
A = 10–4 m2 Emax = 106 V/m C = 15 F k A Cd k C 0 ; d 0 A 15 1012 500 10 6 15 5 = = 8.465 8.86 1012 10 4 8.86 k 8.5 k
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JEE-MAIN-2019 (8th Apr-First Shift)-PCM-23 11.
V(t) = 220 sin(100 t) volt time taken, 1 t 3 sec 100 300 = 3.3 ms
12.
To minimize attenuation, wavelength of carrier waves is close to 1500 nm.
13.
Reynolds number =
14.
vd Volume flow rate = v × r2 100 103 1 v 60 25 104 2 v m /s 3 103 2 10 10 2 Reynolds number = = 2 × 104 103 3 Order 104 2 1 Energy of catapult = Y A 2 1 = Kinetic energy of the ball = mv 2 2 2 1 20 1 Therefore, Y 32 10 6 42 10 2 2 10 2 (20)2 2 42 2 6 2 Y = 3 × 10 Nm
15.
h h ˆ P1 ˆi and P2 j 1 2 Using momentum conservation h ˆ h ˆ i j P P1 P2 = 1 2
P
2
h h 1 2 2
h h 1 2 1 1 1 2 2 2 1 2 h
2
2
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JEE-MAIN-2019 (8th Apr-First Shift)-PCM-24 16.
Net force on particle towards centre of circle is GM2 GM2 FC 2 2 2a 2 a GM2 1 = 2 2 a 2 This force will act as centripetal force. Distance of particle from centre of circle is a . 2 a mv 2 r , FC r 2 2 2 mv GM 1 2 2 a a 2 2 GM 1 v2 1 a 2 2
v2 17.
GM GM 1.35 ; v 1.16 a a
If we approximate the angle 2 as 30° initially then answer will be closer to 57000. but if we solve thoroughly, answer will be close to 55000. So both the answers must be awarded. Detailed solution as following. Exact solution By Snell’s law 1.sin 40° = (1.31) sin 2 .64 64 sin 2 .49 1.31 131 64 64 64 d Now tan 2 13065 114.3 x (131)2 (64)2 Now number of reflections 2 64 64 105 = 114.3 20 10 6 114.3 55991 55000
Approximate solution By Snell’s law 1.sin 40° = (1.31)sin 2
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JEE-MAIN-2019 (8th Apr-First Shift)-PCM-25
0.64 64 0.49 1.31 131 If assume 2 30° d Tan 300 = x 3d x Now number of reflections 2 105 = 3d 3 20 10 6 3 57735 57000 Sin 2 =
18.
Magnetic moment of coil = NIA ˆj = NI ( r 2 ) ˆj Torque on loop (coil) = M B ˆ = NI( r 2 ) B sin 90o ( k) ˆ = NIr 2B ( k)
19.
If we take the position of ship ‘A’ as origin then positions and velocities of both ships can be given as: v A (30iˆ 50ˆj)km / hr vB 10iˆ km / hr ; rA 0iˆ 0ˆj rB (80iˆ 150 ˆj) km Time after which distance between them will be minimum r v t BA 2BA ; v BA Where rBA (80iˆ 150 ˆj) km VBA 10iˆ (30iˆ 50ˆj) ( 40iˆ 50 ˆj) km / hr ˆ (80iˆ 150ˆj) ( 40i 50h) t= 2 40i 50hˆ
= 20.
3200 7500 10700 hr hr 2.6 hrs 4100 4100
The direction of propagation of an EM wave is direction of E B. ˆi ˆj Bˆ Bˆ kˆ
E E 6 B B C 3 108 B = 2 × 10–8 T along z direction. C
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JEE-MAIN-2019 (8th Apr-First Shift)-PCM-26 21.
Suppose ‘m’ gram of water evaporates then, heat required Qreq = mLv Mass that converts into ice = (150 – m) So, heat released in this process Qrel = (150 – m) Lf Now, Qrel = Qreq (150 – m) Lf = mLv M(Lf + Lv) = 150 Lf
m
150L f ; m = 20 g Lf Lv
22.
Energy released for tension n = 2 to n = 1 of hydrogen atom 1 1 E 13.6Z2 2 2 n1 n2 Z = 1, n1 = 1, n2 = 2 1 1 E 13.6 1 2 2 1 2 3 E 13.6 eV 4 + For He ion z = 2 (A) n = 1 to n = 4 15 1 1 E 13.6 22 2 2 13.6 eV 4 1 4 (B) n = 2 to n = 4 3 1 1 E 13.6 22 2 2 13.6 eV 4 1 4 (C) n = 2 to n = 5 1 21 1 E 13.6 22 2 2 13.6 eV 5 25 2 (D) n = 2 to n = 5 1 5 1 E 13.6 22 2 2 13.6 eV 3 9 2
23.
When red is replace with green 1st digit changes to 5 so new resistance will be 500 .
24.
9 = Vz + VR1 VZ = 5.6 V VR1 9 5.6 VR1 3.4 IR1
VR1 R
3.4 ; IR1 = 17 mA 200
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JEE-MAIN-2019 (8th Apr-First Shift)-PCM-27 Vz VR2 IR2 (R2 )
5.6 IR2 ; IR2 7 mA 800 Iz = (17 – 7) mA = 10 mA R
25.
M 0r(2rdr) 0
0 2 R 3 3
R
I0 (MOI about COM)
0r(2rdr) r 2 0
0 2 R 5 5
By parallel axis theorem I = I0 + MR2 0 2R5 0 2R3 8 R2 0 2R5 5 3 15 8 2 = MR 5
=
26.
Dimension of
0 0
[0] = [M–1L–3T4A2] [0] = [MLT–2A–2] 1
Dimension of
0 M1L3 T 4 A 2 2 0 MLT 2 A 2
= [M–2L–4T6A4]1/2 = [M–1L–2T3A2]
27.
E1 E2 E3 2R1 R 2 2R1 Eeq 1 1 1 2R1 R 2 2R1
2 2 1 2
28.
4 2 1 2
4 2 5 10 = 3.3 1 3 3 2 2
As given in the first condition: Both conducting spheres are shown. kQ kQ Vin Vout r1 r2
1 1 = kQ V r1 r2
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JEE-MAIN-2019 (8th Apr-First Shift)-PCM-28 In the second condition: Shell is now given charge –4Q. kQ 4kQ kQ 4kQ Vin Vout r2 r2 r2 r1 =
kQ kQ r1 r2
1 1 = kQ V r1 r2 Hence, we also obtain that potential difference does not depend on charge of outer sphere. P. d. remains same. 29.
Given
a1 1 a2 3 2
I a 1 Ratio of intensities, 1 1 I2 a2 9 2
2 Imax I1 I2 1 3 Now, 4 Imin I1 I2 1 3
30.
qE 5 10 6 2000 mg 2 10 3 10 1 tan = tan1 (0.5) 2 tan
PART B – CHEMISTRY 31.
For cubic unit cell, only FCC has octahedral and tetrahedral voids. 1 ZB = 4, ZA = 4 2,Zo 8 4 Formula = A2B2O8 = AB 2 O 4
32.
Electrophilic substitution reaction takes place.
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JEE-MAIN-2019 (8th Apr-First Shift)-PCM-29 33.
For isoelectronic species the size is compared by nuclear charge. 1 Size Z Nuclear Charge
34.
Basic strength order (CH3CH2)2NH > CH3CH2NH2 > NH3 More the number of +I groups, higher is the basic strength.
35.
Since CN– and NH3 are strong field ligands, low spin complexes are formed. Complex Configuration No. of unpaired electrons 3 0 [V(CN)6]4– 3 t e 2g g
2+
t 42g eg0
2
[Ru(NH3)6]3+
t52g eg0
1
[Fe(CN)6]4–
t 62g eg0
0
[Cr(NH3)6]
Magnetic moment is directly proportional to number of unpaired electrons. 36.
m-cresol is the right answer. 37.
B2H6 3H2O 2H3BO3 3 H2 B2H6 3 O2 B 2O3 3H2O
38.
Sm3+(4f5) = yellow colour, other ions have stable electron configurations with half filled or full-filled electron configuration.
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JEE-MAIN-2019 (8th Apr-First Shift)-PCM-30 H
39.
O - CH3
OCH3
Br SN 2
Conc.HBr excess
CH = CH2
CH = CH2 OH
OH HBr
CH - CH3
CH3Br
CH = CH2
Br OH
Br
40.
neq. CaCO3 = neqCa(HCO3)2 + neqMg(HCO3)2 W 0.81 0.73 or, 2 2 2 100 162 146 w = 1.0 Volume of water = 100 mL Mass of water = 100 g 1.0 Hardness = 106 = 10000 ppm 100 OH
41.
(neq = Number of equivalent)
Neutral ve phenolic group is absent FeCl 3
CH3 C - C2H5 O
42.
x
RMgX vereaction Grignard Re agent
Acidic H C
NaOH I
2 vereaction Iodoform
are present
O
Test Fehling' s solution
ve no CHO group ispresent
y
r = K[A] [B] 0.045 = K(0.05)x (0.05)y 0.090 = K(0.10)x (0.05)y 0.72 = K(0.20)x (0.10)y Dividing (1) by (2) we get
. . . . (1) . . . . (2) . . . . (3)
x
0.045 0.05 x 1 0.090 0.10 Dividing (2) by (3)
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JEE-MAIN-2019 (8th Apr-First Shift)-PCM-31 x
y
0.090 0.10 0.05 y2 0.720 0.20 0.10 Hence, r = K[A][B]2 43.
Fridel-craft acylation. –Cl group is an ortho & para directing 44.
Plastics are non-biodegradable.
45.
Reduction followed by substitution reaction T2
46.
H n Cp.m dT 3
T1
1000
23 0.01T dT
300
3 23 1000 300
0.01 1000 2 300 2 2
= 61950 J 62 kJ 47.
48.
Hydration enthalpy depends upon ionic potential (charge/size). As ionic potential increases hydration enthalpy increases. q hydH0 r
x Kp1/n m Taking log from both sides
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JEE-MAIN-2019 (8th Apr-First Shift)-PCM-32 x 1 logK logP m n 1 2 slope n 3 x Kp2/3 m log
49.
Ptotal= X A PA0 XB PB0 0.5 400 0.5 600 500 mmHg Now, mole fraction of A in vapour P 0.5 400 YA A 0.4 and mole fraction of B in vapour Ptotal 500 YB = 1 – 0.4 = 0.6
50.
For strongest oxidising agent, standard reduction potential should be highest. Peroxy oxygen (–O – O–) is reduced to oxide (O 2–) in the change
51.
According to Aufbau principle, the energy sequence is 3p < 3d < 4p < 4d
52.
53.
Gabriel phthalimide synthesis:
For branched chain RX, elimination reaction takes place.
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JEE-MAIN-2019 (8th Apr-First Shift)-PCM-33 54.
According to the first law of thermodynamics q = U – w
55.
Both nitrogen & oxygen are donating atoms.
56.
The priority of COOH is higher that OH. COOH is the functional group.
3-Hydorxy-4-methylpentanoic acid 57.
n-factors of KMnO4 = 5, n-factor of FeSO4 = 1 n-factors of FeC2O4 = 3, Fe2(SO4)3 does not react n-factors of Fe2(C2O4)3 = 6, neq KMnO4 = neq[FeC2O4 + Fe2(C2O4)3 + FeSO4] or, x 5 = 1 3 + 1 6 + 1 1 x=2
58.
The upper stratosphere consists of ozone (O3), which protect us from harmful ultraviolet (UV) radiations coming from sun. The layer get depleted by CFC’s
59.
60.
Ellingham diagram which are the curves of the graph between G and T helps in predicting the feasibility of thermal reduction of ores.
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JEE-MAIN-2019 (8th Apr-First Shift)-PCM-34
PART C – MATHEMATICS 61.
Vector perpendicular to plane containing the vectors ˆi ˆj kˆ d ˆi 2ˆj 3kˆ is parallel to vector ˆi ˆj kˆ 1 1 1 ˆi 2ˆj kˆ 1 2 3 Required magnitude of projection 2iˆ 3 ˆj kˆ . ˆi 2ˆj kˆ ˆi 2ˆj kˆ
2 6 1 6
3
6
3 2
62.
We have dy dy 1 2y 1 1 dx dx P 2 t 2 , t 2t 1 t 2 9 1 P , 4 2 So, shortest distance 9 2 7 4 4 2 4 2
63.
x 1
2
1 0 x 1 i, 1 i
n
n 1 i 1 n (least natural number) = 4
64. 4th place
2nd place
8th place
6th place
(even places) 3! 6! Number of such numbers 4C3 180 2! 2! 4!
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JEE-MAIN-2019 (8th Apr-First Shift)-PCM-35 5x 5x x 2 sin cos 2 2 dx 2x dx x x sin 2 sin cos 2 2 2 sin3x sin 2x dx sin x sin
65.
3 sin x 4 sin3 x 2 sin x cos x dx sin x
3 4 sin2 x 2cos x dx 3 2 1 cos 2x 2cos x dx 1 2cos 2x 2cos x dx
x sin 2x 2 sin x c 66.
AP OP AO 4 2
h2 k 1 h2 k 2 1 4 2
h2 k 1 h2 k 2 3 2
h2 k 1 9 h2 k 2 6 h2 k 2 2k 8 6 h2 k 2 k 4 3 h2 k 2 k 2 16 8k 9 h2 k 2
2
2
9h 8k 8k 16 0 Locus of P is 9x 2 8y 2 8y 16 0 67.
and 2 4 4 3 4 5 5 If cos then tan and if sin then tan 5 3 13 12 (since here lies in the first quadrant) 0
Now tan 2 tan 4 5 63 3 12 4 5 16 1 tan .tan 1 . 3 12 tan tan
68.
cos sin A sin cos cos sin cos sin A2 sin cos sin cos
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JEE-MAIN-2019 (8th Apr-First Shift)-PCM-36 cos 2 sin 2 sin 2 cos 2 cos 2 sin 2 cos sin A3 sin 2 cos 2 sin cos cos 3 sin 3 sin 3 cos 3 cos 32 sin 32 0 1 Similarly A 32 sin 32 cos 32 1 0 cos 32 0 and sin32 1 32 4n 1 , n I 2 4n 1 , n I 64 for n 0 64
69.
70.
1 x f x loge , x 1 1 x 2x 1 1 2x 2 2x f n 2 1 x 1 2x 1 x2 x 12 1 x 2n n 2f x x 12 1 x 3 1 cos x sin x 2 Consider cot 1 2 1 3 sin x sin x 2 2 sin x 3 cot 1 cos x 3 cot 1 tan x tan1 tan x 3 2 3 x x ; 0 x 2 3 6 6 x 7 x ; x 2 3 2 6 6
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JEE-MAIN-2019 (8th Apr-First Shift)-PCM-37
2 x ; 0 x 6 6 2y 2 7 x ; 6x2 6
2 6 x . 1 ; 0 x 6 dy 2 dx 7 2 x . 1 ; x 6 6 2
x 2 x
71.
x 2
x
2
x 2
x 4 20 2
4 x 20
x 2 20
x 2 2 (not possible) or
x 2 1
x 2 1, 1 x 3,1 x 9, 1 Sum = 10 sin2 x 2 2 1 cos x x lim x 0 1 cos x x2
72.
73.
1
2
. 2 2 1 2
4
2
2. 20C0 5. 20 C1 8. 20C2 11. 20C3 ...... 62. 20C 20 2 20
3r 2 20Cr r 0
20
20
3 r . 20Cr 2 r 0
20
Cr
r 0
20 20 3 r 19Cr 1 2.220 r 0 r 60.219 2.220 225
74.
dy 2x 1 2 y 2 dx x 1 x2 1
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JEE-MAIN-2019 (8th Apr-First Shift)-PCM-38
I.F. e
x2 1
n x 2 1
So, general solution is y. x 2 1 tan1 x c As y 0 0
c 0
1
tan x x2 1 a, y 1 32 1 1 a a 4 16
y x
As,
75.
2 x.cos x g f x n f x n 2 x.cos x /4 2 x.cos x 2 x.cos x I n dx 2 x.cos x 2 x .cos x 0 /2
0 dx 0 log 1 e
0
76.
Required Area 1
x 2 3x dx Area of 0
rectangle PQRS 11 59 8 6 6
77.
Let 7 observations be x1, x 2 , x3 , x 4 , x 5 , x 6 , x 7 7
x 8 xi 56
………..(1)
i 1
Also 2 16 1 7 16 xi2 x 7 i1
2
16
1 7 2 xi 64 7 i1
7 xi2 560 ……..(2) i1 Now, x1 2, x 2 4, x 3 10, x 4 12, x 5 14
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JEE-MAIN-2019 (8th Apr-First Shift)-PCM-39
x6 x 7 14, (from (1)) and x 26 x 27 100 (from (2)) 2
x 26 x 72 x 6 x 7 2x 6 x 7 x 6 x 7 48
78.
Now, MP. 10iˆ 7ˆj kˆ 0
1 2 Length of perpendicular 1 49 PM 0 4 4 50 25 5 , 4 4 2 which is greater than 3 but less than 4.
79.
The contrapositive of a statement p q is ~ q ~ p Here, p : your are born in India q : you are citizen of India So, contrapositive of above statement is “If you are not a citizen of India, then you are not born in India”.
80.
S A sum of numbers between 100 and 200 which are divisible by 7. S A 105 112 ..... 196 14 SA 105 196 2107 2 SB Sum of numbers between 100 and 200 which are divisible by 13. 8 SB 104 117 ..... 195 104 195 1196 2 SC Sum of numbers between 100 and 200 which are divisible by 7 and 13.
SC 182
H.C. F. (91, n) > 1 S A SB SC 3121 81.
f x 9x 4 12x3 36x 2 25 f x 36x 3 36x 2 72x
36x x 2 x 2
36x x 1 x 2
Point of minima 2, 1 S1 Point of maxima 0 S2 82.
6
x x3 1
x x3 1
6
2 3 2 6 C0 x 6 6C2 x 4 x3 1 6C4 x 2 x 3 1 6C6 x3 1
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JEE-MAIN-2019 (8th Apr-First Shift)-PCM-40
2 6 C0 x 6 6C2 x7 6 C2 x 4 6C4 x 8 6C4 x 2 2 6C4 x5 x 9 1 3x 6 3x3 Sum of coefficient of even powers of x 2 1 15 15 15 1 3 24 15 3t t 5 15 3t 15 3t t or t 5 5 15 15 t or t 8 2 15 15 st So, P , 1 quadrant 8 8 15 15 nd or P , II Quadrant 2 2
83.
Now,
84.
4a2 b 2 8 ………..(1) dy 4x 2 dx 1,2 y
4a 1 b 2 b 8a b2 64a2
68a2 8 , a2
85.
86.
2 17
3 1 cos , tan 5 3 4 tan 3 4 1 9 tan 3 3 4 1 13 1 , 3 3 9 sin 5 10 9 sin1 5 10 The required plane is 2x y 4 y 2z 4 0 it passes through (1, 1, 0) 2 1 4 1 4 0 3 3 0 1 xyz0
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JEE-MAIN-2019 (8th Apr-First Shift)-PCM-41
87.
p
n n , but 4 n 1,2,3, 4,5 2 2
Length of chord AB 2 16
n2 2
64 2n2 (say) For n 1, 2 62
n 2, 2 56 n 3, 2 46 n 4, 2 32 n 5, 2 14 Required sum = 62 + 56 + 46 + 32 + 14 = 210 88.
For non – trivial solution D=0 1 c c c
1
c 0 2c 3 3c 2 1 0
c
c
1 2
c 1 2c 1 0 Greatest value of c is
89.
1 2
x f x f 2 x ' x f x f ' 2 x
………(1)
Since f " x 0 f x is increasing x 0,2 Case – I : When x 2 x x 1 ' x 0 x 1,2 x is increasing on 1, 2 Case – II : When x 2 x x 1 ' x 0 x 0, 1 x is decreasing on (0, 1)
90.
P A B
P A B P B
PA P B
(as A B P A B P A )
P A B P A
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FIITJEE Solutions to JEE(Main)-2019 Test Date: 8th April 2019 (Second Shift)
PHYSICS, CHEMISTRY & MATHEMATICS Paper - 1 Time Allotted: 3 Hours
Maximum Marks: 360
Please read the instructions carefully. You are allotted 5 minutes specifically for this purpose.
Important Instructions: 1.
The test is of 3 hours duration.
2.
This Test Paper consists of 90 questions. The maximum marks are 360.
3.
There are three parts in the question paper A, B, C consisting of Physics, Chemistry and Mathematics having 30 questions in each part of equal weightage. Each question is allotted 4 (four) marks for correct response.
4.
Out of the four options given for each question, only one option is the correct answer.
5.
For each incorrect response 1 mark i.e. ¼ (one-fourth) marks of the total marks allotted to the question will be deducted from the total score. No deduction from the total score, however, will be made if no response is indicated for an item in the Answer Box.
6.
Candidates will be awarded marks as stated above in instruction No.3 for correct response of each question. One mark will be deducted for indicating incorrect response of each question. No deduction from the total score will be made if no response is indicated for an item in the answer box.
7.
There is only one correct response for each question. Marked up more than one response in any question will be treated as wrong response and marked up for wrong response will be deducted accordingly as per instruction 6 above.
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JEE-MAIN-2019 (8th Apr-Second Shift)-PCM-2
PART – A (PHYSICS) 1.
A solid sphere and solid cylinder of identical radii approach an incline with the same linear velocity (see figure). Both roll without slipping all throughout. The two climb h maximum heights hsph and hcyl on the incline. The radio sph is given by: hcyl
(A) 1 (C) 2.
4 5 14 (D) 15
(B) 2 5
In the circuit shown, a four wire potentiometer is made of a 400 cm long wire, which extends between A and B. The resistance per unit length of the potentiometer wire is r = 0.01 /cm. If an ideal voltmeter is connected as shown with jockey J at 50 cm from end A, the expected reading of the voltmeter will be:
(A) 0.75V (C) 0.25V
(B) 0.20V (D) 0.50V
3.
A parallel plate capacitor has 1F capacitance. One of its two plates is given + 2C charge and the other plate, +4C charge. The potential difference developed across the capacitor is: (A) 3V (B) 1V (C) 5V (D) 2V
4.
If Surface tension (S), Moment of Inertia (I) and Planck’s constant (h), were to be taken as the fundamental units, the dimensional formula for linear momentum would be: (A) S1 / 2 I1 / 2 h0 (B) S1 / 2 I3 / 2 h1 (C) S3 / 2 I1 / 2 h0 (D) S1 / 2 I1 / 2 h1
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JEE-MAIN-2019 (8th Apr-Second Shift)-PCM-3 5.
Two magnetic dipoles X and Y are placed at a separation d, with their axes perpendicular to each other. The dipole moment of Y is twice that of X. A particle of charge q is passing through their mid-point P, at angle = 450 with the horizontal line as shown in the figure. What would be the magnitude of force on the particle at that instant? (d is much larger than the dimensions of the dipole)
M (A) 0 q 3 4 d / 2
(B) 0
2M (C) 0 q 3 4 d / 2
(D)
M 2 0 q 3 4 d / 2
6.
A damped harmonic oscillator has a frequency of 5 oscillations per second. The amplitude drops to half its value for every 10 oscillations. The time it will take to drop to 1 of the original amplitude is close to: 1000 (A) 10s (B) 100s (C) 50s (D) 20s
7.
A convex lens (of focal length 20 cm) and a concave mirror, having their principal axes along the same lines, are kept 80 cm apart from each other. The concave mirror is to the right of the convex lens. When an object is kept at a distance of 30 cm to the left o the convex lens, its image remains at the same position even if the concave mirror is removed. The maximum distance of the object for which this concave mirror, by itself would produce a virtual image would be: (A) 20 cm (B) 10 cm (C) 30 cm (D) 20 cm
8.
A cell of internal resistance r drives current through an external resistance R. The power delivered by the cell to the external resistance will be maximum when: (A) R = 0.001 r (B) R = 1000 r (C) R = 2r (D) R = r
9.
In a simple pendulum experiment for determination of acceleration due to gravity (g), time taken for 20 oscillations is measured by using a watch of 1 second least count. The mean value of time taken comes out to be 30s.. The length of pendulum is measured by using a meter scale of least count 1 mm and the value obtained is 55.0 cm. The percentage error in the determination of g is close to: (A) 0.7 % (B) 3.5% (C) 6.8% (D) 0.2%
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JEE-MAIN-2019 (8th Apr-Second Shift)-PCM-4 10.
The temperature, at which the root mean square velocity of hydrogen molecules equals their escape velocity from the earth is closest to: [Boltzmans Constant kB 1.38 1023 J / K Avogadro number NA 6.02 10 26 / kg Radius of Earth: 6.4 10 6 m Gravitation acceleration on Earth = 10 ms-2] (A) 800k (B) 104 K 5 (C) 3 10 (D) 650 K
11.
A particle starts from origin O from rest and moves with a uniform acceleration along the positive x – axis. Identify all figures that correctly represent the motion qualitatively. (a = acceleration, v = velocity, x = displacement, t = time) (a)
(b)
(c)
(d)
(A) a, b, c (C) b, c 12.
13.
(B) a (D) a, b, d
The election field in a region is given by E Ax B ˆi where E is in NC-1 and x in meters. The values of constants are A = 20 SI unit and B = 10 SI unit. If the potential at x =1 is V1 and that at x = – 5 is V2 then V1 – V2 is: (A) 320 V (B) – 48 V (C) – 520 V (D) 180 V
Young’s moduli of two wires A and B are in the ratio 7 : 4. Wire A is 2 m long and has radius R. Wire A is 2 m long and has radius R. Wire B is 1.5 m long and has radius 2 mm. If the two wires stretch by the same length for a given load, then the value of R is close to: (A) 1.3 mm (B) 1.5 mm (C) 1.7 mm (D) 1.9 mm
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JEE-MAIN-2019 (8th Apr-Second Shift)-PCM-5
14.
A body of mass m1 moving with an unknown velocity of v1ˆi undergoes a collinear collision with a body of mass m2 moving with a velocity v 2 ˆi . After collision m1 and m2 move with velocities of v ˆi and v ˆi respectively. 3
4
If m2 = 0.5 m1 and v3 = 0.5 v1 then v1 is: v (A) v 4 2 2 (C) v 4 v 2
v2 4 (D) v 4 v 2
(B) v 4
40
Ca and 16 O is close to: (B) 5 (D) 1
15.
The ratio of mass densities of nuclei of (A) 0.1 (C) 2
16.
A rectangles solid box of length 0.3 m is held horizontally, with one of its sides on the edge of a platform of height 5.. When released, it slips off the table in a very short time = 0.01 s, remaining essentially horizontal. The angle by which it would rotate when it hits the ground will be (in radians) close to:
(A) 0.5 (C) 0.02
(B) 0.6 (D) 0.28
17.
An electric dipole is formed by two equal and opposite charges q with separation d. The charges the same mass m. It is kept in a uniform electric field E. If it slightly rotated from its equilibrium orientation, then its angular frequency is: qE 2qE (A) (B) 2md md qE qE (C) (D) 2 md md
18.
The magnetic field of an electromagnetic wave is given by: Wb B 1.6 10 6 cos 2 107 z 6 1015 t 2iˆ ˆj 2 2 The associated electric field will be: V (A) E 4.8 102 cos 2 107 z 6 1015 t ˆi 2jˆ m V (B) E 4.8 102 cos 2 107 z 6 1015 t 2ˆj 2tˆ m V (C) E 4.8 102 cos 2 107 z 6 1015 t ˆi 2jˆ m V (D) E 4.8 102 cos 2 107 z 6 1015 t 2iˆ ˆj m
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JEE-MAIN-2019 (8th Apr-Second Shift)-PCM-6
19.
A common emitter amplifier circuit, built using an npn transistor, is shown in the figure. Its dc current gain is 250, RC = 1k and VCC = 10V. What is the minimum base current for VCE to reach saturation?
(A) 7A (C) 10 A 20.
21.
A positive point charge is released from rest at a distance r0 from a positive line charge with uniform density. The speed (v) of the point charge, as a function of instantaneous distance r from line charge, is proportional to
(A) v e r /ro
r (B) v In r0
r (C) v In r0
r (D) v r0
Let A 1 3, A 2 5, and A 1 A 2 5. The value of 2A 1 3A 2 3A 1 2A 2 is:
(A) –106.5 (C) –118.5 22.
(B) 40A (D) 100 A
(B) –112.5 (D) –99.5
In the figure shown, what is the current (in Ampere) drawn from the battery? You are given R1 15,R2 10, R3 20, R4 5,R5 25, R6 30, E 15V
(A) 13/24 (C) 9/32
(B) 7/18 (D) 20/3
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JEE-MAIN-2019 (8th Apr-Second Shift)-PCM-7 23.
In a line of sight radio communication, a distance of about 50 km is kept between the transmitting and receiving antennas. If the height of the receiving antenna is 70m, then the minimum height of the transmitting antenna should be: (Radius of the Earth = 6.4 10 6 m) (A) 32 m (B) 40 m (C) 51 m (D) 20 m
24.
A rocket has to be launched from each in such a way that it never returns. If E is the minimum energy delivered by the rocket launcher, what should be the minimum energy that the launcher should have if the same rocket is to be launched from the surface of the moon? Assume that the density of the earth and the moon are equal and that the earth’s volume is 64 times the volume of the moon. E E (A) (B) 32 16 E E (C) (D) 64 4
25.
A circuit connected to an ac source of emf e = e0 sin (1000t) with t in seconds, gives a phase difference of between the emf e and current i. Which of the following circuits 4 will exhibit this? (A) RC circuit with R = 1 k and C = 1 F (B) RL circuit with R = 1 k and L = 10 mH (C) RL circuit with R = 1 k and L = 1 mH (D) RC circuit with R = 1 k and C = 10 F
26.
A nucleus A, with a finite de – broglie wavelength A, undergoes spontaneous fission into two nuclei B and C of equal mass. B flies in the same direction as that of A, while C flies in the opposite direction with a velocity equal to half of that of F. The de – Broglie wavelength B and C of B and C are respectively : (A) A , 2 A (B) 2 A , A (C) A , A (D) A , A 2 2
27.
Calculate the limit of resolution of a telescope objective having a diameter of 200 cm, if it has to detect light of wavelength 500 nm coming from a star. (A) 457.5 10 9 radian (B) 610 10 9 radian (C) 305 109 radian
(D) 152.5 10 9 radian
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JEE-MAIN-2019 (8th Apr-Second Shift)-PCM-8 28.
Two very long, straight and insulated wires are kept at 900 angle from each other In xy – plane as shown in the figure.
These wires carry currently of equal magnitude I, whose directions are shown in the figure. The met magnetic field at point P will be: I 0I (A) 0 xˆ yˆ (B) zˆ 2d d I (C) Zero (D) 0 xˆ yˆ 2d 29.
The given diagram shows four processes i.e., isochoric, isothermal and adiabatic. The correct assignment of the processes, in the same order is given by:
(A) a d c b (C) d a c b 30.
(B) a d b c (D) d a b c
A uniform rectangular thin sheet ABCD of mass M has length a and breadth b, as shown in the figure. If the shaded portion HBGO is cut off, the coordinates of the centre of mass of the remaining portion will be:
5a 5b (A) , 3 3 3a 3b (C) , 4 4
2a 2b (B) , 3 3 5a 5b (D) , 12 12
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JEE-MAIN-2019 (8th Apr-Second Shift)-PCM-9
PART –B (CHEMISTRY) 31.
The statement that is INCORRECT about the interstitial compound is: (A) they have metallic conductivity (B) they have high melting points (C) they are chemically reactive (D) they are very hard
32.
The major product of the following reaction is:
33.
34.
(A)
(B)
(C)
(D)
The major product of the following reaction is:
(A)
(B)
(C)
(D)
If p is the momentum of the fastest electron ejected from a metal surface after the irradiation of light having wavelength , then for (A)5 p momentum of the photoelectron, the wavelength of the light should be: (Assume kinetic energy of ejected photoelectron to be very high in comparison to work function)
3 4 4 (C) 9 (A)
1 2 2 (D) 3 (B)
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JEE-MAIN-2019 (8th Apr-Second Shift)-PCM-10 35.
The ion that has sp3d2 hybridization for the central atom is: (A) IF6 – (B) ICI4 – (C) ICI2 –
36.
37.
(D) BrF2 –
The major product obtained in the following reaction is:
(A)
(B)
(C)
(D)
k
k
1 2 For a reaction scheme. A B C if the rate of formation of B is set to be zero then the concentration of B is given by
k1 A k 2
38.
(A)
(B) k1 k 2 A
(C) k1k 2 A
(D)
k1 k 2 A
For the following reactions, equilibrium constants are given: 52 S s O2 g SO2 g ;K1 10 129 2S s 3O2 g 2SO3 g;K 2 10
The equilibrium constant for the reaction 2SO2 g O2 g 2SO3 g is: (A) 1077 (C) 10181 39.
(B) 1025 (D) 10154
Calculate the standard cell potential (in V) of the cell in which following reaction takes place:
Fe2 aq Ag aq Fe3 aq Ag s Given that:
E 0 Ag'/ Ag xV E 0Fe2 /Fe yV E 0Fe3 /Fe zV (A) x – z (C) x – y
(B) x + y – z (D) x + 2y – 3z
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JEE-MAIN-2019 (8th Apr-Second Shift)-PCM-11
40.
41.
The covalent alkaline earth metal halide X CI,Br,I is: (A) BeX2 (C) SrX2
(B) CaX2 (D) MgX2
The structure of Nylon – 6 is: (A)
(B)
(C)
(D)
42.
The IUPAC symbols for the element with atomic number 119 would be: (A) uun (B) uue (C) unh (D) une
43.
Which of the following compounds will show the maximum ‘enol’ content? (A) CH3COCH3 (B) CH3COCH2CONH2 (C) CH3COCH2COCH3 (D) CH3COCH2COOC2H5
44.
0.27 g of a long chain fatty acid was dissolved in 100 cm3 of hexane. 10 mL of this solution was added dropwise to the surface of water in a round watch glass. Hexane evaporates and a monolayer is formed. The distance from edge to centre of the watch glass is 10 cm. What is the height of the monolayer? [Density of fatty acid = 0.9 g cm-3, = 3] (A) 10-4 m (B) 10-8 m -2 (C) 10 m (D) 10-6 m
45.
The percentage composition of carbon by mole in methane is: (A) 80% (B) 20% (C)75% (D) 25%
46.
The Mond process is used for the: (A) Extraction of Mo (C) Purification of Zr and Ti
(B) Extraction of Zn (D) Purification of Ni
47.
Which one of the following alkenes when treated with HCI yields majorly an anti Markovnikov product? (A) H2N – CH = CH2 (B) F3C – CH = CH2 (C) CH3 O – CH = CH2 (D) CI – CH = CH2
48.
5 moles of an ideal gas at 100 K are allowed to undergo reversible compression till its temperature becomes 200 K. If C V 28 JK 1 mol1 , calculate U and pV for this process. (R = 8.0 J K–1 mol-1) (A) U 2.8kJ; pV 0.8kJ (B) U 14 kJ; pV 4kJ (C) U 14kJ; pV 18kJ
(D) U 14kJ; pV 0.8J
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JEE-MAIN-2019 (8th Apr-Second Shift)-PCM-12 49.
The calculated spin only magnetic moments (BM) of the anionic and cationic species of
Fe H2O 6 and Fe CN 6 , respectively, are: 2 (A) 0 and 5.92 (C) 0 and 4.9
(B) 2.84 and 5.92 (D) 4.9 and 0
50.
Fructose and glucose can be distinguished by: (A) Benedict’s test (B) Fehling’s test (C) Barfoed’s test (D) Seliwanoff’s test
51.
The major product in the following reaction is:
52.
(A)
(B)
(C)
(D)
Polysubstitution is a major drawback in: (A) Friedel Craft’s alkylation (C) Acetylation of aniline
(B) Reimer Tiemann reaction (D) Friedel Craft’s acylation
53.
The strength of 11.2 volume solution of H2 O 2 is: [Given that molar mass of H = 1 g mol-1 and O = 16 g mol-1] (A) 3.4% (B) 1.7% (C) 13.6% (D) 34%
54.
For the solution of the gases w, x, y and z in water at 298 K, the Henrys law constants (KH) are 0.5, 2 35 and 40 kbar, respectively. The correct plot for the given data is:
(A)
(B)
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JEE-MAIN-2019 (8th Apr-Second Shift)-PCM-13
(C)
55.
56.
(D)
Among the following molecules / ions C 22 ,N22 ,O 22 ,O 2 which one is diamagnetic and has the shortest bond length? (A) O 22
(B) C 22
(C) O2
(D) N22
The major product obtained in the following reactions is”
(A)
(B)
(C)
(D)
57.
The maximum prescribed concentration of copper in drinking water is: (A) 0.5 ppm (B) 3 ppm (C) 5 ppm (D) 0.05 ppm
58.
The correct statement about ICI5 and ICI4 is: (A) ICI5 is trigonal bipyramidal and ICI4 is tetrahedral (B) ICI5 is square pyramidal and ICI4 is square planar (C)ICI5 is square pyramidal and ICI4 is tetrahedral (D) Both are isostructural
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JEE-MAIN-2019 (8th Apr-Second Shift)-PCM-14 59.
The compound that inhibits the growth of tumors is: (A) trans - Pd CI 2 NH3 2 (B) trans - Pt CI 2 NH3 2 (C) cis - Pd CI 2 NH3 2
60.
(D) cis - Pt CI 2 NH3 2
Consider the bcc unit cells of the solid 1 and 2 with the position of atoms as shown below. The radius of atom B is twice that of atom A. The unit cell edge length is 50% more in solid 2 than in 1. What is the approximate packing efficiency in solid 2?
(A) 90% (C) 65%
(B) 75% (D) 45%
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JEE-MAIN-2019 (8th Apr-Second Shift)-PCM-15
PART–C (MATHEMATICS) 61.
Let S
x, y : y
2
x,0 and A is area of the regions S . If for a
, 0 4, A : A 4 2 : 5 then equals: 1
1
2 3 (A) 4 5
2 3 (B) 2 5 1
1
4 3 (C) 4 25
4 3 (D) 2 25
x
62.
x
Let f x g t dt, were g is a non zero even function. If f(x+5) =g(x) , then 0
f t dt 0
equals 5
(A)
x5
g t dt
(B) 2
x 5 x 5
(C)
g t dt
g t dt 5 5
(D) 5
5
g t dt
x5
63.
Suppose that the points (h, k), (1, 2) and (–3, 4) lie on the line L1. If a line L2 passing k through the points (h, k) and (4, 3) is perpendicular to L1, then equals: h 1 1 (A) (B) 7 3 (C) 3 (D) 0
64.
Let f x a x a 0 be written as f x f1 x f2 x , where f1 x is an even function and f2 x is an odd function. Then f1 x y f1 x y equals
65.
(A) 2f1 x f2 y
(B) 2f1 x f1 y
(C) 2f1 x y f2 x y
(D) 2f1 x y f1 x y
Let f : 1,3 R be defined as x x , 1 x 1 f x x x , 1 x 2 x x, 2x3 where [t] denotes the greatest integer less than or equal to t. Then, ƒ is discontinuous at: (A) four or more points (B) only one point (C) only two points (D) only three points
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JEE-MAIN-2019 (8th Apr-Second Shift)-PCM-16
66.
3 i i 1 , then 1 iz z5 iz8 2 2
If z
9
(A) –1 (C) (–1 + 2i)9 67.
(B) 1 (D) 0
Which one of the following, statements is not a tautology: (A) p v q p v q (B) p v q p (D) p ^ q p v q
(C) p p v q 68.
2y . If the x2 curve passes through the centre of the circle x 2 y 2 2x 2y = 0, then its equation is:
Given that the slope of the tangent to a curve y = y(x) at any point (x, y) is
(B) x loge y 2 x 1
(A) x loge y x 1 2
(C) x loge y 2 x 1 69.
is equal to:
(D) x loge y 2 x 1
If f 1 1,f ' 1 3 , then the derivative of f f f x f x (A) 33 (C) 9
2
at x = 1 is:
(B) 15 (D) 12
70.
The number of four – digit numbers strictly greater than 4321 that can be formed using the digits 0, 1, 2, 3, 4, 5 (repetition of digits is allowed) is: (A) 360 (B) 288 (C) 310 (D) 306
71.
If a point R (4, y, z) lies on the line segment joining the points P (2, –3, 4) and Q (8, 0, 10), then the distance of R from the origin is: (A) 53 (B) 6 (C) 2 14
(D) 2 21
72.
If the system of linear equations x 2y kz 1 2x y z 2 3x y kz 3 Has a solution (x, y, z) 0, then (x, y) lies on the straight line whose equation is: (A) 3x – 4y – 1 = 0 (B) 4x – 3y – 4 = 0 (C) 4x – 3y – 1 = 0 (D) 3x – 4y – 4 = 0
73.
A student score the following marks in five tests : 45, 54, 41, 57, 43. His score is not known for the sixth test. If the mean score is 48 in the six tests, then the standard deviation of the marks in six tests is: 10 100 (A) (B) 3 3 100 10 (C) (D) 3 3
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JEE-MAIN-2019 (8th Apr-Second Shift)-PCM-17 74.
The vector equation of the plane through the line of intersection of the planes x + y + z = 1 and 2x + 3y + 4z = 5 which is perpendicular to the plane x – y + z = 0 is: (A) r ˆi kˆ 2 0 (B) r. ˆi kˆ 2 0 (C) r. ˆi kˆ 2 0 (D) x ˆi kˆ 2 0
75.
If the lengths of the sides of a triangle are in A.P. and the greatest angle is double the smallest, then a ratio of lengths of the sides of this triangle: (A) 4:5:6 (B) 5:6:7 (C) 3:4:5 (D) 5:9:13
76.
In an ellipse, with centre at the origin, if the difference of the lengths of major axis and minor axis is 10 and one of the foci is at (0, 5 3 ), then the length of its latus rectum is: (A) 6 (B) 5 (C) 8 (D) 10
77.
The minimum number of times one has to toss a fair coin so that the probability; of observing at least one head is at least 90% is: (A) 2 (B) 3 (C) 4 (D) 5
78.
If the eccentricity of the standard hyperbola passing, through the point (4, 6) is 2, then the equation of the tangent to the hyperbola at (4, 6) is: (A) 2x – 3y + 10 = 0 (B) x – 2y + 8 = 0 (C) 2x – y – 2 = 0 (D) 3x – 2y = 0
79.
The number of integral values of m for which the equation 1 m2 x 2 2 1 3m x 1 8m 0 has no real root is:
(A) infinitely many (C) 3 20
80.
Then sum
1
k 2
k
(B) 2 (D) 1 is equal to:
k 1
11 219 21 (C) 2 20 2
(A) 2
81.
11 220 3 (D) 2 17 2
(B) 1
ˆ for some real x. Then a b r is possible if: Let a 3iˆ 2ˆj xkˆ and b ˆi ˆj k,
(A) r 5 (C)
3 2
3 3 r3 2 2
3 3 r 5 2 2 3 (D) 0 r 2
(B) 3
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JEE-MAIN-2019 (8th Apr-Second Shift)-PCM-18
82.
Then tangent and the normal lines at the point
3,1 to the circle x 2 y 2 4 and the x
– axis form a triangle. The area of this triangle (in square units) is: 1 4 (A) (B) 3 3 1 2 (C) (D) 3 3 83.
The tangent to the parabola y2 = 4x at the point where it intersects the circle x2 + y2 = 5 in the first quadrant, passes through the point: 1 4 3 7 (A) , (B) , 3 3 4 4 1 1 1 3 (C) , (D) , 4 2 4 4
84.
