# 369physics-12_sqp 2020_solutions.pdf

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Solutions with Marking Scheme Section ‘A’ 1. (a) φ =

q (for one face) 6ε 0

2. (b) Conductor 3. (a) 1W 4. (c) 12.0kJ 5. (a) speed 6. (d) virtual and inverted 7. (a) straight line 8. (d) 60° 9. (b) work function 10. (b) third orbit 11. 45° OR vertical 12. 2 H 13. double 14. 1.227 Å 15. 60° 16. Difference in initial mass energy and energy associated with mass of products. OR Total kinetic energy gained in the process. 17. Increases 18. No/8 19. 0.79 eV 20. Diodes with band gap energy in the visible spectrum range can function as LED OR LEDs are extremely energy efficient and consume upto 90% less power than instead/incandescent bulbs. So, by using LEDs of incandescent lamp, there is large decrease in power costs.

1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1

Section ‘B’ 21. When electric field E is applied on conductor force acting on free electrons →

F = −e E

m a = −e E

−e E a = m Average thermal velocity of electron in conductor is zero (ut)av = 0 Average velocity of electron in conductors in t (relaxation time) = vd (drift velocity) vd = (ut)av + at −eE τ vd = 0 + m →

1

−e E τ vd = m →

1

OSWAAL CBSE Sample Question Papers, Physics, Class - XII

2

C2= 2F

22. 6V

1F=C4

C1=1F

2F=C3 C5= 2F

C2 and C3 are in series 1 1 1 = + =1 C' 2 2 C' = 1mf C' and C4 are in || C" = 1 + 1 = 2mf C" and C5 are in parallel 1 1 1 ⇒ C''' = 1mf = + 2 2 C ''' C"' and C1 are in parallel Ceq = 1 + 1 = 2mf Energy Stored 1 1 U = CV2 = × 2 × 10–6 × 62 2 2 = 36 × 10–6J 23. Gain in KE of particle = Qv 1 mp vp2 = Kp = qp Vp 2 Vp = Va = V 1 mα vα2 = Ka = qa Va  2 Divide (ii) by (i), ma va2 qα = = 2 mp vp2 qp 1

1

1

.... (i)

... (ii) 1

1

v2α 2 mp mp × 2 1 = vp2 = 4 mp × 1 = 2 mα × 1 va : vp = 1 : 2 24. “The angle of incidence at which the reflected light is completely plane polarized, is called as Brewster’s angle (iB) ght)

ed li

Incident Ray

PPL

iB i B

lect (Ref

90°

Rare

r

ht

ed act

Denser

lig

fr

Re

At i = iB, reflected beam is ^ to refracted beam \ iB + r = 90° ⇒ r = 90 – iB Using Snell’s law sin i = m sin r

1

1

Solutions

3

sin iB sin iB sin (90 − iB ) = m ⇒ cos iB = m m = tan iB 25. Work function w = 2.14eV (a) Threshold frequency w = hn0 φ 2.14 × 1.6 × 10 −19 n0 = = h 6.62 × 10 −34 = 5.17 × 1014 Hz (b) As kmax = eV0 = 0.6eV Energy of photon E = kmax + w = 0.6eV + 2.14eV = 2.74eV hc Wave length of photon l = E

1

6.62 × 10 −34 × 3 × 10 −8 2.74 × 1.5 × 10 −19  = 4530Å 26. =

1 Vn

+e

electron

rn

centripetal force = electrostatic attraction mvn2 1 e2 = 2 rn 4 πε0 rn 1 e2 mvn2 = 4 πε0 rn 1 as mvnrn = n. 2π nh put in (i) 2πmrn 2 2 2 n h 1 e m. 2 2 2 = 4 πε m r 4π 0 rn n ε n2 h 2 rn = 0 2 πme

1 ...(i)

vn =

Energy of electron in n = 2 is – 3.4eV \ energy in ground state = – 13.6eV KE = – TE = +13.6eV x x En = 2 Þ – 3.4eV = 2 n 2

1

OR

⇒ energy in ground state x = – 13.6eV. PE = 2TE = – 2×13.6eV = – 27.2eV

1 1

OSWAAL CBSE Sample Question Papers, Physics, Class - XII

4 27.

