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11-Heat and thermodynamics HEAT AND THERMODYNAMICS Heat Heat is that form of energy. Zeroth Law Of Thermodynamics And Temperature If a system A is in thermal equilibrium with system B and the system B is in thermal equilibrium with system C, then systems A and C are in thermal equilibrium with each other.
Temperature scales Relation between Celsius, Kelvin and Fahrenheit Scale: C K 273 F 32 5 5 9
Example 1: The electrical resistance of pure platinum increases linearly with increasing temperature. This property is used in a Platinum resistance thermometer. The relation between R (Resistance at K) and R0 (Resistance at 0K) is given as
R Ro 1 α θ θo θ
where = temperature coefficient of resistance. Now, if a Platinum resistance thermometer reads 00 Celsius when its resistance is 80 and 100o when its resistance is 90, find the temperature at which its resistance is 86. Solution:
Using the given relationship, we have 90 80Ω80Ω100
86 80Ωα[80Ω][θ] . . .
. . . (i)
(ii)
Where is the desired temperature. Taking the ratio of (i) & (ii)
CALORIMETRY Principle of Calorimetry
When two objects having different temperatures are brought in contact, heat flows from the hot object to the cold object. If the system is sufficiently thermally isolated from its surrounding, the heat lost by the hot object = the heat gained by the cold object.
Specific heat capacity
The amount of heat needed to raise the temperature of unit mass of a substance by 1 is known as its specific heat capacity. If Q amount of heat raises the temperature of m mass of a material by T , then its specific heat capacity is given as: s
Q mΔT
Q msT
Also the amount of heat supplied per unit increase in temperature for any body is known as its heat capacity, c
Q ms . T
Specific Latent Heat
In order to change the state of a substance (from solid to liquid or from liquid to gas) heat has to be supplied to it. During this process temperature remains constant. The amount of heat supplied per unit mass for such a process is known as the Specific Latent Heat of that substance for that process. Example 2: 5 gm of water at 30oC and 5 gm of ice at - 20oC are mixed together in a calorimeter. Find the final temperature of mixture and also the final masses of ice and water. Water equivalent of calorimeter is negligible, specific heat of ice = 0.5 cal/gm oC and latent heat of ice = 80 cal/gm. Solution:
In this case heat is given by water and taken by ice Heat available with water to cool from 30oC to 0oC = ms = 51 30 150 cal . Heat required by 5 gm ice to increase its temperature up to 0 oC ms 50.5 20 50 cal Out of 150 cal heat available, 50 cal is used for increasing temperature of ice from - 20oC to 0oC. The remaining heat 100 cal is used for melting the ice. if mass of ice melted is m gm then m 80 100 m 125 . gm
Thus 1.25 gm ice out of 5 gm melts and the mixture of ice and water is at 0oC. Mechanical Equivalent of Heat
1 calorie is the quantity of heat required to raise the temperature of pure water at 1 atm pressure from 14.5C to 15.5C. When work is completely converted to heat, the quantity of heat produced (Q) is found to be proportional to the quantity of work (W) that was converted into heat WQ Or,
W = JQ, where J is known as the Mechanical Equivalent of Heat. J = 4.2 107 erg cal-1 = 4.2 J/cal
Example 3: A bullet splinter of mass of 10 gm moving with a speed of 400 m/s hits an ice block of mass 990 gm kept on a frictionless floor and gets stuck in it. How much ice will melt if 50% of the lost Kinetic energy goes to ice? (Temperature of ice block = 0oC). Solution:
Velocity of bullet + ice block, V =
10 x400 m/s 1000
V =4 m/s Loss of K.E. = =
1 1 mv 2 m M V 2 2 2
1 1 [0.01(400)21(4)2] = [1600 16]= (1584/2) J 2 2
Heat generated = 1584 / 2 4.2 = 95 Cal Mass of ice melted =
95 cal = 1.2 gm. 80 Cal / gm
Thermal Expansion An increase in the temperature of a body is generally accompanied by an increase in its size. This is known as Thermal Expansion.
1 L L T
The value of linear coefficient of thermal expansion at temperature T can be found by taking limit T 0 . 1 L lim T 0 T L
or
1 dL L dT
For most of the solids the value of is small and nearly independent of T. In such cases the linear dimension of an object at a different temperature is given by: L L o (1 T)
(Here L o is the initial length and T is the change in temperature).
Example 4: A clock with an iron pendulum is made so as to keep correct time at 10oC. Given αiron12 10 6 per oC . How fast or slow does the clock move per day if the temperature rises to 250 C? Given iron = 12 106 per 0C. Solution:
When the pendulum keeps correct time, its period of vibration is 2 sec and so it makes 24 60 60 43200 Vibration /day 2
If length of pendulum at 10oC is 10 and at 25oC is 25 25 10[1 (2510)]
10 [115] g
as T 2 i.e. T i.e. n
1
n is no. of vibrations per sec.
n25 10 n10 25
115
1 / 2
1
15 2
15 n 25 n10 1 1210 6 = 43200[1 0.00009] = 43196 .12 2
That is the clock makes (43200 43196.12) = 3.88 vibration loss per day. That is clock losses 3.88 2 = 7.76 sec per day. Area and Volume expansion
If the temperature of a two-dimensional object (lamina) is changed, its area changes. If the coefficient of linear expansion of the material of lamina is small and constant, then its final area is given by A A o 1 T , where A o is the initial area, T is the change in temperature and
is the area coefficient of thermal expansion.
It can be shown that 2 . Similar relation also holds for the volume of a three-dimensional object V Vo 1 T
(Here 3 is known as the coefficient of volume expansion). For most solids, ~ 106/C
Thermal Expansion in liquids: The experimental measurement of for a liquid becomes slightly difficult due to expansion of the container, when a liquid is heated in a container. The initial level of the liquid falls due to expansion of the container. But afterwards it rises due to faster expansion of the liquid.
The actual increase in the the apparent + The increase in the volume of = volume of the liquid increase in the the container. volume of liquid liquid = apparent + container , for liquids, is of the order of 104/C
Anomalous expansion of water: If the temperature of water is increased from 0oC, it contracts until the temperature reaches 4oC and expands thereafter. In other words the density of water is highest at 4oC.
Example 5: A glass vessel of volume Vo is completely filled with a liquid and its temperature is raised by T . What volume of the liquid will overflow? Coefficient of linear expansion of glass = g and coefficient of volume expansion of the liquid = . Solution:
Volume of the liquid over flown = Increase in the volume of the liquid increase in the volume of the container = Vo (1 T) Vo Vo (1 g T) Vo
= Vo T( g ) = Vo T( 3 g )
( 3 )
Thermal Stress When a rod of length L is held between two rigid supports and the temperature of rod is increased by T , the rigid supports prevent the rod from expanding. This causes compressive stress in the rod. As the length of the rod remains unchanged, Thermal expansion = Mechanical compression LT
FL AY
F/A Y L / L
Thermal stress F / A YT , which is compressive.
KINETIC THEORY OF GASES Fluids consist of molecules which are in random motion, which collide with each other and with the walls of the container. The intermolecular separation in a gas is larger by an order of magnitude than the intermolecular separation in a liquid. The intermolecular separation in a gas increases as the temperature increases, and decreases with increasing density. it is experimentally observed that most gases follow a universal equation of state pV = n RT Gas molecules are relatively free: the interactions between them are small. This means that the total energy of the gas molecules is mostly kinetic. A simple model of a gas, with point particles representing molecules and their collisions with the walls of the confining vessel causing the pressure exerted by the gas leads to some important conclusions:
(a)
The pressure exerted by an ideal gas is given by p=
1 2 mc rms , where = number of molecules per unit volume, 3
m = mass of each molecule, crms = the rms speed of each molecule.
(b)
The total internal energy of an ideal gs is
1 2
U = Nf k B T where N = number of molecules in the gas, f = number of
degree of freedom, T = absolute temperature of the gas, kB = R NA = Boltzmann’s constant.
(C)
The rms speed is given by crms=
3RT , where M represents the molar mass of the gas (i.e. mass M
of 1 mole of the gas)
(d)
The pressure of an ideal gas is given by p =
2 U tran , where Utran is the translational energy of the molecules of 3 V
the gas.