The height of a right circular cylinder of maximum volume inscribed in a sphere of radius 3 is: (A) 3 (B) 6 2 (C) 2 3 (D) 3 3 1 1 1 Let the numbers 2, b, c be in an A.P and A 2 b c . If det A 2,16 then ce lies 4 b 2 c 2
85.
86.
in the interval (A) 3,2 22/4
(B) 2 23/4 , 4
(C) {2,3)
(D) [4, 6]
Let f : R R be a differentiable function satisfying f’’(3) + f’(2) = 0. Then 1
1 f 3 x f 3 x lim is equal to: x 1 f 2 x f 2 (A) e2 (C) e
87.
If
x
dx 3
6 2/3
1 x
xf x 1 x 6
function f(x) is equal to: 1 (A) 2 2x 1 (C) 3 2x
1 3
(B) 1 (D) e–1
C where C is a constant of integration, then the
1 2x 3 3 (D) 2 x
(B)
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JEE-MAIN-2019 (8th Apr-Second Shift)-PCM-19 88.
89.
90.
Two vertical poles of heights, 20 m and 80m stand a apart on a horizontal plane. The height (in meters) of the point of intersection of the lines joining the top of each pole to the foot of the other, from his horizontal plane is: (A) 18 (B) 12 (C) 16 (D) 15 1 1 If the fourth term in the binomial expansion of 1 log x x 12 x 10 1, then the value of x is: (A) 104 (B) 100 (C) 103 (D) 10
6
is equal to 200, and x >
If three distinct number a, b, c are in G.P. and the equations ax2 + 2bc + c = 0 and dx2 + 2ex + f = 0 have a common root, then which one of the following statements is correct? d e f (A) , , are in A.P (B) d, e, f are in A.P a b c d e f (C) , , are in G.P (D) d, e, f are in G.P a b c
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JEE-MAIN-2019 (8th Apr-Second Shift)-PCM-20
JEE (Main) – 2019 ANSWERS PART A – PHYSICS 1. 5. 9. 13. 17. 21. 25. 29.
D B C C B C D D
2. 6. 10. 14. 18. 22. 26. 30.
C D B C A C D D
3. 7. 11. 15. 19. 23. 27.
B B D D B A C
4. 8. 12. 16. 20. 24. 28.
A D D A C B C
34. 38. 42. 46. 50. 54. 58.
C B B D D D B
64. 68. 72. 76. 80. 84. 88.
B D B B A C C
PART B – CHEMISTRY 31. 35. 39. 43. 47. 51. 55. 59.
C B D C B B B D
32. 36. 40. 44. 48. 52. 56. 60.
B B A D B A D A
33. 37. 41. 45. 49. 53. 57.
C A C B C A B
PART C – MATHEMATICS 61. 65. 69. 73. 77. 81. 85. 89.
C 62. A A 66. A A 70. C D 74. C C 78. C A 82. D D 86. B All options are incorrect
63. 67. 71. 75. 79. 83. 87. 90.
B A C A A B B A
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JEE-MAIN-2019 (8th Apr-Second Shift)-PCM-21
HINTS AND SOLUTIONS PART A – PHYSICS 1.
For solid sphere 1 1 2 v2 mv 2 mR 2 2 mghsph 2 2 5 R For solid cylinder 1 1 1 v2 mv 2 mR 2 2 mghcyl 2 2 2 R hsph 7 / 5 14 hcyl 3 / 2 15
2.
Resistance of wire AB = 400 × 0.01 = 4 3 i 0.5 A 6 Now voltmeter reading = i (Resistance of 50 cm length) = (0.5 A) (0.01 × 50) = 0.25 volt
3.
4.
Charges at inner plates are 1 C and –1 C. Potential difference across capacitor q 1C 1 10 6 C 1V = c 1F 1 10 6 Farad
p = k saIbhc Where k is dimensionless constant MLT–1 = (MT–2)a(ML2)b(ML2T–1)c a+b+c=1 2b + 2c = 1 –2a – c = –1 1 1 a b c 0 2 2 S1/2II/2h0
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JEE-MAIN-2019 (8th Apr-Second Shift)-PCM-22 5.
M 2M B1 2 0 ; B2 0 3 3 4 (d / 2) 4 (d / 2) B1 = B2 Bnet is at 45° ( = 45°)
Velocity of charge and Bnet are parallel so by F q(b B) force on charge particle is zero. 6.
A = A0e–t A A 0 after 10 oscillations 2 After 2 seconds A0 A 0 e (2) ; 2 = e2 2 n2 n2 2 ; 2 –t A = A0e A n2 n 0 t ; n1000 t A 2 6n10 3n10 2 t ; t n2 n2 t = 19.931 sec t 20 sec
7.
Image formed by lens 1 1 1 1 1 1 ; v u f v 30 20 v = +60 cm If image position does not change even when mirror is removed it means image formed by lens is formed at centre of curvature of spherical mirror. Radius of curvature of mirror = 80 – 60 = 20 cm. Focal length of mirror f = 10 cm for virtual image, object is to be kept between focus and pole. Maximum distance of object from spherical mirror for which virtual image is formed, is 10 cm.
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JEE-MAIN-2019 (8th Apr-Second Shift)-PCM-23
8.
E r R Power generated in R P = i2R E2R P (r R)2 dP For maximum power 0 dR (r R)2 1 R 2(r R) E2 0 (r R)4 Current i =
r=R 9.
T
30 sec 1 T sec, 20 20 L = 55 cm L = 1 mm = 0.1 cm 2 4 L g= T2 percentage error in g is g L 2T 100 100% g T L 0.1 = 55
10.
v rms
3RT m
1 2 20 100% 6.8%. 30 20
v escape 2gRe
v rms v escape
3RT 2gRe m 3 1.38 10 23 6.02 10 26 10 103 104 k 2 11.
Given initial velocity u = 0 and acceleration is constant At time t v = 0 + at v = at 1 Also x = 0(t) + at 2 2 1 2 x at 2
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JEE-MAIN-2019 (8th Apr-Second Shift)-PCM-24
Graph (a), (b) and (d) are correct. 12.
E (20x 10)iˆ 1
V1 V2 (20x 10)dx 5
V1 V2 (10x 2 10x)15 V1 – V2 = 10(25 – 5 – 1 – 1) V1 – V2 = 180 V
13.
Given: YA 7 YB 4
LA = 2m
AA = R2
LB = 1.5 m
AB = (2mm))2
F Y A L given F and are same
AY is same L
A A YA A B YB LA LB 7 ( R2 ) YB 2 4 (2 mm) YB 2 1.5 R = 1.74 mm
14.
Applying linear momentum conservation m1v1ˆi m 2 v 2 ˆi m1v 3 ˆi m 2 v 4 ˆi m1v 1 0.5 m1v 2 m1(0.5 v1 ) 0.5 m1v 4 0.5 m1v1 = 0.5 m1(v4 – v2) v1 = v 4 – v2
15.
Mass densities of all nuclei are same so their ratio is 1.
16.
Angular impulse = change in angular momentum. t = L m 2 mg 0.01 2 3 3g 0.01 3 10 0.01 1 = = 0.5 rad/s 2 2 0.3 2
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JEE-MAIN-2019 (8th Apr-Second Shift)-PCM-25 Time taken by rod to hit the ground 2h 2 5 t = 1 sec. g 10 In this time angle rotate by rod = t = 0.5 × 1 = 0.5 radian 17.
Moment of inertia 2 md2 d (I) m 2 2 2 Now by = I md2 (qE) (d sin ) = 2 2qE sin for small md
2qE md Angular frequency 18.
19.
If we use that direction of light propagation will be along E B. Then (A) option is correct. Magnitude of E = CB E = 3 × 108 × 1.6 × 10–6 × 5 E = 4.8 × 102 5 E and B are perpendicular to each other E B = 0 Either direction of E is ˆi 2ˆj or ˆi 2ˆj from given option Also wave propagation direction is parallel to E B which is kˆ E is along ( ˆi 2ˆj) At saturation state, VCE becomes zero 10V iC = 10 mA 1000 i Now current gain factor C iB iB
20.
2qE md
10 mA = 40 A 250
1 mV 2 q(Vf Vi ) 2 E 20r FIITJEE Ltd., FIITJEE House, 29-A, Kalu Sarai, Sarvapriya Vihar, New Delhi -110016, Ph 46106000, 26569493, Fax 26513942 website: www.fiitjee.com.
JEE-MAIN-2019 (8th Apr-Second Shift)-PCM-26 r n 0 20 r 1 q r0 mv 2 n 2 2 0 r
V
r v n r0
21.
A 1 3, A 2 5, and A1 A 2 5. 2 2 A 1 A 2 A 1 A 2 2 A 1 A 2 cos
3 cos 10 2A 1 3A 2 3A 1 2A 2 2 2 = 6 A1 9A1 A 2 4A 1 A 2 6 A 2
= –118.5 22.
25 45 25 90 160 30 3 3 3 E 15 3 9 I amp. Req 160 32 Req 15
23.
Range =
2RhT 2RhR
50 103 2 6400 103 hT 2 6400 103 70 By solving hT = 32 m.
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JEE-MAIN-2019 (8th Apr-Second Shift)-PCM-27 24.
Minimum energy required (E) = – (Potential energy of object at surface of earth) GMm GMm R R Now Mearth = 64 Mmoon 4 4 R3e 64 Rm3 Re = 4Rm 3 3 E M R 1 4 Now moon moon earth E earth Mearth Rmoon 64 1
25.
Emoon
E 16
and = 100 rad/s 4 Reactance (X) = Resistance (R) Now by checking option. Given phase difference =
Option (A) R = 1000 and XC
1 = 104 10 100 6
Option (B) R = 103 and XL = 10 × 10–3 × 100 = 1 Option (C) R = 103 and XL = 10–3 100 = 10–1 Option (D) R = 103 and XC
1 = 103 10 10 6 100
26.
Let mass of B and C is m each. By momentum conservation mv 2mv 0 mv 2 v = 4v0 PA = 2mv0 pB = 4mv0 pc = 2mv0 h De-Broglie wavelength p h h h A ; B ; C 2mv 0 4mv 0 2mv 0
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JEE-MAIN-2019 (8th Apr-Second Shift)-PCM-28
27.
Limit of resolution of telescope =
1.22 D
1.22 500 10 9 = 305 × 10—9 radian 200 10 2
28.
Magnetic field at point P i ˆ i ˆ Bnet 0 ( k) 0 (k) =0 2d 2d
29.
isochoric Process d Isobaric Process a Adiabatic slope will be more than isothermal so Isothermal Process b Adiabatic Process c Order d a b c
30.
a M 3a M x 2 4 4 M M 4 a 3a 3a 5a 2 16 4 16 5a = 3 3 12 4 4 b M 3b M 5b y= 2 4 4 M 12 M 4
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JEE-MAIN-2019 (8th Apr-Second Shift)-PCM-29
PART B – CHEMISTRY 31.
In case of interstial compounds there is presence of small atoms (or impurity) in the lattice of metal. So interstial compounds are hard, high melting point and chemically inert.
32. O
O
t BuO t BuOH
Cl
Cl
O
H
OH
O
O-H
33.
Reaction involves free radical chlorination followed by hydrolysis. (a) h Cl2 Cl Cl (b)
Cl
CH3 Cl Cl
Cl
CH2 Cl Cl
(c)
(d)
34.
H O
Cl
2 Cl CH2Cl
Cl
2 2 CH2OH Cl
(e)
CH2 HCl
Cl /H O O
CH2Cl
CH2OH
CHO
1 1 m 2u2 1 P 2 mu2 2 2 m 2 m hc 1 P2 Wo K.E 1 Wo 1 2m
K.E
hc 1 1.5P Wo K.E 2 Wo 2 2 m
2
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JEE-MAIN-2019 (8th Apr-Second Shift)-PCM-30 On solving 4 2 1 9 Since the K.E is very high is comparison to work function, then we can assume that K.E + Wo = K.E 35.
[ICl4]– sp3d2 [IF6]– sp3d2 [ICl2]– sp3d [BrF2]– sp3d
36.
37.
Applying steady state of approximation d B K1 A K 2 B dt O K 1 A K 2 B K1 A B K2
38.
2SO2 g O2 g 2SO3 g 2
K eq
SO3 2 O2 SO2
K 2 10129 1025 K1 10104 Oxidation
39.
Fe
2
aq Ag aq Fe3 aq Ag s E0Cell 3 Reduction
Given:
E0 Ag'/ Ag x
1
E0Fe2 /Fe y
2
E0Fe3 /Fe z
3
Using equation:
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JEE-MAIN-2019 (8th Apr-Second Shift)-PCM-31 G0 nFE0 G01 Fx G02 2Fy G03 3Fz Fe2 2e Fe 2Fy Fe3 3e Fe 3Fz
Fe2 Fe3 e 2Fy 3Fz
Ag e Ag Fx 0 GTotal 2Fy 3Fz Fx FECell
E0Cell x 2y 3z 40.
The % covalent character can be predicted by using Fajan’s rule % of covalent character Be
2
Mg Ca Sr
1 cationic radius
2
2
2
Size
41.
The direct monomer of Nylon-6 is caprolactum which polymerises to give Nylon-6 as follows: H N
O 533 K N2 inert
Nylon-6 42
43.
The IUPAC name of element having atomic number 119 is “Ununennium”. So its symbol is uue. O
O
C
C
H3C CH2 CH3 Due to presence of active methylene group and stabilization of enol by intramoelcular H bond forming 6 membered ring structure.
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JEE-MAIN-2019 (8th Apr-Second Shift)-PCM-32 44.
Mass of fatty acid = 0.027 g Radius of plate = 10 cm Density of fatty acid = 0.9 g/cm3 0.027 Volume of fatty acid = 0.03 cm3 0.9 Area of plate = r2 = 3 102 = 300 cm2 Volume 0.03 Height of fatty acid = 104 cm 106 m Area 300
45.
In CH4 Mole of carbon nC = 1 Mole of hydrogen = nH = 4 % of nC =
46.
nC 1 100 100 20% nC nH 5
The mond’s process is used for refining of nickel as per following reaction.
Ni CO g Ni CO 4 Ni s 4CO g 330350K 450 470K
Impure
47.
Pure
In this case antimarkonikov product will be formed as major product because carbocation formed at a double bonded carbon having lesser number of H atom will be unstable due to presence of an electron withdrawing group (CF3) attached to it. HCl
F3C CH CH2 CF3 CH CH3 CF3 CH2 CH2 min or
major
Cl
CF3 CH2 CH2 CF3 CH2 CH2 Cl (major product)
48.
∆U = nCvm ∆T = 5 28 100 = 14 kJ ∆PV = nR∆T = 5 8 100 = 4 kJ
49.
[Fe(CN)6]4– Fe2 3d6 t 62geg0
[Fe(H2O)6]2+ Fe2 3d6 t 42geg2
n n 2
n n 2 24 4.9BM
= 0 50.
Barfoed’s test → Detecting presence of monosaccharides Fehling’s test → For aldehdyes Benedict’s test → For reducing sugars Seliwanoff’s test → Differentiate between aldose and ketose
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JEE-MAIN-2019 (8th Apr-Second Shift)-PCM-33 NH2
51. N
NH2 N
N
N
CH3 I
Base
N
N
N
H (Adenine)
N
CH3
The nitrogen atom marked with * contain high electron density. It is attached to one Hatom which can be removed by base OH– and the CH3 group gets attached there. 52.
Polysubstitution is a major drawback of Friedal craft’s alkylation. It is so because activating behaviour of benzene ring increases with increase in number of alkyl group on benzene ring.
53.
2H2O2 2H2O 68 g/L
O
2 22.4 of O 2 STP
11.2 L of O2 is given by 34 g of H2O2 present in 1 L of H2O2 solution. 34 So % strength of H2O2 = 100 3.4g / 100mL 3.4% 1000 54.
From Henry’s law Pgas = KH.Xgas Pgas = KH 1 XH O
2
Pgas = KH – KH XH O 2
Comparing this equation with straight line equation y = mx + c 55.
2 2 * 2 2 * 2 2 2 2 C 2 1s 1s 2s 2s 2p x 2Py 2Pz 10 4 B.O 3 diamagnetic 2 2 2 * 2 2 * 2 2 2 2 1 * 1 N2 1s 1s 2s 2s 2p x 2Py 2Pz 2p x 2Py 10 6 B.O 2 paramagnetic 2 2 2 * 2 2 * 2 2 2 2 2 * 2 O2 1s 1s 2s 2s 2Pz 2p x 2Py 2p x 2Py 10 8 B.O 1 diamagnetic 2 2 * 2 2 * 2 2 2 2 1 * 1 O2 1s 1s 2s 2s 2Pz 2p x 2Py 2p x 2Py 10 6 B.O 2 paramagnetic 2
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JEE-MAIN-2019 (8th Apr-Second Shift)-PCM-34 56.
The reaction involves carbylamine reaction of 1o amine with CHCl3/KOH followed by reduction with Pd/C/H2. NC NH2 CHCl3 /KOH
H CN
CN
O
O Pd/C/H2
H N - CH3
CH2NH2 OH 57.
Maximum prescribed concentration of copper in drinking water is 3 ppm. Above this concentration water becomes toxic.
58.
Cl
Cl
I
ICl4 sp3 d2
Cl
Cl Cl
Cl
Cl
I
ICl5 sp3d2 Cl
59.
Cl
cis–[PtCl2(NH3)2] is used in chemotherapy to inhibits the growth of tumors 4
60.
r 3 Vol. occupied by atom 3 % packing efficiency 100 3 100 Vol. of unit cell a Let radius of corner atom is r and radius of central atom is 2r So, 3a 2 2r 2r 6r 6r 2 3r 3 Now a
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JEE-MAIN-2019 (8th Apr-Second Shift)-PCM-35
% P.E =
4 3
4 3
4
3
r 3 3 2r
r 3 8r 3 8 3 3 r3
2 3r
3
100
100
4 9r 3 3 8 3 3 r3
100 90.6% 90%
PART C – MATHEMATICS
61.
S 2 x dx 0
S S 4
4 3/ 2 3
2 3/2 2 3/2 5 4 5 (, 0)
1/3
4 4 25
62.
Since g x is even with f 0 0 f x is odd function g x f x 5 g x f x 5 g x f x 5 Replace x by x 5 f x g x 5 x
x
f t dt g t 5 dt 0
0
x 5
g t dt 5
5
g t dt
x 5
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JEE-MAIN-2019 (8th Apr-Second Shift)-PCM-36 63.
Equation of L1 is x 2y 5 and equation of L2 is 2x y 5 Their point of intersection is (3, 1) k 1 h 3
(–3, 4) (h, k)
(1, 2) (4, 3)
L1 L2 x
64.
x
x
a a a a and f2 x 2 2 f1 x y f1 x y
x
f1 x
1 xy a a x y a x y a x y 2 1 ax ay a y a x a y a y 2 ax a x ay a y 2. 2 2
2f1 x f1 y
65.
x 1 x 1,0 x 0,1 x f x x 1,2 2x x 2 x 2, 3 f x is discontinuous at x 0,1 i
66.
3 i z e6 2
8 9
1 iz z iz 1 e e e 5
i /2
i /6
i5 /6
ei /2ei8 /6
11 i 1 ei 2 /6 ei5 /6 e 6
9
9
9
1 3 i /3 i e 2 2
9
ei3 1
67. (A) p q p ~ q ~ p q p ~ q ~ p ~ q p ~ q T FIITJEE Ltd., FIITJEE House, 29-A, Kalu Sarai, Sarvapriya Vihar, New Delhi -110016, Ph 46106000, 26569493, Fax 26513942 website: www.fiitjee.com.
JEE-MAIN-2019 (8th Apr-Second Shift)-PCM-37
(B) p q p ~ p q p ~ p ~ q p ~ p p ~ q =T (C) ~ p p q ~ p p q T
(D) ~ p q ~ p q ~ p ~ q ~ p q ~ p T T
68.
dy 2y dx x 2 2 nC x Passes through (1, 1) 0 2 n C 2 n y 2 x xn y 2 x 1 n y
69.
2
f f f f x f x
dy f ' f f x f ' f x f ' x 2f x f ' x dx Put x = 1 dy 27 6 33 dx
70.
Starting with 5 63 216 Starting with 44 62 36 Starting with 45 62 36 Starting with 43 18 (Not using 0, 1, 2) Starting with 432 = 4 Total = 310
71.
Equation of PQ is
x2 y3 z4 6 3 6 R (4, y, z) lies on this 1 y3 z4 3 3 6 R 4, 2, 6
QR 16 4 36 2 14
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JEE-MAIN-2019 (8th Apr-Second Shift)-PCM-38 72.
For infinitly many solution 1 2 k 2
1
1 0
3
1 k
1 2 Also consider x 2y k 1 and 2x y z 2 2x 4y z 2 2x y z 2 4x 3y 4 k
73.
41 45 54 57 43 x = 48 6 x 48 1 2 482 412 452 542 572 432 482 6 14024 2 2304 6 100 3
AM
74.
P1 : x y z 1 P2 :2x 3y 4z 5 Required plane is P1 P2 0
1 2 x 1 3 y 1 4 z 1 5 which is perpendicular to x y z 0
…..(i)
1 2 1 3 1 4 0
1 3 (i) x z 2 0 r . ˆi kˆ 2 0
75.
Given 2b a c Let A , B 3, C 2 2 sinB sin A sin C 2 sin3 sin sin 2
2 3 4 sin2 1 2cos 2
8 cos 2cos 3 0 3 cos 4 sin A :sinB :sin C
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JEE-MAIN-2019 (8th Apr-Second Shift)-PCM-39
1: 4 3 sin2 :2 cos 5 6 1: : 4 4 4:5:6 76.
be 5 3 b2 e2 75 b a b a 75 b a 15 b 10, a 5 LR
2a2 5 b
1 9 n 2 10 1 1 10 2n 2n 10 n4
77.
1
78.
Let equation of hyperbola be
x2 y2 1 a 2 b2
passes through (4, 6) 16 36 2 2 1 (i) a b b2 Also e2 1 2 b2 3a 2 a From (i) and (ii) a2 4, b2 12 Equation
(ii)
x2 y2 1 4 12
Tangent at (4, 6) is x
y 1 2
Or 2x y 2 79.
2
D 4 1 3m 4 1 m2 1 8m
2
4 1 9m 6m 1 8m m2 8m3
2
8m 4m 4m 1 2
8m 2m 1 0 Infinitely many values of m.
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JEE-MAIN-2019 (8th Apr-Second Shift)-PCM-40 20
k k 2 k 1 1 2 3 20 S 2 3 ..... 20 2 2 2 2 1 1 2 20 S 3 ....... 21 2 2 2 2 2 1 1 1 1 1 20 S 2 3 ...... 20 21 2 2 2 2 2 2 1 1 1 20 2 2 20 21 1 2 2 21 1 21 2 11 s 2 19 2
80.
S
81.
ˆi ab 3
ˆj
kˆ
2
x
1 1 1 2 x ˆi 3 x ˆj 5kˆ a b 2 x 2 x 19
2
82.
x 1/ 2
2
191/4
5 3 2
1 3 Equation of PQ is
Slope of OP
y 1 3 x 3
P
3, 1
y 3x 4 4 Q , 0 3 2 Area 3
O
Q
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JEE-MAIN-2019 (8th Apr-Second Shift)-PCM-41 83.
x 2 4x 5 x 5, x 1 P 1, 2 Tangent at P is y x 1
P (1, 2)
3 7 4 , 4 lies on this.
84.
h 2 3 cos 6 cos , r 3 sin
V r 2h 4 sin2 6 cos
54 sin2 cos dv 0 2 sin cos sin3 0 d 2 sin 3 sin3 0 sin cos
85.
3
h
2 3
1 3
h
6 3
1
1
1
2
b
c
2
c2
4 b
1 log xr
C2 C2 C1, C3 C3 C1 1 2
0
0
b2
c 2
2
4 b 4 c2 4 b 2 c 2
1
1
b2 c2
A b 2 c 2 c b 2,b,c are in AP 2,2 d, 2 2d A d 2d d 2d3 2, 16 d3 1, 8
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JEE-MAIN-2019 (8th Apr-Second Shift)-PCM-42 2d 2, 4 2 2d 4, 6
86.
1 Form f 3 x f 2x f 3 f 2 k lim x 0 x 1 f 2 x f 2 f ' 3 x f ' 2 x lim x 0 1 f 2 x f 2 x f ' 2 x 0 ek 1
87.
I
dx 3
x 1 x6
2/3
dx 1 x7 1 6 x
2/3
dx dt x7 6 1/3 1 dt 1 1 I 2/ 3 1 6 C 6 t 2 x
Put 1 x 6 t
1/3
6 1 1 x x f x 1 x6 2 2 x 1 f x 3 2x
88.
89.
1/3
h 20 h 80 , y x xy x h y h xy , 20 x 80 x h h 1 20 80 100 h 1 1600 h 16
C
80 x–y
h y
3 6 C3 x 1 log x . x1/4 200 2 1 3 1 logx 2
x4
10
1 3 1 log10 x .log10 x 1 4 2
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JEE-MAIN-2019 (8th Apr-Second Shift)-PCM-43 6t 2 5t 4 0, t log10 x D0 So no real solution All options are incorrect
90.
b2 ac Also roots of ax 2 2bx c 0 are equal b x , common root a 2 b b d 2e 0 a a db2 2aeb fa2 0, b2 ac dac 2eab fa2 0 dc 7eb fa 0 Dividing by ac d 2e f 0 a b c d f e 2. a c b
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FIITJEE Solutions to JEE(Main)-2019 Test Date: 9th April 2019 (First Shift)
PHYSICS, CHEMISTRY & MATHEMATICS Paper - 1 Time Allotted: 3 Hours
Maximum Marks: 360
Please read the instructions carefully. You are allotted 5 minutes specifically for this purpose.
Important Instructions: 1.
The test is of 3 hours duration.
2.
This Test Paper consists of 90 questions. The maximum marks are 360.
3.
There are three parts in the question paper A, B, C consisting of Physics, Chemistry and Mathematics having 30 questions in each part of equal weightage. Each question is allotted 4 (four) marks for correct response.
4.
Out of the four options given for each question, only one option is the correct answer.
5.
For each incorrect response 1 mark i.e. ¼ (one-fourth) marks of the total marks allotted to the question will be deducted from the total score. No deduction from the total score, however, will be made if no response is indicated for an item in the Answer Box.
6.
Candidates will be awarded marks as stated above in instruction No.3 for correct response of each question. One mark will be deducted for indicating incorrect response of each question. No deduction from the total score will be made if no response is indicated for an item in the answer box.
7.
There is only one correct response for each question. Marked up more than one response in any question will be treated as wrong response and marked up for wrong response will be deducted accordingly as per instruction 6 above.
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JEE-MAIN-2019 (9th Apr-First Shift)-PCM-2
PART –A (PHYSICS) 1.
In the density measurement of a cube, the mass and edge length are measured as (10.000.10) kg and (0.100.01)m, respectively. The error in the measurement of density is: (A) 0.10 kg/m3 (B) 0.31 kg/m3 3 (C) 0.07 kg/m (D) 0.01 kg/m3
2.
The total number of turns and cross-section area in solenoid is fixed. However, its length L is varied by adjusting the separation between windings. The inductance of solenoid will be proportional to: (A) 1/L (B) L (C) 1/L2 (D) L2
3.
The figure shows a Young’s double slit experimental setup. It is observed that when a thin transparent sheet of thickness t and refractive index is put in front of one of the slits, the central maximum gets shifted by a distance equal to n fringe widths. If the wavelength of light used is , t will be:
4.
(A)
2nD a 1
(B)
nD a 1
(C)
2D a 1
(D)
D a 1
A system of three charges are placed as shown in the figure:
If D >> d, the potential energy of the system is best given by: 1 q2 qQd 1 q2 qQd (A) (B) 4 o d D2 4 o d 2D2 (C)
5.
1 4 o
q2 2qQd D2 d
(D)
1 4 o
q2 qQd 2 D d
A moving coil galvanometer has resistance 50 and it indicates full deflection at 4mA current. A voltmeter is made using this galvanometer and a 5 k resistance. The maximum voltage, that can be measured using this voltamenter, will be close to: (A) 15 V (B) 20 V (C) 10 V (D) 40 V FIITJEE Ltd., FIITJEE House, 29-A, Kalu Sarai, Sarvapriya Vihar, New Delhi -110016, Ph 46106000, 26569493, Fax 26513942 website: www.fiitjee.com.
JEE-MAIN-2019 (9th Apr-First Shift)-PCM-3 6.
If ‘M’ is the mass of water that rises in a capillary tube of radius ‘r’, then mass of water which will rise in a capillary tube of radius ‘2r’ is: M (A) 4 M (B) 2 (C) M (D) 2 M
7.
An NPN transistor is used in common emitter configuration as an amplifier with 1 k load resistance. Signal voltage of 10 mV is applied across the base-emitter. This produces a 3 mA change in the collector current and 15 A change in the base current of the amplifier. The input resistance and voltage gain are: (A) 0.67 k, 300 (B) 0.67 k, 200 (C) 0.33 k, 1.5 (D) 0.33 k, 300
8.
An HCl molecule has rotational, translational and vibrational motions. If the rms velocity of HCl molecules in its gaseous phase is , m is its mass and kB is Boltzmann constant, then its temperature will be: 2
m (A) 3kB
2
m (C) 5kB
9.
10.
2
m (B) 7kB
2
m (D) 6k B
2x N The electric field of light wave is given as E 10 3 cos 2 6 1014 t x . This 7 5 10 C light falls on a metal plate of work function 2eV. The stopping potential of the photoelectrons is: (A) 0.48 V (B) 2.48 V (C) 0.72 V (D) 2.0 V
Following figure shows two processes A and B for a gas. If QA and QB are the amount of heat absorbed by the system in two cases, and UA and UB are changes in internal energies, respectively, then:
(A) QA = QB; UA = UB (C) QA < QB; UA < UB
(B) QA > QB; UA = UB (D) QA > QB; UA > UB
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JEE-MAIN-2019 (9th Apr-First Shift)-PCM-4 11.
A uniform cable of mass ‘M’ and length ‘L’ is placed on a horizontal surface such that its th 1 part is hanging below the edge of the surface. To lift the hanging part of the cable n upto the surface, the work done should be: MgL (A) nMgL (B) 2n2 2MgL MgL (C) (D) 2 n n2
12.
The following bodies are made to roll up (without slipping) the same inclined plane from R a horizontal place: (i) a ring of radius R, (ii) a solid cylinder of radius and (iii) a solid 2 R sphere of radius . If, in each case, the speed of the center of mass at bottom of the 4 incline is same, the ratio of the maximum heights they climb is: (A) 10 : 15 : 7 (C) 4 : 3 : 2
13.
(B) 14: 15: 20 (D) 2 : 3 : 4
A signal A cos t is transmitted using 0 sin 0 t as carrier wave. The correct amplitude modulated (AM) signal is: (A) 0 sin 0 1 0.01A sin t t A A sin 0 t sin 0 t 2 2 (C) 0 sin 0 t A cos t
(B) 0 sin 0 t
(D) 0 A cos t sin 0 t 14.
A concave mirror for face viewing has focal length of 0.4 m. The distance at which you hold the mirror from your face in order to see your image upright with a magnification of 5 is: (A) 0.16 m (B) 1.60 m (C) 0.24 m (D) 0.32 m
15.
Determine the charge on the capacitor in the following circuit:
(A) 200 C (C) 10 C 16.
(B) 60 C (D) 2 C
A rectangular coil (Dimension 5 cm 2 cm) with 100 100 turns, carrying a current of 3 A in the clock-wise direction, is kept centered at the origin and in the X-Z plane. A magnetic field of 1 T is applied along X-axis. If the coil is tilted through 45o about Z-axis, then the torque on the coil is: (A) 0.42 Nm (B) 0.55 Nm (C) 0.27 Nm (D) 0.38 Nm
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JEE-MAIN-2019 (9th Apr-First Shift)-PCM-5 17.
A stationary horizontal disc is free to rotate about its axis. When a torque is applied on it, its kinetic energy as a function of , where is the angle by which it has rotated, is given as k2. If its moment of inertia is then the angular acceleration of the disc is: k k (A) (B) 2 k 2k (C) (D) 4
18.
A simple pendulum oscillating in air has period T. The bob of the pendulum is completely 1 immersed in a non-viscous liquid. The density of the liquid is th of the material of the 16 bob. If the bob is inside liquid all the time, its period of oscillation in this liquid is: 1 1 (A) 2T (B) 2T 10 14 (C) 4T
19.
1 15
(D) 4T
For a given gas at 1 atm pressure, rms speed of the molecules is 200 m/s at 127oC. At 2 atm pressure and at 227oC, the rms speed of the molecules will be: (A) 80 m/s (B) 80 5 m / s (C) 100 m/s
20.
1 14
(D) 100 5 m / s
The magnetic field of a plane electromagnetic wave is given by: B B0 i cos kz t B1j cos kz t where B0 = 3 10–5T and B1 = 2 10–6T. The rms value of the force experienced by a stationary charge Q = 10–4=C at z = 0 is closet to: (A) 0.9 N (B) 0.6 N (C) 0.1 N (D) 3 10–2N
21.
The stream of a river is flowing with a speed of 2 km/h. A swimmer can swim at a speed of 4 km/h. What should be the direction of the swimmer with respect to the flow of the river to cross the river straight? (A) 60o (B) 90o o (C) 120 (D) 150o
22.
A rigid square loop of side ‘a’ and carrying current 2 is laying on a horizontal surface near a long current 1 wire in the same plane as shown in figure. The net force on the loop due to the wire will be:
(A) Repulsive and equal to (C) Zero
0 1 2 2
0 1 2 3 (D) Repulsive and equal to 0 1 2 4
(B) Attractive and equal to
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JEE-MAIN-2019 (9th Apr-First Shift)-PCM-6 23.
Taking the wavelength of first Balmer line in hydrogen spectrum ( n = 3 to n = 2) as 660 nm , the wavelength of the 2nd Balmer line (n =4 to n = 2) will be (A) 889.2 nm (B) 642.7 nm (C) 448.9 nm (D) 388.9 nm
24.
A solid sphere of mass ‘M’ and radius ‘a’ is surrounded by a uniform concentric spherical shell of thickness 2a and mass 2M. The gravitational field at distance ‘3a’ from the centre will be: 2GM 2GM (A) (B) 3a 2 9a 2 GM GM (C) (D) 2 9a 3a2
25.
A string is clamed at both the ends and it is vibrating in its 4th harmonic. The equation of the stationary wave is Y =0.3 sin(0.157x) cos(200t). The length of the string is: (all quantities are in S units) (A) 80 m (B) 60 m (C) 40 m (D) 20 m
26.
A ball is thrown vertically up (taken as +z-axis) from the ground. The correct momentumheight (p-h) diagram is:
(A)
(B)
(C)
(D)
27.
A capacitor with capacitance 5 F is charged to 5 C. If the plates are pulled apart to reduce the capacitance to 2 F, how much work is done? (A) 3.75 10–6 J (B) 2.55 10–6 J (C) 6.25 10–6 J (D) 2.16 10–6 J
28.
A body of mass 2 kg makes an elastic collision with a second body at rest and continues to move in the original direction but with one fourth of its original speed. What is the mass of the second body? (A) 1.5 kg (B) 1.2 kg (C) 1.8 kg (D) 1.0 kg
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JEE-MAIN-2019 (9th Apr-First Shift)-PCM-7 29.
The pressure wave, P = 0.01 sin[1000t – 3x]NM–2, corresponds to the sound produced by a vibrating blade on a day when atmospheric temperature is 0o C . On some other day when temperature is T, the speed of sound produced by the same blade and at the same frequency is found to be 336 ms–1. Approximate value of T is: (A) 12o C (B) 11o C o (C) 15 C (D) 4o C
30.
A wire of resistance R is bent to form a square ABCD as shown in the figure. The effective resistance between E and C is: (E is mid-point of arm CD)
1 R 16 3 (C) R 4
(A)
(B)
7 R 64
(D) R
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JEE-MAIN-2019 (9th Apr-First Shift)-PCM-8
PART –B (CHEMISTRY) 31.
Magnesium powder burns in air to give: (A) MgO and Mg(NO3)2 (C) MgO only
(B) MgO and Mg3N2 (D) Mg(NO3)2 and Mg3N2
32.
The number of water molecule(s) not coordinated to copper ion directly in CuSO45H2O, is: (A) 1 (B) 3 (C) 2 (D) 4
33.
The one that will show optical activity is: (en = ethane-1, 2-diamine)
34.
(A)
(B)
(C)
(D)
Consider the van der Waals constants, a and b, for the following gases. Gas a/(atm dm6mol–2) –2
3
Ar
Ne
Kr
Xe
1.3
0.2
5.1
4.1
–1
b/(10 dm mol ) 3.2 1.7 1.0 Which gas is expected to have the highest critical temperature? (A) Ar (B) Xe (C) Kr (D) Ne
5.0
35.
Among the following, the molecule expected to be stabilized by anion formation is: (A) F2 (B) C2 (C) O2 (D) NO
36.
Liquid ‘M’ and liquid ‘N’ form an ideal solution. The vapour pressures of pure liquids ‘M’ and ‘N’ are 450 and 700 mm Hg, respectively at the same temperature. Then correct statement is: (xM = mole fraction of ‘M’ in solution; xN = mole fraction of ‘N’ in solution; yM = mole fraction of ‘M’ in vapour phase; yN = mole fraction of ‘M’ in vapour phase) x y x y (A) M M (B) M M xN y N xN y N x y (C) xM yM xN yN) (D) M M xN y N FIITJEE Ltd., FIITJEE House, 29-A, Kalu Sarai, Sarvapriya Vihar, New Delhi -110016, Ph 46106000, 26569493, Fax 26513942 website: www.fiitjee.com.
JEE-MAIN-2019 (9th Apr-First Shift)-PCM-9 37.
Among the following the set of parameters that represents path functions, is: (a) q + w (b) q (c) w (d) H – TS (A) (b) and (c) (B) (b), (c) and (d) (C) (a), (b) and (c) (D) (a) and (d)
38.
The degenerate orbitals of Cr H2O 6 (A) dxz and dyz
39.
40.
are: (B) dx2 y 2 and dxy
(C) dyz and dz2
(D) dz2 and dxz
The aerosol is a kind of colloid in which: (A) solid is dispersed in gas (C) liquid is dispersed in water
(B) gas is dispersed in solid (D) gas is dispersed in liquid
The given plots represent the variation of the concentration of a reactant R with time for two different reactions (i) and (ii). The respective orders of the reactions are:
(A) 1, 1 (C) 0, 1 41.
3
(B) 0, 2 (D) 1, 0
The major product of the following reaction is: CH2CH3
i alkaline KMnO4 ii H3O CH2CHO
(A)
CH2COOH
(B)
COOH
(C)
42.
COCH3
(D)
The organic compound that gives following qualitative analysis is: Test Inference (a) Dil. HCl Insoluble (b) NaOH solution Soluble (c) Br2/water Decolourization
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JEE-MAIN-2019 (9th Apr-First Shift)-PCM-10 OH
OH
(A)
(B) NH2
NH2
(C)
43.
(D)
Aniline dissolved in dilute HCl is reacted with sodium nitrite at 0oC . This solution was added dropwise to a solution containing equimolar mixture of aniline and phenol in dil. HCl. The structure of the major product is: (A)
N
N
(C)
N
N
NH
NH2
(B)
N
N
(D)
N
N
44.
Excessive release of CO2 into the atmosphere results in: (A) formation of smog (B) depletion of ozone (C) global warming (D) polar vortex
45.
The ore that contains the metal in the form of fluoride is: (A) saphalerite (B) malachite (C) magnetite (D) cryolite
46.
The correct IUPAC name of the following compound is
OH
O
NO 2
Cl CH3
(A) 5-chloro-4-methyl-1-nitrobenzene (C) 3-chloro-4-methyl-1-nitrobenzene
(B) 2-methyl-5-nitro-1-chlorobenzene (D) 2-chloro-1-methyl-4-nitrobenzene
47.
The major product of the following reaction is : LiAlH4 CH3CH CHCO2CH3 (A) CH3CH2CH2CHO (B) CH3CH = CHCH2OH (C) CH3CH2CH2CO2CH3 (D) CH3CH2CH2CH2OH
48.
The element having greatest difference between its first and second ionization energies, is (A) Ca (B) K (C) Ba (D) Sc
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JEE-MAIN-2019 (9th Apr-First Shift)-PCM-11 49.
For a reaction, N2(g) + 3H2(g) 2NH3(g); identify dihydrogen (H2) as a limiting reagent in the following reaction mixtures. (A) 14g of N2 + 4g of H2 (B) 28g of N2 + 6g of H2 (C) 56g of N2 + 10g of H2 (D) 35g of N2 + 8g of H2
50.
The major product of the following reaction is : i DCI 1equiv. CH3C CH 2 DI (A) CH3CD(I)CHD(Cl) (C) CH3CD2CH(Cl)(I)
(B) CH3C(I)(Cl)CHD2 (D) CH3CD(Cl)CHD(I)
51.
The osmotic pressure of a dilute solution of an ionic compound XY in water is four times that of a solution of 0.01 M BaCl2 in water. Assuming complete dissociation of the given ionic compounds in water, the concentration of XY (in mol L–1) in solution is (A) 4 10–4 (B) 6 10–2 –4 (C) 16 10 (D) 4 10–2
52.
For any given series of spectral lines of atomic hydrogen, let V Vmax Vmin be the difference in maximum and minimum frequencies in cm–1. the ration VLymann / VBalmer is (A) 5 : 4 (C) 9 : 4
53.
(B) 4: 1 (D) 27 : 5
The major product of the following reaction is:
(A)
(B)
(C)
(D)
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JEE-MAIN-2019 (9th Apr-First Shift)-PCM-12 54.
55.
The increasing order of reactivity of the following compounds towards aromatic electrophilic substitution reaction is
(A) D < B < A < C (C) D < A < C < B
(B) A < B < C < D (D) B < C < A < D
C60, an allotrope of carbon contains: (A) 20 hexagons and 12 pentagons. (C) 18 hexagons and 14 pentagons.
(B) 12 hexagons and 20 pentagons. (D) 16 hexagons and 16 pentagons
56.
The correct order of the oxidation states of nitrogen in NO, N2O, NO2 and N2O3 is: (A) N2O < N2O3 < NO < NO2 (B) NO2 < NO < N2O3 < N2O (C) NO2 < N2O3 < NO < N2O (D) N2O < NO < N2O3 < NO2
57.
The standard Gibbs energy for the given cell reaction in kJ mol –1 at 298 K is: Zn s Cu2 aq Zn2 aq Cu s ,Eo 2 V at 298 K
[Faraday’s constant F = 96500 C mol–1] (A) -192 (C) -384 58.
(B) 384 (D) 192
The major product of the following reaction is: 1.PBr3 OH 2. KOH alc.
O OH
(A)
(B) O
O
(C)
(D) HO
59.
O
Which of the following statements is not true about sucrose? (A) The glycosidic linkage is present between C1 of -glucose and C1 of -fructose (B) It is a non-reducing sugar (C) It is also named as invert sugar (D) On hydrolysis it produces glucose and fructose
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JEE-MAIN-2019 (9th Apr-First Shift)-PCM-13 60.
Match the catalysis(Column – I) with products (Column-II) Column-I Column-II Catalyst Product (a) V2O5 (i) Polyethylene (b) TiCl4/Al(Me)3 (ii) Ethanal (c) PdCl2 (iii) H2SO4 (d) Iron oxide (iv) NH3 (A) (a)-(iii); (b)-(iv); (c)-(i); (d)-(ii) (B) (a)-(iv); (b)-(iii); (c)-(ii); (d)-(i) (C) (a)-(ii); (b)-(iii); (c)-(i); (d)-(iv) (D) (a)-(iii); (b)-(i); (c)-(ii); (d)-(iv)
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JEE-MAIN-2019 (9th Apr-First Shift)-PCM-14
PART–C (MATHEMATICS) 61.