p-type semiconductor

n-type semiconductor 1

1. Density of holes >> density of electron

1. density of electron >> density of holes

2. Formed by doping trivalent impurity

2. formed by doping pentavalent impurity

Energy band diagram for p-type

Energy band diagram of n-type semiconductor CB Donor energy level

CB Eg

Acceptor energy Level

Eg

VB

VB

1 (Any two)

OR hc Energy of photon E = λ

6.62 × 10 −34 × 3 × 10 8 eV = 2.06eV 6000 × 10 −9 × 1.6 × 10 −19 As E < Eg (2.8eV), so photodiode cannot detect this photon.

=

1 1

Section ‘C’ 28. Principle of potentiometer, when a constant current flows through a wire of uniform area of crosssection, the potential drop across any length of the wire is directly proportional to the length. Let resistance of wire AB be R1 and its length be ‘l’ then current drawn from driving cell –

I =

V and hence R + R1

1

P.D. across the wire AB will be

VAB = IR1 =

V el × R + R1 a

Where ‘a’ is area of cross-section of wire AB \

1

Ve VAB = R + R a = constant = k ( 1) l

Where R increases, current and potential difference across wire AB will be decreased and hence potential gradient ‘k’ will also be decreased. Thus the null point or balance point will shift to right (towards B) side. 1 29.

Idl

r

a I

O

N

x

r

dBy

dBy

P N1

dB dBx M dB

1 According to Biot-Savart’s law, magnetic field due to a current element is given by →

where r =

µ 0 I dl× r = dB 4 π r 2 →

x 2 + a2

Solutions

5

\

dB =

µ 0 Idl sin 90° 4 π x 2 + a 2

½

And direction of dB is ^ to the plane containing I dl and r . →

Resolving dB along the x – axis and y – axis.

dBx = dB sinf dBy = dB cosf

Taking the contribution of whole current loop we get

Bx == ∫ dBx

µ0

∫ dB sin φ Idl + a2

a

=

∫ 4π x

Bx =

µ0 Ia 2 4 π ( x + a 2 )B / 2

=

µ 0 Ia × 2 πa 4 π x 2 + a2 3 / 2

And \

(

x + a2

∫ dl

cos φ ∫ dB=

Bx 2 + By 2

Bp =

2

½

)

By == ∫ dBy

0

µ0 2IAπ 4 π x 2 + a2 3 / 2

Bp = Bx =

\

2

(

)

µ 2mπ Bp = 0 4 π x 2 + a2 3 / 2 →

(

)

 →  1  m = I A   

For centre x = 0

→  I  µ 2I πa 2 µ B0 = 0   = 0 4 π a3  2a  →

in the direction of m . 30.  Resonant frequency for LCR circuit is given by 1 v0 = 2π LC 1 = −6 2 × 3 . 14 3 × 27 × 10 = 17.69Hz or w0 = 2pv0 = 111 rad/s.  Quality factor of resonance

1 L ωL ω0 Q = = 0 = R C R 2∆ω

\

Q =

1 7.4

3 = 45.0 27 × 10 −6

1

1

OSWAAL CBSE Sample Question Papers, Physics, Class - XII

6

To improve sharpness of resonance circuit by a factor 2, without reducing w0, reduce R to half of its value i.e. R = 3.7Ω 1 A 31. G i

B

i

R

i

r = 39.4° 45°

B

C

1 Two conditions for TIR – (a) Light must travel from denser to rarer medium (b) i > ic 1  sin ic = µ \

 1   = 46°  1.39 

(ic)Red = sin–1 

 1   = 44.8° (ic)Green = sin–1   1.42 

 1   = 43°  1.48 

(ic)Blue = sin–1 

1

 Angle of incidence at face AC is 45° which is more than the critical angle for Blue and Green colours therefore they will show TIR but Red colour will refract to other medium. 1 32. Resolving power (R.P) of an astronomical telescope is its ability to form separate images of two neighbouring astronomical objects like stars etc. D 1 R.P. = = where D is diameter of objective lens and l is wavelength of light used. 1 1 . 22 λ dθ D = 2.54 m l = 6000Å 1.22 λ Limit of resolution dq = 1 D

1.22 × 6000 × 10 −10 m 254 × 10 −2 m = 2.9 × 10–10 OR Basic assumptions in derivation of Lens maker’s formula : (i) Aperture of lens should be small (ii) Lenses should be thin (iii) Object should be point sized and placed on principal axis.