Example6: Find r.m.s speed of Hydrogen molecules at room temperature (=300 k). Solution:
Mass of 1 mole of Hydrogen gas= 2 gm = 2 10-3 kg Vrms = =
3 RT M
3 8.3 300 2 10 3
= 1.93 103 m/s.
2.
When the container contains more than one gas, total pressure exerted by all the gases on the wall is sum of pressures exerted by each gas as it would while filling the container alone. In a way, each gas behaves independent of each other. Thus we have P = P1+P2+P3+……, where P1, P2, P3 are the partial pressures of gases 1, 2 & 3 respectively This is known as Dalton’s Law of partial pressures.
3.
One mole of any gas occupies a volume of 22.4 litre at standard temperature and Pressure which are 273.15 (=0°C) and 1.013105 Pa (=1atm) respectively,
Example 7: 4 gm Hydrogen is mixed with 11.2 litre of He at S.T.P. in a container of volume 20 litre. If the final temperature is 300 K find the pressure. Solution:
4 gm Hydrogen = 2 moles Hydrogen 11.2 He at S.T.P. = 1/2 mole of He P = PH + PHe = (nH+nHe)
8.3 (300 k ) RT = (2+½) V (20 10 3 )m3
= 3.12 105 N/m2.
Internal Energy Internal energy, of any body is sum total of kinetic energies and potential energies of its constituents (at molecular level). In case of an ideal gas, as there are no intermolecular forces, except during collision the possibility of potential energy is ruled out, so it is only kinetic energy. The kinetic energy of the molecules can be of three types. (i)
Translational
(ii)
Rotational
(iii)
Vibrational
In a way, it means that the energy of molecules is shared in various modes. These independent modes of motions are called degrees of freedom. The table given below gives the number of degrees of freedom for various types of molecules at normal temperature.
Nature of motion
Degree of Freedom (f) Translational
Rotational
Vibrational*
Total
(1) Monoatomic
3
0
0
3
(2) Diatomic
3
2
0
5
Poly
Linear
3
2
0
5
Nonlinear
3
3
0
6
Atomicity
* At room temperature the energy associated with vibrational motion is negligibly small in comparison to translational and rotational K.E. Equipartition of Energy According to the Law of equipartition of energy the average K.E. of a molecule is equally shared among different degrees of freedom. The average energy per degree of freedom of a molecule is ½ kT, where k is the Boltzmann’s constant and T is the absolute temperature. Thus, for a monoatomic ideal gas: U (the internal energy) = Also for one mole U =
3 kT 2
3 RT 2
Example 8: Find the average kinetic energy per molecule at temperature T for an equimolar mixture of two ideal gases A and B where A is monoatomic and B is diatomic. Solution:
No. of degrees of freedom per molecule for A = 3 No. of degrees of freedom per molecule for B = 5 Since the mixture is equimolar, the average kinetic energy per molecule will be given by the average of the two values i.e. 35 kT = 4kT 2
where k is Boltzmann’s constant.
Work Done In Different Processes Work done by an enclosed gas on its surroundings is given by W = p dV For different types of processes, we have got different relations between p (Pressure) and V (Volume) and accordingly we have different expressions for work.
(a)
Isochoric Process Here, the volume is constant throughout and therefore the work done by the gas, irrespective of the manner in which pressure varies, is zero Wisochoric = 0
(b)
Isobaric Process In this case, pressure of the gas remains constant throughout the process. Hence, pdV = p V = n RT(n = number of moles) ( T= change in absolute temperature)
(C)
Isothermal process The temperature remains constant throughout the process. Using ideal gas equation, we get, p=
n RT V
Hence, pdV = n RT
V2
V1
(d)
V dV = n RT ln 2 V1 V
Adiabatic Process: pV = Constant = C (say)
p = CV-
W = pdV =
V2
CV
dV
V1
=
p1V1 p 2 V2 = nCv(T2 T1) 1
The work done by a gas can also be evaluated from the p-V diagram of the process. p Area enclosed by the curve in a p-V diagram = work done by the gas
V
First Law of Thermodynamics First law of thermodynamics is simply a re-statement of the principle of conservation of energy for a thermally isolated system. If Q, U & W represent the heat given to the system, change in its internal energy and the work done by the system respectively, the first law of thermodynamics states that,
Q = U + W The heat transferred to the system (Q) is either utilised to do work (W) or increase the internal energy of the system (U). Example 9: 3000 J of heat is given to a gas at constant pressure of 2 105 N/m2. If its volume increases by 10 litres during the process find the change in the internal energy of the gas Q = 3000 J
Solution:
W = P V = (2105 N/m2) (10 10-3m3) = 2 103 J U = Q – W = 3000 2000= 1000 J.
Specific Heat Capacities of Gases S=
1 m
Q T
where Q = amount of heat required for ‘T’ temperature change. m = mass of the gas. In case of gases, the concept of a molar heat capacity is useful. Molar heat capacity is the amount of heat required to raise the temperature of one mole of the gas by one degree. So, if Q amount of heat goes to change the temperature of ‘n’ moles of a gas in a particular process, molar heat capacity ‘C’ can be mathematically given by: C=
1 Q n T
In terms of differentials, C=
1 dQ n dT
Two special cases are:(i)
If volume is kept constant during the process then CV =
1 Q n T V constant
This is the molar heat capacity of the gas at constant volume Note: Since U is independent of the process. U = n Cv T is true for all processes.
(ii)
If pressure remains constant, then Cp =
1 Q n T pconstant
This is the molar heat capacity of the gas at constant pressure
Relation Between Cp and Cv Cp Cv = R This is known as Mayer’s relation.
The Values of Cp and Cv If f is the number of degrees of freedom of a gas molecule then the internal energy of n moles of that gas is given as U = f/2 n RT
U = f/2 n RT = n CvT
Cv = f/2 R
From Mayer’s Relation Cp = Cv + R Cp = (f/2+1)R And the ratio of specific heats = =
Cp Cv
=
f /21 f /2
f 2 2 = 1 f f
Example 10: Find the molar heat capacity of an ideal gas with adiabatic exponent ‘’ for the polytropic process P V = constant. Solution:
We have, from first law of thermodynamics C = Cv +
PdV ndT
(n = number of moles)
We have, P V = constant From Ideal gas equation P V = n RT
Taking ratio, T V1 = Constant Differentiating we get
dV V = T( 1) dT
Putting it in the equation for ‘C’. C = Cv = Cv C=
PV n RT = Cv nT( 1) n T( 1) R 1
R R 1 1
Second law of thermodynamics
(i) Kelvin Statement:- It is impossible to derive a continuous supply of work by cooling a body to a temperature lower than that of the coldest of its surroundings.
(ii) Clausius Statement:-It is impossible for a self acting machine, unaided by an external agency to transfer heat from a body to another at higher temperature.
Reversible Process: A process which can be made to proceed in the reverse direction by variations in its conditions so that all changes occurring in any part of the direct process are exactly reversed in the corresponding part of the reverse process is called a reversible processes. Irreversible Process: A process which can not be made to proceed in the reverse direction is called an irreversible process. Heat Engine: It is a device which continuously converts heat energy into the mechanical energy in a cyclic process.
Efficiency of heat engine:
=
Q work output W Q1 Q 2 = 1 2 Q1 Q1 heat input Q1
Where Q1 is the heat supplied by the source and Q2 is the heat rejected to the sink.
Carnot Engine: It is an ideal heat engine which is based on Carnot's reversible cycle. It works in four steps viz. Isothermal expansion, adiabatic expansion, isothermal compression and adiabatic compression. The efficiency of a Carnot engine is given by =1
Q2 T =1 2 T1 Q1
where T1 and T2 are the temperatures of source and sink respectively. Example 11: The efficiency of a Carnot cycle is 1/6. If on reducing the temperature of the sink by 650C, the efficiency becomes 1/3, find the initial and final temperatures between which the cycle is working. Solution:
Given 1 =
1 1 , 2 = 6 3
If the temperatures of the source and the sink between which the cycle is working are T1 and T2, then the efficiency in the first case will be 1 = 1 -
T2 1 = T1 6
In the second case 2 = 1 Solving T1 = 390 K and
T2 65 T1
=
1 3
T2 = 325 K.