62.
63.
(A) R 1,0
x2 , is surjective, then A is equal to: 1 x2 (B) R–(–1,0)
(C) R–{–1}
(D) [0, ]
If the function f :R–{1, –1}A defined by f x
Let p, q R. if 2 3 is a root of the quadratic equation, x2 + px + q = 0, then: (A) q2 + 4p + 14 = 0 (B) p2 – 4q + 12 = 0 2 (C) p – 4q – 12 = 0 (D) q2 – 4p – 16 = 0 Let 3i j and 2i j 3k . If 1 2 , where 1 is parallel to and 2 is perpendicular to , then 1 2 is equal to: 1 1 (A) (B) 3 i 9 j 5 k 3 i 9 j 5 k 2 2 (C) 3 i 9 j 5 k (D) 3 i 9 j 5 k
64.
The integral
sec
2/3
(A) 3 tan1/3 x C (C) 3 cot 1/3 x C 65.
x cos ec 4/3 x dx is equal to: (Here C is a constant of integration) 3 tan 4/3 x C 4 (D) 3 tan1/3 x C
(B)
A plane passing through the points (0, –1, 0) and (0, 0, 1) and making an angle
with 4
plane y – z + 5 = 0, also passes through the point: (A) 2,1,4 (B) 2, 1, 4
(C)
2, 1, 4
(D) 2, 1, 4
66.
If the tangent to the curve, y = x3 + ax – b at the point (1, –5) is perpendicular to the line, –x + y + 4 = 0, then which one of the following, points lies on the curve? (A) (2, –2) (B) (–2, 2) (C) (–2, 1) (D) (2, –1)
67.
If the standard deviation of the numbers –1, 0, 1, k is (A) 4
5 3
(C) 2 6 68.
(B) (D) 2
5 where k > 0, then k is equal to:
6 10 3
Let f(x) = 15–|x – 10|; x R. then the set of all values of x, at which the function, g(x) = f(f(x)) is not differentiable, is: (A) {5, 10, 15} (B) {10} (C) {5, 10, 15, 20} (D) {10, 15} FIITJEE Ltd., FIITJEE House, 29-A, Kalu Sarai, Sarvapriya Vihar, New Delhi -110016, Ph 46106000, 26569493, Fax 26513942 website: www.fiitjee.com.
JEE-MAIN-2019 (9th Apr-First Shift)-PCM-15 6
69.
2 If the fourth term in the Binomial expansion of xlog8 x x 0 is 20 87, then a value x of x is: (A) 83 (B) 8–2 (C) 8 (D) 82
70.
If f(x) is a non-zero polynomial of degree four, having local extreme points at x = –1, 0, 1; then the set S = {x R; f(x) = f(0)} contains exactly: (A) four irrational numbers (B) four rational numbers (C) two irrational and one rational number (D) two irrational and two rational numbers
71.
For any two statements p and q, the negation of the expression p p q is: (A) p q (C) p q
(B) p q (D) p q
72.
A committee of 11 members is to be formed from 8 males and 5 females. If m is the number of ways the committee is formed with at least 6 males and n is the number of ways the committee is formed with at least 3 females, then: (A) n = m – 8 (B) m + n = 68 (C) m = n = 78 (D) m = n = 68
73.
If the line,
74.
The value of cos2 10o cos10o cos 50o cos2 50o is: 3 3 (A) 1 cos 20o (B) 2 4 3 3 (C) (D) cos 20o 2 4
x 1 y 1 z 2 meets the plane, x + 2y + 3z = 15 at a point P, then the 2 3 4 distance of P from the origin is: 5 (A) (B) 2 5 2 9 7 (C) (D) 2 2
75.
76.
1 1 1 2 1 3 1 n 1 1 78 If ........... , then the inverse of 1 0 1 0 1 0 1 0 1 0 1 12 1 0 (A) (B) 0 1 13 1 1 0 1 13 (C) (D) 1 12 1 0
1 n 0 1 is:
i All the points in the set S : R i 1 lie on a: i (A) straight line whose slope is 1 (B) circle whose radius is 2 (C) straight line whose slope is –1 (D) circle whose radius is 1
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JEE-MAIN-2019 (9th Apr-First Shift)-PCM-16 77.
Let the sum of the first n terms of a non-constant A.P., a1, a2, a3, …… be n n 7 50n A , where A is a constant. If d is the common difference of this A.P., then 2 the ordered pair (d,a50) is equal to: (A) (A, 50 + 46A) (B) (A, 50 + 45A) (C) (50, 50 + 45A) (D)(50, 50 + 46A)
78.
The value of
/2
0
sin3 x dx is: sin x cos x
2 (A) 4 1 (C) 4
79.
If the line y mx 7 3 is normal to the hyperbola (A) (C)
80.
1 2 2 (D) 8
(B)
2
(B)
5 15 2
(D)
The solution of the differential equation x x3 1 2 5 5x 4 3 1 (C) y x 2 5 5x
(A) y
x2 y2 1 , then a value of m is: 24 18
5 2 3 5
dy 2y x 2 x 0 with y(1) = 1, is: dx x2 3 (B) y 2 4 4x 3 2 1 (D) y x 2 4 4x
81.
If one end of a focal chord of the parabola, y2 = 16x is at (1, 4), then the length of this focal chord is: (A) 25 (B) 24 (C) 22 (D) 20
82.
Let S be the set of all values of x for which the tangent to the curve y = f(x) = x3 – x2 – 2x at (x, y) is parallel to the line segment joining the points 1, f 1 and –1, f –1 , then S is equal to: 1 (A) , 1 3 1 (C) ,1 3
1 (B) , 1 3 1 (D) ,1 3
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JEE-MAIN-2019 (9th Apr-First Shift)-PCM-17 83.
Slope of a line passing through P(2, 3) and intersecting the line, x + y = 7 at a distance of 4 units from P, is: 5 1 1 5 (A) (B) 5 1 1 5 7 1 1 7 (C) (D) 7 1 1 7
84.
Let
10
f a k 16 2
10
1 , where the function f satisfies f(x + y) = f(x) f(y) for all natural
k 1
numbers x, y and f(1) = 2. Then the natural number ‘a’ is: (A) 4 (B) 16 (C) 2 (D) 3 85.
Let S = {[–2, 2]: 2 cos2 + 3 sin = 0}. Then the sum of the elements of S is: 13 (A) (B) 2 6 5 (C) (D) 3
86.
If a tangent to the circle x2 + y2 = 1 intersects the coordinate axes at distinct points P and Q, then the locus of the mid-point of PQ is: (A) x2 + y2 – 16x2y2 = 0 (B) x2 + y2 – 2x2y2 = 0 2 2 2 2 (C) x + y – 4x y = 0 (D) x2 + y2 – 2xy = 0
87.
Four persons can hit a target correctly with probabilities
88.
The area (in sq. units) of the region A = {(x, y): x2 y x + 2} is: 31 13 (A) (B) 6 6 9 10 (C) (D) 2 3
89.
Let and be the roots of the equation x2 + x + 1 = 0. Then for y 0 in R, y 1
1 1 1 1 , , and respectively. If all 2 3 4 8 hit at the target independently, then the probability that the target would be hit, is: 25 25 (A) (B) 32 192 7 1 (C) (D) 32 192
y
1
1
y
(A) y(y2 – 3) (C) y3
is equal to:
(B) y3 – 1 (D) y(y2 – 1)
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JEE-MAIN-2019 (9th Apr-First Shift)-PCM-18
90.
2 cos x 1 , x 4 is continuous, then k is If the function f defined on , by f x cot x 1 6 3 k, x 4 equal to: (A) 1 (B) 2 1 1 (C) (D) 2 2
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JEE-MAIN-2019 (9th Apr-First Shift)-PCM-19
JEE (Main) – 2019 ANSWERS PART A – PHYSICS 1. 5. 9. 13. 17. 21. 25. 29.
Bonus B A B D C B D
2. 6. 10. 14. 18. 22. 26. 30.
A D B D C B A B
3. 7. 11. 15. 19. 23. 27.
D A B A B C A
4. 8. 12. 16. 20. 24. 28.
D A B B D A B
34. 38. 42. 46. 50. 54. 58.
C A B D B C B
64. 68. 72. 76. 80. 84. 88.
D A C D B D C
PART B – CHEMISTRY 31. 35. 39. 43. 47. 51. 55. 59.
B B A C B B A A
32. 36. 40. 44. 48. 52. 56. 60.
A A D C B C D D
33. 37. 41. 45. 49. 53. 57.
A A C D B C C
PART C – MATHEMATICS 61. 65. 69. 73. 77. 81. 85. 89.
A A D C A A B C
62. 66. 70. 74. 78. 82. 86. 90.
B or Bonus A C B C D C C
63. 67. 71. 75. 79. 83. 87.
A C C D A D A
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JEE-MAIN-2019 (9th Apr-First Shift)-PCM-20
HINTS AND SOLUTIONS PART A – PHYSICS 1.
2.
m v Maximum % error in will be given by p m L 100% 100% 3 100% …(i) p m L This is not applicable as error is big. m 9.9 min min = 7438 kg/m3 v max (0.11)3 m 10.1 & max max = 13854.6 kg/m3 v min (0.09)3 p = 6416.6 kg/m3 No option is matching. Therefore this question should be awarded bonus. = NBA = LI N 0 nIR2 = LI N N 0 IR 2 LI N and R constant Self inductance (L)
1 1 length
3.
Path difference at central maxima x = ( – 1)t, whole pattern will shift by same amount which will be given by D D n ( 1)t n , according to eh question t d d ( 1) No option is matching, therefore question should be award bonus. Correct option should be (Bonus)
4.
Utotal = Uself of dipole + Uinteraction kq2 kQ = qd d D2
q2 qQd = k 2 D d 5.
G = 50 S = 5000 Ig = 4 × 10–3 V = ig (G + S) V = 4 × 10–3 (50 + 5000) = 4 × 10–3 (5050) = 20.2 volt
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JEE-MAIN-2019 (9th Apr-First Shift)-PCM-21
6.
Height of liquid rise in capillary tube h
2T cos C rg
1 r When radius becomes double height become half h h 2 Now, M = r2h × and M = (2r)2 (h/2) × = 2M. h
7.
Input current = 15 × 10–6 Output current = 3 × 10–3 Resistance out put = 1000 Vinput = 10 × 10–3 Now Vinput = rinput × iinput 10 × 10–3 = rinput × 15 × 10–6 2000 rinput = 0.67 K. 3 Voutput 1000 3 10 3 Voltage gain = = 300 Vinput 10 10 3
8.
According to equipartion energy theorem 1 1 2 m(v rms ) 3 K bT 2 2 2
mv rms T 3k 9.
= 6 × 1014 × 2 f = 6 × 1014 C=f C 3 108 = 5000 Å f 6 1014 12375 Energy of photon = 2.475 eV 5000 From Einstein’s equation KEmax = E – eVs = E – eVs = 2.475 – 2 eVo = 0.475 eV Vo = 0.48 V
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JEE-MAIN-2019 (9th Apr-First Shift)-PCM-22 10.
Initial and final states for both the processes are same, UA = UB Work done during process A is greater than in process B. Because area is more By First law of thermodynamics Q = U + W QA > QB
11.
Mass of the hanging part =
hCOM
M n
L 2n
M L MgL Work done W = mghCOM g 2 n 2n 2n 12.
1 1 m 2 v 2 = mgh 2 R If radius of gyration is k, then k2 2 1 2 v k k solid cylinder R 1 ring h ; 1, 2g Rring R solid cylinder 2
k solid sphere Rsolid sphere
2 5
1 2 H1 : h2 : h3 :: (1 + 1) : 1 : 1 :: 20 :15 : 14 2 5 Therefore most appropriate option is (B) Although which is not in correct sequence. 13.
14.
f f u 40 5 ; u = –32 cm 40 u
m
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JEE-MAIN-2019 (9th Apr-First Shift)-PCM-23 15.
Different potential is shown at different points.
q = eV q = 10F × 20 = 200 C 16.
t MB
NI A B sin 45o = 0.27 Nm 17.
18.
1 2 I k2 2 2k2 2k 2 I I Differentiate (A) wrt time d 2k d dt I dt 2k 2k by (1) I I 2k I Kinetic energy KE =
For a simple pendulum T = 2
…(A)
L gerr
Situation 1: when pendulum is in air geff = g Situation 2:when pendulum is in liquid liquid 1 15g geff g 1 g 1 body 16 16 L 2 15g / 16 T So, T L 2 g T
19.
Vrms
4T 15
3RT Mw
v rms
T
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JEE-MAIN-2019 (9th Apr-First Shift)-PCM-24
v 500 v 5 200 400 200 2 v 100 5 m / s Now,
20.
Maximum electric field E = (B) (C) E0 (3 105 )c ˆj E1 (2 10 6 )c ˆi
Maximum force Fnet 10 4 3 10 8 (3 10 5 )2 (2 10 6 )2 0.9 N Frms
F0 2
= 0.6 N (approx)
21.
For swimmer to cross the river straight 4 sin = 2 1 sin = 30o 2 So, angle with direction of river flow = 90° + = 120°.
22.
F3 & F4 cancel each other. Force on PQ will be F1 = 2B1 a I = I2 0 1 a 2a 0Ii = a= 2a 0I1I2 2 Force on RS will be F2 = I2 B2 a I = I2 0 1 a 22a 0I1I2 = 4 II Net force = F1 – F2 = 0 1 2 repulsion 4
23.
1 1 5R 1 R 2 2 660 2 3 36 1 1 3R 1 R 2 2 4 16 2 Divide equation (1) with (B) 5 16 660 36 3
…(1) …(B)
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JEE-MAIN-2019 (9th Apr-First Shift)-PCM-25
4400 = 488.88 = 488.9 nm 9
24.
We use Gauss’s Law for gravitation g4r2 = (Mass enclosed) 4G 3M4 G GM g= 4 (3a)2 3a2
25.
4th harmonic
; 2 2 2 From equation = 0.157 = 40 ; 2 = 80 m 4
26.
Momentum p = mv
…(1) 2
and for motion under gravity h h
27.
u v 2g
2
…(2)
u2 p 2 / m 2g
Work done = U = Uf – Ui q2 q2 = 2Cr 2Ci
(5 10 6 )2 1 1 6 2 5 10 6 2 10 15 = 106 4 = 3.75 × 10–6 J =
28.
By conservation of linear momentum: v v 2v 0 2 0 mv 2v 0 0 mv 2 2 4 3v 0 mv …(1) 2 Since collision is elastic Vseparation = vapproch v 6 v 0 v 0 m = 1.2 kg 4 5
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JEE-MAIN-2019 (9th Apr-First Shift)-PCM-26 29.
Speed of wave from wave equation (coeffecient of t) v (coeffecient of x) 1000 1000 v ( 3) 3 Since speed of wave 1000 273 So = 3 T 336 T = 277.41 K T = 4.41°C
T
30.
1 8 8 Req 7R R
7R 1 8 56 ; Reql Req 7R 64
PART B – CHEMISTRY 31.
Mg O 2 N2 MgO Mg2N3 Air
32.
[Cu(H2O)4]SO4.H2O
33.
No plane of symmetry 34.
TC
8a 27Rb
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JEE-MAIN-2019 (9th Apr-First Shift)-PCM-27 35.
C2 configuration
1s2 * 1s2 2s2 * 2s2
2py 2 2pz2
2Px
C 2 configuration
1s2 * 1s2 2s2 * 2s2 36.
2py 2 2pz2
2Px1
PMo 450 PNo 700 PM = xM 450 PN = xN 700 YMPT = PM YNPT = PN YMPT xM 450 YNPT xN 700 YM xM 0 : 64 YN xN xM YM xN YN
37.
U = q + w G = H - TS
38.
39.
Fact based (Given in NCERT)
40.
For first order reaction n[R]t = -Kt + n[R]o For zero order reaction [R]t = -Kt + [R]o Where ‘R’ is reactant.
41.
COO
COOH H3O
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JEE-MAIN-2019 (9th Apr-First Shift)-PCM-28
42.
Phenol being acidic insoluble in HCl OH
O Na
NaOH
H2O Soluble in NaOH
OH
OH Br
Br
Br2 /Water
Decolourises Bromine water
Br
43. o
NaNO2 , 0 C NH2
N
NCl
Ph - OH, Ph - NH 2, dil HCl N =N
44.
Fact based (Given in NCERT)
45.
Cryolite is Na3[AlF6]
46.
NH2
NO 2 4 3 2 1
Cl
CH3
47.
LiAlH4 does not reduce double bond it reduces ester in alcohol.
48.
After losing one electron ‘K’ acquires noble gas configuration.
49.
56 g of N2 means 2 mole 2 mole of N 2 requires 6 mole of H2 for complete reaction but available H 2 is 10 g i.e. 5 mole hence H 2 is limiting reagent.
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JEE-MAIN-2019 (9th Apr-First Shift)-PCM-29 Cl
50.
DCI CH3C CH CH3 C CHD
DI Cl
CH3 C CHD2 I
51.
xy 4BaCl2 xy iCRT xy 2 CRT BaCl2 3 0.01 RT RT 12 0.01RT 12 0.01 0.06 2
52.
RH 4 RH 9
VLymann
VBalmer
VLymann VBalmer
9 4
Formula 1 1 V RH 2 2 n1 n2
53. KOH alc
Cl Cl
Cl Free radical polymerisation
n Cl
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JEE-MAIN-2019 (9th Apr-First Shift)-PCM-30 54.
R effect
Hyperconjugation
Me
O - Me
R effect
I effect
CN
Cl
55.
Fact based (Given in NCERT)
56.
N O, N2 O, N O2 , N2 O3
57.
G = -2 96000 2 G = -nFEo G = -2 96000 2 J G = -384 kJ mol–1
2
58.
1
4
3
1.PBr3 OH
Br
2. KOH alc.
O
alc KOH
O
O
59.
C1 of -glucose and C2 of -fructose
60.
V2O5 is used in contact process for H2SO4 TiCl4/Al(Me)3 Ziegler natal catalyst used fin polymerization PdCl2 is used in ethanol formation Iron oxide is used in Haber process for NH3
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JEE-MAIN-2019 (9th Apr-First Shift)-PCM-31
PART C – MATHEMATICS 2
61.
62.
x 1 x2 Range of y :R 1, 0 for surjective function, A must be same as above range. y
In given question p,q R . If we take other root as any real number , then quadratic
equation will be x 2 2 3 x . 2 3 0 Now, we can have none or any of the options can be correct depending upon ' ' . Instead of p,q R it should be p,q Q then other root will be 2 3
p 2 3 2 3 4 and q 2 3 2 3 1 2
p 2 4q 12 4 4 12 16 16 0 Option (B) is correct. 63.
3iˆ ˆj 2iˆ ˆj 3kˆ 1 2 1 3iˆ ˆj , 2 3iˆ ˆj 2iˆ j 3kˆ 2 . 0
3 2 .3 1 0 9 6 1 0 1 2 3 1 1 ˆi ˆj 2 2 1 3 2 ˆi ˆj 3kˆ 2 2
ˆi
ˆj
kˆ
3 1 0 Now, 1 2 2 2 1 3 3 2 2 3 9 9 1 ˆi 0 ˆj 0 kˆ 2 2 4 4 3ˆ 9ˆ 5ˆ i j k 2 2 2 1 3iˆ 9ˆj 5kˆ 2
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JEE-MAIN-2019 (9th Apr-First Shift)-PCM-32
64.
I
dx
sin x
4/3
. cos x
2/3
dx
I
4/3
sin x 2 .cos x cos x sec 2 x I dx 4/3 tan x
put tan x t sec 2 x dx dt dt 3 I 4/3 I 1/3 c t t 3 I c 1/3 tan x 65.
Let ax by cz 1 be the equation of the plane 0 b 0 1 b 1 0 0 c 1 c 1 ab cos a b
1 2
0 1 1
a
2
1 1 0 1 1
2
a 24 a 2 2x y z 1 Now for – sign 2, 2 1 4 1 Hence option (A) 66.
y x3 ax b
1, 5
lies on the curve
5 1 a b a b 6 Also, y ' 3x 2 a
…….(i)
y ' 1, 5 3 a (slope of tangent) this tangent is to x y 4 0 3 a 1 1 a 4 ………(ii) By (i) and (ii) : a 4, b 2
y x3 4x 2 , (2, –2) lies on this curve.
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JEE-MAIN-2019 (9th Apr-First Shift)-PCM-33
2
67.
S.D.
x x
n x 1 0 1 k k x 4 4 4 2
Now
2
2
k k k k 1 4 0 4 1 4 k 4 5 4
2
2
k 5k 2 5 4 2 1 8 16 2 3k 18 4 2 k 24 k 2 6
68.
f x 15 x 10 , x R
f f x 15 f x 10 15 15 x 10 10 15 5 x 10 x 5, 10, 15 are differentiability
point 3
69.
6 2 T4 T31 . xlog8 x 3 x 160 20 87 3 . x 3log8 x x 6 log2 x 8 x 3 log2 x 3 18 2 x 18 log2 x 3 log2 x Let log2 x t
of
non
3
t 2 3t 18 0 t 6 t 3 0 t 6, 3
log2 x 6 x 26 82 log2 x 3 x 23 81 70.
f ' x x 1 x 0 x 1 x3 x
x4 x2 f x 2 4 Now f x f 0
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JEE-MAIN-2019 (9th Apr-First Shift)-PCM-34
x4 x2 2 4 x 0,0, 2 Two irrational and one rational number 71.
~ p ~ p q ~ p ~ ~ p q ~ p p q ~ p p ~ p ~ q c ~ p ~ q ~ p ~ q
72.
73.
Since there are 8 males and 5 females. Out of these 13, if we select 11 persons, then there will be at least 6 males and at least 3 females in the selection. 13 13 13 12 m n 78 2 11 2 Any point on the given line can be 1 2, 1 3, 2 4 ; R Put in plane 1 2 2 6 6 12 15 20 5 15 20 10 1 2 1 Point 2, , 4 2 Distance from origin 4
74.
1 16 1 64 81 16 4 2 2
9 2
1 2 cos2 10o 2cos10o cos 50o 2cos2 50o 2 1 1 cos 20o cos 60o cos 40o 1 cos100o 2 13 cos 20o 2 sin 70o sin 30o 22 13 cos 20o sin70o 22 3 4
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JEE-MAIN-2019 (9th Apr-First Shift)-PCM-35
75.
1 1 1 2 1 2 1 3 1 n 1 1 78 ...... 1 0 1 0 1 0 1 0 1 0 1 0 1 1 2 3 ...... n 1 1 78 1 0 0 1 n n 1 78 n 13, 12 (reject) 2 1 13 we have to find inverse of 0 1 1 13 1 0
i z 1 i z i 1 z circle of radius 1
76.
Let
77.
Sn 50n
n n 7 2
A
Tn Sn Sn 1
50n
n n 7 2
A 50 n 1
n 1n 8 2
A
A 2 n 7n n2 9n 8 2 50 A n 4 d Tn Tn 1 50
50 A n 4 50 A n 5 A T50 50 46A
d, A 50 A,50 46A /2
78.
I
0
sin3 x dx sin x cos x / 4
I
0
sin3 x cos3 x dx sin x cos x
/ 4
1 sin x cos x dx 0
/4
sin2 x x 2 0
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JEE-MAIN-2019 (9th Apr-First Shift)-PCM-36
1 4 4 1 4
79.
x 2 y2 1 a 24 :b 18 24 18 Parametric normal: 24 cos . x 18. y cot 42 42 At x 0; y tan 7 3 (from given equation) 18
3 3 sin 2 5
tan
slope of parametric normal m
80.
24 cos 18 cot
m
4 2 2 sin or 3 5 5
dy 2y x 2 : y 1 1 dx dy 2 y x (LDE in y) dx x
x
2
x dx e2 n x x 2 x4 y. x 2 x. x 2dx C 4 y 1 1
IF e
1 1 3 C C 1 4 4 4 4 x 3 yx 2 4 4 x2 3 y 2 4 4x
1
81.
y 2 4ax 16x a 4 A 1, 4 2. 4. t1 4 t1
1 2
1 length of focal chord a t t
2
2
25 1 4 2 4. 25 2 4
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JEE-MAIN-2019 (9th Apr-First Shift)-PCM-37 82.
f 1 1 1 2 2 f 1 1 1 2 0 m
f 1 f 1
2 0 1 2
1 1 dy 2 3x 2x 2 dx 3x 2 2x 2 1 3x 2 2x 1 0 x 1 3x 1 0
x 1, 83.
1 3
x 2 r cos y 3 r sin 2 r cos 3 r sin 7 r cos sin 2
2 2 1 r 4 2
sin cos 1 sin 2
1 4
3 4 2m 3 1 m2 4 2 3m 8m 3 0 4 7 m 1 7 sin 2
2
1 7 1 7
1 7 1 7
8 2 7 4 7 6 3
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JEE-MAIN-2019 (9th Apr-First Shift)-PCM-38 84.
From the given functional equation: f x 2x x N
16 2
1
2a 1 2a 2 .... 2a10 16 210 1 a
2
10
2 2 2 ..... 2 2a.
16
2. 210 1
1 a 1 2 16 24 a=3
85.
10
2
10
1
2 1 sin2 3 sin 0 2
2sin 3 sin 2 0 2 sin 1 sin 2 0
1 sin ;sin 2 (reject) 2 roots : , 2 , , 6 6 6 6 sum of values 2 86.
Let the mid point be S (h, k) P 2h,0 and Q 0,2k equation of
x y 1 2h 2k PQ is tangent to circle at R (say) PQ :
1
OR 1
2
1 1 2h 2k
2
1
1 1 2 1 2 4h 4k 2 2 x y 4x 2 y 2 0
87.
Let persons be A, B, C, D P (Hit) = 1 – P (none of them hits)
1 P A .P B .P C .P D 1 P A B C D
1 2 3 7 1 . . . 2 3 4 8 25 32
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JEE-MAIN-2019 (9th Apr-First Shift)-PCM-39
88.
x2 y x 2 x 2 y; y x 2 x2 x 2 x2 x 2 0 x 2 x 1 0 x 2, 1 2
Area
9
2
x 2 x dx 2
1
89.
Roots of the equation x 2 x 1 0 are and 2 where , 2 are complex cube roots of unity y 1 2
2
y 2
1
1
y
R1 R1 R2 R3 1 y 2
1
1 2
y
1
1
y
Expanding along R1 , we get
y . y2 D y3 Or If , we get same value or on expansion using 1 , 1 we get 2
value y 3 . 90.
function should be continuous at x
4
lim f x f x 4 4 lim
x 4
lim
x 4
2 cos x 1 k cot x 1 2 sin x k (Using L Hospital Rule) cos ec 2 x
lim 2 sin3 x k x
4
3
1 1 k 2 2 2
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FIITJEE Solutions to JEE(Main)-2019 Test Date: 9th April 2019 (Second Shift)
PHYSICS, CHEMISTRY & MATHEMATICS Paper - 1 Time Allotted: 3 Hours
Maximum Marks: 360
Please read the instructions carefully. You are allotted 5 minutes specifically for this purpose.
Important Instructions: 1.
The test is of 3 hours duration.
2.
This Test Paper consists of 90 questions. The maximum marks are 360.
3.
There are three parts in the question paper A, B, C consisting of Physics, Chemistry and Mathematics having 30 questions in each part of equal weightage. Each question is allotted 4 (four) marks for correct response.
4.
Out of the four options given for each question, only one option is the correct answer.
5.
For each incorrect response 1 mark i.e. ¼ (one-fourth) marks of the total marks allotted to the question will be deducted from the total score. No deduction from the total score, however, will be made if no response is indicated for an item in the Answer Box.
6.
Candidates will be awarded marks as stated above in instruction No.3 for correct response of each question. One mark will be deducted for indicating incorrect response of each question. No deduction from the total score will be made if no response is indicated for an item in the answer box.
7.
There is only one correct response for each question. Marked up more than one response in any question will be treated as wrong response and marked up for wrong response will be deducted accordingly as per instruction 6 above.
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JEE-MAIN-2019 (9th Apr-Second Shift)-PCM-2
PART –A (PHYSICS) 1.
Two coils P and Q are separated by some distance. When a current of 3 A flows through coil P a magnetic flux of 10–3 Wb passes through Q. No current is passed through Q. When no current passes through P and a current of 2 A passes through Q, the flux through P is: (A) 6.67 103 Wb (B) 6.67 10 4 Wb (C) 3.67 10 4 Wb
(D) 3.67 10 3 Wb
2.
A metal wire of resistance 3 is elongated to make a uniform wire of double its previous length. The new wire is now bent and the ends joined to make a circle. If two points on this circle make an angle 600 at the centre, the equivalent resistance between these two points will be: 12 5 (A) (B) 5 3 5 7 (C) (D) 2 2
3.
The resistance of a galvanometer is 50 ohm and the maximum current which can be passed through it is 0.002 A. What resistance must be connected to it in order to convert it into an ammeter of range 0 – 0.5 A? (A) 0.2 ohm (B) 0.002 ohm (C) 0.02 ohm (D) 0.5 ohm
4.
The position of a particle as a function of time t, is given by x t at bt 2 ct 3 where a, b and c are constants. When the particle attains zero acceleration, then its velocity will be: b2 b2 (A) a (B) a 4c c b2 b2 (C) a (D) a 2c 3c
5.
A thin convex lens L (refractive index = 15) is placed on a plane mirror M. When a pin is placed at A, such that OA = 18 cm, its real inverted image is formed at A itself, as shown in figure. When a liquid of refractive index 1 is put between the lens and the mirror, the pin has to be moved to A’. such that OA’ = 27cm, to get its inverted real image at A itself. The value of 1 will be:
(A)
2
(C)
3
4 3 3 (D) 2
(B)
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JEE-MAIN-2019 (9th Apr-Second Shift)-PCM-3 6.
A moving coil galvanometer has a coil with 175 turns and area 1 cm2 It uses a torsion band of torsion constant 10–6 N – m/rad. The coil is placed in a magnetic field B parallel to its plane. The coil deflects by 10 for a current of 1 mA. The value of B (in Tesla) is approximately:(A) 10-3 (B) 10-1 -4 (C) 10 (D) 10-2
7.
A very long solenoid of radius R is carrying current I t kte t k 0 , as a function of time (t 0). counter clockwise current is taken to be positive. A circular conducting coil of radius 2R is placed in the equatorial plane of the solenoid and concentric with the solenoid. The current induced in the outer coil is correctly depicted, as a function of time, by:(A) (B)
(C)
(D)
8.
A massless spring (k = 800 N/m), attached with a mass (500 g) is completely immersed in 1 kg of water. The spring is stretched by 2 cm and released so that it starts vibrating. What would be the order of magnitude of the change in the temperature of water when the vibrations stop completely? (Assume that the water container and spring receive negligible heat and specific heat of mass = 400 J/kg K, specific heat of water = 4184 J/kg K) (A) 10-3 K (B) 10-4 -1 (C) 10 K (D) 10-5 K
9.
A particle P is formed due to a completely inelastic collision of particles x and y having de – Broglie wavelengths x and y respectively. If x and y were moving in opposite directions, then the de – Broglie wavelength of P is: xy (A) x y (B) x y (C)
10.
xy x y
(D) x y
A convex lens of focal length 20 cm produces images of the same magnification 2 when an object is kept at two distances x1 and x2 (x1 > x2) from the lens. The ratio of x1 and x2 is: (A) 5 : 3 (B) 2 : 1 (C) 4 : 3 (D) 3 : 1
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JEE-MAIN-2019 (9th Apr-Second Shift)-PCM-4
11.
Diameter of the objective lens of a telescope is 250 cm. For light of wavelength 600 nm. Coming from a distant object, the limit of resolution of the telescope is close to:(A) 1.5 10 7 rad (B) 2.0 10 7 rad (C) 3.0 10 7 rad
(D) 4.5 10 7 rad
12.
Moment of inertia of a body about a given axis is 1.5 kg m2 Initially the body is at rest. In order to produce a rotational kinetic energy of 1200 J, the angular acceleration of 20 rad/s2 must be applied about the axis of rotation for a duration of: (A) 2 s (B) 5 s (C) 2.5 s (D) 3 s
13.
The logic gate equivalent to the given logic circuit is:
(A) OR (C) NOR 14.
(B) AND (D) NAND
50 W/m2 energy density of sunlight is normally incident on the surface of a solar panel. Some part of incident energy (25%) is reflected from the surface and the rest is absorbed. The force exerted on 1 m2 surface area will be close to c 3 108 m / s (A) 15 10 8 N (C) 10 10 8 N
(B) 35 10 8 N (D) 20 10 8 N
15.
The area of a square is 5.29 cm2 The area of 7 such squares taking into account the significant figures is: (A) 37 cm2 (B) 37.0 cm2 2 (C) 37.03 cm (D) 37.030 cm2
16.
The physical sizes of the transmitter and receiver antenna in a communication system are: (A) Proportional of carrier frequency (B) Inversely proportional to modulation frequency (C) Inversely proportional to carrier frequency (D) Independent of both carrier and modulation frequency
17.
Four point charges –q, +q, +q and –q are placed on y axis at y = -2d, y = -d, y = +d and y = +2d, respectively. The magnitude of the electric field E at a point on the x – axis at x = D, with D > > d, will vary as: 1 1 (A) E (B) E 3 D D 1 1 (C) E 2 (D) E 4 D D
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JEE-MAIN-2019 (9th Apr-Second Shift)-PCM-5 18.
The specific heats, CP and CV of a gas of diatomic molecules, A, are given (in units of J mol-1 K-1) by 29 and 22, respectively. Another gas of diatomic molecules B, has the corresponding values 30 and 21. If they are treated as ideal gases, then:(A) A has one vibrational mode and B has two (B) Both A and B have a vibrational mode each (C) A is rigid but B has a vibrational mode (D) A has a vibrational mode but B has none
19.
The position vector of a particle changes with time according to the relation r t 15t 2 ˆi 4 20t 2 ˆj . What is the magnitude of the acceleration at t = 1? (A) 40 (C) 25
(B) 100 (D) 50
20.
A test particle is moving in a circular orbit in the gravitational field produced by a mass K density r 2 . Identify the correct relation between the radius R of the particle’s orbit r and its period T: (A) T/R2 is a constant (B) TR is constant (C) T2/R3 is a constant (D) T/R is a constant
21.
A particle of mass m is moving with speed 2 v collides with a mass 2m moving with speed v in the same direction. After collision, the first mass is stopped completely while the second one splits into two particles each of mass m, which move at angle 450 with respect to the original direction. The speed of each of the moving particle will be: (A) v / 2 2 (B) 2 2v
(C)
2v
(D) v / 2
4 of its volume submerged. When 5 certain amount of an oil is poured into the bucket, it is found that the block is just under the oil surface with half of its volume under water and half in oil. The density of oil relative to that of water is:(A) 0.5 (B) 0.7 (C) 0.6 (D) 0.8
22.
A wooden block floating in a bucket of water has
23.
Two cars A and B are moving away from each other in opposite directions. Both the cars are moving with a speed of 20 ms-1 with respect to the ground. If an observer in car A detects a frequency 2000 Hz of the sound coming from car B, what is the natural frequency of the sound source of car B? (speed of sound in air = 340 ms-1) (A) 2250 Hz (B) 2060 Hz (C) 2150 Hz (D) 2300 Hz
24.
A wedge of mass M = 4m lies on a frictionless plane. A particle of mass m approaches the wedge with speed v. There is no friction between the particle and the plane or between the particle and the wedge. The maximum height climbed by the particle on the wedge is given by: 2v 2 v2 (A) (B) 7g g (C)
2v 2 5g
(D)
v2 2g
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JEE-MAIN-2019 (9th Apr-Second Shift)-PCM-6
25.
A He+ ion is in its first excited state. Its ionization energy is: (A) 6.04 eV (B) 13.60 eV (C) 54.40 eV (D) 48.36 eV
26.
Two materials having coefficients of thermal conductivity 3K and K and thickness d and 3d, respectively, are joined to form a slab as shown in the figure. The temperatures of the outer surfaces are 2 and 1 respectively 2 1 . The temperature at the interface is:
2 1 2 (C) 1 2 3 3
(A)
1 92 10 10 5 (D) 1 2 6 6
(B)
27.
A thin smooth rod of length L and mass M is rotating freely with angular speed 0 about an axis perpendicular to the rod and passing through its centre. Two beads of mass m and negligible size are at the centre of the rod initially. The beads are free to slide along the rod. The angular speed of the system, when the beads reach the opposite ends of the rod will be:M0 M0 (A) (B) M 3m Mm M0 M0 (C) (D) M 2m M 6m
28.
The parallel combination of two air filled parallel plate capacitors of capacitance C and nC is connected to a battery of voltage, V. When the capacitor are fully charged, the battery is removed and after that a dielectric material of dielectric constant K is placed between the two plates of the first capacitor. The new potential difference of the combined system is: V (A) (B) V Kn n 1 V nV (C) (D) Kn K n
29.
In a conductor, if the number of conduction electrons per unit volume is 8.5 x 1028 m–3 and mean free time is 25fs (femto second), its approximate resistivity is: me 9.1 1031kg (A) 10-5 m (C) 10 7 m
30.
(B) 10-6 m (D) 10-8 m
A string 2.0 m long and fixed at its end is driven by a 240 Hz vibrator. The string vibrates in its third harmonic mode. The speed of the wave and its fundamental frequency is: (A) 320 m/s, 120 Hz (B) 180 m/s, 80 Hz (C) 180m/s, 120 Hz (D) 320 m/s, 80 Hz FIITJEE Ltd., FIITJEE House, 29-A, Kalu Sarai, Sarvapriya Vihar, New Delhi -110016, Ph 46106000, 26569493, Fax 26513942 website: www.fiitjee.com.
JEE-MAIN-2019 (9th Apr-Second Shift)-PCM-7
PART –B (CHEMISTRY) 31.
Increasing order of reactivity of the following compounds for SN1 substitution is: (a) (b)
(c)
32.
(d)
(A) (b) < (c) < (d) < (a) (C) (b) < (a) < (d) < (c)
(B) (a) < (b) < (d) < (c) (D) (b) < (c) < (a) < (d)
The one that is not a carbonate is: (A) bauxite (C) calamine
(B) siderite (D) malachite
33.
During compression of a spring the work done is 10kJ and 2kJ escaped to the surroundings as heat. The change in internal energy, U (in kJ) is: (A) 8 (B) 12 (C) -12 (D) -8
34.
The amorphous form of silica is: (A) quartz (C) cristobalite
35.
(B) kieselguhr (D) tridymite
The major products A and B for the following reactions are, respectively
(A)
(B)
(C)
(D)
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JEE-MAIN-2019 (9th Apr-Second Shift)-PCM-8 36.
Which one of the following about an electron occupying the 1s orbital in a hydrogen atom is incorrect? (The Bohr radius is represented by a0) (A) The electron can be found at a distance 2a0 from the nucleus (B) The probability density of finding the electron is maximum at the nucleus (C) The magnitude of potential energy is double that of its kinetic energy on an average (D) The total energy of the electron is maximum when it is at a distance a0 from the nucleus
37.
The maximum possible denticities of a ligand given below towards a common transition and inner – transition metal ion, respectively, are:
(A) 6 and 8 (C) 8 and 8
(B) 8 and 6 (D) 6 and 6
38.
The correct statements among I to III regarding group 13 element oxides are, I. Boron trioxide is acidic II. Oxides of aluminium and gallium are amphoteric III. Oxides of indium and thallium are basic (A) I, II and III (B) II and III only (C) I and III only (D) I and II only
39.
Among the following species, the diamagnetic molecule is (A) O2 (B) NO (C) B2 (D) CO
40.
The peptide that gives positive ceric ammonium nitrate and carbylamine tests is: (A) Lys - Asp (B) Ser – Lys (C)Gln - Asp (D) Asp – Gln
41.
Assertion : For the extraction of iron, haematite ore is used Reason : Haematite is a carbonate ore of iron (A) Only the reason is correct. (B) Both the assertion and reason are correct and the reason is the correct explanation for the assertion. (C) Only the assertion is correct (D) Both the assertion and reason are correct, but the reason is not the correct explanation for the assertion
42.
10 mL of 1mM surfactant solution forms a monolayer covering 0.24 cm2 on a polar substrate. If the polar head is approximated as cube, what is its edge length? (A) 2.0 pm (B) 2.0 nm (C)1.0 pm (D) 0.1 nm
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JEE-MAIN-2019 (9th Apr-Second Shift)-PCM-9 43.
Consider the given plot of enthalpy of the following reaction between A and B. A+BC+D Identify the incorrect statement
(A) C is the thermodynamically stable product (B) Formation of A and B from C has highest enthalpy of activation (C) D is kinetically stable product (D) Activation enthalpy to form C is 5kJ mol-1 less than that to form D 44.
At a given temperature T, gases Ne, Ar, Xe and Kr are found to deviate from ideal gas RT behaviour. Their equation of state is given as P at T. Here, b is the van der V b Waals constant. Which gas will exhibit steeper increase in the plot of Z (compression factor) versus P? (A) Ne (B) Ar (C) Xe (D) Kr
45.
A solution of Ni(NO3)2 is electrolysed between platinum electrodes using 0.1 Faraday electricity. How many mole of Ni will be deposited at the cathode? (A) 0.20 (B) 0.05 (C). 0.10 (D) 0.15
46.
In the following reaction
Carbonyl compound + MeOH acetal Rate of the reaction is the highest for: (A) Acetone as substrate and methanol in stoichiometric amount (B) Proponal as substrate and methanol in stoichiometric amount (C) Acetone as substrate and methanol in excess (D) Propanal as substrate and methanol in excess HCI
47.
The structures of beryllium chloride in the solid state and vapour, phase, respectively, are: (A) chain and dimeric (B) chain and chain (C)dimeric and dimeric (D) dimeric and chain
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JEE-MAIN-2019 (9th Apr-Second Shift)-PCM-10 48.
49.
Which of the following potential energy (PE) diagrams represents the SN1 reaction?
(A)
(B)
(C)
(D)
The major product of the following reaction is:
(A)
(B)
(C)
(D)
50.
The maximum number of possible oxidation states of actinoids are shown by (A) berkelium (Bk) and californium (Cf) (B) nobelium (No) and lawrencium (Lr) (C) actinium (Ac) and thorium (Th) (D) neptunium (Np) and plutonium (Pu)
51.
Molal depression constant for a solvent is 4.0 K Kg mol-1. The depression in the freezing point of the solvent for 0.03 mol kg–1 solution of K2SO4 is: (Assume complete dissociation of the electrolyte) (A) 0.12 K (B) 0.36 K (C) 0.18 K (D) 0.24 K
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JEE-MAIN-2019 (9th Apr-Second Shift)-PCM-11 52.
53.
Noradrenaline is a/an (A) Neurotransmitter (C) Antihistamine
(B) Antidepresessant (D) Antacid
Hinsberg’s reagent is: (A) C6H5SO2CI
(B) C6H5COCI
(C) SOCI2
(D) COCI2
54.
The layer of atmosphere between 10 km to 50 km above the sea level is called as: (A) troposphere (B) mesosphere (C)stratosphere (D) thermosphere
55.
HF has highest boiling point among hydrogen halides, because it has: (A) lowest dissociation enthalpy (B) strongest van der Waals’ interactions (C) strongest hydrogen bonding (D) lowest ionic character
56.
What would be the molality of 20% (mass/mass) aqueous solution of KI? (molar mass of KI = 166 g mol–1) (A) 1.08 (B) 1.48 (C)1.51 (D) 1.35
57.
Which of the following compounds is a constituent of the polymer?
(A) Formaldehyde (C) Methylamine 58.