=

1

1

Solutions

7

1 Suppose we have a thin lens of material of refractive index n2, placed in a medium of refractive index n1, let O be a point object placed on principle axis then for refraction at surface ABC we get image at I1 , n2 n1 n2 − n1 − ... (1) \ = v' u R1 But the refracted ray before goes to meet at I1 falls on surface ADC and refracts meet at I2 finally; hence I1 works as a virtual object 2nd refracting surface n1 − n2 n1 n2 ... (2) \ − = R2 v v' Equation (1) + (2)

 1 1  n1 n1  = (n2 – n1)  − − R R   1 2 v u  1 1  1 1  \ − = (n21 – 1)  −  R1 R2  v u If u = ∞, v = f  1 1  1  −  = (n – 1) 21 f  R1 R2  which is lens maker’s formula. 238 237 1 33. 92 U →91 Pa +1 H + Q

....(3)

....(4) 1

Q = [MU – MPa – MH] c2 = [ 238.05079 – 237.05121 – 1.00783] u× c2

1

MeV u = – 7.68MeV

1

= – 0.00825u × 931.5

\ Q < 0 ; therefore it can’t proceed spontaneously. We will have to supply energy of 7.68 MeV to nucleus to make it emit proton.

1 238 92

U

34. Circuit Diagram D1

A

Vo Y

C

RL

x

1

B D2

One possible answer: Change the connection of R from point C to point B. Now no current flowing through D2 in the second half. 1 mark for any correct diagram

2 marks for correct explanation

2

OSWAAL CBSE Sample Question Papers, Physics, Class - XII

8

Section ‘D’ 35.(a)

1

According the Gauss’s law – 1 → → {q} . ds = E ∫ ε0 1 → → → → → → E . ds + E . ds + E . ds = ∫ 1 ∫ 2 ∫ 3 ε0 ( λL ) ∫Eds1cos0° + ∫Eds2cos90°+ ∫Eds3cos90° = E∫ds1 =

1

λL ε0

E × 2πrL= E =

λL ε0

λ 2 πε0 r

λL ε0

→ λ  r E = 2 πε0 r

1

(b)

E a

nL a a

a

nR X

Ex = ax = 400x Ey = Ez = 0 Hence flux will exist only on left and right faces of cube as Ex ≠ 0 → →  = 1 {qin} = f  EL .a2( n L) + ER a2 n R ε0 −EL .a 2 + a 2 ER = fNet fNet = – (400a)a2 + a2 (400 × 2a) = – 400a3 + 800a3 = 400a3 = 400 × (0.1)3

1

Solutions

9

0.4 Nm2c-1 1 ( qin )  fNet = ε0

1

fNet =

\ qin = e0fNet

= 8.85 × 10–12 × 0.4 = 3.540×10–12C OR (a) Definition of electrostatic potential – SI unit J/c or Volt. Deduction of expression of electrostatic potential energy of given system of charges :

U=

1 1 2

1  q1 q2 q1 q3 q2 q3  + +   r13 r23  4 πε 0  r12

(b) V = 0V

36. For forward motion from x = 0 to x = 2b. The flux fB linked with circuit SPQR is

K

S

1

L P V Q

R

x = 2b x=0

x=b

The induced emf is,

fB = Blx 0 ≤ x < b = Blb b ≤ x < 2b

e=

1

−dφB dl e = – Bln 0 ≤ x < b e = 0 b ≤ x < 2b 1 When induced emf is non-zero, the current I in the magnitude; e Blv I= = r r The force required to keep arm PQ in constant motion is F =IlB. Its direction is to the left. In magnitude B2 l 2 v F = IlB ; 0 ≤ x < b r

The Joule heating loss is

F = 0 ; b ≤ x < 2b PJ = I 2 r B2 l 2 v 2 = 0≤x
1

10

OSWAAL CBSE Sample Question Papers, Physics, Class - XII

= 0 b ≤ x < 2b 1 One obtains similar expressions for the inward motion from x = 2b to x = 0 Bl = K