OBJECTIVE 1:
Calculate the root mean square speed of smoke particles of mass 5 1017 kg in Brownian motion in air at NTP. Boltzmann constant k 1.38 1023 JK 1
(A)
1.5 cm/s
(B)
2.2 cm/s
(C)
2.3 cm/s
(D)
4.4 cm/s
Ans. (a) Solution:
PV =
=
Vrms
m 1 2 RT mVrms M 3 3RT M
3NK T N
3kT
where = mass of one molecule.
Vrms
3 1.38 10 23 273 5 10 17
= 15 103 m / s 1.5 cm/s
2:
During an experiment an ideal gas is found to obey an additional law VP 2 = constant. The gas is initially at temp T and volume V. What will be the temperature of the gas when it expands to a volume 2V? (A)
T 4 T
(B)
T 2 T
(C)
T 5 T
(D)
T 6 T
Ans. (b) Solution:
According to the given problems VP2 = constant From the gas law PV = nRT
k V nRT V
nR V T K
V1 T1 , i.e, V2 T2
T 2 T
V T 2V T
Q.3-5 We have two vessels of equal volume, one filled with hydrogen and the other with equal mass of Helium. The common temperature is 27oC. 3:
What is the relative number of molecules in the two vessels ? (A)
nH 1 nHe 1
(B)
nH 5 n He 1
(C)
nH 2 n He 1
(D)
nH 3 n He 1
Ans. (C)
4:
If pressure of Hydrogen is 2 atm, what is the pressure of Helium ? (A)
pHe = 2 atm.
(B)
pHe = 3 atm.
(C)
pHe = 4 atm.
(D)
pHe = 1 atm.
Ans. (d)
5:
If the temperature of Helium is kept at 27o C and that of hydrogen is changed, at what temperature will its pressure become equal to that of helium ? The molecular weights of hydrogen and helium are 2 and 4 respectively. (A)
123oC
(B)
140oC
(C)
160oC
(D)
183oC
Ans. (a)
Solution 3-5: 3.
The masses of hydrogen and helium gases in the vessels are equal. This means that the product of the number of molecules and the mass of a molecule must be same for H2 and He gases. Since molecular masses of H2 and He are in the ratio 1: 2, their number of molecules nH and nHe in the vessels must be in the reverse ratio, that is,
4.
The equation of state for one mole of a gas is pV = RT = NkT
nH 2 nHe 1
Where N is Avogadro’s number (no. of molecules in one mole) and k is Boltzmann’s constant. If a gas has n molecules, the equation of state will be pV = nkT For a given volume and a given temperature, we have p n. Since H2 and He have same volume and same temperature (27 oC), we have pH n 2 H pHe nHe 1
Here pH = 2 atm.
5.
pHe = 1 atm.
Again, we have pV = nkT H2 and He have equal volumes. For having equal pressure, we must have nHTH = nHeTHe or
THe n H 2 TH nHe
Here THe = 27 + 273 = 300 K
TH =
1 THe = 150 K 2
= 150 273 = 123oC
6:
A vessel contains a mixture of 7 gm of nitrogen and 11 gm of carbon dioxide at temperature T = 290 K. If pressure of the mixture P = 1 atm, calculate its density (R = 8.31 J/mol k) (A)
2.5 kg/m3
(B)
1.5 kg/m3
(C)
4.5 kg/m3
(D)
7.5 kg/m3
Ans. (b) Solution:
As molecular weight of N2 and CO2 are 28 and 44, and n
m , M
nN
11 7 1 1/ 4 and n C 44 28 4
So, n nN n C
1 1 1 4 4 2
Now, according to gas law PV = nRT V
nRT 1 8.31 290 = 1.19 10 2 m3 5 P 2 1.01 10
and m = 7+11 = 18 gm = 18 10-3 kg so, =
18 10 3 kg m 1.5kg / m 3 2 3 V 1.19 10 m
Q.7-10.The pressure of a monoatomic gas increases linearly from 2 2 3 4 105 N/m to 8 105 N/m when its volume increases from 0.2 m to 3 0.5 m . Calculate 7:
work done by the gas (A)
2.8×105 J
(B)
1.8×106 J
(C)
1.8×105 J
(D)
1.8×102 J
Ans. (C)
8:
increase in internal energy (A)
U 4.8 105 J
(B)
U 4.8 104 J
(C)
U 6.8 105 J
(D)
U 4.8 106 J
Ans. (a)
9:
amount of heat supplied (A)
8.6 105 J
(B)
12.6 105 J
(C)
6.6.105 J
(D)
10.6 105 J
Ans. (C)
10:
molar heat capacity of the gas [R = 8.31 J/mol k] (A)
20.1 J/molK
(B)
17.14 J/molK
(C)
18.14 J/molK
(D)
20.14 J/molK
Ans. (b)
Solution 7- 10: 7.
Work done by the gas,
W PdV = area under P-v curve
= P1 VF VI
PF
B P
PI
A
1 PF PI VF VI 2 VI
1 = VF VI PF PI 2
=
8.
1 0.5 0.8 4 105 1.8 105 J 2
Change in internal energy of a gas is given by U nC V T
nRT PF VF PI VI 1 1
As the gas is monoatomic, = 5/3 So U
10 5 8 0.5 4 0.2 3 10 5 4 0.8 2 5 3 1
U 4.8 105 J
9.
From 1st law of thermodynamics Q U W
= 4.8 1.8 10 5 6.6. 10 5 J
10.
Molar heat capacity is defined as C
6.6 10 5 8.31 Q Q R 5 nT PF VF PI VI 10 8 0.05 4 0.2
i.e, C =
54.846 17.14 J/molK 3.2
V
VF
Q.11-13.Two moles of Helium gas ( = 5/3) are initially at temperature 27oC and occupy a volume of 20 litres. The gas is first expanded at constant pressure until the volume is doubled. Then it undergoes an adiabatic change until the temperature returns to its initial value. 11:
What are the final volume. (A)
113.13103 m 3
(B)
213.13103 m 3
(C)
313.13103 m 3
(D)
13.13103 m 3
Ans. (A)
12:
What are the final pressure of gas? (A)
0.44105 N / m 2
(B)
0.84105 N / m 2
(C)
0.94105 N / m 2
(D)
0.34105 N / m 2
Ans. (a)
13:
What is the work done by the gas? (Gas constant R = 8.3 T/mole K) (A)
13450J
(B)
14450J
(C)
16450J
(D)
12450J
Ans. (d)
Solution:11. From ideal gas equation PV = nRT initial pressure P 2.49 10 5 N / m 2
P
A
B
PA nRT 2 8.3 300 20 10 3 V
C vA
vB
When volume of gas is doubled at constant pressure, its temperature is also doubled. This process is shown on P-V curve by line AB. The gas then cools to temperature T adiabatically. This is shown by curve BC. The whole process is represented by curve ABC. At point B, pressure PB PA 2.49 10 5 N / m 2 . Volume VB 2VA 4010 3 m3 , Temperature TB 600K.
V
Now from adiabatic equation T V 1 = constant We have TA VA ( 1) TC VC( 1) 1
VC VB
VC 1 / 1 = 23/2 2 VB
TB 600 2 TC 300
Final volume
VC 2 2 VB
2 1.414 40103 113.13103 m3
12.
final pressure PC
13.
2 8.3 300 nRTC 0.44 10 5 N / m2 3 VC 113 .13 10
The work done by gas in isobaric process AB 2.49105 (40 20)103 4980 J
The work done by gas during adiabatic process BC W2
nR T2 T1 2 85.3 300 600 7470 J . 1 1 3
Net work done by gas W W1 W2
4980 7470 12450 J
14-15. When 1 gm of water changes from liquid to vapour phase at constant pressure of 1 atmosphere, the volume increases from 1 cm 3 to 1671 c.c. The heat of vaporization at this pressure is 540 cal/gm. Find 14:
The work done (in J) in change of phase (A)
170.78 Joule
(B)
200.67 Joule
(C)
190. 78 Joule
(D)
168 .67 Joule
(B)
3099.33 J
Ans. (d)
15:
Increase in internal energy of water. (A)
2099.33 J
(C)
4099.33 J
(D)
5099.33 J
Ans. (a)
Solution:14. As the process is isobaric W PdV PVF VI
= 1.01 10 6 1671 1 1688 .7 10 6 erg = 168 .67 Joule
15.