(B) Ammonia (D) N – methyl urea
The correct statements among I to III are: I. Valence bond theory cannot explain the color exhibited by transition metal complexes II. Valence bond theory can predict quantitatively the magnetic properties of transition metal complexes III. Valence bond theory cannot distinguish ligands as weak and strong field ones (A) I and II only (B) I, II and III (C) I and III only (D) II and III only
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JEE-MAIN-2019 (9th Apr-Second Shift)-PCM-12 59.
In an acid – base titration, 0.1 M HCI solution was added to the NaOH solution of unknown strength. Which of the following correctly shows the change of pH of the titration mixture in this experiment?
(A) (a) (C) (d) 60.
(B) (c) (D) (b)
p-Hydroxybenzophenone upon reaction with bromine in carbon tetrachloride gives:
(A)
(B)
(C)
(D)
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JEE-MAIN-2019 (9th Apr-Second Shift)-PCM-13
PART–C (MATHEMATICS) 61.
If the tangent to the parabola y 2 x at a point , , ( > 0) is also a tangent to the ellipse, x 2 2y 2 1, then a is equal to (A) 2 2 1 (C) 2 1
62.
(B) 2 1 (D) 2 2 1
Some identical balls are arranged in rows to form an equilateral triangle. The first row consists of one ball, the second row consists of two balls and so on. If 99 more identical balls are added to the total number of balls used in forming the equilateral triangle, then all these balls can be arranged in a square whose each side contains exactly 2 balls less than the number of balls each side of the triangle contains. Then the number of balls used to form the equilateral triangle is (A) 190 (B) 262 (C) 225 (D) 157 f x
63.
If f : R R is a differentiable function and f 2 6, then lim
x2
2 tdt
x 2
is:
6
(B) 2f ' 2 (D) 24f’ (2)
(A) 0 (C) 12f’ (2) 64.
If the system of equations 2x 3y z 0, x ky 2z 0 and 2x y z 0 has a non – x y z trivial solution (x, y, z), then k is equal to y z x 3 (A) (B) –4 4 1 1 (C) (D) 2 4
65.
The common tangent to the circles x 2 y 2 4 and x 2 y 2 6x 8y 24 0 also passes through the point (A) (–4, 6) (B) (6. –2) (C) (–6, 4) (D) (4, –2)
66.
If the sum and product of the first three term in an A.P. are 33 and 1155, respectively, then a value of its 11th tern is (A) –25 (B) 25 (C) –36 (D) –35
67.
The value of the integral
1
x cot 1 x 1
2
x 4 dx is
0
1 loge 2 4 2 1 (C) loge 2 2 2
(A)
loge 2 2 (D) loge 2 4
(B)
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JEE-MAIN-2019 (9th Apr-Second Shift)-PCM-14 68.
The value of sin10o sin 30o sin50o sin70o is 1 1 (A) (B) 36 32 1 1 (C) (D) 18 16
69.
Let z C be such that z 1 . If w
5 3z , then 5 1 z
(A) 5Im w 1
(B) 4Im w 5
(C) 5 Re w 1
(D) 5 Re w 4 n
70.
If some three consecutive in the binomial expansion of x 1 in powers of x are in the ratio 2 : 15 : 70, then the average of these three coefficient is: (A) 964 (B) 625 (C) 227 (D) 232
71.
If cos x
72.
If the two lines x a 1 y 1 and 2x a 2 y 1 a R 0,1 are perpendicular, then the
dy y sin x 6x, 0 x and y 0, then y is equal to: dx 2 3 6 2 2 (A) (B) 2 4 3 2 2 (C) (D) 2 3 2 3
distance of their point of intersection from the origin is: 2 2 (A) (B) 5 5 2 2 (C) (D) 5 5 73.
A water tank has the shape of an inverted right circular cone, whose semi vertical angle 1 is tan 1 .Water is poured in at a constant rage of 5 cubic meter per minute. Then the 2 rate (in m/min) at which the level of water is rising at the instant when the depth of water in the tank is 10 m is: 2 1 (A) (B) 5 1 1 (C) (D) 10 15
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JEE-MAIN-2019 (9th Apr-Second Shift)-PCM-15 74.
Two poles standing on a horizontal ground are of heights 5m and 10 m respectively. The line joining their tops makes an angle of 15o with ground. Then the distance (in m) between the poles, is 5 (A) 2 3 (B) 5 3 1 2 (C) 5 2 3 (D) 10 3 1
75.
77.
x 2 y 1 z such that BC = 5 units. 3 0 4 Then the area (in sq. units) of this triangle, given that the point A (1, –1, 2) is: (A) 2 34 (B) 34
(D) 5 17
0 2x 2x The total number of matrices A 2y y y ; x, y R, x y for which A T A 3I3 1 1 1 (A) 6 (B) 2 (C) 3 (D) 4
The area in sq. units) of the smaller of the two circles that touch the parabola, y 2 4x at the points (1, 2) and the axis is (A) 4 2 2 (B) 8 3 2 2
(C) 4 3 2 78.
79.
The vertices B and C of a ABC lie on the line,
(C) 6
76.
(D) 8 2 2
a x 1, x 5 If the function f x is continuous at x = 5, then the value of a – b is b x 3, x 5 2 2 (A) (B) 5 5 2 2 (C) (D) 5 5 x If f x x , x R , where [x] denotes the greatest integer function, then: 4 (A) Both lim f x and lim f x exist but are not equal x 4
x 4
(B) lim f x exists but lim f x does not exist x 4
x 4
(C) lim f x exists but lim f x does not exist x 4
x 4
(D) f is continuous at x = 4
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JEE-MAIN-2019 (9th Apr-Second Shift)-PCM-16
80.
If
e
sec x
sec x tan x f x sec x tan x sec x dx e 2
sec x
f x C , then a possible
choice of f x is (A) sec x tan x
1 2
(C) sec x x tan x 81.
(B) x sec x tan x 1 2
(D) sec x tan x
1 2
1 2 2
If m is chosen in the quadratic equation m2 1 x 2 3x m2 1 0 such that the sum of its roots is greatest, then the absolute difference of the cubes of its roots is: (A) 8 3 (B) 4 3 (C) 10 5
(D) 8 5
82.
Two newspaper A and B are published in a city. It is known that 25% of the city populations reads A and 20% reads B while 8% reads both A and B. Further, 30% of those who read A but not B look into advertisements and 40% of those who read B but not A also look into advertisements, while 50% of those who read both A and B look into advertisements. Then the percentage of the population who look into advertisement is: (A) 12.8 (B) 13.5 (C) 13.9 (D) 13
83.
Let P be the plane, which contains the line of intersection of the planes, x y z 6 0 and 2x 3y z 5 0 and it is perpendicular to the xy – plane. Then the distance of the point (0, 0, 256) from P is equal to: (A) 63 5 (B) 205 5 17 11 (C) (D) 5 5
84.
If P q r is false, then the truth values of p, q, r are respectively (A) F, T, T (B) T, F, F (C) T, T, F (D) F, F, F 1 The domain of the definition of the function f x log x3 x is 2 4x (A) 1,2 2, (B) 1,0 1,2 3,
85.
(C) 1,0 1,2 2,
(D) 2, 1 1,0 2,
86.
The sum of the series 1 2 3 3 5 4 7 .... upto 11th term is (A) 915 (B) 946 (C) 945 (D) 916
87.
The mean and the median of the following ten numbers in increasing order 10, 22, 26, y 29, 34, x, 42, 67, 70, y are 42 and 35 respectively, then is equal to x 7 9 (A) (B) 3 4 7 8 (C) (D) 2 3 FIITJEE Ltd., FIITJEE House, 29-A, Kalu Sarai, Sarvapriya Vihar, New Delhi -110016, Ph 46106000, 26569493, Fax 26513942 website: www.fiitjee.com.
JEE-MAIN-2019 (9th Apr-Second Shift)-PCM-17
88.
89.
90.
y2 The area (in sq. units) of the region A x, y : x y 4 is: 2 53 (A) (B) 18 3 (C) 30 (D) 16
If a unit vector r makes angles with ˆi, with ˆj and 0, with kˆ , then a value of 3 4 is 5 5 (A) (B) 12 6 2 (C) (D) 3 4 A rectangle is inscribed in a circle with a diameter lying along the line 3y x 7 . If the two adjacent vertices of the rectangle are (–8, 5) and (6, 5) then the area of the rectangle (in sq. units) is (A) 72 (B) 84 (C) 98 (D) 56
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JEE-MAIN-2019 (9th Apr-Second Shift)-PCM-18
JEE (Main) – 2019 ANSWERS PART A – PHYSICS 1. 5. 9. 13. 17. 21. 25. 29.
B B C A D B B D
2. 6. 10. 14. 18. 22. 26. 30.
B A D D D C B D
3. 7. 11. 15. 19. 23. 27.
A D C C D A D
4. 8. 12. 16. 20. 24. 28.
D D A C D C C
34. 38. 42. 46. 50. 54. 58.
B A A D D C C
64. 68. 72. 76. 80. 84. 88.
C D D D D B B
PART B – CHEMISTRY 31. 35. 39. 43. 47. 51. 55. 59.
C B D D A B C A
32. 36. 40. 44. 48. 52. 56. 60.
A D B C D A C D
33. 37. 41. 45. 49. 53. 57.
A A C B C A A
PART C – MATHEMATICS 61. 65. 69. 73. 77. 81. 85. 89.
C B C B B D C C
62. 66. 70. 74. 78. 82. 86. 90.
A A D C A C B B
63. 67. 71. 75. 79. 83. 87.
C A D B D D A
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JEE-MAIN-2019 (9th Apr-Second Shift)-PCM-19
HINTS AND SOLUTIONS PART A – PHYSICS 1.
Mutual induction q MIp 10–3 = M(3) 1 M 10 3 3 p = MIq p = 6.67 × 10–4 Wb
2.
2 (V Volume of wire) A (V / ) V Final resistance = 3 × (B)2 = 12 Req = 2 || 10 5 = 3 R
3.
Since shunt resistance is connected in parallel with galvanometer, both will have same voltage drop. (0.002) (RG) = (0.5 – 0.002) (rs) rs 0.2
4.
x = at + bt2 – ct3 dx V = a + 2bt – 3ct2 dt dv a 2b 6ct dt Put acceleration = 0 b t 3c b Find V at t 3c b2 V= a 3c
5.
For image to form at object itself, says must retrace their path back to object. Hence must incident on mirror normally. Case 1: Object will be at focus of lens 1 1 1 1 ( 1) f R R 18 R = 18 cm Case 2: Retraction at 1st surface:
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JEE-MAIN-2019 (9th Apr-Second Shift)-PCM-20
1 1.5 1 1.5 27 V1 R
…(i)
2nd retraction:
1.5 1.5 u V1 R
…(ii)
From (i) and (ii) = 6.
7.
4 . 3
MB C = i N A B 106 103 104 175 B 180 B = 10–3 Tesla
d d = (onI) ( R2 ) dt dt dI = ( on R 2 ) dt d (kt e t ) = on R2 dt = –C (t(e–t) (–) + e–t) = –C (e–t) (1 – t)
(C constant)
Clearly, at t = 0 Induced current 0 Also, apply Lenz law to find correct option. 8.
By law of conservation of energy 1 2 kx = (m1s1 + m2s2)T 2 16 10 2 T = 3.65 × 10–5 4384
9.
Conservation of momentum p x p y p final mxvx – myvy = (mx + my) V h h h x y
xy x y
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JEE-MAIN-2019 (9th Apr-Second Shift)-PCM-21 10.
Magnification is 2 If image is real, x1
3f 2
If image is virtual, x2 =
f 2
x1 =3:1 x2
1.22 d 1.22 600 10 9 = 250 10 2 = 2.9 × 10–7 rad.
11.
Limit of resolution =
12.
KE =
13.
Truth table can be formed as
1 2 I 1200 (given) 2 = 40 rad/s = o + t 40 = 0 + (20) t t = 2 sec.
A 0 0 1 1
B 0 1 0 1
Equivalent 0 1 1 1
Hence the equivalent is “OR” gate. 14.
Radiation pressure for 100% reflection = Radiation pressure for 0% reflection =
2I C
I C
2I I Hence, in given case, radiation pressure = (0.25) (0.75) C C I = (1.25) C Force = P × (Area) = 20.83 × 10–8 N 15.
Total Area = A1 + A2 + ……… A7 = A + A + ……….. 7 times = 37.03 m2 Addition of 7 terms all having 2 terms beyond decimal, so final answer must have 2 terms beyond decimal (as per rules of significant digits)
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JEE-MAIN-2019 (9th Apr-Second Shift)-PCM-22
16.
The physical size of antenna of receiver and transmitter both are inversely proportional to carrier frequency.
17.
Electric field at p = 2E1cos1 –2E2cos2 2Kq D 2Kq D = 2 2 2 2 1/2 2 2 2 (d D ) (d D ) [(2d) D ] [(2d) D 2 ]1/ 2 = 2KqD (d2 D2 )3 /2 (4d2 D2 )3/ 2 3/ 2 3/ 2 d2 4d2 1 2 1 2 D D Applying binomial approximation d << D 2KqD 3 d2 3 4d2 = 1 1 D3 2 D2 2D2 2KqD 12 d2 3 d2 = D3 2 D2 2 D2
2KqD = D3
= 18.
9kqd2 D4
For A: Cp 2 29 1 Cv f 22 It gives f = 6.3 6
(3 translational, 2 rotational and 1 vibrational)
For B: Cp 2 30 1 Cv f 21 f = 4.67 5 (3 translational, 2 rotational, no vibrational)
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JEE-MAIN-2019 (9th Apr-Second Shift)-PCM-23
19.
20.
r (15t 2 )iˆ (4 20t 2 )ˆj dr v (30t)iˆ (40t)jˆ dt dv a (30)iˆ (40)jˆ dt a 50
For circular motion of particle: mV 2 mE r GM = m 2 r r
k Where M = (4 x 2 dx) 2 x 0 = 4kr 2 mV G(4 k) m r r V = constant 2R T= V T Constant R
21.
Linear momentum conservation v m 2v 2m v m 0 m 2 2 v 2 2 v. 22.
In 1st situation Vbbg = Vswg Vs b 4 Vb w 5
…(i)
Here Vb is volume of block Vs is submerged volume of block b is density of block w is density of water & Let o is density of oil Finally in equilibrium condition V V Vb b g b o g b w g 2 2 FIITJEE Ltd., FIITJEE House, 29-A, Kalu Sarai, Sarvapriya Vihar, New Delhi -110016, Ph 46106000, 26569493, Fax 26513942 website: www.fiitjee.com.
JEE-MAIN-2019 (9th Apr-Second Shift)-PCM-24 2b = 0 + w 3 o 0.6 w 5 23.
Doppler effect: v uo f (fo ) v us 340 ( 20) 2000 (fo ) 340 ( 20) fo = 2250 Hz
24.
Let mass attains height ‘h’ on wedge and at that time, both attain velocity vf. Conservation of momentum: mv = (m + 4m)Vf …(A) COE: 1 1 mv 2 (m 4m)Vf2 mgh 2 2 From (A) and (B) 2V 2 h 5g
25.
26.
27.
…(B)
z2 T.E. = (13.6) 2 eV n z=n=2 Ionisation energy = –T.E. = 13.6 eV At steady state: q q t 1 t 2 3kA( 2 ) kA( 1 ) d 3d 1 92 10 Conservation of angular momentum about rotation axes: Li = Lf M 2 2 M 2 2 m f o 2 12 12
M f o M 6m
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JEE-MAIN-2019 (9th Apr-Second Shift)-PCM-25 28.
After fully charging, battery is disconnected.
Total chare of the system = CV + nCV = (n + 1)CV After the insertion of dielectric of constant K New potential (common) Total charge VC Total capacitance (n 1)CV (n 1)V = KC nC K n
2m ne2 = 3.34 × 10–8 m
29.
30.
We have: nv f 2 3v 240 = 22 v = 320 m/s Fundamental frequency =
v = 80 Hz. 2
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JEE-MAIN-2019 (9th Apr-Second Shift)-PCM-26
PART B – CHEMISTRY 31.
Stability of carbonium ions involved in the reactions follow the order:
CH2
CH2 H3C >
>
CH - CH2 > CH3CH2
H3C OCH3 Reactivity Stability of carbonium ions 32.
Malachite = CuCO3.Cu(OH)2 Bauxite = Al2O3.2H2O Calamine = ZnCO3 Siderite = FeCO3
33.
Work done on system = +10 kJ Heat escaped = -2kJ U = q + w = 10 – 2 = 8 kJ
34.
Kieselguhr is amorphous form of silica.
35.
O
O I
CN KCN/DMSO SN 2
H2/Pd OH CH2NH2
36.
37.
PE 2 The energy of electrons increases as they are away from the nucleus. The probability of finding the 1s electron may be higher at a0 but the energy is not. The probability density of finding the electron is not zero at any place in the atom. So, the energy may be higher when it is far from a0. T.E K.E
General coordination number of CN– in transition element is 6 and in inner transition element is 8-12. because inner transition metal ions can make available more number of vacant orbitals of nearly same energy than transition metal ions. The high effective nuclear charge of inner-transition metal ions make them form complex with high coordination number.
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JEE-MAIN-2019 (9th Apr-Second Shift)-PCM-27 38.
B2O3 is acidic in nature Al2O3 and Ga2O3 are amphoteric Oxides of In and Tl are basic in nature. Because the metallic character of the elements increases on moving down the group.
39.
CO is diamagnetic in nature due to absence of any unpaired electron.
40.
Due to –OH group of serine it give cerric ammonium nitrate test whereas due to –NH2 group lysine give +ve carbylamine test.
41.
Extraction of Fe is done form haematite ore this is true but reason is wrong as haematite is Fe2O3.
42.
MVml 103 10 105 mole 1000 1000 10–5 NA molecules covering area = 0.24 cm2 0.24 1 ---------------------------------------- = cm2 10 5 NA Moles
0.24 a2 23 10 6 10 a2 = 4 10–20 cm2 a = 2 10–10 cm a = 2 10–12 m a = 2 pm 5
43.
Ea = (D C) = 15 – 0 = 15 kJ mol–1 Ea = (A + B) C = 15 kJ mol–1 Ea = (A + B) D = 10 kJ mol–1 Ea = C (A + B) = 20 kJ mol–1 (high activation enthalpy in the reaction) Activation enthalpy to form C is 5 kJ mol–1 more than that of form D.
44.
P
RT V b
P(V – b) = RT a P 2 V b RT V At high pressure. P(V - b) = RT PV – Pb = RT PV Pb 1 RT RT Pb Z=1+ RT Z > 1, Z b The value of ‘b’ for the gases follows the order Ne < Ar < Kr < Xe
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JEE-MAIN-2019 (9th Apr-Second Shift)-PCM-28
45.
1F deposits 1 g equivalent of Ni or ½ mole of Ni 1 0.1 F will deposit 0.05 moles of Ni 20
46.
Ketones < Aldehyde Rate of Nucleophilic addition reaction Only aldehydes are responsible for formation of acetals. Excess of MeOH is used to drive the reaction towards forward direction.
47.
Solid state – chain Cl Cl
- Be
Be
Be
-
Cl Cl In vapour state dimeric Cl
-
Cl Be
Be - Cl Cl
48.
For SN1 carbocation is formed.
The first transition state should have higher energy than the second transition state. Reaction intermediate is also formed. 49.
Chloroform is used as a solvent in this reaction
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JEE-MAIN-2019 (9th Apr-Second Shift)-PCM-29 50.
Ac +3
Th
Pu +3 +4 +5
U +3 +4 +5
Np Pu Am Cm Bk Cf Es Fm Md +3 +3 +3 +3 +3 +3 +3 +3 +3 +4 +4 +4 +4 +4 +4 +5 +5 +5 +6 +6 +6 +6 +7 +7 Np and Pu has maximum no. of possible oxidation states
No Lr +2 +3
51.
Kf = 4 K Kg mol–1 i=3 Molality = 0.03 Tf = i Kfm = 3(4) (0.03) Tf = 0.36 K
52.
It is an organic chemical in the catecholamine family that functions in the brain and body as hormone and neurotransmitter.
53.
Benzene sulphonyl chloride C6H5SO2Cl
54.
Fact based.
55.
HF has strong hydrogen bond. It has highest boiling point among hydrogen halides. The strong hydrogen bond is due to more difference in electronegativity between F and H atoms.
56.
20% w/w KI Mass of solute(KI) = 20 g Mass of solvent = 100 – 20 = 80 g Molar mass of KI = 38 + 128 = 166 gm solute 20 1000 Molality 1.506 1.51 mw Kg solvent 166 80
57.
58.
(i) VBT does not explain colour exhibited by complex of transition metal because splitting of d-orbitals in explained by CFT. (ii) VBT does not distinguish between strong and weak field complex. (iii) VBT does not give quantitative interpretation of magnetic properties. According to question statement I and III are correct.
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JEE-MAIN-2019 (9th Apr-Second Shift)-PCM-30 59.
The nature of the curve is such that the pH change of the reaction becomes larger for very small change in volume of titrant. O
60.
O Br Br2 /CCl4
HO
HO O C
– R effect, deactivate the ring for EAS. OH +R effect activate the
ring towards EAS
PART C – MATHEMATICS 61.
Equation of tangent to the parabola y 2 x At , is T 0
x 2 x 2 y 2 2 1 y x 2 2 y
1 ,c m 2 2 This is also a tangent to ellipse x 2 2y 2 1
C a 2m2 b 2
1 1 2 2 4 2
2 1 1 2 4 4 2
4 22 1 0
2
2 1 2 2 1 2 2 2 1 2 2 1
62.
n n 1
99 n 2
2
2 n2 n 198 2n2 8n 8 n2 9n 190 0
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JEE-MAIN-2019 (9th Apr-Second Shift)-PCM-31 n 19 n 10 0 n 19 19 20 Number of balls is 190 2
63.
lim
f x
x2 6
2t dt dx {given that f 2 6 } x 2
0 from, so we use L – Hospital Rule 0 f ' x .2f x lim x 2 1 f ' 2 .2f 2 12f ' 2
64.
system of equations has non trival solution 2 3 1 D 0 1
k
2 1
2 0 1
9 2 So equation are 2x 3y z 0 ……….(1) 9 x y 2z 0 ……….(2) 2 2x y z 0 ……….(3) (1) – (3) 4y 2z 0 2y z ………….(4) y 1 z 2 From equation (1) and (4) 2x 3y 2y 0 2x y 0 x 1 z or 4 y 2 x x y z 1 k y z x 2 k
65.
Circle x 2 y2 4 c1 0,0 ; r1 2
and circle x 2 y 2 6x 8y 24 0 c 2 3, 4 ; r2 7 d c 1c 2 5
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JEE-MAIN-2019 (9th Apr-Second Shift)-PCM-32 also d r1 r2 circles touch internally Equation of common tangent S1 S2 0 6x 8y 20 0 3x 4y 10 0 Point (6, –2) satisfy it. 66.
Let the three numbers in A.P. are a – d, a, a + d Given that a d a a d 33 a 11 and a d a a d 1155
a a2 d2 1155
11 121 d2 1155 2
d 16 d 4 If d = 4 then first term a – d = 7 If d = –4 then first term a – d = 15 T11 7 10 4 47 , T11 15 10 4 25 67.
1
I x cot 1 1 x 2 x 4 dx 0
1 1 I x tan1 dx 2 4 0 1 x x x 2 x 2 1 1 1 I x tan dx 2 2 0 1 x x 1
1
I x tan1 x 2 tan 1 x 2 1 dx 0
It x 2 t 2x dx dt 1 1 I tan1 t tan1 t 1 dt 2 0 1 1 1 1 tan1 t dt tan1 t 1 dt 2 0 2 0 1 1 1 1 tan1 t dt tan1 t dt 0 2 2 0 1 1 1 1 tan1 t dt tan1 t dt 2 0 2 1
1
tan1 t dt 0
t .tan1 t
1 t t 1
1
0
0
2
dt
1 1 log 1 t2 0 4 2 1 loge 2 4 2
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JEE-MAIN-2019 (9th Apr-Second Shift)-PCM-33
68.
sin10o sin30o sin50o sin70o sin10o sin30o sin50o sin70o
sin30o sin10o sin 60o 10o sin 60o 10o
1 sin30o sin3 10o 4 1 1 1 2 4 2 1 16
69.
W
5 3z 5 1 z
5w 5wz 5 3z 5w 5 z 3 5w given z 1
R w
5w 5 1 3 5w 5w 5 3 5w
w 1
3 5
1 5
1 5
1
3 w 5
n
Cr 1 2 r 2 Cr 15 n r 1 15 15r 2n 2r 2 ………..(1) 17r 2n 2 n C 15 r 1 3 also given n r Cr 1 70 n r 14 3n 3r 14r 14 ………..(2) 17r 3n 14 Solving (1) and (2) n = 16, r = 2 16 C1 16 C2 16 C3 Average of coefficient 3 16 120 560 3 = 232
70.
Given:
71.
cos x
n
dy y sin x 6x dx
dy y tan x 6x sec x dx
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JEE-MAIN-2019 (9th Apr-Second Shift)-PCM-34
tan x dx e e loge sec x
1 sec x
: solution of equation 1 1 y. 6x sec x. dx sec x sec x
y 3x 2 c sec x
…….(1)
given y 0 3 2 So, 0 3 C 9 2 C 3 Now from (1) y 2 3x 2 sec x 3 At x 6 3y 32 2 2 36 3 2 y 2 3 72.
Two lines are perpendicular m1m2 1
1 2 2 1 a 1 a a3 a 2 2 0 a 1 a 2 2a 2 0
a 1 So lines are
L1 : x 2y 1 0 L 2 :2x y 1 0
Solving these equation we get point of intersection 1 3 P , 5 6 Now distance of P from origin 1 9 2 OP 25 25 5
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JEE-MAIN-2019 (9th Apr-Second Shift)-PCM-35
73.
1 Given tan1 2 1 r tan 2 h h r 2 1 2 V r h 3 1 h3 V 3 4 dv dh 3h2 dt 12 dt dh dh 1 5 100 4 dt dt 5
74.
r
In ABC tan15o
5 x x 5 2 3
h = 10 m
B
5 x
2 3
5 cm
A
15o
10 cm C
5 x
E
D
A (1, –1, 2)
75.
Let any point on given line is D 3 2, 1, 4 Now AD BC D.R. of BC a1 3, b1 0, c1 y D.R. of AD a2 3 3, b2 2, c 2 4 2 a1a 2 b1b 2 c 1c 2 0
B
D
C
x 2 y 1 z 3 0 4
5
3 3 3 0 4 4 2 0 25 17 17 25 Co – ordinate of point D
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JEE-MAIN-2019 (9th Apr-Second Shift)-PCM-36
68 1 25 ,1, 25 576 324 2 AD 4 34 625 25 5 1 Area of ABC BC AD 2 1 2 5 34 2 6 34 76.
A T A 3I3
0 2x 2x 0 2y 2y y y 2x y 1 1 1 2x y 8x 2 0 0 3 0 0 6y 2 0 0 3 0 0 3 0 0 3 8x 2 3 x 8
1 3 0 0 1 0 3 0 1 0 0 3 0 0 3
1 2 4 matrices are possible 6y 2 3 y
77.
Equation of tangent parabola y 2 4x at
to
the
x 1 (1, 2) is 2y 4 2 y x 1 Equation of normal y x 3 Let centre be C (3 – r, r) Now PC2 r 2 2
2
3 r 1 r 2 r
y=x+1
y P (1, 2)
C r
2
(3 – r, r) x
2
2 2 r r2 r 2 8r 8 0 r 42 2 For r 4 2 2 possible So r 4 2 2
(3 – r, 0)
3 r 0
Area r 2 16 8 16 2
8 3 2 2
Not
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JEE-MAIN-2019 (9th Apr-Second Shift)-PCM-37
78.
a x 1, x 5 f x b x 3, x 5 Contributes at x = 5 L.H.L. = R.H.L = f (5) b 5 3 a 5 1
b 5 3 a 5 1 a b 5 2
ab 79.
2 2 5 5
x f x x 4 x lim f x lim x 4 1 3 x 4 x4 4 x lim f x lim x 3 0 3 x4 x 4 4 f 4 3 Continuous at x = 4
80.
e
sec x
sec x tan x f x sec x tan x sec x dx e 2
sec x
f x C
Diff. both side w.r.t.x
esec x sec x tan x f x sec x tan x sec 2 x
esec x .sec x tan xf x esec x f n f ' x sec 2 x tan x sec x f x tan x sec x C
81.
m
2
2
1 x 2 3x m 1 0
3 m 1 2 m 1
2
m2 1 is maximum
m2 1 is minimum m0 3 and 1
3 3 2 2
2
2
4
9 4 9 1
8 5
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JEE-MAIN-2019 (9th Apr-Second Shift)-PCM-38 82.
Let population = 100 n A 25 n B 20 n A B 8
B
A
n A B 12 n A B 17
17
8
12
Now % of the population who look advertisement 30 40 50 17 12 8 100 100 100 13.9 83.
Equation of plane P1 P2 0
x y z 6 2x 3y z 5 0 x 1 2 y 1 3 2 1 6 5 0 This plane is to xy – plane 1 0 So, equation of plane x 2y 11 0 x 2y 11 0 distance of the point (0, 0, 256) from this plane. 0 0 11 11 1 4 5 84.
p q r is false
T F F So, p T, q F and r F 85.
1 log10 4 x2 1 Let f1 and 4 x2 4 x2 0 x 2 f x
x
3
x
f2 log10 x 3 x
x3 x 0 x x 1 x 1 0
x 1, 0 1, 2 x 1, 0 1,2 2,
86.
S 1 2 3 3 5 4 7 ....... upto 11 terms nth term of the series is Tn n 2n 1 11
11
S Tn 2n2 n n 1
n 1
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JEE-MAIN-2019 (9th Apr-Second Shift)-PCM-39
Sn
2n n 1 2n 1
n n 1
6 2 Put n 1 2 1112 23 1112 S11 6 2 S11 946
87.
mean = 42 10 22 26 29 34 x 42 67 70 y 45 10 x y 120 …………(i) and median = 35 34 x 35 x 36 2 from (i) y = 84 y 84 7 x 36 3
88.
y 2 2x …..(i) and x y 4 0 ………(ii) solving (1) and (2) 2 x y 2x x 2 10x 16 0 x 8,2 and y 4, 2 4
A
y2 y 4 dy 2 2 4
y2 y3 A 4y 6 2 4 64 8 A 4 16 1 8 18 6 6 89.
cos2 cos2 cos2 1 1 1 cos2 1 4 2 3 1 cos2 1 4 4 1 cos 2 2 or 3 3
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JEE-MAIN-2019 (9th Apr-Second Shift)-PCM-40 90.
Let vertex C is 6,k then centre of circle
5k 1, 2 It lies on diameter 3y x 7 5k 3 1 7 2 k 1 So, AB = 14 and BC = 6 Area = 14 6 84
D
C (6, k)
A (–8, 5)
B (6, 5)
3y = x + 7
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FIITJEE Solutions to JEE(Main)-2019 Test Date: 10th April 2019 (First Shift)
PHYSICS, CHEMISTRY & MATHEMATICS Paper - 1 Time Allotted: 3 Hours
Maximum Marks: 360
Please read the instructions carefully. You are allotted 5 minutes specifically for this purpose.
Important Instructions: 1.
The test is of 3 hours duration.
2.
This Test Paper consists of 90 questions. The maximum marks are 360.
3.
There are three parts in the question paper A, B, C consisting of Physics, Chemistry and Mathematics having 30 questions in each part of equal weightage. Each question is allotted 4 (four) marks for correct response.
4.
Out of the four options given for each question, only one option is the correct answer.
5.
For each incorrect response 1 mark i.e. ¼ (one-fourth) marks of the total marks allotted to the question will be deducted from the total score. No deduction from the total score, however, will be made if no response is indicated for an item in the Answer Box.
6.
Candidates will be awarded marks as stated above in instruction No.3 for correct response of each question. One mark will be deducted for indicating incorrect response of each question. No deduction from the total score will be made if no response is indicated for an item in the answer box.
7.
There is only one correct response for each question. Marked up more than one response in any question will be treated as wrong response and marked up for wrong response will be deducted accordingly as per instruction 6 above.
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JEE-MAIN-2019 (10th Apr-First Shift)-PCM-2
PART –A (PHYSICS) 1.
A particle of mass m is moving along a trajectory given by x = x0 + a cos 1t y = y0 + b sin 2t The torque, acing on the particle about the origin, at t = 0 is: (A) my 0a12kˆ (B) m( x0b y0 a)12kˆ (C) m( x b2 y a2 )kˆ (D) Zero 0
2
0
1
2.
A stationary source emits sound waves of frequency 500 Hz. Two observers moving along a line passing through the source detect sound to be of frequencies 480 Hz and 530 Hz. Their respective speeds are, in ms–1 (Given speed of sound = 300 m/s) (A) 16, 14 (B) 12, 16 (C) 8, 18 (D) 12, 18
3.
Two particles, of masses M and 2M, moving as shown, with speeds of 10 m/s and 5 m/s, collide elastically at the origin. After the collision, they move along the indicated directions with speeds v1 and v2 respectively. The value of v1 and v2 are nearly: (A) 3.2 m/s and 12.6 m/s (B) 6.5 m/s and 6.3 m/s (C) 6.5 m/s and 3.2 m/s (D) 3.2 m/s and 6.3 m/s
4.
Given below in the left column are different modes of communication using the kinds of waves given in the right column. A. Optical Fibre Communication P. Ultrasound B. Radar Q. Infrared Light C. Sonar R. Microwaves D. Mobile Phones S. Radio Waves From the options given below, find the most appropriate match between entries in the left and the right column. (A) A – S, B – Q, C – R, D – P (B) A – Q, B – S, C – P, D – R (C) A – R, B – P, C – S, D – Q (D) A – Q, B – S, C – R, D – P
5.
In an experiment, the resistance of a material is plotted as a function of temperature (in some range). As shown in the figure, it is a straight line. One may conclude that: 2 2 (A) R(T) R0 e T / T0 R0 T2 2 2 (C) R(T) R0 e T / T0
(B) R(T)
2
(D) R(T) R0 e T0 / T
2
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JEE-MAIN-2019 (10th Apr-First Shift)-PCM-3 6.
A cylinder with fixed capacity of 67.2 lit contains helium gas at STP. The amount of heat needed to raise the temperature of the gas by 20°C is: [Given that R = 8.31 J mol–1 K–1] (A) 350 J (B) 700 J (C) 748 J (D) 374 J
7.
A ball is thrown upward with an initial velocity V0 from the surface of the earth. The motion of the ball is affected by a drag force equal to m2 (where m is mass of the ball, is its Instantneous velocity and is a constant). Time taken by the ball to rise to its zenith is: 1 1 (A) ln 1 V0 (B) ta n1 V0 g g g g (C)
sin1 V0 g g
1
(D)
2 tan1 V0 2 g g 1
8.
A thin disc of mass M and radius R has mass per unit area (r) = kr2 where r is the distance from its centre. Its moment of inertia about an axis going through its centre of mass and perpendicular to its plane is: MR2 MR2 (A) (B) 2 3 MR2 2MR2 (C) (D) 6 3
9.
Two coaxial discs, having moments of inertia I1 and
I1 , area rotating with respectively 2
1 , about their common axes. They are brought in contact 2 with each other and thereafter they rotate with a common angular velocity. If Ef and Ei are the final and initial total energies, then (Ef – Ei) is: I 2 3 (A) 1 1 (B) I112 6 8 2 I I 2 (C) 1 1 (D) 1 1 12 24
angular velocities 1 and
10.
The ratio of surface tensions of mercury and water is given to be 7.5 while the ratio of their densities is 13.6. Their contact angles, with glass, are close to 135° and 0°, respectively. It is observed that mercury gets depressed by an amount h in a capillary tube of radius r1, while water rises by the same amount h in a capillary tube of radius r2. The ratio, (r1/r2), is then close to: (A) 3/5 (B) 4/5 (C) 2/3 (D) 2/5
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JEE-MAIN-2019 (10th Apr-First Shift)-PCM-4 11.
Figure shows charge (q) versus voltage (V) graph for series and parallel combination of to given capacitors. The capacitances are: (A) 40 F and 10 F (B) 50 F and 30 F (C) 60 F and 40 F (D) 20 F and 30 F
12.
Two wires A and B are carrying currents I1 and I2 as shown in the figure. The separation between them is d. A third wire C carrying a current I is to be kept parallel to them at a distance x from A such that the net force acting on it is zero. The possible values of x are: I I2 (A) x 1 d and x d (I1 I2 ) I1 I2 I I (C) x 2 d and x 2 d I1 I2 I1 I2
(B) x
I2d (I1 I2 )
I (D) x 1 I1 I2
I2 d and x d I1 I2
13.
A proton, an electron, and a Helium nucleus, have the same energy. They are in circular orbitals in a plane due to magnetic field perpendicular to the plane. Let rp, re and rHe be their respective radii, then, (A) re > rp = rHe (B) re > rp > rHe (C) re < rp < rHe (D) re < rp = rHe
14.
The value of acceleration due to gravity at Earth’s surface is 9.8 ms–2. The altitude above its surface at which the acceleration due to gravity decreases to 4.9 ms–2, is close to: (Radius of earth = 6.4 × 106 m) (A) 6.4 × 106 m (B) 9.0 × 106 m 6 (C) 2.6 × 10 m (D) 1.6 × 106 m
15.
In the given circuit, an ideal voltmeter connected across the 10 resistance reads 2 V. The internal resistance r, of each cell is: (A) 1 (B) 0.5 (C) 1.5 (D) 0
16.
The electric field of a plane electromagnetic wave is given by E E0 ˆi cos(kz) cos( t) The corresponding magnetic field B is then given by: E E (A) B 0 ˆj sin(kz) cos( t) (B) B 0 kˆ sin(kz) cos( t) C C E E (C) B 0 ˆj cos(kz) sin( t) (D) B 0 ˆj sin(kz) sin( t) C C
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JEE-MAIN-2019 (10th Apr-First Shift)-PCM-5 17.
An npn transistor operates as a common emitter amplifier, with a power gain of 60 dB. The input circuit resistance Is 100 and the output load resistance is 10 k. the common emitter current gain is: (A) 6 × 102 (B) 102 (C) 60 (D) 104
18.
In a meter bridge experiment, the circuit diagram and the corresponding observation table are shown in figure.
Sl. No. 1. 2. 3. 4.
R () 1000 100 10 1
l (cm) 60 13 1.5 1.0
Which of the readings is inconsistent? (A) 4 (C) 2
(B) 3 (D) 1
19.
A transformer consisting of 300 turns in the primary and 150 turns in the secondary gives output power of 2.2 kW. If the current in the secondary coils is 10 A, then the input voltage and current in the primary coil are: (A) 440 V and 5 A (B) 440 and 20 A (C) 220 V and 20 A (D) 220 V and 10 A
20.
A uniformly charged ring of radius 3a and total charge q is placed in xy-plane centered at origin. A point charge q is moving towards the ring along the z-axis and has speed v at z = 4a. The minimum value of v such that it crosses the origin is:
21.
1/2
(A)
2 m
1 q2 5 4 0 a
(C)
2 m
4 q2 15 4 0 a
1/2
(B)
2 m
1 q2 15 4 0 a
(D)
2 m
2 q2 15 4 0 a
1/2
1/2
A current of 5 A passes through a copper conductor (resistivity = 1.7 × 10–8 m) of radius of cross-section 5 mm. Find the mobility of the charges if their drift velocity is 1.1 × 10–3 m/s. (A) 1.8 m2/Vs (B) 1.0 m2/Vs 2 (C) 1.3 m /Vs (D) 1.5 m2/Vs
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JEE-MAIN-2019 (10th Apr-First Shift)-PCM-6 22.
n-moles of an ideal gas with constant volume heat capacity CV undergo an isobaric expansion by certain volume. The ratio of the work done in the process, to the heat supplied is nR nR (A) (B) CV nR CV nR 4nR 4nR (C) (D) CV nR CV nR
23.
A message signal of frequency 100 MHz and peak voltage 100 V is used to execute amplitude modulation on a carrier wave of frequency 300 GHz and peak voltage 400 V. The modulation index and difference between the two side band frequencies are: (A) 4 ; 2 × 108 Hz (B) 4 ; 1 × 108 Hz 8 (C) 0.25 ; 1 × 10 Hz (D) 0.25 ; 2 × 108 Hz
24.
A ray of light AO in vacuum is incident on a glass slab at angle 60° and refracted at angle 30° along OB as shown in the figure. The optical path length of light ray from A to B is : 2b 2b (A) 2a (B) 2a 3 3 (C)
25.
2 3 2b a
(D) 2a + 2b
The displacement of a damped harmonic oscillator is given by x(t) = e–0.1 t cos(10t + ). Here t is in seconds. The time taken for its amplitude of vibration to drop to half of its initial value is close to: (A) 13 s (B) 27 S (C) 4 s (D) 7 s
26.
Two radioactive materials A and B have decay constants 10 and , respectively. If initially they have the same number of nuclei, then the ratio of the number of nuclei of a to that of B will be 1/e after a time: 1 1 (A) (B) 11 10 1 11 (C) (D) 9 10
27.
A 25 × 10–3 m3 volume cylinder is filled with 1 mol of O2 gas at room temperature (300 K). The molecular diameter of O2, and its root mean square speed, are found to be 0.3 nm and 200 m/s, respectively. What is the average collision rate (per second) for an O2 molecule? (A) ~1010 (B) ~1011 12 (C) ~10 (D) ~1013
28.
In a photoelectric effect experiment the threshold wavelength of incident light is 260 nm, 1237 the maximum kinetic energy of emitted electrons will be: Given E (in eV) = (in nm) (A) 15.1 eV
(B) 1.5 eV
(C) 4.5 eV
(D) 3.0 eV
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JEE-MAIN-2019 (10th Apr-First Shift)-PCM-7 29.
One plano-convex and one plano-concave lens of same radius of curvature ‘R’ but of different materials are joined side by side as shown in the figure. If the refractive index of the material of 1 is 1 and that of 2 is 2, then the focal length of the combination is R 2R (A) (B) 2(1 2 ) 1 2 R R (C) (D) 1 2 2 (1 2 )
30.
A moving coil galvanometer allows a full scale current of 10–4 A. A series resistance of 2 M is required to convert the above galvanometer into a voltmeter of range 0 – 5 V. Therefore the value of shunt resistance required to convert the above galvanometer into a ammeter of range 0–10 mA is: (A) 200 (B) 100 (C) 10 (D) 500
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JEE-MAIN-2019 (10th Apr-First Shift)-PCM-8
PART –B (CHEMISTRY) 31.
The species that can have a trans-isomer is: (en = ethane-1,2-diamine, ox = oxalate) (A) [Pt(en)Cl2] (B) [Pt(en)2Cl2]2+ (C) [Cr(en)2(ox)]+ (D) [Zn(en)Cl2]
32.
The major product of the following reaction is
(A)
(B)
(C) CH3CH = CH – CH2NH2
(D)
33.
The synonym for water gas when used in the production of methanol is (A) syn gas (B) natural gas (C) fuel gas (D) laughing gas
34.
The regions of the atmosphere, where clouds form and where we live, respectively, are: (A) Troposphere and Stratosphere (B) Troposphere and Troposphere (C) Stratosphere and Stratosphere (D) Stratosphere and Troposphere
35.
Consider the hydrated ions of Ti2+, V2+, Ti3+ and Sc3+. The correct order of their spin-only magnetic moments is: (A) Ti3+ < T2+ < Sc3+ < V2+ (B) Sc3+ < Ti3+ < Ti2 < V2+ 2+ 2+ 3+ 3+ (C) V < Ti < Ti < Sc (D) Sc3+ < Ti3+ < V2+ < Ti2+
36.