0

L

M

L

M

b

2b

b

0

e

Blv

b

0

b

2b

0

1

Blv Inward

Outward

BlV 2 2

r

2

0

b

2b

bb

0

OR Working principle of cyclotron 1 Diagram 1 Working of cyclotron with explanation 2 Any two applications [CBSE Marking Scheme, 2020] 1 Detailed Answer : Cyclotron : The cyclotron, devised by Lawrence and Livingstone, is a device for accelerating charged particles to high speed by the repeated application of accelerating potentials. Construction : The cyclotron consists of two flat semi-circular metal boxes called 'dees' and are arranged with a small gap between them. A source of ions is located near the mid-point of the gap between the dees (fig.). The dees are connected to the terminals of a radio frequency oscillator, so that a high frequency alternating potential of several million cycles per second exists between the dees. Thus dees act as electrodes. the dees are enclosed in an insulated metal box containing gas at low pressure. The whole apparatus is placed between the poles of a strong electromagnet which provides a magnetic field perpendicular to the plane of the dees. A cyclotron makes use of the principle that the energy of the charged particles can be increased to a high value by making it pass through an electric field repeatedly. The magnetic field acts on the charged particle and makes them move in a circular path inside the dee. Number of times the particle moves from one dee to another it is acted upon by the alternating electric field, and is accelerated by this field, which increases the energy of the particle. Magnetic Pole N S Dee

Dee

Magnetic Pole S

Working : The principle of action of the apparatus is shown in figure. The positive ions produced from a source S at the centre are accelerated by a dee which is at negative potential at that moment. Due to the presence of perpendicular magnetic field the ion will move in a circular path inside the dees. The magnetic field and the frequency of the applied voltages are so chosen that as the ion comes out of a dee, the dees change their polarity (positive becoming negative and vice-versa) and the ion is further accelerated and moves with higher velocity along a circular path of greater radius. The phenomenon

Solutions

11

is continued till the ion reaches at the periphery of the dees where an auxiliary negative electrode (deflecting plate) deflects the accelerated ion on the target to be bombarded. Uses : (i) It is used to bombard nuclei with high energetic particles accelerated by cyclotron and study the resulting nuclear reaction. (ii) It is used to implant ions into solids and modify their properties or even synthesize new materials. Magnetic field out of paper

Deflecting plate

Exit port Charged particle

D1 D2

Electric Oscillator

37. M B B' A

A'

F

N

1

C

Deduction of mirror formula

1 1 1 + = f v u

For a convex mirror f is always +ve. \ f > 0 Object is always placed in front of mirror hence u < 0 (for real object) ⇒

2

1

1 1 1 + = f v u

1 1 1 = − f u v As u < 0 i.e., –ve hence 1 > 0 v

⇒ v > 0 i.e. +ve for all values of u. Image will be formed behind the mirror and it will be virtual for all values of u. OR (a) Ray Diagram : (with proper labelling)

1

1

OSWAAL CBSE Sample Question Papers, Physics, Class - XII

12

B Fe

A' A FO

A

FO

B

Fe

s uO Objective Lens

VO

B'

Eye Piece

D

B

Magnifying power,

m =

v0  D 1 +  u0  fe 

m =

L D 1 +  fo  fe 

1

(b)  m = mo, me = – 30 (virtual, inverted)  fo = 1.25 cm fe = 5.0 cm Let us setup a compound microscope such that the final image be formed at D, then L 25 me = 1 + = 1 + =6 f0 5 and position of object for this image formation can be calculated – 1 1 1 = – ve ue fe

1

1 1 1 − = −25 ue 5  \

6 1 1 1 = + = 25 5 25 ue

−25 = –4.17 cm 6 m = mo × me + vo −30 = = −5 \ vo = –5uo mo = uo 6 ue =

1 1 1 − == vo uo fo

1 1 1 − = = −5uoo uoo 1.25

1 1 −16 = − = −5uo 5uo 1.25

1

uo = – 1.5 cm vo = 7.5 cm Tube length = vo +|ue| = 7.5 cm + 4.17 cm

L = 11.67 cm

1

Object should be placed at 1.5 cm distance from the objective lens.



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