[1 erg = 10-7J]
From 1st law of thermodynamics Q U W Q mL 1 540 cal
= 2268 J, [ 1 cal = 4.2J] so, U Q W = 2268 168.67J = 2099.33 J
16:
A glass flask of volume one litre at 0 o C is filled level full of mercury at this temperature. The flask and mercury are now heated to 100 oC. How much mercury will spill out if coefficient of volume expansion of mercury is 1.82 104 / o C and linear expansion of glass is 0.1 104 / o C respectively? (A)
14.2 c.c.
(B)
15.2 c.c.
(C)
18.2 c.c.
(D)
20.2 c.c.J
Ans. (b) Solution: In case of thermal expansion of liquid, change in volume of liquid relative to container is given by V V L S
Here V = 1 litre = 1000 c.c. S = 3 glass = 0.3 104 / o C So, V 1000 1.82 0.3 10 4 100 0 = 15.2 c.c.
17:
Two cylinders A and B fitted with pistons contain equal amounts of an ideal diatomic gas at 300K. The piston A is free to move, while that of B is held fixed. The same amount of heat is given to the gas in each cylinder. If the rise in temperature of the gas in A is 30K, then the rise in temperature of the gas in B is. (A)
30K
(B)
18K
(C)
50K
(D)
42K
Solution:
For cylinder A. For cylinder B
dQ = nCPdT1 dQ = nCvdT2
nCPdT1 = nCvdT2 From (I) and (II) c v dT2 c v R30
dT2
(c v R)30 cv
5 2
For diatomic gas c v R
dT2 42K .
18:
80 gm of water at 30o C is poured on a large block of ice at 0 o C . The mass of ice that melts is (A)
30 gm
(B)
80 gm
(C)
150 gm
(D)
1600 gm
Solution: Since the block of ice at 0 o C is large, the whole of ice will not melt, hence final temperature is 0 o C .
Q1 = heat given up by water in cooling up to 0O C
= ms 80 1 30 0 = 2400 cal If m gm be the mass of ice melted, then
Q 2 = ML = m 80 Q1 Q2 m 80 2400 or m 30 gm
Here A is correct.
19:
A gas at pressure Po is contained in a vessel. If the masses of all the molecules are halved and their speeds doubled, the resulting pressure would be (A)
4Po
(B)
2Po
(C)
Po
(D)
Po 2
Solution:
P
1 mn 2 V rms where m = mass of one gas molecules 3 V
n = total no. of gas molecules i.e, P m and P Vrms Here m is halved and Vrms is doubled
pressure will be doubled Hence, (B) is correct
20:
The volume V versus temperature T graphs for a certain amount of a perfect gas at two pressures P1 and P2 are shown in the figure. Here (A) P1 < P2
(C) P1 = P2
(B) P1 > P2
(D) can’t be
Solution:
v
For a perfect gas, PV
m RT M
mR RT PM
So, the slope of the graph is Slope
1 P
mR PM
P2 P1 V
O
T
Hence P1 > P2 Hence, (C) is correct
21:
At room temperature the rms speed of the molecules of a certain diatomic gas is found to be 1930 m/s. The gas is (A)
H2
(B)
F2
(C)
O2
(D)
Cl 2
Solution:
M
Vrms 3RT Vrms
2
3RT M
3 x8.31x300 2
(1930)
2.0078x103 kg 2.00 gm
It is molecular weight of hydrogen (H2 ) .
22:
The latent heat of vaporization of water is 2240 J. If the work done in the process of vaporization of 1 gm is 168 J, then increase in internal energy is (A)
2408 J
(B)
2240 J
(C)
2072 J
(D)
1904 J
Solution:
L = 2240 J, m = 1 gm
dW = 168 J dQ = mL = dU + dW or 1 2240 dU 168 dU = 2072 J Hence, (C) is correct
23:
For a gas, y = 1.286. What is the number of degrees of freedom of the moleculas of this gas ? (A)
3
(B)
5
(C)
6
(D)
7
Ans. (d) Solution:
(D) 1
2 2 2 1.286 or 0.286 or 7 n n 0.286
24:
Which of the following temperatures is the highest? (A)
100 K
(B)
–13oF
(C)
–20oC
(D)
–30oC
Solution:
25:
(B ) –13oF is (13+32)o below ice point on F scale.
An ideal gas ( = 1.5) is expanded adiabatically. How many times has the gas to be expanded to reduce the root mean square velocity of molecules 2.0 times (A)
4 times
(B)
16 times
(C)
8 times
(D)
2 times
Ans. (B)
3RT M
Vrms
Solution: Vrms
T
Vrms is to reduce two times i.e, temperature of the gas will have to reduce four times or T 1 T 4
During adiabatic process TV 1 T V 1 1
1 V T 1 4 1.5 1 4 2 16 or, V T
V 16V
Hence, (B) is correct
26:
A thin copper wire of length L increases in length by 1% when heated from 0 o C to 100o C . If a thin copper plate of area 2L L is heated from 0 o C to 100o C , the percentage increase in its area will be (A)
1%
(B)
2%
(C)
3%
(D)
4%
Ans. (b)
L = Lo 1
Solution:
L L o L Lo
= percentage increase in length =
L L o 1
1 100 2
1 Hence 2L 2L o 1 100 2
2L2 2L2o 2L2o
or
A 2L2o
1 100
2
2 100
2 2% 100
Hence, (B) is correct
27:
Gas at pressure Po is contained in a vessel. If the masses of all the molecules are doubled and their speed is halved, the resulting pressure P will be equal to (A)
2Po
(B)
Po/4
(C)
Po
(D)
Po/2
Ans. (d) Solution: P=
Po = 1 3
1 mn 2 v rms 3 v
m n 2 v rms v
where m = 2m, v rms
v rms 2
putting the value
2 P mv rms 2m v 2 rms 1 = 2 2 Po m 4 v rms 2 mv rms
P = Po/2
28:
The molar heat capacity in a process of a diatomic gas if it does a work of Q/4, when Q amount of heat is supplied to it is (A)
2 R 5
(B)
(C)
10 R 3
(D)
5 R 2
6 R 7
Ans. (C) Solution:
5 2
dU = CV dT = R dT dT
2dU 5R
From 1st law of thermodynamics dU = dQ – dW or dU = Q Now molar heat capacity C
Q 3Q 4 4
dQ Q 5QR 10 R dT 3 3Q dU 2 2 4 5R
Hence (C) is correct
29:
For an ideal gas: (A)
the change in internal energy in a constant pressure process from temperature T1 to T2 is equal to nCv (T2 - T1), where Cv is the molar specific heat at constant volume and n the number of moles of the gas.
(B)
the change in internal energy of the gas and the work done by the gas are equal in magnitude in an adiabatic process.
(C)
the internal energy does not change in an isothermal process.
(D)
no heat is added or removed in an adiabatic process.
(A)
A, B
(B)
A, B, C
(C)
A, B, C, D
(D)
A, C
Solution: (C) Change in internal energy depends only on change in temperature since internal energy is a function of state only i.e. dU = nCv,dT. In adiabatic process, dQ = 0, Hence, dU + dW = 0 dU = dW
i.e. magnitude of change in internal energy is equal to magnitude of work done.
30:
Heat is supplied to a diatomic gas at constant pressure. The ratio of Q : U : W is (A)
5:3:2
(B)
5:2:3
(C)
7:5:2
(D)
7:2:5
Ans. (C) Q nC P dT
Solution:
U nC V T
7 nR T , 2
5 nR T , 2
7 CP 2 R 5 C V 2 R
and W Q U nRT Q : U : W 7 : 5 : 2 Hence, C is correct
31:
Two mole of argon are mixed with one mole of hydrogen, then Cp/Cv for the mixture is nearly (A)
1.2
(B)
1.3
(C)
1.4
(D)
1.5
Ans.(C) Solution: fav =
Average degree of freedom 2 3 1 5 11 23 5
mix = 1 +
1 f av
=1+
5 16 = = 1.4 11 11
Hence, C is correct Answer.