Amylopectin is composed of (A) -D-glucose, C1–C4 and C2–C6 linkages (B) -D-glucose, C1–C4 and C2–C6 linkages (C) -D-glucose, C1–C4 and C1–C6 linkages (D) -D-glucose, C1–C4 and C1–C6 linkages
37.
A gas undergoes physical adsorption on a surface and follows the given Freundlich adsorption isotherm equation x kp0.5 m Adsorption of the gas increases with: (A) Increase in p and decrease in T (B) Increase in p and increase in T (C) Decrease in p and increase in T (D) Decrease in p and decrease in T
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JEE-MAIN-2019 (10th Apr-First Shift)-PCM-9 38.
39.
A bacterial infection in an internal would grows as N(t) = N0 exp(t), where the time t is in hours. A dose of antibiotic, taken orally, needs 1 hour to reach the wound. Once it dN reaches there, the bacterial population goes down as 5N2 . What will be the plot of dt N0 vs, t after 1 hour? N
(A)
(B)
(C)
(D)
2
The graph between and r(radial distance) is shown below. This represents:
(A) 1s orbital (C) 2s orbital
(B) 3s orbital (D) 2p orbital
40.
Consider the following table: Gas a/ (k Pa dm6 mol–1) b/ (dm3 mol–1) A 642.32 0.05196 B 155.21 0.04136 C 431.91 0.05196 D 155.21 0.4382 a and b a van der Waals constants. The correct statement about the gases is: (A) Gas C will occupy more volume than gas A ; gas B will be lesser compressible than gas D (B) Gas C will occupy lesser volume than gas A ; gas B will be more compressible than gas D. (C) Gas C will occupy lesser volume than gas A ; gas B will be lesser compressible than gas D (D) Gas C will occupy more volume than gas A; gas B will be more compressible than gas D.
41.
During the change of O2 to O 2 , the incoming electron goes to the orbital: (A) 2p y (B) 2p x (C) * 2p z
(D) * 2p x
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JEE-MAIN-2019 (10th Apr-First Shift)-PCM-10 42.
The principle of column chromatography is (A) Differential absorption of the substances on the solid phase. (B) Differential adsorption of the substances on the solid phase. (C) Gravitational force. (D) Capillary action.
43.
At 300 K and 1 atmospheric pressure, 10 mL of a hydrocarbon required 55 mL of O2 for complete combustion, and 40 mL of CO2 is formed. The formula of the hydrocarbon is: (A) C4H10 (B) C4H6 (C) C4H7Cl (D) C4H8
44.
Three complexes, [CoCl(NH3)5]2+ (I), [co(NH3)5H2O]3+ (II) and [Co(NH3)6]3+ (III) Absorb light in the visible region. The correct order of the wavelength of light absorbed by them is (A) (II) > (I) > (III) (B) (III) > (II) > (I) (C) (I) > (II) > (III) (D) (III) > (I) > (II)
45.
Major products of the following reaction are:
(A)
CH3OH and HCO2H
(C)
46.
(B)
(D)
The major product of the following reaction is:
(A)
(B)
(C)
(D)
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JEE-MAIN-2019 (10th Apr-First Shift)-PCM-11 47.
Match the refining methods (Column I) with metals (Column II). Column I Column II (Refining methods) (Metals) (I) Liquation (a) Zr (II) zone Refining (b) Ni (III) Mond process (C) Sn (IV) Van Arkel Method (d) Ga (A) (I) – (b); (II) – (c); (III) – (d); (IV) – (a) (B) (I) – (b); (II) – (d); (III) – (a); (IV) – (c) (C) (I) – (c); (II) – (a); (III) – (b); (IV) – (d) (D) (I) – (c); (II) – (d); (III) – (b); (IV) – (a)
48.
The increasing order of the reactivity of the following compounds towards electrophilic aromatic substitution reaction is:
(A) (III) < (I) < (II) (C) (II) < (I) < (III)
(B) (III) < (II) < (I) (D) (I) < (III) < (II)
49.
A process will be spontaneous at all temperatures if: (A) H < 0 and S < 0 (B) H < 0 and S > 0 (C) H > 0 and S > 0 (D) H > 0 and S < 0
50.
Consider the following statements (a) The pH of a mixture containing 400 mL of 0.1 M H2SO4 and 400 mL of 0.1 M NaOH will be approximately 1.3. (b) Ionic product of water is temperature dependent. (c) A monobasic acid with Ka = 10–5 has a pH = 5. the degree of dissociation of this acid is 50%. (d) The Le Chatelier’s principle is not applicable to common-ion effect. The correct statements are: (A) (a) and (b) (B) (a), (b) and (c) (C) (b) and (c) (D) (a), (b) and (d)
51.
Ethylamine (C2H5NH2) can be obtained from N-ethylphthalimide on treatment with: (A) CaH2 (B) H2O (C) NaBH4 (D) NH2NH2
52.
The correct order of catenation is: (A) C > Sn > Si Ge (C) Si > Sn > C > Ge
(B) Ge > Sn > Si > C (D) C > Si > Ge Sn
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JEE-MAIN-2019 (10th Apr-First Shift)-PCM-12 53.
The major product of the following reaction is
(A)
(B)
(C)
(D)
54.
Which of the following is a condensation polymer? (A) Nylon 6, 6 (B) Neoprene (C) Buna–S (D) Teflon
55.
Consider the statements S1 and S2: S1: Conductivity always increases with decrease in the concentration of electrolyte. S2: Molar conductivity always increases with decrease in the concentration of electrolyte. The correct option among the following is: (A) S1 is wrong and S2 is correct (B) Both S1 and S2 are wrong (C) S1 is correct and S2 is wrong (D) Both S1 and S2 are correct
56.
Increasing rate of SN1 reaction in the following compounds is
(A) (A) < (B) < (C) < (D) (C) (B) < (A) < (D) < (C)
(B) (A) < (B) < (D) < (C) (D) (B) < (A) < (C) < (D)
57.
The oxoacid of sulphur that does not contain bond between sulphur atoms is: (A) H2S2O3 (B) H2S2O4 (C) H2S2O7 (D) H2S4O6
58.
The isoelectronic set of ions is (A) Li+, Na+, O2– and F– (C) N3–, O2–, F– and Na+
(B) F–, Li+, Na+ and Mg2+ (D) N3–, Li+, Mg2+ and O2–
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JEE-MAIN-2019 (10th Apr-First Shift)-PCM-13
59.
The alloy used in the construction of aircrafts is (A) Mg–Mn (B) Mg-Zn (C) Mg-Al (D) Mg-Sn
60.
At room temperature, a dilute solution of urea is prepared by dissolving 0.60 g of urea in 360 g of water. If the vapour pressure of pure water at this temperature is 35 mm Hg, lowering of vapour pressure will be: (molar mass of urea = 60 g mol–1) (A) 0.027 mmHg (B) 0.031 mmHg (C) 0.028 mmHg (D) 0.017 mmHg
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JEE-MAIN-2019 (10th Apr-First Shift)-PCM-14
PART–C (MATHEMATICS) 61.
If Q(0, –1, –3) is the image of the point P in the plane 3x – y, 1.4z = 2 and R is the point (3, –1, –2), then the area (in square units) of PQR is: 91 (A) (B) 2 13 2 65 91 (C) (D) 2 4
62.
If the coefficients of x2 and x3 are both zero, in the expansion of the expression (1 + ax + bx2) (1 – 3x)t5 in powers of x, then the ordered pair (a, b) is equal to (A) (–54, 315) (B) (28, 861) (C) (28, 315) (D) (–21, 714)
63.
The sum 3 1 5 (13 23 ) 7 (13 23 33 ) ........ 12 12 22 12 22 3 2 Upto 10th term, is (A) 620 (B) 660 (C) 680 (D) 600
64.
If the length of the perpendicular from the point (, 0, ) ( = 0) to the line, x y 1 z 1 3 is , then is equal to: 1 0 1 2 (A) 1 (B) –1 (C) 2 (D) –2
65.
66.
sin (p 1)x sin x , x0 x If f(x) = q , x0 2 xx x , x0 x/2 Is continuous at x = 0, then the ordered pair (p, d) is equal to 3 1 5 1 (A) , (B) , 2 2 2 2 1 3 3 1 (C) , (D) , 2 2 2 2
Which one of the following Boolean expressions is a tautology? (A) (p q) (p~q) (B) (p q) (p~q) (C) (p q) (~p~q) (D) (p q) (p~q)
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JEE-MAIN-2019 (10th Apr-First Shift)-PCM-15
67.
If and are the roots of the quadratic equation, x2 + x sin – 2sin = 0, 0, then 2 12 12 is equal to: 12 ( 12 ) ( )24 212 (sin 8)12 212 (C) (sin 8)6
212 (sin 4)12 26 (D) (sin 8)12
(A)
68.
(B)
ABC is a triangular park with AB = AC = 100 metres. A vertical tower is situated at the mid-point of BC. If the angles of elevation of the top of the tower at A and B are cot– 1 1 (3 2 ) and cos ec (2 2 ) respectively, then the height of the tower (in metres) is: (A) 25 100 (C) 3 3
69.
(B) 10 5 (D) 20
If a > 0 and z
(1 i)2 , has magnitude ai
3 1 i 5 5 1 3 (C) i 5 5
(A)
x
70.
sin cos
If 1 sin
x
1
cos
1
x
x
and
sin 2 cos 2
2 sin 2
x
1
cos 2
1
x
, x 0 ; then
for all 0, : 2 (A) 1 – 2 = –2x3 (C) 1 – 2 = x(cos2 – cos4)
71.
2 , then z is equal to: 5 1 3 (B) i 5 5 1 3 (D) i 5 5
(B) 1 + 2 = –2 (x3 + x –1) (D) 1 + 2 = –2x3
If y = 1f(x) is the solution of the differential equation
dy (tan x y) sec 2 x, dx
x , , such that y(0) = 0, then y is equal to 2 2 4 1 1 (A) e (B) 2 2 e 1 (C) e – 2 (D) 2 e
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JEE-MAIN-2019 (10th Apr-First Shift)-PCM-16 72.
Let f: R R be differentiable at c R and f(c) = 0. If g(x) = f(x) , then at x =c, g is: (A) differentiable if f(c) = 0 (C) not differentiable
(B) differentiable if f(c) 0 (D) not differentiable if f(c) = 0
73.
Let f(x) = x2, x R. For any A R, define g(A) = {x R : f(x) A}. If S = [0, 4], then which one of the following statements is not true? (A) f(g(S) f(S) (B) f(g(S)) = S (C) g(f(S)) S (D) g(f(S)) = g(S)
74.
If the system of linear equations x+y+z=5 x = 2y + 2z = 6 x + 3y + z = u ( R), has infinitely many solutions then the value of + is: (A) 12 (B) 7 (C) 10 (D) 9
75.
The number of 6 digit numbers hat can be formed using the digits 0, 1, 2, 5, 7 and 9 which are divisible by 11 and no digits is repeated, is (A) 36 (B) 60 (C) 72 (D) 48
76.
The line x = y touches a circle at the point (1, 1). If the circle also passes through the point (1, –3), then its radius is (A) 3 2 (B) 3 (C) 2 (D) 2 2
77.
Let f(x) = ex – x and g(x) = x2 – x, x R. Then the set h(x) = (fog) (x) is increasing is: 1 1 (A) 0, 1, (B) 1, 2 2 1 (C) , 0 1, (D) 0, 2
of all x R, where the function 1 2 ,
2
78.
The value of
[sin 2x(1 cos 3x)] dx where [t] denotes the greatest integer function, is: 0
(A) (C) 2
(B) –2 (D) –
79.
If the circles x2 + y2 + 5Kx + 2y + K = 0 and 2(x2 + y2) + 2Kx + 3y – 1 = 0, (KR), intersect at the points P and Q, then the line 4x + 5y – K = 0 passes through P and Q for: (A) exactly one value of K (B) not value of K (C) infinitely many values of K (D) exactly two values of K
80.
If the lines x – 2y = 12 is tangent to the ellipse length of the latus rectum of the ellipse is (A) 12 2 (C) 8 3
x2 y2 9 2 1 at the point 3, , then the 2 a b 2
(B) 9 (D) 5
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JEE-MAIN-2019 (10th Apr-First Shift)-PCM-17 81.
If a1, a2, a3 ………… an are in A.P and a1 + a4 + a7 + …………… + a16 = 114, then a1 + a6 + a11 + a16 is equal to (A) 76 (B) 64 (C) 98 (D) 38
82.
If for some x R, the frequency distribution of the marks obtained by 20 students in a test is Marks 2 3 5 7 2 2 Frequency (x+1) 2x – 5 x – 3x x Then the mean of the marks is: (A) 2.8 (B) 3.2 (C) 2.5 (D) 3.0
83.
If lim x 1
x4 1 x3 k 3 lim 2 , then k is x 1 x k x k 2
3 8 4 (C) 3
8 3 3 (D) 2
(A)
84.
85.
(B)
All the pairs (x, y) that satisfy the inequality 2 1 2 sin x 2 sin x 5 sin2 y 1 also 4 Satisfy the equation: (A) 2 sin x 3 sin y (B) sin x sin y (C) 2sin x = sin y (D) sin x = 2 sin y (n 1)1/3 (n 2)1/3 (2n)1/3 lim .... 4/3 n n4/3 n 4/3 n is equal to: 3 3 (A) (2)4/3 4 4 3 4 (C) (2)4/3 4 3
4 3/4 (2) 3 4 (D) (2)4/3 3
(B)
86.
If a directrix of a hyperbola centered at the origin and passing through the point (4, 2 3 ) is 5x 4 5 and its eccentricity is e, then (A) 4e4 + 8e2 – 35 = 0 (B) 4e4 – 24e2 + 35 = 0 4 2 (C) 4e – 12e – 27 = 0 (D) 4e4 – 24e2 + 27 = 0
87.
The region represented by x y 2 and x y 2 is bounded by a: (A) rhombus of area 8 2 sq. units (C) rhombus of side length 2 units
(B) square of area 16 sq. units (D) square of side length 2 2 units
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JEE-MAIN-2019 (10th Apr-First Shift)-PCM-18 88.
Assume that each born child is equally likely to be a boy or a girl. If two families have two children each, then the conditional probability that all children are girls given that at least two are girls is: 1 1 (A) (B) 10 17 1 1 (C) (D) 12 11
89.
Let A(3, 0, –1), B(2, 10, 6) and C(1, 2, 1) be the vertices of a triangle and M be the midpoint of AC. If G divides BM in the ratio, 2 : 1, then cos(GOA) (O being he origin) is equal to: 1 1 (A) (B) 30 2 15 1 1 (C) (D) 6 10 15
90.
If
(x
2
dx 2x 10)2
f(x) x 1 = A tan1 C 2 3 x 2x 10 Where C is a constant of integration, then: 1 1 (A) A and f(x) (x 1) (B) A and f(x) 9(x 1)2 27 54 1 1 (C) A and f(x) 3(x 1) (D) A and f(x) 3(x 1) 54 81
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JEE-MAIN-2019 (10th Apr-First Shift)-PCM-19
JEE (Mains) – 2019 ANSWERS PART A – PHYSICS 1. 5. 9. 13. 17. 21. 25. 29.
A C D D B B D C
2. 6. 10. 14. 18. 22. 26. 30.
D C D C A B C Bonus
3. 7. 11. 15. 19. 23. 27.
B B A B A D A
4. 8. 12. 16. 20. 24. 28.
B D B D D D B
34. 38. 42. 46. 50. 54. 58.
B D A C B A C
64. 68. 72. 76. 80. 84. 88.
B D A D B B D
PART B – CHEMISTRY 31. 35. 39. 43. 47. 51. 55. 59.
B B C B D D A C
32. 36. 40. 44. 48. 52. 56. 60.
D C D C A D D D
33. 37. 41. 45. 49. 53. 57.
A B C D B A C
PART C – MATHEMATICS 61. 65. 69. 73. 77. 81. 85. 89.
A D B C A A A D
62. 66. 70. 74. 78. 82. 86. 90.
C D D C D A B C
63. 67. 71. 75. 79. 83. 87.
B A C B B B D
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JEE-MAIN-2019 (10th Apr-First Shift)-PCM-20
HINTS AND SOLUTIONS PART A – PHYSICS 1.
2.
F ma m[ a12 cos , tiˆ b22 sin 2 tjˆ ft 0 = ma12 ˆi rt 0 (x0 a)iˆ y ˆj r F my0 a12kˆ
v vo o v vo 1 v 0 530 vo 1 300 = 18 m/s 500
480 vo 1 300 = 12 m/s 500
3.
2MV, cos 30° + Mv2 cos 45° = 10 M cos 30° + 10 cos 45° v v1 3 2 5 3 5 2 …(i) 2 2MV, sin 30° – MV2 sin 45° = –10 M sin 30° + 10 M sin 45° V V1 2 5 5 2 …(ii) 2 5( 3 1) 10 2 17.5 V1 = 6.5 m/s 2.7 3 1 V2 = 6.3 m/s
4.
Factual
5.
1 T 2 ln R(T) 1 1 ln R(To ) T02 T2 ln R(T) = ln R(To) 1 o2 T
R(T) = Ro e
T2 02 T
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JEE-MAIN-2019 (10th Apr-First Shift)-PCM-21
6.
7.
3 Q = nCv T = n RT 2 67.2 3 = 8.31 (20) 22.4 2 748 J dv dt
(g v 2 )
g dv gdt g 2 v
Integrating 0 t and V0 0: V g gt tan1 0 g 1 t tan1 V g 0 g R
8.
R
IDisc (dm)r 2 IDisc ( 2rdr)r 2 0
0
R
IDisc (kr 2 2rdr)r 2
Mass of disc
0
R
R
IDisc 2k r 2dr 0
R
IDisc
r6 2k 6 0 6
IDisc IDisc IDisc
R 2k 6
M 2rdr kr 2 0
R
M 2k r 3 dr 0
r4 M 2k 4
R
0
kR 6 kR 4 R 2 2 r4 M 2k 3 4 2 3 2 M2R 2 ; IDisc MR2 3 3
R
0
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JEE-MAIN-2019 (10th Apr-First Shift)-PCM-22
9.
1 1 I 12 II 12 2 22 4 I 2 9 9 = 1 1 I112 2 8 16 I 3I 5 3I I11 1 1 1 ; I11 1 4 2 4 2 5 1 3I1 25 2 1 ; Ef 1 6 2 2 36 25 2 = I11 48 25 2 2 Ef Ei I112 I2 1 49 48 25 2 = I11 48 25 9 2 2 E f Ei I112 I11 48 16 48 Ei
=
10.
I112 24
2S1 cos r11g 2s cos 2 h 2 r22 g r 2 1 r2 5 h
11.
As q = CV Hence slope of graph will give capacitance. Slope will be more in parallel combination. Hence capacitance in parallel should be 50 F and in series combination must be 8 F. Only in option 40 f and 10 F. Cparallel = 40 + 10 = 50 F 40 10 Cseries = = 8 F 40 10
12.
Net force on wire carrying current I per unit length is 0I1I 0I2I 0 2x 2(d x) I1 I 2 x xd Id x 1 I1 I2
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JEE-MAIN-2019 (10th Apr-First Shift)-PCM-23
13.
r rHe
14.
mv 2mK qB qB rp re
GM GM 2 (R h) 2R2
R + h = 2R h ( 2 1)R 2.6 × 106 m 15.
15 10 2 2r 25 = 8 + 2r 3 i 8 2r 3 2 i R eq 6 8 2r 16 + 4r = 18 r = 0.5 Req
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JEE-MAIN-2019 (10th Apr-First Shift)-PCM-24
16.
17.
E B || v
Given that wave is propagating along positive z-axis and E along positive x-axis. Hence B along y-axis. From Maxwell equation B V E t E E B i.e. and B0 0 Z dt C
Av × = Pgain P 60 10 log10 P0
P 10 6 2
Rout Rin
104 100 = 100
= 2 2 = 104 ;
18.
as x
R(100 )
for (A)
x
1000 (100 60) 667 40
for (B)
x
100 (100 13) 669 13
for (C)
x
10 (100 1.5) 656 98.5
for (D)
x
1 (100 1) 95 1
So reading in serial no. (4) is completely different hence correct answer (A). 19.
Given NP = 300, Ns = 150, P0 = 2200W Is = 10 A P0 = V0I0 2200 = V0 × 10 V0 = 220 V Vi NP Vi = 2 × 220 = 440 V V0 Ns Also, P0 = ViIi 2200 Ii =5A 440
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JEE-MAIN-2019 (10th Apr-First Shift)-PCM-25 20.
Ui + Ki = Uf + Kf kq2 1 kq2 mv 2 3a 16a 2 9a 2 2
1 kq2 1 1 2kq2 mv 2 2 a 3 5 15a v
21.
=
4kq2 15ma
Vd E
E = J 1.1 103
5 25 10 6 1.1 103 25 10 6 = 1.01 m2 / Vs 1.7 10 8 5 1.7 10 8
22.
w = nRT H = (Cv + nR)T nR H Cv nR
23
fm = 100 MHz = 108 Hz, (Vm)0 = 100 V fc = 300 GHz, (Vc)0 = 400 V UBF – LBF = 2fm = 2 × 108 Hz
24.
From Snell’s law 1 sin 60° = sin 30° 3 Optical path = AO + (OB) a b 3 = o cos 60 cos 30o = 2a + 2b
25.
A = A0e–0.1 t =
A0 2
ln2 = 0.1 t t = 10 ln2 = 6.93 7 sec. 26.
N1 = N0e–10t ; 1 N1 e 9 t e N2 9t = 1 1 t= 9
N2 = N0e–t
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JEE-MAIN-2019 (10th Apr-First Shift)-PCM-26
27.
Vav RT
2 2NAP = 2 × 0.3 × 10–9 RT P V V 2 2NA
8 Vrms 3
Vav
200 2 2NA 8 v 3 25 10 3 = 17.68 × 108/sec = 0.1768 × 1010/sec ~ 1010
This answer does not match with JEE – Answer key 28.
K max
hc hc o
K max hc 0 0
380 260 K max (1237) 380 260 = 1.5 eV
29.
For 1st lens
1 1 1 1 1 1 1 f1 1 R R
For 2nd lens
1 1 2 1 1 0 2 f2 1 R R
1 1 1 feq f1 f2 1 1 1 ( 2 1) R feq feq R R 1 2 30.
200 + 10–4 G = 5 G = –ve So, answer is Bonus.
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JEE-MAIN-2019 (10th Apr-First Shift)-PCM-27
PART B – CHEMISTRY 31.
The trans-isomer of [Pt(en)2Cl2]2+
32.
33.
The synonoyms for water gas is syn gas.
34.
Fact based.
35.
36.
Amylopetcin is a homopolymer of a-D-glucose where C1–C4 linkage and C1 – C6 linkage are present.
37.
Increase in pressure leads to the increase in adsorption capacity and the physical adsorption is an exothermic process with the increase in temperature adsorption decrease.
38.
Initially N > N0 and ‘’N” is increasing through first-order kinetics. So N0/N in initial time decreases.
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JEE-MAIN-2019 (10th Apr-First Shift)-PCM-28
But after 1 hr the value of N decreases with a faster rate. So N0/N will increases. 39.
As we know that for s-orbital graph starts from top and no. of radial node = n - -1 For 2s orbital it will = 2 - 0 - 1 = 1.
The graph
40.
is of 2s
Gas A and C have same value of 'b' but different value of 'a' so gas having higher value of 'a' have more force of attraction so molecules will be more closer hence occupy less volume. Gas B and D have same value of 'a' but different value of 'b' so gas having lesser value of 'b' will be more compressible.
41.
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JEE-MAIN-2019 (10th Apr-First Shift)-PCM-29 42.
The principle of column chromatography is differential adsorption of substance and hence option 1 is correct.
43.
44.
As we known that Strong ligand CFSE Eabsorbed
1
absorbed We have [CoCl(NH3)5]2+, [Co(NH3)5H2O]3+ and [CO(NH3)6]3+ (I) (II) (III) III > II> I (as per Eabsorbed) absorbed I > II > III
45.
46. O
O H
OH
H OH
NC
NC O
O
NO I
H
OH
NC
I
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JEE-MAIN-2019 (10th Apr-First Shift)-PCM-30 47.
48.
49.
Liquation is used for Sn. Zone refining is used for Ga. Mond's process is used for Ni. Van arkel process is used for Zr.
Rate of aromatic electrophilic substitution is
At constant P and T and for the process to be spontaneous we should have G = -ve and we know that G = H - TS If H = -ve and S = +ve then at all the temperature the process will be spontaneous.
50.
51.
It is the final step of Gabriel pthalimide synthesis reaction. O
O
C
C - OH
C2H5 NH2
H3O N - C2H5
C O
Product
C - OH O
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JEE-MAIN-2019 (10th Apr-First Shift)-PCM-31 52.
In this order of catenation is asked. Catenation is a self-linking property here and for group 14: Self-linking is through covalent bonding C > Si > Ge Sn In C there is 2p – 2p overlapping further 3p – 3p, 4p-4p and so on and the extent of overlapping is more in 2p-2p > 3p-3p > 4p-4p 5p-5p
53.
It is SN1 reaction mechanism is favourable hence reaction complete via SN1 mechanism
54.
Nylon-6,6 is a condensation polymer of hexamethylene diamine and adipic acid. Buna-S, Teflon and Neoprene are addition polymer.
55.
We know that K Here m = molar conductivity m C K = conductivity C = concentration When concentration increase conductivity always increases. The molar conductivity always increase with the decrease in the concentration.
56.
Rate of SN1 reaction stability of C – I.M
57.
H2S2O7
58.
In this we have to choose isoelectronic set of ions: Isoelectronic species are those which have same no. of electron in total, so option 3 is correct.
59.
For aircraft construction aluminium and its alloys are used, because they are lighter.
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JEE-MAIN-2019 (10th Apr-First Shift)-PCM-32 60.
PART C – MATHEMATICS 61.
MQ
1 12 2 9 1 16
13
26
Q
13 2
PM 26 RQ 9 1 10 13 7 2 2 1 7 91 Ar PQR 26 2 2 2
M
RM 10
R
P
62.
Coefficient of x 2 15C2 9 3a
15C2 9 45a b 0 3
15
C1 b 0
(1) 15
Coefficient of x 27 C3 9a 15C2 3b 15 C1 0 (2) 273 21a b 0 (1) + (2) give a = 28, b = 315 24a 672 0
3 n 1 2 1
3
63.
Tn
1 2
2
23 ..... n3
.... n2
3 n n 1 2 Sn Tn
3 .n. n 1 2 On solving Sn
n n 1n 2 2
S10 660
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JEE-MAIN-2019 (10th Apr-First Shift)-PCM-33
64.
x y 1 z 1 1 0 1 A point on this line is A 0,1, 1 AB.BC 0 1 We get 2 1 1 C , 1, 2 2 2 BC 3
B ,0,
,1, , 1
2
2
1 2 1 2 2 1 2 3 0, 1
1
65.
C
A 0,1, 1
0
x x2 x x 0 x3/2 1 x 1 1 lim x 0 x 2 sin 1 x sin x LHL lim x 0 x p 1 1 p2 For function to be continuous LHL RHL f 0 RHL lim
3 1 p,q , 2 2 66.
From options p q ~ p ~ q p q ~ p q Not a tautology
p q p q p q ~ q p q p ~ q p q ~ q p q p ~ q p q ~ q 67.
tautology Not a tautology Not a tautology
x 2 x sin 2sin 0 sin 2 sin 12
12 12 Now, 24 1 1 24 12 12 FIITJEE Ltd., FIITJEE House, 29-A, Kalu Sarai, Sarvapriya Vihar, New Delhi -110016, Ph 46106000, 26569493, Fax 26513942 website: www.fiitjee.com.
JEE-MAIN-2019 (10th Apr-First Shift)-PCM-34 12
12 2 4 2 4
12
12
2 sin 2 sin 8 sin 212 12 sin 8
68.
Q
cos ec 2 2 cot 3 2 x 3 2 ….(i) h 1 So ….(ii) 4 2 7 10 x From (i) and (ii) h = 20
B
h
P
100
A
69.
z
1 i
2
100
C
2i a i
a2 1 2 z a3 2 5 a 1 2i 3 i z 10 1 3i 5 a i 2
x
70.
x
sin cos
1 sin
x
1
cos
1
x
2
x x 1 sin x sin cos cos sin x cos
x
3
x
sin 2 cos 2
2 sin 2
x
1
cos 2
1
x
x3 1 2 2x 3
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JEE-MAIN-2019 (10th Apr-First Shift)-PCM-35
71.
dy tan x y sec 2 x dx dy y sec 2 x tan x sec 2 x dx dt Let tan x t sec 2 x dx dy t y dt dy y t (Linear differential equation) dt After solving we get ye t e t t 1 c y tan x 1 ce tan x y 0 0 c 1
y tan x 1 e tan x So, y e2 4 72.
g' c lim
f c h f c h
h 0
lim
f c 0
f c h
h f c h f c h lim . h 0 h h h 0
h
0 if f ' c 0 h i.e. g x is differentiable at x c if f ' c 0 lim f ' c h 0
73.
g s 2, 2
f g s 0,4 5 f S 0, 16 f g s f s g f s 4, 4 g s therefore, g f s S 74.
x 3y z u a x y z 5 b x 2y 2z 6 Comparing coefficients we get a b 1 and a 2b 3 a,b 1, 2 So, x 3y z u x 3y 3z u 7, 3
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JEE-MAIN-2019 (10th Apr-First Shift)-PCM-36 75.
Let the six digit number be abcdef for this number to be divisible by 11 a c e b d f must be multiple of 11 possibility is a c c b d f 12 Case I a,c,e 7,5,0 and b,d,f 9,2,1 So, number of numbers 2 2! 3! 24
Case II a,c,e 3,2,1 b,d,f 7,5,0 So number of numbers 3! 3! 36 Total 24 36 60 76.
Equation of circle is given as S L 0 2
2
x 1
y 1 x y 0 passes through (1, –3) 16 4 0 4 2
(1, 1)
2
x 1 y 1 4 x y 0 r 2 2
77.
h x f g x h' x f ' g x g' x and f ' x e x 1
1 g' x h ' x e 1 2x 1 0 h ' x e
g x
x2 x
Case : 1
ex
2
x
1 and 2x 1 0 1 x 0, …..(i) 2 Case : 2 2 ex x 1 and 2x 1 0 x 1, …….(ii) from (i) and (ii) 1 x 0, 1, 2 2
78.
I
sin 2x 1 cos 3x dx 0
2
2I
sin 2x 1 cos 3x sin 2x sin 2x cos 3x dx 0
2
2I
dx 0
2I 2 dx 0
I dx 0
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JEE-MAIN-2019 (10th Apr-First Shift)-PCM-37
79.
80.
81.
1 1 yk 0 2 2 and given line 4x 5y k 0 …..(ii) 1 k 1 2 on comparing (i) and (ii) we get k 10 k No real value of k exist. Equation of common chord 4kx
……..(i)
9 Tangent at 3, 2 3x 9y 1 a2 2b2 Comparing with x 2y 12 3 9 1 2 2 a 4b 12 and b3 3 a6 2b2 Length of latus rectum 9 a
a1 a4 a7 a10 a13 a16 114 6 a1 a16 114 2 a1 a16 38
So a1 a6 a11 a16
4 a1 a16 2
2 38 76
82.
Mean x
xf f
i i i
2
fi x 1 2x 5 x 2 3x x 20
x 3, 4 (rejected) x
x f f
i i
2.8
i
83.
x4 1 lim x 1 x 2 1 ……..(1) x 1 x 1 x 1 3 3 x k k 2 k 2 k2 lim 2 ………….(2) x k x k 2 2k (1) = (2) 8 k 3 lim
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JEE-MAIN-2019 (10th Apr-First Shift)-PCM-38
84.
2
sin2 x 2sin x 5
2
sin x 1
2
. 4 sin
4
2
y
4sin
1
2
y
2
sin x 1 4 2 sin2 y
0
2
2
this is possible only if sin x 1 and sin y 1 1/ 3
85.
n 1n r lim n r 1 n n 1
1/ 3
1 x
dx
0
86.
3 4/3 2 1 4
Let hyperbola be
x2 y2 1 and passes through 4, 2 3 therefore a 2 b2
16 12 1 ……….(i) a2 b2 b2 a 2 e 2 1
4 5 a 16 2 a2 e 5 e 5 On solving (i) and (ii) 4e4 24e2 35 0 x
87.
………(ii)
y
Shown figure is square with side length 2 2
(0, 2) x – y = –2 x+y=2 x
(–2, 0) x + y = –2
(2, 0) x–y=2 (0, –2)
88.
P (Boy) = P (girl)
1 2
Required probability
all four girls Atleast two girls
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JEE-MAIN-2019 (10th Apr-First Shift)-PCM-39 4
1 2 4 4 4 1 1 1 4 4 2 C 3 2 C2 2 1 11
89.
G will be centroid of ABC G 2,4,2 OG 2iˆ 4ˆj 2kˆ OA 3i kˆ OG.OA 1 cos GOA 15 OG OA
90.
dx
x
2
2x 10
2
dx
x 1
21
9
2
Let x 1 3 tan
dx 3 sec 2 d 1 1 1 sin2 cos2 d 1 cos 2 d 27 54 54 2 3 x 1 1 1 x 1 tan C 2 54 3 x 2x 10
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FIITJEE Solutions to JEE(Main)-2019 Test Date: 10th April 2019 (Second Shift)
PHYSICS, CHEMISTRY & MATHEMATICS Paper - 1 Time Allotted: 3 Hours
Maximum Marks: 360
Please read the instructions carefully. You are allotted 5 minutes specifically for this purpose.
Important Instructions: 1.
The test is of 3 hours duration.
2.
This Test Paper consists of 90 questions. The maximum marks are 360.
3.
There are three parts in the question paper A, B, C consisting of Physics, Chemistry and Mathematics having 30 questions in each part of equal weightage. Each question is allotted 4 (four) marks for correct response.
4.
Out of the four options given for each question, only one option is the correct answer.
5.
For each incorrect response 1 mark i.e. ¼ (one-fourth) marks of the total marks allotted to the question will be deducted from the total score. No deduction from the total score, however, will be made if no response is indicated for an item in the Answer Box.
6.
Candidates will be awarded marks as stated above in instruction No.3 for correct response of each question. One mark will be deducted for indicating incorrect response of each question. No deduction from the total score will be made if no response is indicated for an item in the answer box.
7.
There is only one correct response for each question. Marked up more than one response in any question will be treated as wrong response and marked up for wrong response will be deducted accordingly as per instruction 6 above.
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JEE-MAIN-2019 (10th Apr-Second Shift)-PCM-2
PART –A (PHYSICS) 1.
One mole of an ideal gas passes through a process where pressure and volume obey 1 V 2 the relation P Po 1 o . Here Po and Vo are constants. Calculate the change in 2 V the temperature of the gas if its volume change from Vo to 2Vo 1 Po Vo 1 Po Vo (A) (B) 4 R 2 R 5 Po Vo 3 Po Vo (C) (D) 4 R 4 R
2.
A solid sphere of mass M and radius R is divided into two unequal parts. The first part 7M has a mass of and is converted into a uniform disc of radius 2R. The second part is 8 converted into a uniform solid sphere. Let I1 be the moment of inertia of the disc about its axis and I2 be the moment of inertia of the new sphere about its axis. The ratio of I1/I2 is given by: (A) 285 (B) 185 (C) 65 (D) 140
3.
The correct figure that shows, schematically, the wave pattern produced by superposition of two waves of frequencies 9 Hz and 11 Hz,
4.
(A)
(B)
(C)
(D)
In an experiment, brass and steel wires of length 1 m each with areas of cross section 1 mm2 are used. The wires are connected in series and one end of the combined wire is connected to a rigid support and other end is subjected to elongation. The stress requires to produced a new elongation of 0.2 mm is [Given, the Young’s Modulus for steel and brass are respectively 120 109 N/m2 and 60 109 N/m2] (A) 1.8 106 N/m2 (B) 0.2 106 N/m2 6 2 (C) 1.2 10 N/m (D) 4.0 106 N/m2
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JEE-MAIN-2019 (10th Apr-Second Shift)-PCM-3 5.
When heat Q is supplied to a diatomic gas of rigid molecules at constant volume its temperature increases by T. The heat required to produce the same change in temperature, at constant pressure is 3 5 (A) Q (B) Q 2 3 7 2 (C) Q (D) Q 5 3
6.
A bullet of mass 20 g has an initial speed of 1 ms–1 just before it starts penetrating a mud wall of thickness 20 cm. If the wall offers a mean resistances of 2.5 10–2 N, the speed of the bullet after emerging from the other side of the wall is close to (A) 0.7 ms–1 (B) 0.3 ms–1 –1 (C) 0.1 ms (D) 0.4 ms–1
7.
The elastic limit of brass is 379 MPa. What should be the minimum diameter of a brass rod if it is to support a 400 N load without exceeding its elastic limit? (A) 1.00 mm (B) 1.16 mm (C) 0.90 mm (D) 1.36 mm
8.
A plane is inclined at an angle = 30o with respect to the horizontal. A particle is projected with a speed u = 2 ms–1, from the base of the plant, making an angle = 15o with respect to the plane as shown in the figure. The distance from the base at which the particle hits the plane is close to (Take g = 10 ms2)
(A) 18 cm (C) 26 cm
(B) 14 cm (D) 20 cm
9.
The magnitude of the magnetic field at the centre of an equilateral triangular loop of side 1 m which is carrying a current of 10 A is: [Take o = 4 10–7 NA–2] (A) 9T (B) 1T (C) 3T (D) 18T
10.
Two radioactive substances A and B have decay constants 5 and respectively. At t = 0, a sample has the same number of the two nuclei. The time taken for the ratio of the 2
1 number of nuclei to become will be e (A) 1/ (C) 2/
(B) 1/4 (D) 1/2
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JEE-MAIN-2019 (10th Apr-Second Shift)-PCM-4 11.
Two blocks A and B of masses mA = 1 kg and mB = 3 kg are kept on the table as shown in figure. The coefficient of friction between A and B is 0.2 and between B and the surface of the table is also 0.2. The maximum force F that can be applied on B horizontal, so that the block A does not slide over the block B is: [Take g = 10 m/s2]
(A) 8 N (C) 12 N
(B) 16 N (D) 40 N
12.
The formula X = 5YZ2 X and Z have dimensions of capacitance and magnetic field respectively. What are the dimensions of Y in SI units? (A) [M-2 L0 T–4 A–2] (B) [M-3 L-2 T8 A–1] -2 -2 6 3 (C) [M L T A ] (D) [M-1 L-2 T4 A2]
13.
In Li++, electron in first Bohr orbit is excited to a level by a radiation of wavelength . When the ion gets deexcited to the ground state in all possible ways(including intermediate emission) a total of six spectral lines are observed. What is the value of ? (Given: h = 6.63 1034 js; e = 3 108 ms–1) (A) 10.8 nm (B) 11.4 nm (C) 9.4 nm (D) 12.3 nm
14.
The graph shows how the magnification m produced by a thin lens varies with image distance . What is the focal length of the lens used?
b2 ac b 2c (C) a
(A)
15.
a c b (D) c
(B)
A spaceship orbits around a planet at a height of 20 km from its surface. Assuming that only gravitational field of the plant acts on the spaceship. What will be the number of complete revolutions made by the spaceship in 24 hours around the plane? [Given: Mass of plane = 8 1022 kg, Radius of planet = 2 106 m, Gravitational constant G = 6.67 10–11 Mn2/kg2] (A) 9 (B) 11 (C) 13 (D) 17 FIITJEE Ltd., FIITJEE House, 29-A, Kalu Sarai, Sarvapriya Vihar, New Delhi -110016, Ph 46106000, 26569493, Fax 26513942 website: www.fiitjee.com.
JEE-MAIN-2019 (10th Apr-Second Shift)-PCM-5 16.
Light is incident normally on a completely absorbing surface with an energy flux of 25 W cm–2. If the surface has an area of 25 cm2, the maximum transferred to the surface in 40 min time duration will be (A) 6.3 10–4 Ns (B) 3.5 10–6 Ns –3 (C) 5.0 10 Ns (D) 1.4 10–6 Ns
17.
The time dependence of the position of a particle of mass m = 2 is given by r t 2t ˆi 3t 2 ˆj Its angular momentum with respect to the origin at time t = 2 is . (A) -48 kˆ (B) 48( ˆi + ˆj ) (C) 36 kˆ
(D) -34( kˆ - ˆi )
18.
Water from a tap emerges vertically downwards with an initial speed of 1.0 ms–1. The cross-sectional area of the tap is 10–4m2. Assume that the pressure is constant throughout the stream of water and that flow is streamlined. The cross-sectional area of the stream, 0.15 m below the tap would be: (take g = 110 ms–2) (A) 5 10–4 m2 (B) 5 10–5 m2 –5 2 (C) 1 10 m (D) 2 10–5 m2
19.
Space between two concentric conducting spheres of radii a and b (b >a) is filled with a medium of resistivity . The resistance between the two spheres will be 1 1 1 1 (A) (B) 2 a b 2 a b 1 1 1 1 (C) (D) 4 a b 4 a b
20.
A metal coin of mass 5 g and radius 1 cm is fixed to a thin stick AB of negligible mass as shown in the figure. The system is initially at rest. The constant torque, that will make the system rotate about AB at 25 rotations per second is 5 s is close to
(A) 2.0 10–5 Nm (C) 1.6 10–5 Nm
(B) 4.0 10–6 Nm (D) 7.9 10–6 Nm
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JEE-MAIN-2019 (10th Apr-Second Shift)-PCM-6 21.
The figure represents a voltage regulator circuit using a Zener diode. The breakdown voltage of the Zener diode is 6 V and the load resistance is RL= 4k. The series resistance of the circuit is Ri = 1k. If the battery voltage V8 varies from 8 V to 16 V, what are the minimum and maximum values of the current through Zener diode?
(A) 0.5 mA; 0.6 mA (C) 1.5 mA; 8.5 mA
22.
(B) 1 mA; 8.5 mA (D) 0.5 mA; 8.5 mA
A simple pendulum of length L is placed between the plates of a parallel plate capacitor having electric field E, as shown in figure. Its bob has mass m and charge q. the time period of the pendulum is given by
(A) 2
L qE g2 m
(C) 2
L qE g m
2
(B) 2
(D) 2
L qE g m
L g2
q2E 2 m2
23.
In free space, a particle A of charge 1 C is held fixed at a point P. Another particle B of the same charge and mass 4 g is kept at a distance of 1 mm from P. If B is released, then its velocity at a distance of 9 mm from P is 1 9 109 Nm2C2 Take 4o (A) 1.5 102 m/s (B) 2.0 103 m/s (C) 1.0 m/s (D) 3.0 104 m/s
24.
A 2 mW laser operates at a wavelength of 500 nm. The number of photons that will be emitted per second is [Given Planck’s constant h = 6.6 10–34 Js, speed of light c = 3.0 108 m/s] (A) 1 1016 (B) 1.5 1016 16 (C) 2 10 (D) 5 1015
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JEE-MAIN-2019 (10th Apr-Second Shift)-PCM-7 25.
A coil of self inductance 10 mH and resistance 0.1 is connected through a switch to a battery of internal resistance 0.9 . After the switch is closed the time taken for the current to attain 80% of the saturation values is: [take ln 5 = 1.6] (A) 0.016 s (B) 0.324 s (C) 0.002 s (D) 0.103 s
26.
A submarine experiences a pressure of 5.05 106 Pa at a depth of d1 in a sea. When it goes further to a depth of d2, it experiences a pressure of 8.08 106 Pa. Then d2 – d1 is approximately (density of water = 103 kg/m3 and acceleration due to gravity = 10 ms–2) (A) 400 m (B) 500 m (C) 600 m (D) 300 m
27.
A cubical block of side 0.5 m floats on water with 30% of its volume under water. What is the maximum weight that can be put on the block without fully submerging it under water? [Take density of water = 103 kg/m3] (A) 46.3 kg (B) 65.4 kg (C) 30.1 kg (D) 87.5 kg
28.
In a Young’s double slit experiment the ratio of the slit’s width is 4 : 1. The ratio of the intensity of maxima to minima, close to central fringe on the screen will be (A)
4
3 1 : 16
(C) 9 : 1
(B) 25 : 9 (D) 4:1
29.
A source of sound S is moving with the velocity of 50 m/s towards a stationary observer. The observer measures the frequency of the sound as 1000 Hz. What will be the apparent frequency of the source when it is moving away from the observer after crossing him? (Take velocity of sound in air is 350 m/s) (A) 1143 Hz (B) 857 Hz (C) 750 Hz (D) 807 Hz
30.
A square loop is carrying a steady current I and the magnitude of its magnetic dipole moment is m. If this square loop is changed to a circular loop and it carries the same current, the magnitude of the magnetic dipole moment of circular loop will be m 3m (A) (B) 4m 2m (C) (D)
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JEE-MAIN-2019 (10th Apr-Second Shift)-PCM-8
PART –B (CHEMISTRY) 31.
The difference between H and U (H - U), when the combustion of one mole of heptane(I) is carried out a temperature T is equal to (A) -4 RT (B) -3 RT (C) 3 RT (D) 4 RT
32.
The major product obtained in the given reaction is
(A)
(B)
(C)
(D)
33.
The ratio of the shortest wavelength of two spectral series of hydrogen spectrum is found to be about 9. The spectral series are: (A) Lyman and Paschen (B) Brackett and Pfund (C) Paschen and Pfund (D) Balmer and Brackett
34.
The correct order of the first ionization enthalpies is (A) Mn < Ti < Zn < Ni (B) Zn < Ni < Mn < Ti (C) Ti < Mn < Zn < Ni (D) Ti < Mn < Ni < Zn
35.
The correct statements among (a) to (b) are: (a) saline hydrides produce H2 gas when reacted with H2O. (b) reaction of LiAH4 with BF3 leads to B2H6. (c) PH3 and CH4 are electron - rich and electron-precise hydrides, respectively. (d) HF and CH4 are called as molecular hydrides. (A) (c) and (d) only (B) (a), (b) and (c) only (C) (a), (b), (c) and (d) (D) (a), (c) and (d) only
36.
Air pollution that occurs in sunlight is: (A) oxidising smog (C) reducing smog
37.
(B) acid rain (D) fog
For the re action of H2 w i t h I2, the rate constant is 2.5 10-4 dm3 mol–1 s–1 at 327°C and 1.0 dm3 mol–1 s–1 at 527°C. The activation energy for the reaction, in kJ mol–1 is: (R = 8.314 J K–1 mol–1) (A) 72 (B) 166 (C) 150 (D) 59 FIITJEE Ltd., FIITJEE House, 29-A, Kalu Sarai, Sarvapriya Vihar, New Delhi -110016, Ph 46106000, 26569493, Fax 26513942 website: www.fiitjee.com.
JEE-MAIN-2019 (10th Apr-Second Shift)-PCM-9 38.
In chromatography, which of the following statements is INCORRECT for Rf? (A) Rf value depends on the type of chromatography. (B) The value of Rf can not be more than one. (C) Higher Rf value means higher adsorption. (D) Rf value is dependent on the mobile phase.
39.
A hydrated solid X on heating initially gives a monohydrated compound Y. Y upon heating above 373K leads to an anhydrous white powder Z. X and Z, respectively, are: (A) Washing soda and soda ash. (B) Washing soda and dead burnt plaster. (C) Baking soda and dead burnt plaster. (D) Baking soda and soda ash.
40.
The INCORRECT statement is: (A) the spin-only magnetic moments of [Fe(H2O)6]2+ and [Cr(H2O)6]2+ are nearly similar. (B) the spin-only magnetic moment of [Ni(NH3)4(H2O)2]2+ is 2.83 BM. (C) the gemstone, ruby, has Cr3+ ions occupying the octahedral sites of beryl. (D) the color of [CoCl(NH3)5]2+ is violet as it absorbs the yellow light.
41.
Which of these factors does not govern the stability of a conformation in acyclic compounds? (A) Torsional strain (B) Angle strain (C) Steric interactions (D) Electrostatic forces of interaction
42.
The correct statement is: (A) zincite is a carbonate ore (B) aniline is a froth stabilizer (C) zone refining process is used for the refining of titanium (D) sodium cyanide cannot be used in the metallurgy of silver
43.
For the reaction, 2 SO2 g O2 g 2 SO3 g H = –57.2 kJ mol–1 and KC = 1.7 1016 Which of the following statement is INCORRECT? (A) The equilibrium constant is large suggestive of reaction going to completion and so no catalyst is required. (B) The equilibrium will shift in forward direction as the pressure increase. (C) The equilibrium constant decreases as the temperature increases. (D) The addition of inert gas at constant volume will not affect the equilibrium constant.
44.
The increasing order of nucleophilicity of the following nucleophiles is : (a) CH3CO2 (b) H2O (c) CH3SO3 (A) (b) < (c) < (a) < (d) (C) (d) < (a) < (c) < (b)
(d) OH (B) (a) < (d) < (c) < (b) (D) (b) < (c) < (d) < (a)
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JEE-MAIN-2019 (10th Apr-Second Shift)-PCM-10 45.
The correct match between Item-I and Item-II is:
(A) (a) (III), (b)(I), (c)(II), (d)(IV) (B) (a)(IV), (b)(II), (c)(I), (d)(III) (C) (a)(II), (b)(IV), (c)(I), (d)(III) (D) (a)(III), (b)(I), (c)(IV), (d)(II) 46.
The major product 'Y' in the following reaction is:Ph
CH3
i SOCl
NaOCl 2 X ii Aniline Y
O
47.
(A)
(B)
(C)
(D)
Points I, II and III in the following plot respectively correspond to (Vmp : most probable velocity)
(A) Vmp of N2 (B) Vmp of H2 (C) Vmp of O2 (D) Vmp of N2
(300K); Vmp of H2(300K); V mp (300K); Vmp of N2(300K); Vmp (400K); Vmp of N2(300K); Vmp (300K); Vmp of O2(400K); Vmp
of O2(400K) of O2(400K) of H2(300K) of H2(300K)
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JEE-MAIN-2019 (10th Apr-Second Shift)-PCM-11 48.
The highest possible oxidation states of uranium and plutonium, respectively, are (A) 6 and 4 (B) 7 and 6 (C) 4 and 6 (D) 6 and 7
49.
Compound A (C9H10O) shows positive iodoform test. Oxidation of A with KMnO4/KOH gives acid B(C8H6O4). Anhydride of B is used for the preparation of phenolphthalein. Compound A is:-
50.
The noble gas that does NOT occur in the atmosphere is: (A) He (B) Ra (C) Ne (D) Kr
51.
The pH of a 0.02 M NH4Cl solution will be [given Kb (NH4OH) = 10–5 and log 2 = 0.301] (A) 2.65 (B) 5.35 (C) 4.35 (D) 4.65
52.
The crystal field stabilization energy (CFSE) of [Fe(H 2 O) 6 ]Cl 2 and K 2 [NiCl 4 ], respectively, are :(A) –0.4o and –0.8t (B) –0.4o and –1.2t (C) –2.4o and –1.2t (D) –0.6o and –0.8t
53.
The correct option among the following is: (A) Colloidal particles in lyophobic sols can be precipitated by electrophoresis. (B) Brownian motion in colloidal solution is faster the viscosity of the solution is very high. (C) Colloidal medicines are more effective because they have small surface area. (D) Addition of alum to water makes it unfit for drinking.
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JEE-MAIN-2019 (10th Apr-Second Shift)-PCM-12
54.
Which one of the following graphs between molar conductivity (m) correct? (A) (B)
(C)
55.
versus
C is
(D)
1 g of non-volatile non-electrolyte solute is dissolved in 100g of two different solvents A and B whose ebullioscopic constants are in the ratio of 1 : 5. The ratio of the elevation Tb A in their boiling points, is : Tb B (A) 5 : 1 (C) 1 : 5
(B) 10 : 1 (D) 1 : 0.2
56.
Which of the following is NOT a correct method of the preparation of benzylamine from cyanobenzene? (A) (i) HCl/H2O (ii) NaBH4 (B) (i) LiAIH4 (ii) H3O+ (C) (i) SnCl2 + HCl(gas) (ii) NaBH4 (D) H2/Ni
57.
The number of pentagons in C60 and trigons (triangles) in white phosphorus, respectively, are: (A) 12 and 3 (B) 20 and 4 (C) 12 and 4 (D) 20 and 3
58.
The minimum amount of O2(g) consumed per gram of reactant is for the reaction : (Given atomic mass : Fe = 56, O = 16, Mg = 24, P = 31, C = 12, H = 1) (A) C3H8(g) + 5O2(g) 3 CO2(g) + 4 H2O(l) (B) P4(s) + 5O2(g) P4O10(s) (C) 4Fe(s) + 3O2(g) 2 Fe2O3(s) (D) 2 Mg(s) + O2(g) 2 MgO(s)
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JEE-MAIN-2019 (10th Apr-Second Shift)-PCM-13 59.
Number of stereo centers present in linear and cyclic structures of glucose are respectively (A) 4 & 5 (B) 5 & 5 (C) 4 & 4 (D) 5 & 4
60.
The major product 'Y' in the following reaction is:
(A)
(B)
(C)
(D)
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JEE-MAIN-2019 (10th Apr-Second Shift)-PCM-14
PART–C (MATHEMATICS) x 2 ax b 3 , then a + b is equal to x 1 x 1 (A) 5 (B) 1 (C) -4 (D) -7
61.
If lim
62.
The sum of the real roots of the equation x 6 1 2
3x
x 3 0 is equal to
3
2x
x2
(A) -4 (C) 6 63.
(B) 0 (D) 1
Lines are drawn parallel to the line 4x – 3y + 2 = 0 at a distance
3 from the origin. Then 5
which one of the following points lies on any of these lines? 1 2 1 1 (A) , (B) , 4 3 4 3 1 1 1 2 (C) , (D) , 4 3 4 3 x , x R, x 3 at a point (, ) (0, 0) on it is x 3 parallel to the line 2x + 6y – 11 = 0 then (A) |2 + 6| = 11 (B) |2 + 6| = 19 (C) |6 + 2| = 19 (D) |6 + 2| = 9
64.
If the tangent to the curve y
65.
The distance of the point having position vector ˆi 2ˆj 6kˆ from the straight line passing through the point (2, 3, -4) and parallel to the vector 6iˆ 3ˆj 4kˆ is
2
(A) 7
(B) 4 3
(C) 2 13
(D) 6
66.
If the line ax + y = c, touches both the curves x2 + y2 = 1 and y 2 4 2 x , then |c| is equal to 1 (A) (B) 2 2 1 (C) (D) 2 2
67.
Let f(x) = loge(sinx), (0 < x < ) and g(x) = sin–1(e–x), (x 0). If is a positive real number such that a = (fog)’() and b = (fog)(), then (A) a2 + b – a = 22 (B) a2 – b – a = 0 (C) a2 – b – a = 1 (D) a2 + b + a = 0 FIITJEE Ltd., FIITJEE House, 29-A, Kalu Sarai, Sarvapriya Vihar, New Delhi -110016, Ph 46106000, 26569493, Fax 26513942 website: www.fiitjee.com.
JEE-MAIN-2019 (10th Apr-Second Shift)-PCM-15 68.
If 5x + 9 = 0 is the directrix of the hyperbola 16x2 – 9y2 = 144, then its corresponding focus is 5 (A) (5, 0) (B) ,0 3 (C) (-5, 0)
69.
y y If cos–1x – cos–1 2 , where -1 x 1, – 2 y 2, x 2 , then for all x, y, 4x2 – 4xy
cos + y2 is equal to (A) 4 sin2 - 2x2y2 (C) 2 sin2 70.
5 (D) ,0 3
(B) 4 cos2 + 2x2y2 (D) 4 sin2
The locus of the centres of the circles, which touch the circle, x2 + y2 = 1 externally, also touch the y-axis and lie in the first quadrant is (A) x 1 2y, y 0 (B) y 1 4x, x 0 (C) x 1 4y, y 0
(D) y 1 2x, x 0
71.
Let a1, a2, a3, …… be and A.P with a6 = 2. Then the common difference of this A.P., which maximizes the product a1a4a5 is 3 8 (A) (B) 2 5 2 6 (C) (D) 3 5
72.
The smallest natural number n, such that the coefficient of x in the expansion of n
2 1 n x 3 is C23 is x (A) 38 (C) 23
(B) 58 (D) 35
73.
Suppose that 20 pillars of the same height have been erected along the boundary of a circular stadium. If the top of each pillar has been connected by beams with the top of all its non-adjacent pillars, then the total number of beams is (A) 210 (B) 180 (C) 170 (D) 190
74.
A spherical iron ball of radius 10 cm is coated with a layer of ice of uniform thickness that melts at a rate of 50 cm3/min. When the thickness of the ice is 5 cm, then the rate at which the thickness (in cm/min) of ice decreases is 1 5 (A) (B) 36 6 1 1 (C) (D) 9 18
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JEE-MAIN-2019 (10th Apr-Second Shift)-PCM-16 75.
If both the means and the standard deviation of 50 observations x1, x2, ………, x50 are equal to 16, then the mean of (x1 - 4)2, (x2 - 4)2, …., (x50 - 4)2 is (A) 400 (B) 380 (C) 525 (D) 480
76.
The number of real roots of the equation 5 + |2x - 1| = 2 x(2 x - 2) is (A) 4 (C) 2
(B) 3 (D) 1
77.
The tangent and normal to the ellipse 3x2 + 5y2 = 32 at the point P(2, 2) meet the x-axis at Q and R, respectively. Then the area(in sq. units) of the triangle PQR is 34 68 (A) (B) 15 15 14 16 (C) (D) 3 3
78.
A perpendicular is drawn from a point on the line
79.
The integral
x 1 y 1 z to the plane x + y + z 2 1 1 = 3 such that the foot of the perpendicular Q also lies on the plane x – y + z = 3. Then the co-ordinates of Q are (A) (2, 0, 1) (B) (-1, 0, 4) (C) (1, 0, 2) (D) (4, 0, -1)
/3
/ 6
sec 2/3 x cos ec 4/3 x dx is equal to
(A) 35/6 – 32/3 (C) 37/6 – 35/6
(B) 35/3 – 31/3 (D) 34/3 – 31/3
80.
The angles A, B and C of a triangle ABC are in A.P and a : b = 1 : the area (in sq. cm) of this triangle is 4 (A) 2 3 (B) 3 2 (C) 4 3 (D) 3
3 . If c = 4 cm, then
81.
Let a, b and c be in G.P with common ratio r, where a 0 and 0 < r
1 . If 3a, 7b and 2
15c are the first three terms of an A.P., then the 4th term of this A.P is 2 7 (A) a (B) a 3 3 (C) 5a (D) a 82.
Let y = y(x) be the solution of the differential equation dy y tan x 2x x 2 tan x, x , , such that y(0) = 1. Then dx 2 2 (A) y ' y ' (B) y ' y ' 2 2 4 4 4 4 (C) y y 2 4 4
2 (D) y y 2 4 4 2
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JEE-MAIN-2019 (10th Apr-Second Shift)-PCM-17 1 2 and units from the planes 4x – 2y 3 3 + 4z + = 0 and 2x – y + 2z + = 0, respectively, then the maximum value of + us equal to (A) 15 (B) 13 (C) 5 (D) 9
83.
If the plane 2x – y + 2z + 3 = 0 has the distances
84.
The area(in sq. units) of the region bounded by the curves y = 2x and y = |x +1| in the first quadrant is 3 3 (A) (B) loge 2 2 2 3 1 1 (C) (D) 2 loge 2 2
85.
Let be a real number for which the system of linear equations x+y+z=6 4x + y - z = - 2 3x + 2y – 4z = -5 Has indefinitely many solutions. Then is a root of the quadratic equation (A) 2 - – 6 = 0 (B) 2 - 3 – 4 = 0 2 (C) + 3 – 4 = 0 (D) 2 + – 6 = 0
86.
The sum 3 3 3 3 3 3 3 3 3 1 2 1 2 3 1 2 3 ..... 15 1 1 ...... 1 2 3 ...... 15 is equal to 1 2 1 2 3 1 2 3 ..... 15 2 (A) 620 (B) 1860 (C) 1240 (D) 660
87.
Minimum number of times a fair coin must be tossed so that the probability of getting at least one head is more than 99% is (A) 8 (B) 6 (C) 7 (D) 5
88.
The negation of the Boolean expression ~ s(~rs) is equivalent to (A) sr (B) ~s ~r (C) r (D) s r
89.
If
5 x2
x e
2
dx g x e x c , where c is a constant of integration, then g(-1) is equal to
(A) -1 (C) 90.
(B) 1 5 2
(D)
1 2
If z and w are two complex numbers such that |zw| = 1 and arg(z) – arg(w) = (A) zw i (C) zw
(B) zw
1 i 2
, then 2
1 i 2
(D) zw i
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JEE-MAIN-2019 (10th Apr-Second Shift)-PCM-18
JEE (Mains) – 2019 ANSWERS PART A – PHYSICS 1. 5. 9. 13. 17. 21. 25. 29.
D C C A A D A C
2. 6. 10. 14. 18. 22. 26. 30.
D A D D B D D C
3. 7. 11. 15. 19. 23. 27.
A B B B D None D
4. 8. 12. 16. 20. 24. 28.
None D B C A D C
34. 38. 42. 46. 50. 54. 58.
D C B A C B C
64. 68. 72. 76. 80. 84. 88.
C C A D A C D
PART B – CHEMISTRY 31. 35. 39. 43. 47. 51. 55. 59.
A C A A D B C A
32. 36. 40. 44. 48. 52. 56. 60.
C A C A D A A C
33. 37. 41. 45. 49. 53. 57.
A B B D A A C
PART C – MATHEMATICS 61. 65. 69. 73. 77. 81. 85. 89.
D A D C B D A C
62. 66. 70. 74. 78. 82. 86. 90.
B B D D A B A D
63. 67. 71. 75. 79. 83. 87.
A C B A C B C
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JEE-MAIN-2019 (10th Apr-Second Shift)-PCM-19
HINTS AND SOLUTIONS PART A – PHYSICS 1.
n = 1 mole 1 V 2 P = Po 1 o ; PV = nRT = RT 2 V RT P V V2 RT Po 1 o 2 V 2V PV Vo2 V 2 Po T o 1 = V R 2V 2 R 2V 2 P V2 1 1 T o (2Vo Vo ) o R 2 2Vo Vo Vo2 V o 2 P V2 1 1 T o (2Vo Vo ) o R 2 2Vo Vo
=
Po R
Po Vo2 (1 2) Vo R 2 2Vo P V 3 Po Vo = o Vo o = R 4 4 R
=
2.
7M 2 (ZR) 7M 4R 2 7MR2 8 I1 2 28 4 2 2 2 2 MR 2M R MR I2 5 82 58 4 80 2 I1 7MR 80 140 I2 4MR 2
3.
By looking into graph.
4.
=1M A = 10–6 M2 F Stress = A Stress Stress = Y
Ys = 120 × 109
Ys = 120 × 109
F 9
YB = 60 × 10
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JEE-MAIN-2019 (10th Apr-Second Shift)-PCM-20
F AY
1 2
1F 2F 0.2 10 3 AY1 AY2
F 0.2 10 3 A Y1 Y2 0.2 10 3 0.2 10 3 109 120 1 1 1 2 9 120 10 60 109 0.2 10 6 120 = 8 106 3 =
5.
Q = CVT Q = CPT C 2 7 Q = P Q 1 Q Q CV 5 5
6.
2.5 × 10–2 × 0.2 =
7.
8.
1 20 10 3 V 2 12 2 5 × 10–3 = 10 × 10–3 (1 –V2) 1 1 1 1 – V2 = ; V2 = ; V= = 0.7 2 2 2
400 379 106 2 d 4 4 400 106 d2 = 0.336 × 10–6 × 4 379 d 2 0.336 10 3 M 1.16 mm
T
2 u sin g cos
1 g sin T 2 2 ucos 2u sin gsin 4u2 sin2 = gcos 2 g2 cos2
R ucos T
=
u2 sin2 u2 sin 1 cos 2 g cos g cos2
1 u2 sin 3 2 = 1 2 2 3 g cos 10 2 4
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JEE-MAIN-2019 (10th Apr-Second Shift)-PCM-21
= = 9.
4 10 3
4 5 3
8 3 1 30 2
8 8 3 8 8( 3 1) = 20 cm 30 30 30
i = 10 A =1m o = 4 × 10–7 B=
N A2
i C
o i
3 4 3 2 i 3 4 10 7 10 3 = o = 20 3 107 2 2 1 = 3 T
10.
11.
1 e t 5 t 2 e 1 t 2 MA = 1 kg, MB = 3 kg AB 0.2 B = 0.2 Fmax = (MA + MB) × 0.2 × 10 + (MA + MB) × 0.2 ×
A B
F
10 = 4 × 2 + 4 × 2 = 16 12.
13.
X = 5YZ2 X Y= = M–3 L–2 T8A4 2 5Z
hc 1 13.6 eV(g) 1 16 1240 eV 15 9 13.6 eV 16 1240 16 = 10.8 nm 15 9 13.6
h=4
h=1
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JEE-MAIN-2019 (10th Apr-Second Shift)-PCM-22
14.
f
m
b c C
c b
a
15.
b
V
mV 2 GMm 2 r r GM V r
n
VT GM T 2r r 2r
GM T 6.67 1011 8 1022 T = r 3 2 (202 104 )3 2 =
16.
17.
24 3600 2 3.14
24 3600 6.67 8 1011 24 3600 = 11 3 12 (202) 10 2 3.14 1242.8 78.51
W 25 10 4 W / m2 cm2 P = 25 × 25 ; W = 625 W hc dn P dt h dn P 625 F dt C 3 108 625 40 60 Momentum = = 5 × 10–3 Ns 3 108 I 25
v 2iˆ 6 ˆj At t = 2 v 2iˆ 12ˆj P mv 4i 24 ˆj At t = 2 r 4iˆ 12ˆj
ˆi ˆj kˆ L r P 4 12 0 4 24 0
= 4( 24) 4 12 kˆ = –96 48 kˆ = ( ) 48 kˆ
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JEE-MAIN-2019 (10th Apr-Second Shift)-PCM-23
18.
10 4 1 (1)2 2 10 0.15 A A
10 4 = 5 × 10–5 2 b
19.
R a
=
dx 4 x 2
1 1 4 a b
20.
m = 5 × 10–3 kg, r = 10–2 m = 25 × 2 rad/5 = 50 rad/sec t I I 5mr 2 t 4 t 3 5 5 10 10 4 50 = 45 25 = 106 = 2 × 10–5 4
21.
Vbreakdwon = 6V, RL = 4k, Ri = 1 k 6 iL = 103 1.5 10 3 = 1.5 mA 4 ii = 2 × 10–3 i = i1 – iL = 0.5 mA – minimum current
Ri ii
i RL
VB iL
ii = 10 × 10–3 = 10 mA imax = 8.5 mA 22.
T 2
L q2E 2 g 2 M 2
+ + + + + + + + + + + +
– – – – – E – – – – – – m – q qE – – – – Mg –
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JEE-MAIN-2019 (10th Apr-Second Shift)-PCM-24 23.
qA = 1 c ; qB = 1 c,
mB = 4 × 10–9 kg, rAB = 10–3 m
1 1 1 MB V 2 k qA qB 3 2 9 10 3 10
1 8 4 109 V 2 9 109 10 6 10 6 103 2 9 8 V 2 109 4 109 2 hc dn dt dn 2 103 dt hc 2 10 3 500 10 9 = 6.6 10 34 3 108 1000 = 1014 = 5 × 1015 6.6 3
24.
2 103
25.
L = 10 × 10–3 H, r1 = 0.1 i = 1 e t/2
L = 10
–2
r1 = 0.1
isaturation = 80% isaturation = 0.8 0.8 = 1 e t/2
t/2
0.8 = 1 e ; e t/2 0.2 et/L = 5 t = L ln 5 = 10 × 10–3 × 1.6 = 16 × 10–3
r2 = 0.9
26.
P1 = 5.05 × 106 ; P2 = 8.08 × 106 P2 – P1 = g(d2 – d1) 3.03 106 d2 d1 = 3.03 × 102 = 303 103 10
27.
0.3 3 = 3
××
kg m3 m + 3 = 3
0.3 3
= 300
M = 3(w – ) = (5)3 {1000 – 300} = 700 × (5)3 = 87.5 kg 28.
IMax 9 IMin 1
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JEE-MAIN-2019 (10th Apr-Second Shift)-PCM-25
29.
fa
V fo = 1000 Hz V Vs
s
V = 50 m/s
V fo V Vs fa V Vs 350 50 300 3 fa V Vs 350 50 400 4 3 fa 1000 = 750 Hz 4 fa
30.
m = I2 I4 2 m 4 m 4 m m
m =
2r = 4
r
2
r 2
4 2 42 = 2
PART B – CHEMISTRY 31.
32.
33.
1 1 1 RH 2 2 Z2 2 n1 n2 1 1 1 RH 1 1 Z2 1 n1 n2 As for shortest wavelength both n1 and n12 are
1 9 n11 2 1 n12
Now if n11 = 3 and n1 is 1 it will justify the statement hence Lyman and Paschen is correct.
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JEE-MAIN-2019 (10th Apr-Second Shift)-PCM-26
34.
As Zn is fully filled and left to right in group IP increases.
35.
36.
Fact based
37.
38.
Rf value can’t measure the extent of adsorption.
39.
40.
In gemstone, ruby has Cr3+ ion occupying the octahedral sites of aluminium oxide (Al2O3) normally occupied by Al3+ion.
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JEE-MAIN-2019 (10th Apr-Second Shift)-PCM-27 41.
Angle strain govern stability in cyclic compound.
42.
Fact based.
43.
In option (B)- ng is –ve therefore increase in pressure will bring reaction in forward direction. In option (C)- as the reaction is exothermic therefore increase in temperature will decrease the equilibrium constant. In option (D)- Equilibrium constant changes only with temperature. Hence, option (B), (C) and (D) are correct therefore option (1) is incorrect choice.
44.
45
(a) (b) (c) (d)
High density polythene Polyacrylonitrile Novolac Nylon 6
(III) (I) (IV) (II)
Ziegler-Natta Catalyst Peroxide catalys t Acid or base catalyst Condensation at high temperature & pressure
46.
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JEE-MAIN-2019 (10th Apr-Second Shift)-PCM-28 47.
48.
The highest oxidation state of U and Pu is 6+ and 7+ respectively.
49.
50.
Fact based.
51.
52.
CFSE = [-0.4n t2g + 0.6 n eg] o
53.
In electrophoresis precipitation occurs at the electrode which is oppositely charged therefore (A) is correct. FIITJEE Ltd., FIITJEE House, 29-A, Kalu Sarai, Sarvapriya Vihar, New Delhi -110016, Ph 46106000, 26569493, Fax 26513942 website: www.fiitjee.com.
JEE-MAIN-2019 (10th Apr-Second Shift)-PCM-29
54.
Both NaCl and KCl are strong electrolytes and as Na+(aq.) has less conductance than K+ (aq.) due to more hydration therefore the graph of option (B) is correct.
55.
56.
Benzylamine will not give cyanobenzene with HCl/H2O & NaBH4.
57.
Refer structure of C60 & P4 P
P
P P
58.
4 mol of Fe require 3 32 gram 1 3 32 1 mol of Fe require 0.428 g 56 4 56
59.
60.
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JEE-MAIN-2019 (10th Apr-Second Shift)-PCM-30
PART C – MATHEMATICS 2
61.
62.
63.
x ax b 5 x 1 ……..(i) 1 a b 0 . ……..(ii) 2a5 a b 7 lim
x 1
By expansion, we get 5x 3 30x 30 5x 0 5x3 35x 30 0 x3 7x 6 0 , All roots area real So, sum of roots = 0 Y
Required line is 4x 3y 0
3 5 5 3 So, required equation of line 4x 3y 3 0 and 4x 3y 3 0
is
(–2, 0) X
1 2 (1) 4 3 3 0 4 3 4x – 3y + 2 = 0
64.
dy 2 3 2 dx , 2 3
Given that : 2 3 1 2 2 3 3
0, 3
1 . 2 6 2 19
65.
0 0
AP.n AD 61 n
P (–1, 2, 6)
PD AP 2 AD 2 110 61 7
A (2, 3, –4)
D
n 6iˆ 3 ˆj 4kˆ
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JEE-MAIN-2019 (10th Apr-Second Shift)-PCM-31
66.
Tangent to y 2 4 2 x is y mx
2 /m 1 m2
2 it is also tangent to x 2 y 2 1 m
1 m 1
Tangent will be y x 2 or y x 2 compare with y ax C 67.
a 1 and C 2 fog x x fg b
fg x ' 1 fg ' 1 a 68.
x2 y2 1 9 16
16 5 9 3 corresponding focus will be (–ae, 0) i.e. 5, 0 .
a 3, b 4 and e 1
69.
y 2 y cos cos1 x cos1 cos 2 cos1 x cos1
x
y y2 1 x2 1 cos 2 4 2
xy y2 2 cos 1 x 1 2 4 y2 x2 xy cos 1 cos2 sin2 4
70.
h2 k 2 h 1
x 2 y 2 x 2 1 2x y 2 1 2x y 1 2x; x 0
71.
Let a is first term and d is common difference then, a 5d 2 (given) ……(1) f d 2 5d 2 2d 2 d
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JEE-MAIN-2019 (10th Apr-Second Shift)-PCM-32
2 8 , 3 5 8 f " d 0 at d 5 8 d 5 f ' x 0 d
n
72.
Tr 1 nCr x 2n2r . x 3r r 0
2n 5r 1 2n 5r 1 for r 15, n 38 smallest value of n is 38.
73.
74.
75.
Total cases = number of diagonals in 20 sided polygon. 20C2 20 170
4 3 10 h 103 3 dV 2 dh 4 10 h dt dt 2 dh 50 4 10 5 dt dh 1 cm dt 18 min V
Mean
x 50
i
16
Standard deviation
256 2
e
x
2 i
50
2
16
2 i
50
New mean
x
i
4
2
x
2 i
16 50 8 x i
50 50 256 2 16 8 16 400
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JEE-MAIN-2019 (10th Apr-Second Shift)-PCM-33 76.
Let 2x t 5 t 1 t 2 2t
t 1 t 2 2t 5
gt f t From the graph So, number of real root is 1.
77.
3x 2 5y 2 32 dy 3 dx 2,2 5 3 16 x 2 Q , 0 5 3 5 4 Normal : y 2 x 2 R , 0 3 5 1 68 Area is QR 2 QR 2 15 Tangent : y 2
78.
79.
Let point P on the line is 2 1, 1, foot of perpendicular Q is given by x 2 1 y 1 z 2 3 1 1 1 3 Q lies on x y z 3 and x y z 3 x z 3 and y 0 2 3 y 0 1 0 3 Q is (2, 0, 1) 1 I dx 2/3 cos x sin1/3 x.sin x
tan2/3 x .sec 2 x.dx tan2 x sec 2 x .dx tan x t, sec 2 x dx dt 4/3 tan x dt t 1/3 3 t 1/3 tan4/3 1/ 3
1 3 tan x 3 1/3
tan x
1/3
/3
I
/6
3
1
1/3
1/3
3
3
37/6 3 5/6
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JEE-MAIN-2019 (10th Apr-Second Shift)-PCM-34
80.
81.
, by sine Rule 3 1 sin A 2 A 30o ,a 2, b 2 3, c 4 1 2 3 2 2 3 sq.cm 2 B
b ar c ar 2 3a,7b and 15c are in A.P. 14b 3a 15c 14 ar 3a 15 ar 2
14r 3 15r 2 15r 2 14r 3 0
r
3r 1 5r 3 0
1 3 , 3 5
1 1 Only acceptable value is r , because r 0, 3 2 7 2 c .d 7b 3a 7ar 3a a 3a a 3 3 2 15 2 4th term 15c a a aa 3 9 3 82.
dy y tan x 2x x 2 tan x dx tan x dx I.F. e eln.sec x sec x
y.sec x 2x x 2 tan x sec x.dx 2
2x sec x dx x sec x .tan x dx
y sec x x 2 sec x y x 2 cos x y 0 0 1
1
2
y x cos x 1 2 y 2 4 16 2 1 y 2 4 16 y ' x 2x sin x 1 y ' 2 4 2
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JEE-MAIN-2019 (10th Apr-Second Shift)-PCM-35
1 y ' 2 4 2 y ' y' 2 4 4 83.
4x 2y 4z 6 0 6 16 4 16 6 2 8, 4 3 4 4 1 3 2 5, 1
6 1 6 3
2 3
Maximum value of 13 .
84.
Required Area 1
x 1 2 dx x
0
1
x2 2x x ln 2 0 2 1 2 1 1 0 0 ln 2 ln 2 2 3 1 2 ln 2 85.
D0 1 1
1
4
0 3
3 2 4
13 23 .....n3 1 15.16 . 2 2 n 1 1 2 ..... n 15 n n 1 60 2 n 1 15
86.
Sum
15
n 1
n n 1 n 2 n 1 6
60
15.16.17 60 620 6
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JEE-MAIN-2019 (10th Apr-Second Shift)-PCM-36 n
87.
99 1 1 100 2 n
1 1 2 100 n7
88.
~ ~ s ~ s s r ~ s
s r s ~ s s r s r 89.
Let x 2 t 2xdx dt 1 2 t 1 2 t t .e dt t .e 2t.e t , dt 2 2 1 t 2 .e t t.e t 1.e t .dt 2 t2 t 2e t te t e t t 1 e t 2 2
x4 2 x 2 1 e x C 2 for k 0 1 5 g 1 1 1 2 2 90.
z . w 1
z. w e
i 2
z re
i 2
i 2
z.w e
.ei e
1 and w ei r
i 2
i
i
.e i e 2 i
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FIITJEE Solutions to JEE(Main)-2019 Test Date: 12th April 2019 (First Shift)
PHYSICS, CHEMISTRY & MATHEMATICS Paper - 1 Time Allotted: 3 Hours
Maximum Marks: 360
Please read the instructions carefully. You are allotted 5 minutes specifically for this purpose.
Important Instructions: 1.
The test is of 3 hours duration.
2.
This Test Paper consists of 90 questions. The maximum marks are 360.
3.
There are three parts in the question paper A, B, C consisting of Physics, Chemistry and Mathematics having 30 questions in each part of equal weightage. Each question is allotted 4 (four) marks for correct response.
4.
Out of the four options given for each question, only one option is the correct answer.
5.
For each incorrect response 1 mark i.e. ¼ (one-fourth) marks of the total marks allotted to the question will be deducted from the total score. No deduction from the total score, however, will be made if no response is indicated for an item in the Answer Box.
6.
Candidates will be awarded marks as stated above in instruction No.3 for correct response of each question. One mark will be deducted for indicating incorrect response of each question. No deduction from the total score will be made if no response is indicated for an item in the answer box.
7.
There is only one correct response for each question. Marked up more than one response in any question will be treated as wrong response and marked up for wrong response will be deducted accordingly as per instruction 6 above.
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JEE-MAIN-2019 (12th Apr-First Shift)-PCM-2
PART –A (PHYSICS) 1.
The value of numerical aperature of the objective lens of a microscope is 1.25. If light of
wavelength 5000 A is used, the minimum separation between two points, to be seen as distinct, will be : (A) 0.48 m (B) 0.38 m (C) 0.24 m (D) 0.12 m 2.
The resistive network shown below is connected to a D.C. source of 16 V. The power consumed by the network is 4 Watt. The value of R is:
(A) 8 (B) 16 3.
The figure shows a square loop L of side 5 cm which is connected to a network of resistances. The whole set up is moving towards right with a constant speed of 1 cms–1. At some instant, a part of L is in a uniform magnetic field of 1 T, perpendicular to the plane of the loop. If the resistance of L is 1.7 , the current in the loop at that instant will be close to :
(A) 115 A (C) 60 A 4.
(B) 6 (D) 1
(B) 170 A (D) 150 A
A circular disc of radius b has a hole of radius a at its centre (see figure). If the mass per unit area of the disc varies as 0 , then the radius of gyration of the disc about its axis r passing through the centre is : (A)
ab 3
(B)
a 2 b 2 ab 3
(C)
ab 2
(D)
a 2 b 2 ab 2
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JEE-MAIN-2019 (12th Apr-First Shift)-PCM-3 5.
A man (mass = 50 kg) and his son (mass = 20 kg) are standing on a frictionless surface facing each other. The man pushes his son so that he starts moving at a speed of 0.70 ms–1 with respect to the man. The speed of the man with respect to the surface is : (A) 0.47 ms–1 (B) 0.28 ms–1 –1 (C) 0.14 ms (D) 0.20 ms–1
6.
Two identical parallel plate capacitors, of capacitance C each, have plates of area A, separated by a distance d. The space between the plates of the two capacitors, is filled with three dielectrics, of equal thickness and dielectric constants K1 , K 2 and K 3 . The first capacitor is filled as shown in fig. I, and the second one is filled as shown in fig. II. If these two modified capacitors are charged by the same potential V, the ratio of the energy stored in the two, would be ( E1 refers to capacitor (I) and E2 to capacitor (II)) :
E1 K1K 2K 3 E 9K 1K 2K 3 (B) 1 E 2 (K1 K 2 K 3 )(K 2K 3 K 3K 1 K1K 2 ) E 2 (K1 K 2 K 3 )(K 2K 3 K 3K 1 K1K 2 ) E (K K 2 K 3 )(K 2K 3 K 3K1 K1K 2 ) E (K K 2 K 3 )(K 2K 3 K 3K1 K1K 2 ) (C) 1 1 (D) 1 1 E2 9K 1K 2K 3 E2 K1K 2K 3
(A)
7.
Two moles of helium gas is mixed with three moles of hydrogen molecules (taken to be rigid). What is the molar specific heat of mixture at constant volume? (R = 8.3 J/mol K) (A) 17.4 J/mol K (B) 15.7 J/mol K (C) 19.7 J/mol K (D) 21.6 J/mol K
8.
The stopping potential V0 (in volt) as a function of frequency () for a sodium emitter, is shown in the figure. The work function of sodium, from the data plotted in the figure, will be : (Given : Planck's constant (h) = 6.63 × 10–34 Js, electron charge e = 1.6 × 10–19 C) (A) 1.82 eV (B) 1.66 eV (C) 2.12 eV (D) 1.95 eV
9.
At 40°C, a brass wire of 1 mm is hung from the ceiling. A small mass, M is hung from the free end of the wire. When the wire is cooled down from 40ºC to 20ºC it regains its original length of 0.2 m. The value of M is close to : (Coefficient of linear expansion and Young's modulus of brass are 10–5/ºC and 1011 N/m2, respectively; g = 10 ms–2) (A) 0.5 kg (B) 9 kg (C) 0.9 kg (C) 1.5 kg
10.
A magnetic compass needle oscillates 30 times per minute at a place where the dip is 45°, and 40 times per minute where the dip is 30º. If B1 and B2 are respectively the total magnetic field due to the earth at the two places, then the ratio B1/B2 is best given by : (A) 3.6 (B) 1.8 (C) 2.2 (D) 0.7 FIITJEE Ltd., FIITJEE House, 29-A, Kalu Sarai, Sarvapriya Vihar, New Delhi -110016, Ph 46106000, 26569493, Fax 26513942 website: www.fiitjee.com.
JEE-MAIN-2019 (12th Apr-First Shift)-PCM-4 11.
An excited He+ ion emits two photons in succession, with wavelengths 108.5 nm and 30.4 nm, in making a transition to ground state. The quantum number n, corresponding 1240 eV to its initial excited state is (for photon of wavelength , energy E :) (in nm) (A) n = 4 (B) n = 6 (C) n = 5 (D) n = 7
12.
When M1 gram of ice at –10ºC (specific heat = 0.5 cal g–1°C–1) is added to M2 gram of water at 50ºC, finally no ice is left and the water is at 0ºC. The value of latent heat of ice, in cal g–1 is: 50M2 5M2 (A) 5 (B) 5 M1 M1 50M2 5M1 (C) (D) 50 M1 M2
13.
A shell is fired from a fixed artillery gun with an initial speed u such that it hits the target on the ground at a distance R from it. If t1 and t2 are the values of the time taken by it to hit the target in two possible ways, the product t1t2 is: (A) 2R / g (B) R / 4g (C) R / g (D) R/2g
14.
The transfer characteristic curve of a transistor, having input and output resistance 100 and 100 k respectively, is shown in the figure. The Voltage and Power gain, are respectively :
15.
(A) 5 10 4 , 2.5 106
(B) 5 10 4 , 5 106
(C) 5 10 4 , 5 105
(D) 2.5 104 , 2.5 106
Shown in the figure is a shell made of a conductor. It has inner radius a and outer radius b, and carries charge Q. At its centre is a dipole P as shown. In this case : (A) Surface charge density on the inner surface of the shell is zero everywhere. (B) Electric field outside the shell is the same as that of a point charge at the centre of the shell. (C) Surface charge density on the inner surface is uniform and (Q / 2) equal to 4a2 (D) Surface charge density on the outer surface depends on P
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JEE-MAIN-2019 (12th Apr-First Shift)-PCM-5 16.
Which of the following combinations has the dimension of electrical resistance ( 0 is the permittivity of vacuum and 0 is the permeability of vacuum)? 0 (A) (B) 0 0 0 (C)
17.
0 0
0 0
(D)
A galvanometer of resistance 100 has 50 divisions on its scale and has sensitivity of 20 A / division. It is to be converted to a voltmeter with three ranges, of 0–2 V, 0–10 V and 0–20 V. The appropriate circuit to do so is : (A)
(B)
(C)
(D)
18.
In a double slit experiment, when a thin film of thickness t having refractive index is introduced in front of one of the slits, the maximum at the centre of the fringe pattern shifts by one fringe width. The value of t is ( is the wavelength of the light used) : (A) (B) 2( 1) ( 1) 2 (C) (D) (2 1) ( 1)
19.
A progressive wave travelling along the positive x-direction is represented by y(x, t) = A sin (kx – t + ). Its snapshot at t = 0 is given in the figure:
For this wave, the phase is : (A) (C)
(B) 2
2
(D) 0
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JEE-MAIN-2019 (12th Apr-First Shift)-PCM-6 20.
21.
A uniform rod of length is being rotated in a horizontal plane with a constant angular speed about an axis passing through one of its ends. If the tension generated in the rod due to rotation is T(x) at a distance x from the axis, then which of the following graphs depicts it most closely?
(A)
(B)
(C)
(D)
A point dipole p p0 x is kept at the origin. The potential and electric field due to this dipole on the y-axis at a distance d are, respectively : (Take V = 0 at infinity)
p
p (A) , 4 0 d2 4 0 d3 p p (C) , 4 0 d2 4 0 d3
22.
p (B) 0, 4 0 d3 p (D) 0, 4 0 d3
An electromagnetic wave is represented by the electric field E E0 n sin[t (6y 8z)]. Taking unit vectors in x, y and z directions to be i, j, k , the direction of propogation s , is
: 3 j 4k (A) S 5 4k 3 j (C) S 5
23.