32:
An ideal gas is taken through the cycle A BC A as shown in figure. If the net heat supplied to the gas in the cycle is 5J, the work done by the gas in the process CA is, (A) -5J
(B) -10J
(C) -15J
(D) -20J
Solution:
(A) Qnet Wnet WCA WAB
5 WCA 10 (2 1) 5 10 WCA WCA 5
33:
When an ideal gas at pressure P, temperature T and volume V is isothermally compressed to a V/n, its pressure becomes Pi . If the gas is compressed adiabatically to V/n, its pressure becomes P a. The ratio Pi / Pa is (A)
1
(B)
n
(C)
n
(D)
n 1-
Solution:
(D) For isothermal process, PV = constant. Therefore
PiVi = PV or Pi
V PV or Pi = nP n
……(i)
For adiabatic process, PV = constant. Therefore Pa (Va) = PV
or
(P, V, T)
V Pa PV n
or
Pa = nP
…..(ii)
P
V Pn , ,T ' n
From (i) and (ii) we get Pi n n (1 ) Pa n
34:
V Pi , ,T n
v
When an ideal monatomic gas is heated at constant pressure, the fraction of heat energy supplied which increases the internal energy of the gas is (A)
2/5
(B)
3/5
(C)
3/7
(D)
3/4
Solution:
U
(B) U
w 1
nRT 1
w nRT
Q U 1
Q W U
U ( 1) Q U U 1 1 Q
U 1 3 [for monoatomic gas = 5/3 ] Q 5
35:
A monatomic ideal gas, initially at temperature T1 is enclosed in a cylinder fitted with a frictionless piston. The gas is allowed to expand adiabatically to a temperature T2 by releasing the piston suddenly. If L1 and L2 are the lengths of the gas column before and after expansion respectively, then T1/T2 is given by (A)
(L1/L2)2/3
(B)
L1/L2
(C)
L2/L1
(D)
(L2/L1) 2/3
Solution:
(D) TV 1 constant
Initial position T1(L1A)–1 = constant Final position T2 (L2A)–1 = constant
T L 1 1 T2 L2
1
1
T1 L2 T2 L1 L 2 L1
5 / 3 1
[for monoatomic gas
L 2 L1
36:
1
= 5/3]
2/3
An ideal mono atomic gas at 300K expands adiabatically to twice its volume. What is the fine temperature (A)
189K
(B)
289K
(C)
30Kj
(D)
Non of these
Solution:
(A) TV 1 constant
1 TV T2V2 1 1 1
300 V11 T2 2V1
1
300 T2 21 T2
37:
300 300 300 300 5 / 31 2 / 3 189K 1 2 2 2 1.587
What will be P-V graph corresponding to the P-T graph (process AB) for an ideal gas shown in figure
B
C
P
A
D
T
(A)
Hyperbolic
(B)
Circle
(C)
Straight line
(D)
Elliptical
Solution: (A)
38:
Starting with the same initial conditions, an ideal gas expands from volume V1 to V2 in three different ways. The work done by the gas is W1 if the process is purely isothermal, W2 if purely isobaric and W3 if purely adiabatic. Then (A)
W2 > W1 >W3
(B)
W2 > W3 >W1
(C)
W1 > W2 >W3
(D)
W1 > W3 >W2
Solution:
(A) Isobaric
P Isothermal Adiabatic V
39: of =
One mole of argon is heated using PV5/2 = = const. By which amount heat is obtained by the process when the temperature change by T = -26K. (A)
100J
(B)
200J
(C)
108J
(D)
208J
(C)
Here n = 1
Solution: C= C=
R R 1 1 x
For
polytropic process
R R 5 / 3 1 3 1 2
3 Q nCT 1 R 2R 26 108J 2
40: =
3 moles of an ideal monoatmic gas performs a cycle as shown in the fig. The gas temperatures T1 = = 400K, T2= = 800K, T3 = = 2400K, T4 = 1200K. What will be the net work done.
P
B
A
C D T
(A)
20J
(B)
20000J
(C)
200J
(D)
2000J
Solution:
(B) W AB = CCD = 0, because process is isochoric.
WAD = nRT = 3R(TA – TB) = 3 8.31 (2400 – 800) = 39884
Total work done W AD + W BC = 39888 – 19944 = 20 103J = 19944 20 103J
41:
How much heat is absorbed by the system in going through the process shown in the fig. (consider that value is taken in SI system)
900 P
100 800
V
400
(A)
20.4104 J
(B)
30.4104 J
(C)
21.4104 J
(D)
25.12104 J
Solution:
(D) Work done = Area of PV diagram
=
P2 P1 V2 V1 4
=
800 400 4
= 800 100 3.14 = 8 104 3.14 = 25.12 104
42:
3000J of heat is given to a gas at constant pressure of 2105 N/m2. If its volume increases by 10 litres during the process, what will be the change in the internal energy of the gas (A)
1000J
(B)
100J
(C)
200J
(D)
2000J
Solution: (A) Q 3000J P = 2 105 = Vi Vf = (Vi + 10 10–3) W = Pdv W = 2 105 10 10-3 = 2 103
Q W U 3000 = 2 103 + U
Q 1000
43:
A gas at atmospheric pressure is contained in a cylinder of volume 80 litre. When it is compressed adiabatically to 20 litre its pressure rises to 7 atm. What will be the ratio of specific heats of the gas (A)
1.33
(B)
1.4
(C)
1.67
(D)
1.5
Pi = 1atm = 1 105 N/m2
Solution: (B) Vi = 80 10 – 3m3 Vf = 20 10–3 m3 Pf = 7atm = 7 105 N/m2 PiVi = PfVf
1 105 (80 10–3) =(7 105)(20 10–3)
80 10 3 7 3 20 10
4
7
log 4 log7 log7 1.40 log 4
44:
A gas consisting of rigid diatomic molecules was initially under standard conditions. Then gas was compressed adiabatically to one fifth of its initial volume. What will be the mean kinetic energy of a rotating molecule in the final state? (A)
1.44 J
(B)
4.55J
(C)
787.98 10–23
(D)
757.310-23J
Solution: (C)
1.4
TV 1 constant T1V1 -1 = T2 V2 -1 7
(300)V1
7/5 – 1
V 5 = T2 1 5
1
T2
300 V12 / 5 1 V12 / 5 5
2/5
300 300 52 / 5 300 (1.903) = 571 2 / 5 5
Mean kinetic energy of rotating molecules = KT = 1.38 10 – 23 571 KT = 787.98 10–23
45:
Immediately after the explosion of an atom bomb, the ball of fire produced has a radius of 100m and a temperature 105K . What will be the approximate temperature when the ball expands adiabatically to a radius of 1000m (suppose mono atomic gas is there) (A)
1000K
(B)
100K
(C)
105 (10–3)2/3
(D)
200K
Solution:
Vi =
(C) r = 100 m
4 (100)3 3
Ti = 105 K after explosion r = 1000 m, Vf = 1 TV Tf Vf 1 i i
V Tf Ti i Vf
1
5
3 4 3 (100) Tf = 105 3 4 (1000)3 3
1
= 105 (10–3)2/3 = 1.05 K
46:
Which of the following is false? (A)
Enthalpy is a path function.
(B)
Work is a path function.
(C)
Heat is a path function.
4 (1000)3 3
(D) Solution:
47:
Energy is a state function (C)
A gas mixture consists of 32 gram of oxygen and 36 gram of Ar a temperature T. Neglecting all vibration modes, the total internal energy of the system is (A) 4RT
(B) 8RT
(C) 9RT
(D) 11RT
Solution: (D) F 5 3 n RT 2 RT 4 RT 11RT 2 2 2
48:
Energy
=
A mono atomic gas is supplied heat Q very slowly keeping the pressure constant. The work done by the gas is (A)
2/5 Q
(B)
3/5 Q
(C)
Q /5
(D)
2/3 Q
Solution:
(D) For monatomic gas
U 1 Q 3
or U
Q 3
From the first law of thermodynamics Q = U + W
2 W Q 3 49:
Which of the following parameters does not characterise the thermodynamic state of matter (A)
Work
(B)
Pressure
(C)
Temperature
(D)
Volume
Solution:
50:
(A) Work
Which of the following is correct (A)
For an isothermal change PV = = constant
(B)
For a isothermal process, the change in internal energy must be equal to the work done
P V For an adiabatic change 2 2 , where is the ratio of the P1 V1 two specific heats
(C)
(D)
In an adiabatic process external work done must be equal to the heat entering the system
Solution:
51:
(A)
An ideal gas goes through cyclic process ABC and following (P vs T) curve is obtained. This process can be represented by
B P C
A
T
B
B
P
P C
A
A
C V
V
(A)
(B)
B B
P
P
C
A
V
(C)
C
A
V
(D)
1 . Process B C is isobaric V and C A adiabatic. Slope of adiabatic > slope of isothermal.