4 j 3k (B) S 5 3i 4j (D) S 5
A submarine (A) travelling at 18 km/hr is being chased along the line of its velocity by another submarine (B) travelling at 27 km/hr. B sends a sonar signal of 500 Hz to detect A and receives a reflected sound of frequency . The value of is close to : (Speed of sound in water = 1500 ms–1) (A) 499 Hz (B) 502 Hz (C) 504 Hz (D) 507 Hz
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JEE-MAIN-2019 (12th Apr-First Shift)-PCM-7
24.
The trajectory of a projectile near the surface of the earth is given as y = 2x – 9x2. If it were launched at an angle 00 with speed v0 then (g = 10 ms–2) : 1 5 2 3 1 1 (A) 0 cos1 (B) 0 cos1 and v 0 ms and v 0 ms 3 5 5 5 2 3 1 (C) 0 sin1 and v 0 ms 5 5
1 5 1 (D) 0 sin1 and v 0 ms 3 5
25.
A thin ring of 10 cm radius carries a uniformly distributed charge. The ring rotates at a constant angular speed of 40 rad s–1 about its axis, perpendicular to its plane. If the magnetic field at its centre is 3.8 × 10–9 T, then the charge carried by the ring is close to ( 0 4 107 N / A 2 ). (A) 2 × 10–6 C (B) 7 × 10–6 C –5 (C) 4 × 10 C (D) 3 × 10–5 C
26.
The truth table for the circuit given in the fig is :
(A)
(C)
27.
A B Y
A B Y
0
0
1
0
0
0
0
1
0
0
1
0
1 1
0 1
0 0
1 1
0 1
1 1
(B)
A B Y
A B Y
0
0
1
0
0
1
0
1
1
0
1
1
1 1
0 1
1 1
1 1
0 1
0 0
(D)
A sample of an ideal gas is taken through the cyclic process abca as shown in the figure. The change in the internal energy of the gas along the path ca is –180 J. The gas absorbs 250 J of heat along the path ab and 60 J along the path bc. The work done by the gas along the path abc is : (A) 120 J (B) 100 J (C) 140 J (D) 130 J
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JEE-MAIN-2019 (12th Apr-First Shift)-PCM-8 28.
A concave mirror has radius of curvature of 40 cm. It is at the bottom of a glass that has water filled up to 5 cm (see figure). If a small particle is floating on the surface of water, its image as seen, from directly above the glass, is at a distance d from the surface of water. The value of d is close to : (Refractive index of water = 1.33) (A) 13.4 cm (B) 8.8 cm (C) 6.7 cm (D) 11.7 cm
29.
To verify Ohm's law, a student connects the voltmeter across the battery as, shown in the figure. The measured voltage is plotted as a function of the current, and the following graph is obtained: If V0 is almost zero, identify the correct statement: (A) The potential difference across the battery is 1.5 V when it sends a current of 1000 mA. (B) The emf of the battery is 1.5 V and the value of R is 1.5 (C) The emf of the battery is 1.5 V and its internal resistance is 1.5 (D) The value of the resistance R is 1.5
30.
A person of mass M is, sitting on a swing of length L and swinging with an angular amplitude 0. If the person stands up when the swing passes through its lowest point, the work done by him, assuming that his centre of mass moves by a distance ( < < L), is close to : (A) Mg(1 0 2 ) (B) Mg(1 02 ) (C) Mg
2 (D) Mg 1 0 2
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JEE-MAIN-2019 (12th Apr-First Shift)-PCM-9
PART –B (CHEMISTRY) 31.
The mole fraction of a solvent in aqueous solution of a solute is 0.8.The molality (in mol kg–1) of the aqueous solution is (A) 13.88 10–2 (B) 13.88 10–1 (C) 13.88 (D) 13.88 10–3
32.
The major products of the following reaction are:
(A)
(B)
(C)
(D)
33.
The correct statement among the following is (A) (SiH3 )3 N is planar and less basic than (CH3 )3 N. (B) (SiH3 )3 N is planar and more basic than (CH3 )3 N. (C) (SiH3 )3 N is pyramidal and less basic than (CH3 )3 N. (D) (SiH3 )3 N is pyramidal and more basic than (CH3 )3 N.
34.
The complex ion that will lose its crystal field stabilization energy upon oxidation of its metal to +3 state is
2
(A) Ni(phen)3
2
(C) Co(phen)3
2
(B) Zn(phen)3
2
(D) Fe(phen)3
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JEE-MAIN-2019 (12th Apr-First Shift)-PCM-10 35.
The correct set of species responsible for the photochemical smog is : (A) NO, NO2, O3 and hydrocarbons (B) CO2, NO2 SO2 and hydrocarbons (C) N2, NO2 and hydrocarbons (D) N2, O2, O3 and hydrocarbons
36.
The major product(s) obtained in the following reaction is/ are:
37.
38.
(A)
(B)
(C)
(D)
The major product of the following addition reaction is: Cl2 /H2O H3 C CH CH2 (A)
(B)
(C)
(D)
The correct sequence of thermal stability of the following carbonates is : (A) BaCO3 SrCO3 CaCO3 MgCO3 (B) BaCO3 CaCO3 SrCO3 MgCO3 (C) MgCO3 CaCO3 SrCO3 BaCO3
(D) MgCO3 SrCO3 CaCO3 BaCO3
39.
The group number, number of valence electrons and valency of an element with atomic number 15, respectively, are: (A) 15, 5 and 3 (B) 15, 6 and 2 (C) 16, 5 and 2 (D) 16, 6 and 3
40.
Peptization is a : (A) process of converting soluble particles to form colloidal solution (B) process of converting precipitate into colloidal solution (C) process of converting a colloidal solution into precipitate (D) process of bringing colloidal molecule into solution
41.
The metal that gives hydrogen gas upon treatment with both acid as well as base is: (A) iron (B) magnesium (C) zinc (D) mercury
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JEE-MAIN-2019 (12th Apr-First Shift)-PCM-11 42.
The electrons are more likely to be found:
(A) in the region a and b (C) only in the region a 43.
(B) in the region a and c (D) only in the region c
An example of a disproportionation reaction is : (A) 2KMnO 4 K 2MnO 4 MnO2 O2 (B) 2NaBr Cl2 2naCl Br2 (C) 2 CuBr CuBr2 Cu (D) 2MnO 4 10 16H 2Mn2 5 2 8 H2O
44.
5 moles of AB2 weigh 125 10–3 kg and 10 moles of A2B2 weigh 300 10–3 kg. The molar mass of A(MA) and molar mass of B(MB) in kg mol–1 are : (A) MA = 50 10–3 and MB = 25 10–3 (B) MA = 10 10–3 and MB = 5 10–3 –3 –3 (C) MA = 5 10 and MB = 10 10 (D) MA = 25 10–3 and MB = 50 10–3
45.
The basic structural unit of feldspar, zeolites, mica and asbestos is :
(A)
(C)
SiO2
(B)
(SiO3 )2
(D)
(SiO 4 )4
46.
Enthalpy of sublimation of iodine is 24 cal g–1 at 200°C. If specific heat of I2(s) and I2 (vap) are 0.055 and 0.031 cal g–1K–1 respectively, then enthalpy of sublimation of iodine at 250°C in cal g–1 is : (A) 2.85 (B) 11.4 (C) 5.7 (D) 22.8
47.
Which of the following is a thermosetting polymer? (A) PVC (B) Bakelite (C) Buna-N (D) Nylon 6
48.
An element has a face-centred cubic (fcc) structure with a cell edge of a. The distance between the centres of two nearest tetrahedral voids in the lattice is: a (A) (B) 2a 2 3 (C) a (D) a 2
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JEE-MAIN-2019 (12th Apr-First Shift)-PCM-12 49.
An organic compound 'A' is oxidized with Na2O2 followed by boiling with HNO3.The resultant solution is then treated with ammonium molybdate to yield a yellow precipitate. Based on above observation, the element present in the given compound is : (A) Phosphorus (B) Sulphur (C) Nitrogen (D) Fluorine
50.
The idea of froth floatation method came from a person X and this method is related to the process Y of ores. X and Y, respectively, are: (A) washer woman and concentration (B) fisher woman and concentration (C) fisher man and reduction (D) washer man and reduction
51.
In the following reaction; xA yB d[A] d[B] log10 log10 0.3010 dt dt ‘A’ and ‘B’ respectively can be : (A) C2H2 and C6H6 (C) N2O4 and NO2
(B) n-Butane and Iso-butane (D) C2H4 and C4H8
52.
Glucose and Galactose are having identical configuration in all the positions except position. (A) C–4 (B) C–3 (C) C–5 (D) C–2
53.
Which of the following statements is not true about RNA? (A) It has always double stranded -helix structure (B) It is present in the nucleus of the cell (C) It controls the synthesis of protein (D) It usually does not replicate
54.
What is the molar solubility of Al(OH)3 in 0.2 M NaOH solution? Given that, solubility product of Al(OH)3 = 2.4 10–24 : (A) 3 10 19 (B) 12 10 21 23 (C) 12 10 (D) 3 10 22
55.
The increasing order of the pKb of the following compound is :
(A)
(B)
(C)
(D)
(A) (c) < (a) < (d) < (b) (C) (a) < (c) < (d) < (b)
(B) (b) < (d) < (a) < (c) (D) (b) < (d) < (c) < (a)
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JEE-MAIN-2019 (12th Apr-First Shift)-PCM-13 56.
Given : Co3 e Co 2 ; E 1.81V Pb4 2e Pb2 ; E 1.67 V Ce 4 e Ce3 ; E 1.61V Bi3 3e Bi ; E 0.20V Oxidizing power of the species will increase in the order (A) Ce4 Pb 4 Bi3 Co 3 (B) Co3 Pb 4 Ce 4 Bi3 3 4 4 3 (C) Bi Ce Pb Co (D) Co3 Ce4 Bi3 Pb4
57.
But-2-ene on reaction with alkaline KMnO4 at elevated temperature followed by acidification will give : (A) One molecule of CH3CHO and one molecule of CH3COOH (B)
2 molecules of CH3CHO
(C)
2 molecules of CH3COOH
(D) 58.
The major product of the following reaction is :
(A)
(B)
(C)
(D)
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JEE-MAIN-2019 (12th Apr-First Shift)-PCM-14 59.
60.
Complete removal of both the axial ligands (along the z-axis) from an octahedral complex leads to which of the following splitting patterns? (relative orbital energies not on scale).
(A)
(B)
(C)
(D)
An ideal gas is allowed to expand from 1 L to 10 L against a constant external pressure of 1bar. The work done in kJ is: (A) – 9.0 (B) – 0.9 (C) – 2.0 (D) + 10.0
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JEE-MAIN-2019 (12th Apr-First Shift)-PCM-15
PART–C (MATHEMATICS) 61.
Let P be the point of intersection of the common tangents to the parabola y2 = 12x and the hyperbola 8x2 – y2 = 8. If S and S' denote the foci of the hyperbola where S lies on the positive x-axis then P divides SS' in a ratio: (A) 2:1 (B) 13:11 (C) 5:4 (D) 14:13
62.
If three of the six vertices of a regular hexagon are chosen at random, then the probability that the triangle formed with these chosen vertices is equilateral is : 3 1 (A) (B) 10 5 1 3 (C) (D) 10 20
63.
Let f : R R be a continuously differentiable function such that f(2) = 6 and f '(2)
f (x)
6
4t 3 dt (x 2)g(x), then lim g(x) is equal to : x2
(A) 24 (C) 12 64.
1 . If 48
(B) 18 (D) 36
2x3 1 x 4 x dx is equal to : (Here C is a constant of integration)
The integral
x3 1 1 (A) loge C 2 x2
(C) loge
x3 1 x
C
2
x3 1 1 (B) loge C 2 x3
(D) loge
x3 1 x2
C
65.
The coefficient of x18 in the product (1 x)(1 x)10 (1 x x 2 )9 is : (A) 84 (B) 126 (C) –126 (D) –84
66.
The number of ways of choosing 10 objects out of 31 objects of which 10 are identical and the remaining 21 are distinct, is : (A) 220 (B) 220 + 1 21 (C) 2 (D) 220 – 1
67.
Let a random variable X have a binomial distribution with mean 8 and variance 4. If k P(X 2) 16 , then k is equal to : 2 (A) 17 (B) 137 (C) 1 (D) 121
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JEE-MAIN-2019 (12th Apr-First Shift)-PCM-16
68.
The equation z i z 1 ,i 1, represents : 1 2 (C) a circle of radius 1
(A) a circle of radius
69.
(B) the line through the origin with slope 1 (D) the line through the origin with slope –1
If m is the minimum value of k for which the function f(x) x kx x 2 is increasing in the interval [0,3] and M is the maximum value of f in [0, 3] when k = m, then the ordered pair (m, M) is equal to : (A) (5, 3 6 ) (B) (4, 3 2) (C) (3, 3 3)
(D) (4, 3 3 )
70.
If the data x1, x2, ...., x10 is such that the mean of first four of these is 11, the mean of the remaining six is 16 and the sum of squares of all of these is 2,000; then the standard deviation of this data is : (A) 2 2 (B) 2 (C) 4 (D) 2
71.
If the area (in sq. units) of the region then a – b is equal to : 10 (A) 3 8 (C) 3
72.
73.
(x, y) : y
2
4x, x y 1, x 0, y 0 is a 2 b,
(B) 6 (D)
2 3
1 Consider the differential equation, y 2 dx x dy 0 . If value of y is 1 when x = 1, y then the value of x for which y = 2, is : 3 1 1 (A) e (B) 2 2 e 3 1 5 1 (C) (D) 2 2 e e
If and are the roots of the equation 375x2 – 25x – 2 = 0, then n
n
lim r lim r is equal to :
n
r 1
1 12 7 (C) 116
(A)
n
r 1
29 358 21 (D) 346
(B)
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JEE-MAIN-2019 (12th Apr-First Shift)-PCM-17 74.
75.
76.
77.
78.
79.
A 2 m ladder leans against a vertical wall. If the top of the ladder begins to slide down the wall at the rate 25 cm/ sec., then the rate (in cm/sec.) at which the bottom of the ladder slides away from the wall on the horizontal ground when the top of the ladder is 1 m above the ground is : 25 (A) 25 (B) 3 25 (C) 25 3 (D) 3 5 5 The number of solutions of the equation 1 + sin4x = cos23x, x , is : 2 2 (A) 3 (B) 4 (C) 5 (D) 7
dy d2 y If ey + xy = e, the ordered pair , 2 at x = 0 is equal to : dx dx 1 1 1 1 (A) , 2 (B) , 2 e e e e 1 1 1 1 (C) , 2 (D) , 2 e e e e 12 3 The value of sin1 sin 1 is equal to : 13 5 33 63 (A) cos1 (B) sin1 65 65 9 56 (C) cos1 (D) sin1 2 2 65 65
If the normal to the ellipse 3x2 + 4y2 = 12 at a point P on it is parallel to the line, 2x + y = 4 and the tangent to the ellipse at P passes through Q(4, 4) then PQ is equal to : 157 5 5 (A) (B) 2 2 221 61 (C) (D) 2 2 Let a 3i 2j 2k and b i 2j 2k be two vectors. If a vector perpendicular to both the vectors a b and a b has the magnitude 12 then one such vector is: (A) 4 2i 2j k (B) 4 2i 2j k
(C) 4 2i 2j k 80.
(D) 4 2i 2j k
The equation y = sinx sin(x + 2) – sin2(x+1) represents a straight line lying in : (A) first, third and fourth quadrants (B) first, second and fourth quadrants (C) third and fourth quadrants only (D) second and third quadrants only
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JEE-MAIN-2019 (12th Apr-First Shift)-PCM-18
81.
1 x2 3 For x 0, , let f(x) x, g(x) tan x and h(x) = . If (x) ((hof )og)(x), then 1 x2 2 is equal to : 3 11 (A) tan (B) tan 12 12 5 7 (C) tan (D) tan 12 12
82.
If the angle of intersection at a point where the two circles with radii 5 cm and 12 cm intersect is 90°, then the length (in cm) of their common chord is : 13 120 (A) (B) 2 13 13 60 (C) (D) 5 13
83.
If
84.
Let Sn denote the sum of the first n terms of an A.P.. If S4 = 16 and S6 = –48, then S10 is equal to : (A) –410 (B) –260 (C) –320 (D) –380
85.
86.
87.
2 0
cot x dx m( n),thenm n is equal to cot x cos ecx 1 (A) 1 (B) 2 1 (C) (D) –1 2
5 2 1 If B = 0 2 1 is the inverse of a 3 3 matrix A, then the sum of all values of for 3 1 which det (A) + 1 = 0, is : (A) 0 (B) –1 (C) 1 (D) 2 x 2 y 1 z 1 intersects the plane 2x + 3y – z + 13 = 0 at a point P and 3 2 1 the plane 3x + y + 4z = 16 at a point Q, then PQ is equal to : (A) 2 14 (B) 14 (C) 2 7 (D) 14
If the line
j k and i k is If the volume of parallelepiped formed by the vectors i j k, minimum, then is equal to : 1 (A) 3 (B) 3 1 (C) (D) 3 3 FIITJEE Ltd., FIITJEE House, 29-A, Kalu Sarai, Sarvapriya Vihar, New Delhi -110016, Ph 46106000, 26569493, Fax 26513942 website: www.fiitjee.com.
JEE-MAIN-2019 (12th Apr-First Shift)-PCM-19
88.
2 3 If A is a symmetric matrix and B is a skew-symmetrix matrix such that A + B = , 5 1 then AB is equal to : 4 2 4 2 (A) (B) 1 4 1 4 4 2 (C) 1 4
4 2 (D) 1 4
89.
If the truth value of the statement p (~ q r) is false (F), then the truth values of the statement p, q, r are respectively : (A) T, T, F (B) F, T, T (C) T, F, T (D) T, F, F
90.
For x R,let [x] denote the greatest integer x, then the sum of the series 1 1 2 1 1 1 99 3 3 100 3 100 3 100 (A) –135 (B) –153 (C) –133 (D) –131
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JEE-MAIN-2019 (12th Apr-First Shift)-PCM-20
JEE (Mains) – 2019 ANSWERS PART A – PHYSICS 1. 5. 9. 13. 17. 21. 25. 29.
C D B A C D D C
2. 6. 10. 14. 18. 22. 26. 30.
A B D A B A D A
3. 7. 11. 15. 19. 23. 27.
B A C B A B D
4. 8. 12. 16. 20. 24. 28.
B B A D D A B
34. 38. 42. 46. 50. 54. 58.
D C B D A D C
64. 68. 72. 76. 80. 84. 88.
C B C C C C B
PART B – CHEMISTRY 31. 35. 39. 43. 47. 51. 55. 59.
C A A C B D C B
32. 36. 40. 44. 48. 52. 56. 60.
C C B C A A C B
33. 37. 41. 45. 49. 53. 57.
A B C D A A C
PART C – MATHEMATICS 61. 65. 69. 73. 77. 81. 85. 89.
C A D A D A C A
62. 66. 70. 74. 78. 82. 86. 90.
C A B D B B A C
63. 67. 71. 75. 79. 83. 87.
B B B C A D Bonus
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JEE-MAIN-2019 (12th Apr-First Shift)-PCM-21
HINTS AND SOLUTIONS PART A – PHYSICS 1.
Numerical aperature of the microscope is given as 0.61 NA d Where d = minimum sparaton between two points to be seen as distinct d
0.61 (0.61) (5000 10 m10 ) = 2.4 × 10–7 m NA 1.25 = 0.24 m
2.
162 4 8R R = 8 P
3.
Since it is a balanced wheatstone bridge, its equivalent resistance =
4 3
= Bv = 5 × 10–4 V So total resistance 4 R 1.7 3 3 i 166 A 170 A R 4.
dI = (dm)r2 = ( dA)r2 = 0 2dr r 2 = (0 20r2dr r b
I = DI 0 2r 2 dr a
b3 a3 = 0 2 3 m = dm dA b
= 0 2 dr a
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JEE-MAIN-2019 (12th Apr-First Shift)-PCM-22
5.
0 = 50V1 – 20V2 and V1 + V2 = 0.7 V1 = 0.2 6.
30 AK1 d 3 AK C2 0 2 d 30 AK 3 C3 d C1
1 1 1 1 Ceq C1 C2 C3
Ceq
30 AK 1K 2K 3 d(K1K 2 K 2K 3 K 3K1 )
…(i)
0K1A 3d 0K 2 A C2 3d K A C3 0 3 3d C1
Ceq C1 C2 C3
=
0 A (K1 K 2 K 3 ) 3d
…(ii)
Now, 1 C V2 9K1K 2K 3 E1 2 eq 1 E2 (K1 K 2 K 3 ) (K 1K 2 K 2K 3 K 3K1 ) Ceq V 2 2
7.
8.
n1f1 n2 f2 2 3 3 5 21 n1 n2 5 5 fR 21 R Cv = = 17.4 J/mol K 5 5 2 fmix
hv = + ev0 h v0 e e v0 is zero for = 4 × 1014 Hz
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JEE-MAIN-2019 (12th Apr-First Shift)-PCM-23
hv e e = h 6.63 10 34 4 1014 = = 1.66 eV 1.6 10 19 0
9.
Ay Mg Mg = (Ay)T = 2 It is closest to 9.
10.
1 B1 cos 45o 2 I o f1 B1 cos 45 f2 B2 cos 30o
f1
11.
12.
1 1 1 R 2 2 z2 n m 1 1 1 R 2 2 22 1085 n m m=2 n=5 Heat lost = Heat gain
M2 × 1 × 50 = M1 × 0.5 × 10 + M1Lf
Lf
= 13.
1 B 2 cos 30o 2 I B1 0.7 B2
f2
50M2 5M1 M1 50M2 5 M1
Range will be same for time t1 and t2, so angles of projection will be ‘’ & ‘90° – ’
t1
2u sin 2u sin(90o ) u2 sin 2 t2 and R g g g
t1t 2
4u2 sin cos 2 2u2 sin cos g2 g g 2R = g
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JEE-MAIN-2019 (12th Apr-First Shift)-PCM-24
14.
I R 5 10 3 Vgain C out = 103 6 100 10 IB Rin 1 108 5 10 4 20 I Pgain C (Vgain ) Ib 5 103 = (5 10 4 ) 6 100 10 = 2.5 × 106
15.
Total charge of dipole = 0, so charge induced on outside surface = 0. But due to non uniform electric field of dipole, the charge induced on inner surface is non zero and non uniform. So, for any observer outside the shell, the resultant electric field is due to Q uniformly distributed on outer surface only and it is equal to. E
16.
KQ r2
[0] = M–1 L–3 T4 A2 [0] = M L T–2 A–2 [R] = M L2 T–3 A–2
[R] 0 0 17.
20 × 50 × 10–6 = 10–3 Amp. 2 V1 3 100 R1 10 1900 = R1 10 V2 3 (2000 R 2 ) 10 R2 = 8000 20 V3 3 10 103 R3 = 10 × 103 R3 10
18.
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JEE-MAIN-2019 (12th Apr-First Shift)-PCM-25 x = ( – 1)t = 1 for one maximum shift t 1 19.
y = A sin (kx – wt + ) at x = 0, t = 0 and slope is negative = 20.
T
x
x x
dm 2 x
x
xx
m dx2 x T
m2 2 ( x2 ) 2 m 2 2 T ( x2 ) 2
=
21.
22.
V=0
KP E 3 r p = 40 d3
E E0nˆ sin( t (6y 8z) = E0nˆ sin(t k r ) where r xiˆ yjˆ zkˆ and k r = 6y – 8z k 6 ˆj 8kˆ direction of propagation sˆ kˆ 3ˆj 4kˆ = 5
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JEE-MAIN-2019 (12th Apr-First Shift)-PCM-26
23.
f0 = 500 Hz frequency received by B again = 1500 (B) (A) & 7.5 m/s 5 m/sec 1500 7.5 1500 5 f2 f0 = 502 Hz 1500 5 1500 7.5 24.
25.
26.
Equation of trajectory is given as y = 2x – 9x2 …(A) Comparing with equation: g y = x tan – x2 …(B) 2 2 2u cos We get, tan = 2 1 cos 5 g Also, =9 2 2u cos2 10 25 u 2 ; u2 2 9 1 29 5 5 u m/s 3
0i 0 q 2R 2R 2 q = 3 × 10–5 C B
C=A+B and y = A.C
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JEE-MAIN-2019 (12th Apr-First Shift)-PCM-27 27.
28.
Light incident from particle P will be reflected at mirror.
R 20 cm 2 1 1 1 20 ; v1 cm v u f 3 This image will act as object for light getting refracted at water surface. 20 35 So, object distance d 5 cm 3 3 Below water surface. After refraction, final image is at 35 1 d d 2 1 3 4 / 3 35 8.75 cm 4 8.8 cm u 5cm, f m
29.
V = E – Ir When V = V0 = 0 0 = E – Ir E=r When I = 0, V = E = 1.5 V r = 1.5 30.
Angular momentum conservation MV0L = MV1(L – )
L V1 V0 L wg + wp = KE
mg w p
1 m V12 V02 2
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JEE-MAIN-2019 (12th Apr-First Shift)-PCM-28
w p mg
L 2 1 L
1 mV02 2
1 = mg mV02 2
2 L 1 1 L
Now, << L By, Binomial approximation 1 = mg mV02 2
= mg
2 L 1 1 L
1 2 mV02 2 L
W P = mg mV02
L
g Here, V0 = maximum velocity = × A = ( L) L 0
So, w p mg m 0 gL
2
= mg 1 0
2
L
PART B – CHEMISTRY
31.
Xsolvent = 0.8 = 8/10 NTotal= 10, nsolutent = 8, nsolute = 2 Wt of solvent = 8 18 2 1000 Molality 8 18
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JEE-MAIN-2019 (12th Apr-First Shift)-PCM-29 32.
33. 34.
p-d bonding in N(SiH3)3. When Fe2+ oxidizes to Fe3+ 3d X
35.
Smog photochemical is NO, NO2, O3.
36.
37.
38.
Smaller the size of cation, more polarisability high thermal decomposition of carbonates.
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JEE-MAIN-2019 (12th Apr-First Shift)-PCM-30
39.
15 1s2, 2s2, 2p6, 3s2, 3p3 Period- III, Group-V
40.
Peptization is process to form collidal solution by adding peptizing agent.
41.
HCl Zn ZnCl2 H2 NaOH Na 2 ZnO2 H2
42.
At a & c, probability of finding electron is maximum 43.
2 Cu Cu2 Cu
44.
Mol. wt is of 1 mol AB2 A + 2B = 25 A2B2 2A + 2B = 30
45.
Basic unit is silicate (SiO 4 )4
46.
CP
47.
Thermosetting are cross-linked polymer.
48.
Two nearest tetrahedral voids are along one body-diagonal.
49.
Canary yellow ppt comes in test of PO34 ion.
50.
Concentration(on dressing), i.e. washing of ore.
51.
H2 H1 T2 T1
1 dA 1 dB x dt y dt d A x dB log10 log10 log y dt dt
x
log y log 0.3
x 2 y
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JEE-MAIN-2019 (12th Apr-First Shift)-PCM-31 6
52.
CH2OH O OH
5
Glucose:
4
OH
1 2
OH 3
OH 6
They differ at C-4
CH2OH
OH Galactose:
4
O OH
OH 3
1 2
OH 53.
Not always double stranded -helix.
54.
3 Al OH3 Al 3OH
2.4 10
-24
s (3s 0.2) = s(3s + 0.2)3
55.
CH3O–, CH3–, H is electron donating, But –NO2 is electron withdrawing group.
56.
Lower the standard reduction potential, more the ability to get reduced higher the oxidizing power.
57.
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JEE-MAIN-2019 (12th Apr-First Shift)-PCM-32 58.
59.
Ligand filed exert mass repulsion along x, y axis as compared to Z-axis so dx 2 y 2 and dxx will have increase in energy y
60.
W = - Pext(V2 – V1)
PART C – MATHEMATICS 61.
Tangents y 2 12x y 2x
3 m
x2 y2 1 y mx m2 8 1 8 Common tangent gives 3 m m2 8 m4 8m2 9 0 m 3 1 y 3x 1 P ,0 3 y 3x 1
x2 y2 1 1 8 e3 ae 3 S 3,0
S ' 3, 0
P divides SS’ in 5 : 4. 62.
Only two equilateral triangle are possible A1A 3 A 5 and A 2 A 5 A 6
2 2 1 6 C3 20 10
A4 A5
A3
A6
A2 A1
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JEE-MAIN-2019 (12th Apr-First Shift)-PCM-33 f x
4t 3 dt
63.
lim
6
x2
lim
x2 4.f 3 x .f ' x
1 4f 2 f ' 2 18 x2
3
64.
65.
2x 3 1 x 4 x dx 1 2x 2 2 x1 dx x x 1 x2 t x 1 2x x 2 dx dt dt t n t C 1 n x 2 C x 9
10
1 x 1 x 1 x x 2 3 9
1 x 1 x 2
9
66.
C6 84
Since
21
C0 ..... 21C10 21C11 ....... 21C21 221
but we have to select 10 objects and
67.
21
21
C0 ..... 21C10 21 C11 ....... 21C21
C0 ..... 21C10 220
np 8 npq 4 1 1 q p 2 2 n 16 16
1 p x r 16 Cr 2 16 C0 16 C1 16 C 2 p x 2 216 137 16 2
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JEE-MAIN-2019 (12th Apr-First Shift)-PCM-34 68.
z i z 1 gives y x
69.
f x x kx x 2 f ' x
3kx 4x 2
2 kx x 2 For increasing f x 0
3kx 4x 2 0 4x 2 3kx 0 3k 4x x 0 4 3k 3 0 4 k4
kx x 2 0 x 2 kx 0 x x k 0 so x 0, 3 ve x 3 minimum value of k is m 4
f x x kx x 2
3 4 3 32 3 3, M 3 3 70.
x1 ...... x 4 44 x 5 ...... x10 96
x 14, xi 140
x Variance
2 i
2
x 4
n Standard deviation = 2 71.
x, y : y
2
4x, x y 1, x 0, y 0
32 2
A
2 x dx
0
2 x 3/2
1 1 3 2 2 2
1 3 2 2
y2 = 4x
32 2
1 0 2 22 2 22 3/2 2 8 2 10 3 3 8 10 a ,b 3 3 a b 6 72.
1
y 2 dx xdy
O
32 2
1
dy y
dx x 1 2 3 dy y y
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JEE-MAIN-2019 (12th Apr-First Shift)-PCM-35 1
IF e
y 2 dy
1
e
1
e y .x e y . 1 y
1 y
xe e
C x
73.
1 y
1 dy C y3
1 y
e C y
1 e
3 1 when y = 2 2 e
375x 2 25x 2 0 25 2 , 375 375 2 ...upto inf inite terms 2 ..........upto inf inite terms
74.
1 1 1 12
dy x2 y2 4 25 dt dx dy x y 0 dt dt dx 3 1 25 0 dt dx 25 cm / sec dt 3 4
A 25 cm/s 2
y
B O
x
2
75.
1 sin x cos 3x sin x 0 and cos3x 1 0,2, 2, ,
76.
ey xy e differentiate w.r.t. x dy dy ey x y0 dx dx dy dy ey x y0 dx dx dy dy 1 x e y y, again differentiate w.r.t.x dx dx 0,1 e d2 y dy y dy d2 y dy dy .e . x. 0 dx 2 dx dx dx 2 dx dx 2 d2 y dy y dy x ey 0 .e 2 dx 2 dx dx
ey .
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JEE-MAIN-2019 (12th Apr-First Shift)-PCM-36
d2 y 1 1 e 2 0 dx 2 e1 e 2 dy 1 2 2 dx e e
77.
12 3 sin1 sin1 13 5
sin1 x 1 y 2 y 1 x 2
33 56 56 sin1 cos1 sin1 65 65 2 65 78.
3x 2 4y 2 12 x2 y2 1 4 3 x 2 cos , y 3 sin
Equation of normal is
a2 x b2 y a 2 b2 x1 y1
2x sin 3 cos y sin cos 2 Slope tan 2 tan 3 3 Equation of tangent is it passes through (4, 4) 3x cos 2 3 sin y 6 12 cos 8 3 sin 6 2
1 3 cos ,sin 120o 2 2
Hence point P is 2cos120o , 3 sin120o
79.
2 P 1, , Q 4, 4 2 5 5 PQ 2 ab ab
ˆi
ˆj
3
1
kˆ
2 1 2 2 3 2
2
2 8iˆ 8ˆj 4kˆ
2iˆ 2ˆj kˆ Required vector 12 3
4 2iˆ 2ˆj kˆ
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JEE-MAIN-2019 (12th Apr-First Shift)-PCM-37
80.
2y 2 sin x sin x 2 2 sin2 x 1
2y cos 2 cos 2x 2 1 cos 2x 2 cos 2 1
2y 2 sin2 y sin2
81.
1 2
1 0 2
f x x, g x tan x,h x
1 x2 1 x2
fog x tan x
hofog x h
tan x
1 tan x 1 tan x
tan x 4 x tan x 4 tan tan tan 12 3 4 3 12 11 tan tan 12 12 82.
Let length of common chord 2x 25 x 2 144 x 2 13 12 5 After solving x 13 120 2x 13 / 2
83.
0
/2
0
/ 2
12
5 x x 5
12
cot x dx cot x cos ec x
cos x cos x 1
1
1 2 sec 0
2
x 1 2 x 2cos2 2
2cos2
x dx 2
x2 x tan 2 0
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JEE-MAIN-2019 (12th Apr-First Shift)-PCM-38
1 2 2 mn 1 84.
m
1 , n 2 2
2 2a 3d 16 3 2a 5d 48 2a 3d 8 2a 5d 16 d 12 S10 5 44 9 12 320
85.
B 5 5 2 2
2 2 2 25 1 A 0 2 12 0 Sum = 1 86.
x 2 y 1 z 1 3 2 1 x 3 2, y 2 1, z 1 Intersection with plane 2x 3y z 13 0 2 3 2 3 2 1 1 13 0 13 13 0 1 P 1, 3, 2 Intersection with plane 3x y 4z 16 3 3 2 2 1 4 1 16 1 Q 5, 1,0 PQ 62 42 22 56 2 14
1 1 87.
Volume of paralleopiped 0 1
0 1
f 3 1 Its graphs as follows
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JEE-MAIN-2019 (12th Apr-First Shift)-PCM-39
2.0 1.5 1.0 0.5
1
1 3
3 where 1.32 For minimum value of volume of paralelopiped and corresponding value of ; the minimum value is zero, cubic always has at least one real root. Hence answer to the question must be root of cubic 3 1 0 . None of the options satisfies the cubic. Hence Question must be Bonus.
88.
A A ',B B ' 2 3 A B ……….(1) 5 1 2 5 A ' B ' 3 1 2 5 A B ……….(2) 3 1 After adding equation (1) and (2) 2 4 0 1 A , B 4 1 1 0 4 2 AB 1 4
89.
P ~ q r ~ p ~ q r
~ p F p T ~ q F q T ~ r F r F
90.
1 1 1 1 66 1 67 1 99 3 3 100 .... 3 100 3 100 ... 3 100 133 1 67
2 33
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FIITJEE Solutions to JEE(Main)-2019 Test Date: 12th April 2019 (Second Shift)
PHYSICS, CHEMISTRY & MATHEMATICS Paper - 1 Time Allotted: 3 Hours
Maximum Marks: 360
Please read the instructions carefully. You are allotted 5 minutes specifically for this purpose.
Important Instructions: 1.
The test is of 3 hours duration.
2.
This Test Paper consists of 90 questions. The maximum marks are 360.
3.
There are three parts in the question paper A, B, C consisting of Physics, Chemistry and Mathematics having 30 questions in each part of equal weightage. Each question is allotted 4 (four) marks for correct response.
4.
Out of the four options given for each question, only one option is the correct answer.
5.
For each incorrect response 1 mark i.e. ¼ (one-fourth) marks of the total marks allotted to the question will be deducted from the total score. No deduction from the total score, however, will be made if no response is indicated for an item in the Answer Box.
6.
Candidates will be awarded marks as stated above in instruction No.3 for correct response of each question. One mark will be deducted for indicating incorrect response of each question. No deduction from the total score will be made if no response is indicated for an item in the answer box.
7.
There is only one correct response for each question. Marked up more than one response in any question will be treated as wrong response and marked up for wrong response will be deducted accordingly as per instruction 6 above.
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JEE-MAIN-2019 (12th Apr-Second Shift)-PCM-2
PART – A (PHYSICS) 1.
Figure shown a DC voltage regulator circuit, with a Zener diode of breakdown voltage = 6V. If the unregulated input voltage varies between 10 V to 16 V, then what is the maximum Zener current?
(A) 2.5 mA (C) 7.5 mA 2.
Three particles of masses 50 g, 100 g and 150 g are placed at the vertices of an equilateral triangle of side 1 m (as shown in the figure). The (x, y) coordinates of the centre of mass will be :
(A) (C)
3.
(B) 3.5 mA (D) 1.5 mA
3 7 m, m 7 12 3 5 m, m 4 12
7 (B) m, 12 7 (D) m, 12
3 m 8 3 m 4
The ratio of the weights of a body on the Earth's surface to that on the surface of a 1 planet is 9 : 4. The mass of the planet is th of that of the Earth. If 'R' is the radius of the 9 Earth, what is the radius of the planet ? (Take the planets to have the same mass density) R R (A) (B) 3 4 R R (C) (D) 9 2
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JEE-MAIN-2019 (12th Apr-Second Shift)-PCM-3 4.
A block of mass 5 kg is (i) pushed in case (A) and (ii) pulled in case (B), by a force F = 20 N, making an angle of 30º with the horizontal, as shown in the figures. The coefficient of friction between the block and floor is m = 0.2. The difference between the accelerations of the block, in case (B) and case (A) will be : (g = 10 ms–2)
(A) 0.4 ms2 (C) 0 ms2
(B) 3.2 ms2 (D) 0.8 ms2
5.
In the given circuit, the charge on 4 F capacitor will be : (A) 13.4 C (B) 24 C (C) 9.6 C (D) 5.4 C
6.
A diatomic gas with rigid molecules does 10 J of work when expanded at constant pressure. What would be the heat energy absorbed by the gas, in this process? (A) 40 J (B) 30 J (C) 35 J (D) 25 J
7.
A Carnot engine has an efficiency of 1/6. When the temperature of the sink is reduced by 62ºC, its efficiency is doubled. The temperatures of the source and the sink are, respectively (A) 62ºC, 124ºC (B) 99ºC, 37ºC (C) 37ºC, 99ºC (D) 124ºC, 62ºC
8.
Consider an electron in a hydrogen atom, revolving in its second excited state (having radius 4.65 Å). The de-Broglie wavelength of this electron is : (A) 12.9 Å (B) 9.7 Å (C) 6.6 Å (D) 3.5 Å
9.
A small speaker delivers 2 W of audio output. At what distance from the speaker will one detect 120 dB intensity sound? [Given reference intensity of sound as 10–12W/m2] (A) 30 cm (B) 10 cm (C) 40 cm (D) 20 cm
10.
A system of three polarizers P1, P2, P3 is set up such that the pass axis of P3 is crossed with respect to that of P1. The pass axis of P2 is inclined at 60º to the pass axis of P3. When a beam of unpolarized light of intensity I0 is incident on P1, the intensity of light transmitted by the three polarizers is I. The ratio (I0/I) equals (nearly) : (A) 10.67 (B) 1.80 (C) 5.33 (D) 16.00
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JEE-MAIN-2019 (12th Apr-Second Shift)-PCM-4 11.
Let a total charge 2Q be distributed in a sphere of radius R, with the charge density given by r(r) = kr, where r is the distance from the centre. Two charges A and B, of –Q each, are placed on diametrically opposite points, at equal distance, a, from the centre. If A and B do not experience any force, then : 3R 1 (A) a 1 (B) a 2 4 R 2 4 (C) a 8
12.
1
4
(D) a R / 3
R
An electron, moving along the x-axis with an initial energy of 100 eV, enters a region of magnetic field B 1.5 10 3 T kˆ at S (See figure). The field extends between x = 0 and
x = 2 cm. The electron is detected at the point Q on a screen placed 8 cm away from the point S. The distance d between P and Q (on the screen) is : (electron's charge = 1.6 × 10–19 C, mass of electron = 9.1 × 10–31 kg)
(A) 12.87 cm (C) 1.22 cm
(B) 2.25 cm (D) 11.65 cm
13.
Two particles are projected from the same point with the same speed u such that they have the same range R, but different maximum heights, h1 and h2. Which of the following is correct? (A) R2 = 4 h1h2 (B) R2 = 2 h1h2 2 (C) R = 16 h1h2 (D) R2 = h1h2
14.
A particle is moving with speed v x along positive x-axis. Calculate the speed of the particle at time t = t(assume that the particle is at origin at t = 0). b2 (A) b 2 (B) 2 b2 b2 (C) (D) 4 2
15.
One kg of water, at 20ºC, is heated in an electric kettle whose heating element has a mean (temperature averaged) resistance of 20 . The rms voltage in the mains is 200 V. Ignoring heat loss from the kettle, time taken for water to evaporate fully, is close to : [Specific heat of water = 4200 J/kg ºC), Latent heat of water = 2260 kJ/kg] (A) 3 minutes (B) 10 minutes (C) 22 minutes (D) 16 minutes
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JEE-MAIN-2019 (12th Apr-Second Shift)-PCM-5 16.
Half lives of two radioactive nuclei A and B are 10 minutes and 20 minutes, respectively. If, initially a sample has equal number of nuclei, then after 60 minutes, the ratio of decayed numbers of nuclei A and B will be : (A) 9 : 8 (B) 1 : 8 (C) 8 : 1 (D) 3 : 8
17.
In an amplitude modulator circuit, the carrier wave is given by, C(t) = 4 sin (20000 t) while modulating signal is given by, m(t) = 2 sin (2000 t). The values of modulation index and lower side band frequency are: (A) 0.5 and 9 kHz (B) 0.3 and 9 kHz (C) 0.5 and 10 kHz (D) 0.4 and 10 kHz
18.
A uniform cylindrical rod of length L and radius r, is made from a material whose Young's modulus of Elasticity equals Y. When this rod is heated by temperature T and simultaneously subjected to a net longitudinal compressional force F, its length remains unchanged. The coefficient of volume expansion, of the material of the rod, is (nearly) equals to : (A) 9F/(r2YT) (B) F/(3r2YT) (C) 3F/(r2YT) (D) 6F/(r2YT)
19.
Consider the LR circuit shown in the figure. If the switch S is closed at t = 0 then the L amount of charge that passes through the battery between t = 0 and t is : R
EL 7.3R2 7.3 EL (C) R2
EL 2.7R 2 2.7 EL (D) R2
(A)
(B)
20.
Two sources of sound S1 and S2 produce sound waves of same frequency 660 Hz. A listener is moving from source S1 towards S2 with a constant speed u m/s and he hears 10 beats/s. The velocity of sound is 330 m/s. Then, u equals: (A) 15.0 m/s (B) 10.0 m/s (C) 5.5 m/s (D) 2.5 m/s
21.
A plane electromagnetic wave having a frequency n = 23.9 GHz propagates along the positive z-direction in free space. The peak value of the electric field is 60 V/m. Which among the following is the acceptable magnetic field component in the electromagnetic wave? (A) B 2 10 7 sin 1.5 10 2 x 0.5 1011 t ˆj (B) B 60 sin 0.5 103 x 1.5 1011 t kˆ (C) B 2 10 7 sin 0.5 10 2 z 1.5 1011 t ˆi (D) B 2 107 sin 0.5 103 z 1.5 1011 t ˆi
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JEE-MAIN-2019 (12th Apr-Second Shift)-PCM-6 22.
Find the magnetic field at point P due to a straight line segment AB of length 6 cm carrying a current of 5 A. (See figure) (0 = 4p × 10–7 N-A–2)
(A) 2.0 × 105 T (C) 2.5 × 105 T
(B) 3.0 × 105 T (D) 1.5 × 105 T
23.
The electron in a hydrogen atom first jumps from the third excited state to the second excited state and subsequently to the first excited state. The ratio of the respective wavelengths, 1/2, of the photons emitted in this process is : (A) 20/7 (B) 7/5 (C) 9/7 (D) 27/5
24.
A smooth wire of length 2r is bent into a circle and kept in a vertical plane. A bead can slide smoothly on the wire. When the circle is rotating with angular speed w about the vertical diameter AB, as shown in figure, the bead is at rest with respect to the circular ring at position P as shown. Then the value of 2 is equal to:
3g 2r
(B) g 3 / r
(C) 2g / r
(D) 2g / r 3
(A)
25.
A solid sphere, of radius R acquires a terminal velocity 1 when falling (due to gravity) through a viscous fluid having a coefficient of viscosity . The sphere is broken into 27 identical solid spheres. If each of these spheres acquires a terminal velocity, 2, when falling through the same fluid, the ratio (1/2) equals : (A) 27 (B) 1/27 (C) 9 (D) 1/9
26.
The number density of molecules of a gas depends on their distance r from the origin as, 4 n r n0 e r . Then the total number of molecules is proportional to : (A) n03/4 (C) n01/4
(B) n03 (D) n0 1/2
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JEE-MAIN-2019 (12th Apr-Second Shift)-PCM-7
27.
A transparent cube of side d, made of a material of refractive index 2, is immersed in a liquid of refractive index 1(1 < 2). A ray is incident on the face AB at an angle q(shown in the figure). Total internal reflection takes place at point E on the face BC.
The q must satisfy : (A) sin1
1 2
(B) sin1
22 1 12
(C) sin 1
1 2
(D) sin1
22 1 12
28.
A tuning fork of frequency 480 Hz is used in an experiment for measuring speed of sound () in air by resonance tube method. Resonance is observed to occur at two successive lengths of the air column, 1 30 cm and 2 70 cm. Then n is equal to : (A) 332 ms1 (B) 338 ms1 1 (C) 384 ms (D) 379 ms1
29.
A spring whose unstretched length is has a force constant k. The spring is cut into two pieces of unstretched lengths 1 and 2 where, 1 = n 2 and n is an integer. The ratio k1/k2 of the corresponding force constants, k1 and k2 will be: 1 (A) n (B) 2 n 1 (C) n2 (D) n
30.
A moving coil galvanometer, having a resistance G, produces full scale deflection when a current Ig flows through it. This galvanometer can be converted into (i) an ammeter of range 0 to I0 (I0 > Ig) by connecting a shunt resistance RA to it and (ii) into a voltmeter of range 0 to V(V = GI0) by connecting a series resistance RV to it. Then, Ig R (A) R AR V G2 and A R V I0 Ig Ig R (B) R AR V G and A R V I0 Ig
2
2
Ig I0 Ig R (C) R AR V G and A I I R V Ig 0 g
2
2
Ig I0 Ig R (D) R A R V G2 and A I R I I V g 0 g
2
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JEE-MAIN-2019 (12th Apr-Second Shift)-PCM-8
PART –B (CHEMISTRY) 31.
What will be the major product when m-cresol is reacted with propargyl bromide (HC C – CH2Br) in presence of K2CO3 in acetone
(A)
(B)
(C)
(D)
32.
Which one of the following is likely to give a precipitate with AgNO3 solution? (A) CH2 = CH – Cl (B) CHCl3 (C) (CH3)3CCl (D) CCl4
33.
The INCORRECT match in the following is: (A) G° < 0, K > 1 (B) G° < 0, K < 1 (C) G° = 0, K = 1 (D) G° > 0, K < 1
34.
Benzene diazonium chloride on reaction with aniline in the presence of dilute hydrochloric acid gives : (A)
(B)
(C)
(D)
35.
The INCORRECT statement is: (A) LiNO3 decomposes on heating to give LiNO2 and O2. (B) Lithium is least reactive with water among the alkali metals. (C) LiCl crystallises from aqueous solution as LiCl.2H2O. (D) Lithium is the strongest reducing agent among the alkali metals.
36.
The C–C bond length is maximum in (A) graphite (C) diamond
(B) C70 (D) C60
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JEE-MAIN-2019 (12th Apr-Second Shift)-PCM-9 37.
Consider the following reactions :
‘A’ is: (A) CH2 = CH2 (C) CH CH 38.
(B) CH3 – C CH (D) CH3 – C C – CH3
In which one of the following equilibria, Kp Kc? (A) 2NO g (B) 2C s O2 g N2 g O2 g 2CO g
(C) NO2 g SO2 g NO g SO3 g (D) 2HI g H2 g I2 g 39.
The coordination numbers of Co and Al in [Co(Cl)(en)2]Cl and K3[Al(C2O4)3], respectively, are (en = ethane-1,2-diamine) (A) 6 and 6 (B) 5 and 3 (C) 3 and 3 (D) 5 and 6
40.
Which of the given statements is INCORRECT about glycogen? (A) It is present in some yeast and fungi (B) It is present in animal cells (C) Only -linkages are present in the molecule (D) It is a straight chain polymer similar to amylase
41.
The primary pollutant that leads to photochemical smog is: (A) acrolein (B) nitrogen oxides (C) ozone (D) sulphur dioxide
42.
The ratio of number of atoms present in a simple cubic, body centered cubic and face centered cubic structure are, respectively : (A) 1 : 2 : 4 (B) 4 : 2 : 3 (C) 4 : 2 : 1 (D) 8 : 1 : 6
43.
Thermal decomposition of a Mn compound (X) at 513 K results in compound Y, MnO2 and a gaseous product. MnO2 reacts with NaCl and concentrated H2SO4 to give a pungent gas Z. X, Y and Z, respectively. (A) K2MnO4, KMnO4 and SO2 (B) K3MnO4, K2MnO4 and Cl2 (C) K2MnO4, KMnO4 and Cl2 (D) KMnO4, K2MnO4 and Cl2
44.
An ' Assertion' and a 'Reason' are given below. Choose the correct answer from the following options. Assertion (A): Vinyl halides do not undergo nucleophilic substitution easily. Reason (R): Even though the intermediate carbocation is stabilized by loosely held p-electrons, the cleavage is difficult because of strong bonding. (A) Both (A) and (R) are correct statements but (R) is not the correct explanation of (A) (B) Both (A) and (R) are wrong statements (C) Both (A) and (R) are correct statements and (R) is the correct explanation of (A) (D) (A) is a correct statement but (R) is a wrong statement. FIITJEE Ltd., FIITJEE House, 29-A, Kalu Sarai, Sarvapriya Vihar, New Delhi -110016, Ph 46106000, 26569493, Fax 26513942 website: www.fiitjee.com.
JEE-MAIN-2019 (12th Apr-Second Shift)-PCM-10 45.
The molar solubility of Cd(OH)2 is 1.84 × 10–5 M in water. The expected solubility of Cd(OH)2 in a buffer solution of pH = 12 is : (A) 6.23 × 1011 M (B) 1.84 × 109 M 2.49 (C) 10 9 M (D) 2.49 × 1010 M 1.84
46.
The compound used in the treatment of lead poisoning is: (A) EDTA (B) Cis-platin (C) D-penicillamine (D) desferrioxime B
47.
The pair that has similar atomic radii is: (A) Ti and Hf (C) Sc and Ni
(B) Mn and Re (D) Mo and W
48.
25 g of an unknown hydrocarbon upon burning produces 88 g of CO2 and 9 g of H2O. This unknown hydrocarbon contains. (A) 24g of carbon and 1 g of hydrogen (B) 22g of carbon and 3 g of hydrogen (C) 18g of carbon and 7 g of hydrogen (D) 20g of carbon and 5 g of hydrogen
49.
The decreasing order of electrical conductivity of the following aqueous solutions is: 0.1 M Formic acid (a) 0.1 M Acetic acid (b) 0.1 M Benzoic acid (c) (A) a > c > b (B) c > a > b (C) c > b > a (D) a > b > c
50.
Among the following, the energy of 2s orbital is lowest in: (A) K (B) Na (C) H (D) Li
51.
In the following skew conformation of ethane, H'–C–C–H" dihedral angle is :
(A) 58° (C) 149° 52.
(B) 120° (D) 151°
The correct statement is: (A) leaching of bauxite using concentrated NaOH solution gives sodium aluminate and sodium silicate (B) the Hall-Heroult process is used for the production of aluminium and iron (C) the blistered appearance of copper during the metallurgical process is due to the evolution of CO2 (D) pig iron is obtained from cast iron
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JEE-MAIN-2019 (12th Apr-Second Shift)-PCM-11 53.
NO2 required for a reaction is produced by the decomposition of N2O5 in CCl4 as per the equation 2N2O5 (g) 4NO 2 g O 2 g The initial concentration of N2O5 is 3.00 mol L–1 and it is 2.75 mol L–1 after 30 minutes. The rate of formation of NO2 is : (A) 1.667 × 102 mol L1 min1 (B) 4.167 × 103 mol L1 min1 (C) 8.333 × 103 mol L1 min1 (D) 2.083 × 103 mol L1 min1
54.
The correct name of the following polymer is:
(A) Polyisoprene (C) Polyisobutane 55.
56.
(B) Polytert-butylene (D) Polyisobutylene
Heating of 2-chloro-1-phenylbutane with EtOK/EtOH gives X as the major product. Reaction of X with Hg(OAc)2/H2O followed by NaBH4 gives Y as the major product. Y is: (A)
(B)
(C)
(D)
The IUPAC name of the following compound is :
(A) 3, 5-dimethyl-4-propylhept-1-en-6-yne (B) 3-methyl-4-(1-methylprop-2-ynyl)-1-heptene (C) 3-methyl-4-(3-methylprop-1-enyl)-1-heptyne (D) 3, 5-dimethyl-4-propylhept-6-en-1-yne 57.
Among the following, the INCORRECT statement about colloids is : (A) The range of diameters of colloidal particles is between 1 and 1000 nm (B) The osmotic pressure of a colloidal solution is of higher order than the true solution at the same concentration (C) They can scatter light (D) They are larger than small molecules and have high molar mass
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JEE-MAIN-2019 (12th Apr-Second Shift)-PCM-12 58.
In comparison to boron, berylium has: (A) greater nuclear charge and greater first ionisation enthalpy (B) lesser nuclear charge and lesser first ionisation enthalpy (C) greater nuclear charge and lesser first ionisation enthalpy (D) lesser nuclear charge and greater first ionisation enthalpy
59.
A solution is prepared by dissolving 0.6 g of urea (molar mass = 60 g mol–1) and 1.8 g of glucose (molar mass = 180 g mol–1) in 100 mL of water at 27ºC. The osmotic pressure of the solution is : (R = 0.08206 L atm K–1 mol–1) (A) 8.2 atm (B) 1.64 atm (C) 4.92 atm (D) 2.46 atm
60.
The temporary hardness of a water sample is due to compound X. Boiling this sample converts X to compound Y. X and Y, respectively, are : (A) Mg(HCO3)2 and MgCO3 (B) Ca(HCO3)2 and CaO (C) Mg(HCO3)2 and Mg(OH)2 (D) Ca(HCO3)2 and Ca(OH)2
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JEE-MAIN-2019 (12th Apr-Second Shift)-PCM-13
PART–C (MATHEMATICS) 61.
62.
The general solution of the differential equation (y2 – x3) dx – xydy = 0 (x 0) is : (where c is a constant of integration) (A) y2 + 2x3 + cx2 = 0 (B) y2 – 2x3 + cx2 = 0 2 2 3 (C) y + 2x + cx = 0 (D) y2 – 2x2 + cx3 = 0 ˆ b 2iˆ ˆj kˆ and c ˆi 2ˆj 3kˆ . Then Let R and the three vectors a ˆi ˆj 3k, the set S :a,b and c are coplanar
(A) Contains exactly two numbers only one of which is positive (B) is empty (C) Contains exactly two positive numbers (D) is singleton 63.
Let (0, /2) be fixed. If the integral
tan x tan
tan x tan dx
A(x) cos2 + B(x) sin2 + C, where C is a constant of integration, then the functions A(x) and B(x) are respectively: (A) x and loge sin x (B) x and loge cos x (C) x and loge sin x
(D) x and loge sin x
64.
Let S be the set of all R such that the equation, cos2 x + sin x = 2 – 7 has a solution. Then S is equal to : (A) [3, 7] (B) R (C) [2, 6] (D) [1, 4]
65.
Let f(x) 5 x 2 and g(x) x 1 , x R . If f(x) attains maximum value at and g(x) attains minimum value at , then lim
x 1 x 2 5x 6
x
3 2 1 (C) 2
(A)
66.
Let z C with Im(z) = 10 and it satisfies (A) n = 40 and Re(z) = 10 (C) n = 40 and Re(z) = – 10
x 2 6x 8 3 (B) 2 1 (D) 2
is equal to :
2z n 2i 1 for some natural number n. Then : 2z n (B) n = 20 and Re(z) = 10 (D) n = 20 and Re(z) = – 10 6
67.
1 x8 2 3 The term independent of x in the expansion of 2x 2 is equal to: x 60 81 (A) 36 (B) 36 (C) 108 (D) 72
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JEE-MAIN-2019 (12th Apr-Second Shift)-PCM-14
68.
If the area (in sq. units) bounded by the parabola y2 =4x and the line y = x, > 0, is then is equal to : (A) 48
(B) 4 3
(C) 2 6
(D) 24
1 , 9
69.
If [x] denotes the greatest integer x, then the system of linear equations [sin ] x + [– cos ] y = 0 [cot ] x + y = 0 2 7 (A) have infinitely many solutions if , and has a unique solution if , 2 3 6 2 7 (B) have infinitely many solutions if , , 2 3 6 2 7 (C) has a unique solution if , and have infinitely many solutions if , 2 3 6 2 7 (D) has a unique solution if , , 2 3 6
70.
For an initial screening of an admission test, a candidate is given fifty problems to solve. 4 If the probability that the candidate can solve any problem is , then the probability that 5 he is unable to solve less than two problems is : 164 1 (A) 25 5
(C)
54 4 5 5
48
201 1 (B) 5 5
49
(D)
49
316 4 25 5
48
71.
If a1, a2, a3, ……. are in A.P. such that a1 + a7 + a16 = 40, then the sum of the first 15 terms of this A.P. is: (A) 200 (B) 280 (C) 150 (D) 120
72.
An ellipse, with foci at (0, 2) and (0, –2) and minor axis of length 4, passes through which of the following points? (A) 2, 2 (B) 2,2 2
(C) 1,2 2 73.
(D)
2,2
The length of the perpendicular drawn from the point (2, 1, 4) to the plane containing the lines r ˆi ˆj ˆi 2jˆ kˆ and r ˆi ˆj ˆi ˆj 2kˆ is:
(A) (C)
1 3 1 3
(B)
3
(D) 3
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JEE-MAIN-2019 (12th Apr-Second Shift)-PCM-15
74.
Let A, B and C be sets such that A B C. Then which of the following statements is not true? (A) If (A – C) B, then A B (B) If (A – B) C, then A C (C) (C A) (C B) = C (D) B C
75.
If , and are three consecutive terms of a non-constant G.P. such that the equations x2 + 2x + = 0 and x2 + x – 1 = 0 have a common root, then ( + ) is equal to : (A) (B) 0 (C) (D)
76.
The tangents to the curve y = (x – 2)2 –1 at its points of intersection with the line x – y = 3, intersect at the point : 5 5 (A) ,1 (B) , 1 3 2 5 5 (C) ,1 (D) , 1 2 2
77.
The angle of elevation of the top of vertical tower standing on a horizontal plane is observed to be 45° from a point A on the plane. Let B be the point 30 m vertically above the point A. If the angle of elevation of the top of the tower from B be 30°, then the distance (in m) of the foot of the tower from the point A is: (A) 15 1 3 (B) 15 3 3
(C) 15 3 3
(D) 15 5 3
78.
A triangle has a vertex at (1, 2) and the mid points of the two sides through it are (–1, 1) and (2,3). Then the centroid of this triangle is : 7 1 (A) 1, (B) ,1 3 3 1 1 5 (C) ,2 (D) , 3 3 3
79.
A person throws two fair dice. He wins Rs. 15 for throwing a doublet (same numbers on the two dice), wins Rs.12 when the throw results in the sum of 9, and loses Rs. 6 for any other outcome on the throw. Then the expected gain/loss (in Rs.) of the person is : 1 (A) loss (B) 2 gain 4 1 1 (C) gain (D) loss 2 2
80.
If
20
C1 22
20
C3 32
(A, ) is equal to: (A) (420, 18) (C) (420, 19)
20
C3 22 ....... 202
20
C20 A 2 , then the ordered pair
(B) (380, 18) (D) (380, 19)
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JEE-MAIN-2019 (12th Apr-Second Shift)-PCM-16
81.
x sin x cos x The derivative of tan 1 , with respect to , where x 0, is: 2 sin x cos x 2 1 (A) 2 (B) 2 2 (C) 1 (D) 3
82.
A circle touching the x-axis at (3, 0) and making an intercept of length 8 on the y-axis passes through the point : (A) (3, 5) (B) (1, 5) (C) (3, 10) (D) (2, 3)
83.
The equation of a common tangent to the curves, y2 = 16x and xy = –4 is : (A) x – 2y + 16 = 0 (B) 2x – y + 2 = 0 (C) x + y + 4 = 0 (D) x – y + 4 = 0
84.
A plane which bisects the angle between the two given planes 2x – y + 2z – 4 = 0 and x + 2y + 2z – 2 = 0, passes through the point: (A) (1, 4, 1) (B) (2, 4, 1) (C) (2, 4, 1) (D) (1, 4, 1)
85.
A value of such that
1
dx
x x 1 log
e
(A) (C) 86.
1 2
(B) 2
1 2
(D) 2
A group of students comprises of 5 boys and n girls. If the number of ways, in which a team of 3 students can randomly be selected from this group such that there is at least one boy and at least one girl in each team, is 1750, then n is equal to: (A) 24 (B) 28 (C) 27 (D) 25 1 cos2
87.
A value of (0, /3), for which
2
cos 2
cos
18 7 (C) 36
sin2
4 cos 6
2
4 cos 6
1 sin 2
sin
0, is:
1 4 cos 6
9 7 (D) 24
(A)
88.
9 8 is:
(B)
A straight line L at a distance of 4 units from the origin makes positive intercepts on the coordinate axes and the perpendicular from the origin to this line makes an angle of 60° with the line x + y = 0. Then an equation of the line L is : (A) 3 1 x 3 1 y 8 2 (B) 3x y 8
(C) x 3y 8
(D)
3 1 x
3 1 y 8 2
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JEE-MAIN-2019 (12th Apr-Second Shift)-PCM-17
89.
lim x 0
(A) 2 (C) 3 90.
x 2 sin x x 2 2 sin x 1 sin2 x x 1
is: (B) 6 (D) 1
The Boolean expression (p (q)) is equivalent to: (A) (p) q (B) p q (C) p q (D) q p
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JEE-MAIN-2019 (12th Apr-Second Shift)-PCM-18
JEE (Mains) – 2019 ANSWERS PART A – PHYSICS 1. 5. 9. 13. 17. 21. 25. 29.
B B C C A C C D
2. 6. 10. 14. 18. 22. 26. 30.
D C A B C D A B
3. 7. 11. 15. 19. 23. 27.
D B C C B A D
4. 8. 12. 16. 20. 24. 28.
D B C A D D C
34. 38. 42. 46. 50. 54. 58.
B B A A A D D
64. 68. 72. 76. 80. 84. 88.
C D D D A B D
PART B – CHEMISTRY 31. 35. 39. 43. 47. 51. 55. 59.
C A D D D C A C
32. 36. 40. 44. 48. 52. 56. 60.
C C D D A A A C
33. 37. 41. 45. 49. 53. 57.
B B B D A A B
PART C – MATHEMATICS 61. 65. 69. 73. 77. 81. 85. 89.
B C A B C A B A
62. 66. 70. 74. 78. 82. 86. 90.
B C C A C C D C
63. 67. 71. 75. 79. 83. 87.
C B A D D D B
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JEE-MAIN-2019 (12th Apr-Second Shift)-PCM-19
HINTS AND SOLUTIONS PART A – PHYSICS 1.
Maximum current will flow from zener if input voltage is maximum. When zener diode works in breakdown state, voltage across the zener will remain same. Vacross 4k 6V
6 6 A mA 4000 4 Since input voltage = 16 V Potential difference across 2K = 10V 10 Current through 2k = 5mA 2000 Current through zener diode = (Is – IL) = 3.5 mA Current through 4K =
2.
The co-ordinates of the centre of mass 1 3 ˆ 0 150 i j 100 ˆi 2 2 rcm 300 7 ˆ 3ˆ rcm i j 12 4 7 3 Co-ordinate , m 12 4
3.
Since mass of the object remains same Weight of object will be proportional to ‘g’ (acceleration due to gravity) Given: Wearth 9 gearth Wplanet 4 gplanet Also, gsurface
GM (M is mass planet, G is universal gravitational constant, R is radius of R2
planet)
2 2 2 Rplanet 9 GMearth Rplanet Mearth Rplanet 9 2 2 2 4 GMplanet Rearth Mplanet Rearth R earth
Rplanet
Rearth R 2 2
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JEE-MAIN-2019 (12th Apr-Second Shift)-PCM-20 4.
N1 = 60 10 3 0.2 60 a1 5 a1 – a2 = 0.8
N2 = 40 10 3 0.2 40 a2 5
5.
So total charge flow = q = 5.4 F × 10V = 54 F The charge will be distributed in the ratio of capacitance q1 2.4 4 q2 3 5 9X = 54 C X = 6 C Charge on 4 F capacitor will be = 4X = 4 × 6 C = 24 C 6.
7 R 2 Since gas undergoes isobaric process For a diatomic gas, Cp =
7 7 Q = n RT (nRT) = 35 J 2 2 7.
Efficiency of Carnot engine = 1
Tsink Tsource
Given,
T Tsink 1 5 1 sink 6 Tsource Tsource 6
…(i)
Also,
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JEE-MAIN-2019 (12th Apr-Second Shift)-PCM-21
T 62 2 62 1 1 sink 6 Tsource Tsource 6
…(ii)
Tsource = 372 K = 99°C
5 372 = 310 K = 37°C 6 (Note: Temperature of source is more than temperature of sink) Also, Tsink
8.
2rn = nn 2(4.65 10 10 ) 3 3 3 = 9.7 Å
9.
Loudness of sound is given by I dB 10log (I is intensity of sound, I0 is reference intensity of sound) I0 I 120 10 log I0 2 I = 1 W/m P 2 Also, I 2 4r 4r 2 2 I r m = 0.399 m 4 2 40 cm
10.
Since unpolarised light falls on P1
I0 . 2 Pass axis of P2 will be at an angle of 30° with P1 Intensity of light transmitted from I 3I P2 0 cos2 30o 0 2 8 Pass axis of P3 is at an angle of 60° with P2 Intensity of light transmitted from 3I 3I P3 0 cos2 60o 0 8 32 I0 32 10.67 I 3 Intensity of light transmitted from P1
11.
E 4 a
E
2
0
kr 4r 2 dr e0
k 4 a 4 4 4 0
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JEE-MAIN-2019 (12th Apr-Second Shift)-PCM-22 R
2Q kr 4r 2 dr 0
2Q k R 4 1 QQ 40 (2a)2 R = a81/4 QE
12.
mv qB
R
2m (K.E.) qB
=
2 9.1 10 31 (100 1.6 10 19 ) 1.6 1019 1.5 10 3 R = 2.248 cm 2 QU sin ; tan 2.248 TU 2 QU 1.026 6 QU = 11.69 R=
PU = R(1 – cos ) = 1.22 d = QU + PU 13.
For same range angle of projection will be will be & 90 – u2 2 sin cos R g h1
u2 sin2 g
h2
u2 sin2 (90 ) g
R2 16 h1h2 14.
v b x dv b dx bv ; a dt 2 x dt 2 x
b b x a
2 x
;
dv b2 a ; dt 2
v
b2 2
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JEE-MAIN-2019 (12th Apr-Second Shift)-PCM-23 15.
Q=P×t Q = mcT + mL V2 P = rms R 4200 × 80 + 2260 × 103 =
(200)2 t 20
t = 1298 sec t 22 min 16.
NOA NOA 6 2 2t/t1/ 2 Number of nuclei decayed N 63NOA = NOA OA 6 2 64 NOB NOB NB NOBet t/t1/ 2 3 2 2 Number of nuclei decayed N 7N = NOB OB OB 3 2 8 Since, NOA = NOB Ratio of decayed numbers of nuclei 63 NOA 8 9 A&B= 64 7NOB 8 NA NOA et
17.
Modulation index is given by A 2 m m 0.5 Ac 4 & (a) carrier wave frequency is given by = 2fc = 2 × 104 fc = 1 kHz lower side band frequency fc – fm 10 kHz – 1 kHz = 9 kHz
18.
Length of cylinder remains unchanged F F so A Compressive A Thermal F YT ( is linear coefficient of expansion) r 2 F YTr 2 The coefficient of volume expansion = 3 F 3 YTr 2
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JEE-MAIN-2019 (12th Apr-Second Shift)-PCM-24
19.
q Idt L /R
Rt L 1 e dt 0 EL 1 EL q 2 ; q R e 2.7R2
q
20.
E R
f = 660 Hz, v = 330 m/s
v u v u f1 f ; f2 f v v f f2 f1 [v u (v u)] v f 10 = f2 – f1 = [2u] v u = 2.5 m/s 21.
22.
23.
Magnetic field when electromagnetic wave propagates in +z direction. B = B0 sin (kz – t) where 60 B0 2 10 7 8 3 10 2 k 0.5 103 = 2f = 1.5 × 1011
0I 2 sin 4d d = 4 cm 3 sin = 5 B
1 1 1 1 1 1 R 2 2 R 2 2 ; 1 4 3 nf ni 1 1 1 7 1 R R 2 2 ; 1 2 3 9 16 2 5 = R 49 5 1 20 36 7 2 7 9 16
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JEE-MAIN-2019 (12th Apr-Second Shift)-PCM-25
24.
r N sin = m 2 2 N cos = mg r2 tan 2g r 2
25.
3r 2
r2 2g
;
…(i) …(ii)
2
2g 3r
We have
VT
2 r2 (0 )g 9
VT r2 Since mass of the sphere will be same 4 4 R3 27 r 3 3 3 r
26.
R 3
v1 R2 9 v2 r2
Given number density of molecules of gas as a function of r is n(r) = n0 er
4
Total number of molecule =
4
r 2 n(r)dV = n0 e 4r dr 0
0
Number of molecules is proportional to n0–3/4 27.
1 2 1 sin = 2 sin (90° – C) 2 2 1 12 2 sin 1 sin c =
sin1
12 22 12
For TIR sin1
u22 1 12
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JEE-MAIN-2019 (12th Apr-Second Shift)-PCM-26 28.
v = 2f(l2 – l1) v = 2 × 480 × (70 – 30) × 10–2 v = 960 × 40 × 10–2 v = 38400 × 10–2 m/s v = 384 m/s
29.
k1
30.
When galvanometer is used an ammeter shunt is used in parallel with galvanometer.
C 1 C k2 2 k1 C 2 1 2 k 2 1C n 2 n
IgG = (I0 – Ig)RA Ig RA G I0 Ig When galvanometer is used as a voltmeter, resistance is used in series with galvanometer.
Ig(G + Rv) = V = GI0 (given V = GI0) (I0 Ig )G RV Ig Ig R R AR V G & A R V I0 Ig
2
2
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JEE-MAIN-2019 (12th Apr-Second Shift)-PCM-27
PART B – CHEMISTRY 31.
O
CH3 Above anion act as nucleophile for SN2 attack on CH C – CH2 – Br and acetone acting as polar aprotic solvent. 32.
(CH3)3CCl will give Cl– and most stable carbocation. Hence (CH3)3CCl likely to give a precipitate with AgNO3 solution.
33.
G° = -2.303 RT log Keq
34.
According to NCERT C – N coupling take place, when diazonium ion is treated with aniline.
35.
1 2LiNO 3 Li2O 2NO 2 O2 2
36.
Since carbon in diamond is sp3 hybridized and its C – C bond order is 1. In graphite and fullerene there is both C – C and C = C in conjugation, hence there is partial double bond character between carbon atoms.
37.
Ag2O CH3 C C H CH3 C C Ag
(A) Hg2+/H+
White PPt.
O H3C
C
CH3
(B) NaBH4 OH ZnCl2 CH3 C CH3 Turbidity within 5 minutes Conc.HCl
(C) H
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JEE-MAIN-2019 (12th Apr-Second Shift)-PCM-28 38.
If ng = 0 KP = KC If ng 0, KP KC Hence (B) is correct answer.
39.
en and C2 O24 are a bidentate ligand. So coordination number of [Co(Cl)(en)2]Cl is 5 and K3[Al(C2O4)3] is 6
40.
Glycogen is a multibranched polysaccharide.
41.
NO2 and Hydrocarbons are primary precursors of photochemical smog.
42.
No. of atoms in simple cubic = 1, bcc = 2 & fcc = 4
43.
44.
Due to resonance there is partial double bond character between carbon and chlorine, hence it do not undergoes nucleophilic substitution reaction.
45.
Ksp of Cd(OH)2 = 4s3 = 4 × (1.84 × 10–5)3 If pH = 12 pOH = 2 [OH–] = 10–2 M Ksp = [Cd2+] [OH–]2 4 × (1.84 ×10–5)3 = [Cd2+] [OH–]2 3 4 1.84 10 15 2+ [Cd ] = 10 4 2+ Cd = 4 × 6.22 × 10–11 = 2.49 × 1010 M
46.
(A) EDTA (ethylene diamine tetra acetate) is used for lead poisoning (B) Cis platin is used as a anti cancer drug (C) D-penicillamine is used for copper poisoning (D) desferrioxime B is used for iron poisoning
47.
Due to lanthanoide contraction Mo & W have similar atomic radii.
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JEE-MAIN-2019 (12th Apr-Second Shift)-PCM-29 48.
49.
Stronger the acidic strength greater will be its electrical conductivity. Ka value of formic acid > benzoic acid > acetic acid.
50.
Greater the nuclear charge, stronger will be the attraction, hence lower will be energy of 2s
51.
Dihedral angle is then angle between bond pairs present on adjacent atoms.
52.
Since NaOH is a strong base hence it reacts with Al2O3 and SiO2 to form salts.
53.
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JEE-MAIN-2019 (12th Apr-Second Shift)-PCM-30 54.
Isobutylene on polymerization will form given polymer. OH
55. Ph
alc.KOH
Ph X
Cl
Hg OAc 2 ,H2O NaBH4
Ph
OMDM follow Markonikovs addition rule. 56.
According to IUPAC rules, select the largest chain including functional group, if alkene and alkyne are present at equivalent position then priority is given to alkene.
57.
Definitions and property of colloidal will explain & solve above question..
58.
Since boron has higher nuclear charge because it has greater atomic number and lower 1st I.E. then beryllium due to fully filled s-orbital.
59.
60.
18 6 CRT .0821 300 60 180 = 0.2 .082 300 = 4.926 atm.
Mg HCO3 2 aq Mg OH2 2CO2
PART C – MATHEMATICS 61.
y
2
x3 dx xy dy 0
y 2 x 3 xy
x0
dy 0 dx
dy y2 x3 dx dy 1 2 y y x 2 ……..(i) dx x Let y 2 dy d 2y dx dx Putting this value in equation (i) 1 d 1 x2 2 dx x d 2 2x2 (ii) dx x 2 dx 1 I.F. e x e 2 n x 2 x Sol. of equation (ii) 1 1 3 2x 2 dx C x 2 2x C x2 or, xy
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JEE-MAIN-2019 (12th Apr-Second Shift)-PCM-31
y 2 2x3 cx 2 y 2 2x 3 cx 2 0 Hence correct answer is option B. 62.
a.b.c 0 3 1 2 1 0 2 3
3 2 1 2 6 3 4 0 2
2
3 2 6 12 3 0 32 18 0 2 6 0 not possible for real S is empty set 63.
tan x tan
sin x
tan x tan dx sin x dx Let, x t sin t 2 dt cos 2 dt cot t sin 2 dt sin t t.cos 2 n sin t .sin 2 C
x cos 2 ln sin x .sin 2 C Hence the correct answer is option (C) 64.
Given, cos 2x 2 sin x 2 7
1 2sin2 x sin x 2 7 2sin2 x sin x 2 8 0 sin x
2 8 2 8
4 8 sin x 4 8 8 sin x , 4 4 sin x 2 (Not possible) For solution 2 8 1 1 4 4 2 8 4 4 2 12 2, 6
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JEE-MAIN-2019 (12th Apr-Second Shift)-PCM-32 65.
f x 5 x 2 f x attains maximum value when x 2 0 x 2 g x x 1 g x attains minimum value of x 1 Lim
x 1 x 2 5x 6
x
Lim
x 2 6x 8 x 1 x 2 x 3
x 2 x 4 2 1 2 3 1 2 2 4 x 2
66.
Let z x 10i 2z n given 2i 1 2z n 2 x 10i n 2i 1 2 x 10i n
2x n 20i 2i 1 2x n 20i Comparing real and imaginary part 2x n 2 20 2x n and 20 2 2x n 20 2x n 40 2x n and 20 4x 2n 20 4x 40 and 4x 2n 40 x 10 and 40 2n 40 n 40 n 40 and Re z 10
67.
1 x8 2 3 2x 2 x 60 81
6
1 independent of x in 60 6 6 1 2 3 2 3 8 2x Term of x in 2x x2 8 x3
Term independent of x will be
6
Tr 1
3 in 2x 2 2 will be Tr 1 6Cr 2x 2 x
6 r
3 2 x
r
r
6Cr 26 r 1 3r x122r 2r Case I : For term independent of x, 12 4r 0 r 3 T4 6C3 23 33 6 20 23 33 Case II : For term of x 8 12 4r 8 4r 20 r 5
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JEE-MAIN-2019 (12th Apr-Second Shift)-PCM-33 T6 6C5 .21 1 .35 . x 8
Required Answer
1 1 20 23 33 6 2 1 35 60 81
72 36 36 Hence the correct answer is option (B).
68.
y 2 4x and y x 2 x 2 4x x 0 and x
4
y
4/
Area
4x x dx
0
4/
y x
1 9 4/
x3/2 x2 1 2 9 3 / 2 0 2 0 3/4 x 4 x 16 1 22 3/2 3 2 9 32 8 1 3 9 8 1 3 9 24
x
69.
sin x cos y 0 cot x y 0
……..(1) ………(2)
Case I
2 When , 2 3 3 sin , 1 2 1 1 cos , 0 cos 0, 2 2 1 cot , 0 3 sin 0 cos 0 cot 1 Equation (1) and (2) will 0x 0y 0 system will have infinitely many solution x y 0 Case II When
7 1 , sin , 0 6 2
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JEE-MAIN-2019 (12th Apr-Second Shift)-PCM-34
3 cos 1, 2
cot
3,
sin 1, cos 1 cot 1,2,3,...... x y 0 Ix y 0 I 1,2,..... It will have unique solution in all cases x 0, y 0 70.
Total problems = 50 4 P (Solving) 5 1 P (Not solving) 5 P (unable to solve less than two problems) = P (not solving one problem) + P (not solving zero problem) 0 50 1 49 1 4 1 4 50 C0 50 C1 5 5 5 5 50 49 4 4 50 50. 5 5.5 49 50 49 4 4 10. 5 5 49
4 4 10 5 5
71.
54 4 . 5 5
49
a1,a 2 ,......an are in A.P. a1 a7 a16 40 a a 6d a 15d 40 3a 21d 40 40 a 7d 3 15 515 2a 14d 2 15 a 7d
15
40 3
= 200
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JEE-MAIN-2019 (12th Apr-Second Shift)-PCM-35 72.
Given 2a 4 and 2be 4 a 2, be 2
b2e2 4 b2 a2 4 b2 8 equation of ellipse x2 y2 1 4 8 Clearly option (D) satisfy the given curve. 73.
Equation of plane containing both lines is x 1 y 1 z 1
2
1 0
1
1
2
x 1 4 1 y 11 2 z 1 2 0 3 x 1 3 y 1 3z 0 x 1 y 1 z 0 x y z 0 distance from point (2, 1, 4) is
74.
2 1 4 12 12 12
3
For A C, A C B But A B option A is NOT true Let x Cx C A C B x C A and x C B x C or x A
x C or
and
x B
x C or x A B xC or x C A B C) xC C A C B C Now
x C x C A
(as A=C
B
(1) and
x C B x C A C B C C A C B from (1) and (2) C C A C B option B is true
(2)
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JEE-MAIN-2019 (12th Apr-Second Shift)-PCM-36 Let x A and x B x A B
x C (as A B C ) Let x A and x B x A B x C (as A B C ) Hence x A x C AC Option C is true As C A B B C A B As A B B C Hence the correct answer is option (A)
75.
, , are in G.P.
x 2 2 x 0 and x 2 x 1 0 have a common roots. Both roots will be common. 2 1 1 1 2 2 2 76.
xy3 0 ……..(i) will be chord of contact of parabola Let the required point is P x1,y1 chord
of
contact
for
point
P
x x1 y y1 xx1 4 3 2 2 y y1 2x1x 4x 4x1 6 As equation (i) and (ii) are same line 2x1 4 1 4x1 y1 6 1 1 3 2x1 4 1
x1
5 2
4x1 y1 6 3 10 y1 9 0 y 1 1
5 Hence correct answer is , 1 which is option (D). 2
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is
JEE-MAIN-2019 (12th Apr-Second Shift)-PCM-37 77.
AB 30m NP
In ANM tan 45o
MN 1 AN M
MN AN PM MN 30 AN 30
P
PM AN 30 In BPM tan 30 PB AN
30o
B
o
1 AN 30 AN 3
45o N
AN 3AN 30 3
AN
30 3 3 1
30 3
A
15 3 3
3 1
2
78. A (1, 2)
E (–1, 1) F (2, 3)
B (x2, y2)
C (x3, y3)
2 1 y 2 1, 2 1 2 2
x3 1 y 2 2 and 3 3 2 2
x 2 3, y 2 0
x 3 3, y3 4
B 3, 0
C 3, 4
x x 2 x 3 y1 y 2 y 3 , Centroid 1 3 3 1 3 3 2 D 4 1 , 3 , 2 3 3
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JEE-MAIN-2019 (12th Apr-Second Shift)-PCM-38 79. Coin Probability
+15 6 36
+12 4 36
–6 26 36
6 36 4 Probability of sum of 9 36 26 Other probability 36 6 4 26 Expected gain/loss 15 12 6 36 36 36 90 48 156 1 36 36 36 2 Hence correct answer is option (D). Probability of doublet
80.
20
20C0 20C1 20C2 x 2 ....... 20 C20 x 20 ……….(i) Differential equation w.r.t. x 19 20 1 x 20 C1.1 2. 20C2 x ....... 20 20C20 x19 …………..(ii) Multiply equation (2) by x 19 20x 1 x 20 C1x 2. 20C2 x 2 ...... 20 20C20 x 20 ……….(iii)
1 x
Differential equation 3 w.r.t. x 19 18 20 1 x 19x 1 x 1. 20C1 22 . 20C 2 x ...... 20 2 Put x 1 in equation (iv) 20 219 19.218 12 20C1 22 20 C2 ...... 20 2 20 C20
18
20 2
20
C 20 x19 …….(iv)
18
2 19 20 21 2
420 218 A 420, 18 Hence correct Option is (A). 81.
sin x cos x Given, y tan1 sin x cos x tan x 1 y tan1 tan x 1 1 tan x y tan1 1 tan x y tan1 tan x 4 0 x 2
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JEE-MAIN-2019 (12th Apr-Second Shift)-PCM-39
x 0 2 x0 4 4 y x tan1 tan x x x , 4 2 2 dy 1 yx 2 4 x 1 d 2 2
82.
2
2
Equation of required circle will be x 3 y r r 2
x 2 6x 9 y 2 2r y r 2 r 2 x 2 y 2 6x 2ry 9 0
………(1) 2
r f
Length of y intercept 2 f c 2
82 r 9 16 r 2 9 r=5 So equation of required circle will be x 2 y 2 6x 10y 9 0 two circles
x 2 y 2 6x 10y 9 0 ………..(2) x 2 y 2 6x 10y 9 0 ………..(3) Given option (C) i.e. (3, 10) satisfy equation (3). 83.
4 ……(i) is always tangent to y 2 16x m If it is tangent to the xy 4 y mx
4 x mx 4 m 2 2 m x 4x 4m m2 x 2 4x 4m m2 x 2 4x 4m 0 for tangent D = 0 16 16m3 0 m 1 put in equation (i) yx4 So the correct answer is option (D) 84.
Equation of angle bisectors x 2y 2z 2 2x y 2z 4 2 2 2 2 1 2 2 22 1 22
……..(1)
Case I: take positive sign 2x y 2z 4 x 2y 2z 2
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JEE-MAIN-2019 (12th Apr-Second Shift)-PCM-40
x 3y 2 0 ………(2) Case II : take negative sign 2x y 2z 4 x 2y 2z 2 2x y 2z 4 x 2y 2z 2 3x y 4z 6 0 ………(3) Option (B) satisfy equation (3) 1
85.
dx
x x 1 log
e
1
9 8
x 1 x dx log
x x 1
e
1
dx x
1
9 8
dx 9 loge x 1 8 1
x 9 loge loge x 1 8
2 1 2 9 loge log loge 2 2 2 1 8 2 1 2 1 9 log loge 8 2 2 2 2
2 1 9 4 1 8 8 42 4 1 9 4 2 4
322 32 8 36 2 36 4 2 4 8 0 2 2 0 2 1 0 1, 2 Hence the correct answer is option (B). 86.
Given 5 boys and n girls Total ways of farming team of 3 Members under given condition 5 C1 . nC2 5C2 . nC1
5C1. nC2 5C2 . nC1 1750
5n n 1 2 n n 1
10n 1750
2n 350 2 n2 3n 700 n2 3n 700 0 n 25
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JEE-MAIN-2019 (12th Apr-Second Shift)-PCM-41
87.
0, 3 1 cos2 sin2 4 cos 6 2 2 cos 1 sin 4 cos 6 0 2 2 cos sin 1 4 cos 6
R2 R 2 R1,R3 R3 R1 1 cos2 sin2 4 cos 6
1
1
0
1
0
1
0
C1 C1 C2
sin2 4 cos 6
2 0 1
1
0
0
1
0
expanding along first column 2 1 0 1 4 cos 6 0
2 4 cos 6 0 1 cos 6 2 2 6 3 9 88.
OP = 4 given OP makes 60o with x y 0 Let slope of OP = m m1 tan60o 1 m m 1 3 or 3 m 1 m 1 3m 3 or m 1 3 3m
m
m
3 1 3 1 or m 1 3 3 1 3 1 3 1
tan
or m
3 1
3 1 3 1
or tan
3 1
3 1 3 1 equation of line x cos y sin P
3 1 x
3 1 y 8 2
or
3 1 x
3 1 y 8 2
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JEE-MAIN-2019 (12th Apr-Second Shift)-PCM-42
89.
Lim x 0
x 2 sin x x 2 2 sin x 1 sin2 x x 1
x 2 sin x x 2 sin x 1 sin2 x x 1 x 2 sin x Lim 2 .2 x 0 x 2 sin x sin2 x x Applying L’H Rule 2. 1 2 cos x Lim x 0 2x 2 cos x 2 sin x cos x 1 Lim x 0
2
x 2 2 sin x 1 sin2 x x 1
2 3
2 2 1 Hence the correct answer is option (A).
90.
~ p ~ q p q Hence the correct answer is option (C).
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