Solution:(B) Process A to B isothermal. Then P
52:
A container contain 0.1 mol of H2 and 0.1 mol of O2 , If the gases are in thermal equilibrium then (A)
Only the average kinetic energy of the molecule of H2 and O2 is same.
(B)
Average speed of the molecule of H2 and O2 is same.
(C)
Only the specific heat at constant pressure of two gases is same.
(D) energy gases.
(d) The specific heat at constant pressure and the kinetic are same for both the
Solution: (D) The specific heat at constant pressure (Cp) is the amount of heat required to raise the temperature of one gram through 1°C when the pressure of the gas is kept constant. Again, the mean kinetic energy per molecule (3/2)kT depends only upon temperature. Clearly both the specific heats at constant pressure and mean kinetic energy are depending on the temperature which is again same for the two gases.
53:
Two systems are in thermal equilibrium. The quantity which is common for them is (A)
Heat
(B)
Momentum
(C)
Temperature
(D)
Specific heat
Solution:
54:
Mean molecular weight is defined as (A)
the number of free particles per positron mass
(B)
the number of free particles per electron mass
(C)
the number of free particles per neutral mass
(D)
the number of free particles per photon mass
Solution:
55:
(C)
(D)
Which one of the following statements is true about a gas undergoing an adiabatic change
(A)
The temperature of the gas remains constant
(B)
The pressure of the gas remains constant
(C)
The volume of the gas remains constant
(D)
The gas is completely insulated from the surroundings
Solution:
56:
If an ideal gas is allowed to expand adiabatically, the work done is equal to (A)
The loss in heat
(B)
The loss in internal energy
(C)
The gain in internal energy
(D)
The gain in enthalpy
Solution:
57:
(A)
Isothermal
(B)
Adiabatic
(C)
Isobaric
(D)
Isochoric
(A)
zero
(B)
infinite
(C)
positive
(D)
negative
(B)
The internal energy of the system remains constant when it undergoes (A)
a cyclic process
(B)
an adiabatic process
(C)
an isothermal process
(D)
an isobaric process
Solution:
60:
(A)
Specific heat of a gas undergoing adiabatic changes is
Solution:
59:
(B)
For the Boyle’s law to hold, the necessary condition is
Solution:
58:
(D)
(C)
The first law of thermodynamics incorporates the concepts of
(A)
conservation of energy
(B)
conservation of heat
(C)
conservation of work
(D)
equivalence of heat and work
Solution:
(D)
12-Transference of heat
SYLLABUS Thermal expansion of solids, liquids and gases; Calorimetry, latent heat; Heat conduction in one dimension; Elementary concepts of convection and radiation; Newton’s law of cooling; bulk modulus of gases; Blackbody radiation: absorptive and emissive powers; Kirchhoff’s law, Wien’s displacement law, Stefan’s law, Specific heat of a liquid using calorimeter (M and E)
1.
Thermal Expansion
When a body is heated, it expands in terms of length, area & volume and temperature rises. In a solid, molecules can only have thermal agitation (random vibrations). As temperature of a body increases, the vibrations of molecules will become fast and due to this the rate of collision among neighbouring molecules increases, it develops a thermal stress in the body and due to this the intermolecular separation increases which results in thermal expansion of body. At ordinary temperatures, the atoms in a solid oscillate about their equilibrium position with an amplitude of approximately 10–11 m. The average spacing between the atoms is about 10–10 m. As the temperature of solid increases, the atoms oscillate with greater amplitudes, as a result the average separation between them increases. Consequently the object expands. In the similar way the block diagram shown in Fig. explains the way how thermal expansion takes place. Temperature of Body Increases
Themal vibration energy of molecules increases
Rate of collision between neighbouring molecules increases
Interatomic separation increases
Thermal exp ansion takes place
Thermal expansion of a substance can be classified in three broad categories, these are (i) Linear Expansion (ii) Superficial Expansion and (iii) Cubical Expansion or Volume Expansion (i) Linear Expansion: Consider a rod of length l1 at a temperature T1. Let it be a heated to a temperature T2 and the increased length of the rod be l2 then l2 l1 (1 t) Coefficient of linear expansion and t T2 T1 Example 1. The density of substance at 0C is 10 g cm-3 and at 100C, its density is 9.7g cm-3. The coefficient of linear expansion of substance is d0 dt (1 3t) Solution: 10 9.7(1 3 100) 0.3 10 1 1 or 0.0001/ C or 9.7 300 9.7 300
(ii) Superficial Expansion (expansion in surface area): If A1 is the area of solid at T1 C and A2 is the area at T2C . Then A2 A1(1 t ) Coefficient of superficial (areal) expansion and t T2 T1
(iii) Volume expansion (Part A: expansion in solids): If V1 is the volume of solid at T1C and V1 is the volume at T2 C then V2 V1 1 t coefficient of cubical (volume) expansion and t T2 T1
Note: For isotopic solids: 2, 3
Relation between expansion coefficient are 6 3, 2
As temperature increases, density of solid decreases. If d1 is the density at T1C, d2 is the density at T2C then d1 d2 = or d2 =d1 1- t where t T2 T1 1+t (iv) Expansion of gases (Part B: expansion in gases): Pressure coefficient of a gas is the ratio of increase in pressure for 1C rise in temperature to the pressure at 0C, provided the volume of the gas is kept constant. p =
pt po po t
Where p = pressure coefficient Pt = pressure at tC Po= pressure at 0C Volume coefficient of a gas is similarly defined as (pressure being kept constant) V Vo v = t Vo t Where v = volume coefficient Vt = volume at tC Vo= volume at 0C Experiments have shown the value of p (or v) is the same for all gases and equal to 1/273 per degree celsius, i.e. 1 1 perC or per K, p = v = 273 273 Where K stands for absolute of kelvin temperature. Example 2. A grid iron pendulum consists of 5 iron rods and 4 brass rods. What will be the length of each brass rod if the length of each iron rod is 1 m ? (Fe= 12 10-6/C, brass = 20 10-6/C) Solution: There are 5 iron rods and 4 brass rods in the pendulum. The number of iron in one half of the pendulum including the central rod is 3. The number of brass rods is 2. Since the pendulum is compensated, lbrass liron or
2l2 2 t 3l11t or l2
3l11 2 2
3 1 12 10 6 m 0.9m 2 20 106
2.
ERROR in measurament DUE TO THERMAL EXPANSION
2.1
HEATING A METALLIC SCALE
A metallic scale (linear) expands in length when heated. As a result all the markings are displaced from their usual (correct) positions. A reading of l unit on a heated scale is equivalent to an actual length of l 1 t , where is coefficient of linear expansion of material of scale, and t is the temperature of the heated scale. If the reading is x, actual length x(1 t) actual length reading (1 t)
2.2
1
2
3 Scale
1
2
3 Heated scale
DIFFERENCE OF LENGTHS OF TWO RODS
Consider two rods 1 and 2 of lengths l1 and l2. Let they be heated through a temperature t . If l’1 and l’2 are their expanded lengths, then: l’2 l2 1 2 t where 2 is coefficient of linear expansion of rod 2 l’1 l1 1 1t where 1 is coefficient of linear expansion of rod 1 Since the difference of length of two rods is constant l '2 l '1 l 2 l1 l11 l 22
Example3. Two rods of length l1 and l2 are made of materials whose coefficients of linear expansion are 1 and 2 respectively. If the difference between the two lengths is independent of temperature, then relation between length and linear expansion coefficients. Solution: l1 l11t l2 l22 t
l1 l2 0
2.3
l11t l22 t or
l1 1 l2 2
TIME PERIOD OF PENDULUM
Time period (T) of a sample pendulum of length l is given by l T 2 g If there is a rise in temperature by t, length of the pendulum increases and hence the time period increase. As a result the clock slows down. If l0 be the length of the pendulum and corresponding time period be T0 then l T0 2 0 g If the pendulum be heated by t (rise in temperature), the new time period T1 is:
T1 2
l1 g
l 1 t T1 l 1 0 ln T0 l0
T1 1 as is very small 1 t 1 t T0 2 T T 1 T 1 1 0 t t T0 2 T0 2 The above relation gives the time lost per second by the pendulum clock. If t is the fall in temperature, same equation will give the time gained by the clock per second as its oscillation will become faster due to reduction in its length.
Example 4:
A clock with a metallic pendulum is 5 second fast each day at a temperature of 15 o C and 10 seconds slow each day at a temperature of 30 o C . Find coefficient of linear expansion for the metal.
Solution:
Time lost or gained per second by a pendulum clock is given by 1 ( Here T is difference of temperature) t T 2 Here temperature is higher then graduation temperature thus clock will loose time and if it is lower then graduation temperature will gain time. 1 t T 86400 [as 1 day 86400sec.] 2 If graduation temperature of clock is T0 then we have. At 15C, clock is gaining time, thus 1 5 = T0 15 86400 ……(1) 5 At 30C clock is loosing time, thus 1 ……(2) 10 = 30 T0 86400 5 Dividing equation (2) by (1), we get 2(T0 – 15) = (30 – T0) or T0 = 20C Thus from equation (1) 1 5 = 20 15 86400 2 2.31 105 C
2.4
STRESS IN OBJECTS DUE TO THERMAL EXPANSION
If cross–sectional area of wire is A and F be the tension developed in wire due to stretching. Thus the stress developed in the wire due to this tension F is given as Force Strees = = F/A …..(1) Area Strain produced in wire due to its elastic properties is L Strees = T …..(2) L If young’s modulus of the material of wire is Y, we have
Y
strees strain
F/A T or F YA T …..(3) Equation - (3) gives the expression for tension in the wire due to decrease in its temperature by T. This result gives the tension in wire if initially wire is just taught between clamps. If it already has some tension in it then this expression will give the increment in tension in the wire.
or
Y
Example 5. A uniform metal rod of 2 m m2 cross-section is heated from 0C to 20C. The coefficient of liner expansion of the rod is 12 10-6 per C, Y=1011 N/m2. The energy stored per unit volume of the rod is solution: Energy per unit volume 1 1 1 stress strain (Yt)(t) Y 2t 2 2 2 2 12 11 10 144 10 400 J m3 2 288 10 J m3 2880 J m3
3.
SPECIFIC HEAT
When heat energy flows into a substance, the temperature of the substance usually rises. The heat required to raise the temperature of unit mass of a body through 1 oC or ( 1 oK ) is called specific heat capacity or simply specific heat of the material of the body. If Q heat changes the temperature of mass m by T. 1 Q … (1) c m T The SI unit of specific heat is J/kg K. Heat is so frequently measured in calories, therefore the practical unit cal/g C is also used quite often. The specific heat capacity of water is approximately 1 cal/g C. From Eq. (1), we can define the specific heat of a substance as “the amount of energy needed to raise the temperature of unit mass of that substance by 1C (or 1 K)”. A closely related quantity is the Molar heat capacity C. It is defined as, 1 dQ … (2) C n dT where n is the number of moles of the substance. If M is the molecular mass of the m substance, then n = were m is the mass of the substance and, M M dQ C= … (3) m dT The SI units of C is J / mole K. Key points: (a) It depends on nature of material of body. Dulong and petit has found formula for elemental solids that (with few exceptions such as carbon) Atomic weight Specific heat 6 cal / oC So, heavier the element lesser will be the specific heat, i.e., CHg < CCu< CAl
Specific heat of a substance also depends on temperature (particularly at low temperatures) the variation of specific heat with temperature for wateris shown in Fig (A) for metals in Fig (B). This temperature dependence of specific heat is usually neglected. 1.008
6
1.004
4
Molar Sp. heat
Sp. heat cal / gm C
(b)
1.000
0.996
40
60
80
100
100
Temp. in C
Fig. (A) Water
(e)
(f)
(g) (h)
(i)
300 200 Temp. in K
400
500
Fig. (B) Metals
Specific heat also depends on the state of substance, i.e., solid, liquid or gas. e.g., specific heat of solid copper will be different from that of liquid copper. In case of water Specific heat of Ice (solid) Water (liquid) Steam (gas)
(d)
2
0 20
(c)
c T3
In cal/g C 0.5 1 0.47
In SI units, i.e., J/kg K 2100 4200 1970
If a substance is undergoing change of state which takes place at constant temperature (called isothermal change) , specific heat 1 dQ 1 dQ [as T = 0] c m 0 m dT i.e., specific heat of a substance at its melting pint or boiling point or isothermal change is infinite. Specific heat is found to be maximum for hydrogen (3.5 cal/gm C) then for water (1 cal/gm C = 4200 J/kg K). For all other substances specific heat is lesser than 1 cal/gm C and is minimum for radon and actinium (= 0.22 cal/gam C) If the temperature of a body changes without transfer of heat with the surroundings (adiabatic change) as in shaking a liquid or compressing a gas, Q 0 c= =0 [as Q = 0] mT mT i.e., specific heat of a substance, when it undergoes adiabatic change, is zero. Specific heat of a substance can also be negative. Negative specific heat means that in order to raise the temperature, a certain quantity of heat is to be withdrawn from the body. Specific heat of saturated water vapours is negative. When specific heats are measured, the values obtained are also found to depend on the conditions of the experiment. In general measurements made at constant pressure are different from those at constant volume. For solids and liquids this difference is very small and usually neglected. The specific heat of gases are quite different under constant pressure condition (cp) to constant volume condition (cv). As by definition c = (Q/m T), heat required to change the temperature of m gm of a substance through T: Q = mc T and as T = (Q/mc), greater the specific heat of a substance lesser will be the change in temperature for a given mass when same amount of heat is supplied. Now as specific heat of water is very large (1 cal/g C), by absorbing or releasing large
amount of heat its temperature changes by small amounts. This is why, it is used in hot water bottles or as coolant in radiators. This is also how the sea moderates the climate of nearby coastal land.
4.
WATER EQUIVALENT
Water–equivalent of a body is the mass of water which when given same amount of heat as to the body, changes the temperature of water through same range as that of the body, i.e., W = (m c) gm The unit of water equivalent W is gm while its dimensions [M]. Units and dimensions of some physical–quantities used in heat are given below in a tabular form. S. No. 1. 2. 3.
Physical quantity Heat Specific–heat Molar sp. heat
Symbol Q c C
Dimensions
1 –1
]
4. 5. 6.
5.
Latent heat Thermal capacity Water–equivalent
L Tc W
[ML2T–2] [L2T–2–1] [ML2T–2– [L2T–2] [ML2T–2–1] [M]
Units SI Joule J/kg K J/mol K
CGS calorie cal/gm C cal/mol C
J/kg J/K kg
cal/gm cal/C gm
PHASE CHANGES AND LATENT HEAT
Suppose that we slowly heat a cube of ice whose temperature is below 0C at atmospheric pressure, what changes do we observe in the ice? Initially we find that its temperature increases according to equation Q = mc(T2 – T1). Once 0C is reached, the additional heat does not increase the temperature of the ice. Instead, the ice melts and temperature remains at 0C. The temperature of the water then starts to rise and eventually reaches 100C, whereupon the water vaporizes into steam at this same temperature. During phase transitions (solid to liquid or liquid to gas) the added heat causes a change in the positions of the molecules relative to one another, without affecting the temperature. The heat necessary to change a unit mass of a substance from one phase to another is called the latent heat (L). Thus, the amount of heat required for melting and vaporizing a substance of mass m are given by, Q = mL … (1) For a solid-liquid transition, the latent heat is known as the latent heat of fusion (Lf) and for the liquid-gas transition, it is known as the latent heat of vaporization (Lv).
6.
PRINCIPLE OF CALORIMETRY
When two bodies (one being solid and other liquid or both being liquid) at different temperatures are mixed, heat will be transferred from body at higher temperature to body at lower temperature till both acquire same temperature. The body at higher temperature releases heat while body at lower temperature absorbs it, so that: Heat lost = Heat gained, i.e. principle of calorimetry represents the law of conservation of heat energy. While, using this principle always keep in mind that:
(a)
Temperature of mixture (T) is always lower temperature (TL) and higher temperature (TH), i.e., TL T TH i.e. the temperature of mixture can never be lesser than lower temperatures (as a body cannot be cooled below the temperature of cooling body) and greater than higher temperature (as a body cannot be heated above the temperature of heating body when there is no chemical reaction).
(b)
When temperature of a body changes, the body releases heat if its temperature falls and absorbs heat when its temperature rises. The heat released or absorbed by a body of mass m is given by: Q = mc T Where c is specific heat of the body and T change in its temperature in C or K.
(c)
When state of a body changes, change of state takes place at constant temperature [m.pt. or b.pt.] and heat released or absorbed is given by Q = mL Where L is latent heat. Heat is absorbed if solid converts into liquid (at m.pt.) or liquid converts into vapours (at b.pt.) and is released if liquid converts into solid or vapours converts into liquid.
HEAT TRANSFER Heat can be transferred from one place to the other by any of three possible ways: conduction, convection and radiation. In the conduction, convection processes, a medium is necessary for the heat transfer. Radiation, however, does no have this restriction. This is also the fastest mode of heat transfer, in which heat is transferred from one place to the other in the form of electromagnetic radiation. (i) Conduction Figure shows a rod whose ends are in thermal contact with a hot reservoir at temperature T1 and a cold reservoir at temperature T2. The sides of the rod are covered with insulating medium, so the transport of heat T1 T2 is along the rod, not through the sides. The molecules at the hot reservoir have greater vibrational energy. This T1 Q energy is transferred by collisions to the atoms at the end (Hot) face of the rod. These atoms in turn transfer of heat through a substance in which heat is transported without direct mass transport is called conduction. Most metals use another, more effective mechanism to conduct heat. The free electrons, which move throughout the metal can rapidly carry energy from the hotter to cooler regions, so metals are generally good conductors of heat. The presence of ‘free’ electrons also causes most metals to be good electrical conductors. A metal rod at 5C feels colder than a piece of wood at 5C because heat can flow more easily from your hand into the metal. Heat transfer occurs only between regions that are at different temperature, and the rate dQ . This rate is also called the heat current, denoted by H. Experiments of heat flow is dt show that the heat current is proportional to the cross-section area A of the rod and to the
T2 (Cold)
temperature gradient
dT , which is the rate of change of temperature with distance along dx
the bar. In general dQ dT H= kA dt dx
… (1)
dQ dT a positive quantity since is negative. The dt dx constant k, called the thermal conductivity is measure of the ability of a material to conduct heat. A substance with a large thermal conductivity k is a good heat conductor. The value of k depends on the temperature, increasing slightly with increasing temperature, but k can be taken to be practically constant throughout a substance if the temperature difference between its ends is not too great. Let us apply Eq. (1) to a rod of length L and constant cross sectional area A in which a steady state has been reached. In a steady state the temperature at each point is constant in time. Hence, dT T1 T2 dx Therefore, the heat Q transferred in time t is T T Q kA 1 2 t …(2) L
The negative sign is used to make
Thermal Resistance (R) Eq. (2) in differential form can be written as dQ T T …(3) H dt l / kA R Here, T = temperature difference (T.D) and l = thermal resistance of the rod. R kA (ii) Convection Although conduction does occur in liquids and gases also, heat is transported in these media mostly by convection. In this process, the actual motion of the material is responsible for the heat transfer. Familiar examples include hot-air and hot-water home heating systems, the cooling system of an automobile engine and the flow of blood in the body. You probably have warmed your hands by holding them over an open flame. In this situation, the air directly above the flame is heated and expands. As a result, the density of this air decreases and then air rises. When the movement results from differences in density, as with air around free, it is referred to as natural convection. Air flow at a beach is an example of natural convection. When the heated substance is forced to move by a fan or pump, the process in called forced convection. If it were not for convection currents, it would be very difficult to boil water. As water is heated in a kettle, the heated water expands and rises to the top because its density is lowered. As the same time, the denser, cool water at the surface sinks to the bottom of the kettle and is heated. Heating a room by a radiator is an example of forced convection. Ingen Hausz Experiment:
Ingen Hausz provided a method to compare the thermal conductivities of different materials. He took wax coated rods of different materials but of the same area. One end of the rods is kept in a hot water bath and the other end is kept at the temperature of surroundings. If 1, 2, 3 . . . represent the lengths upto which the wax has melted and K1, K2, K3 . . . are their thermal conductivities, then K1 21
or
K2 22
= . . . = constant
K 1 21 K 2 22
Example-6. A wall is made of two equally thick layers A and B of different materials. The thermal conductivity of A is twice that of B. In the steady state, the temperature difference across the wall is 36C. The temperature difference across the layer A will be Solution: K constant A K 1 B 2K 2 1 A 36 12C 2 1
(iii) Radiation The third means of energy transfer is radiation which does not require a medium. The best known example of this process is the radiation from sun. All objects radiate energy continuously in the form of electromagnetic waves. 1 2 Energy 2 0E The rate at which an object radiates energy is proportional to the fourth power of its absolute temperature. This is known as the Stefan’s law and is expressed in equation form as P = AeT4 Here P is the power in watts (J/s) radiated by the object, A is the surface area in m2, e is a fraction between 0 and 1 called the emissivity of the object and is a universal constant called Stefan’s constant, which has the value = 5.67 10–8 W/m2-K4 BLACK BODY RADIATION (i)
Perfectly black body
A body that absorbs all the radiation incident upon it and has an emissivity equal to 1 is called a perfectly black body. A black body is also an ideal radiator. It implies that if a black body and an identical another body are kept at the same temperature, then the black body will radiate maximum power as is obvious from equation p = eat4 also. Because e = 1 for a perfectly black body while for any other body e < 1. Materials like black velvet or lamp black come close to being ideal black bodies, but the best practical realization of an ideal black body is a small hole leading into a cavity, as this Absorbs 98% of the radiation incident on them. (ii) Absorptive power ‘a’ “It is defined as the ratio of the radiant energy absorbed by it in a given time to the total radiant energy incident on it in the same interval of time.” energy absorbed a energy incident As a perfectly black body absorbs all radiations incident on it, the absorptive power of a perfectly black body is maximum and unity. (iii) Spectral absorptive power ‘a’ The absorptive power ‘a’ refers to radiations of all wavelengths (or the total energy) while the spectral absorptive power is the ratio of radiant energy absorbed by a surface to the radiant energy incident on it for a particular wavelength . It may have different values for different wavelengths for a given surface. Let us take an example, suppose a = 0.6, a = 0.4 for 1000 Å and a = 0.7 for 2000 Å for a given surface. Then it means that this surface will absorbs only 60% of the total radiant energy incident on it. Similarly it absorbs 40% of the energy incident on it corresponding to 1000 Å and 70% corresponding to 2000 Å. The spectral absorptive power a is related to absorptive power a through the relation
a a d 0
(iv) Emissive power ‘e’ (Don’t confuse it with the emissivity e which is different from it, although both have the same symbols e). “For a given surface it is defined as the radiant energy emitted per second per unit area of the surface.” It has the units of W/m2 or J/s–m2. For a black body e=T4. (v) Spectral emissive power ‘e’ “It is emissive power for a particular wavelength .” Thus,
e e d 0
Kirchhoff’s law: “According to this law the ratio of emissive power to absorptive power is same for all surfaces at the same temperature.” T
T
1
2
Perfectly black body
T
Fig.
e1 e e 2 Perfectly black body a1 a2 a but (a)black body = 1 and (e)black body = E (say) e Then, cons tan t E a for any surface Similarly for a particular wavelength , e E a for any body Hence,
Here
E = emissive power of black body at temperature T = T4 From the above expression, we can see that e a i.e., good absorbers for a particular wavelength are also good emitters of the same wavelength. COOLING BY RADIATION Consider a hot body at temperature T placed in an environment at a lower temperature T0. The body emits more radiation than it absorbs and cools down while the surrounding absorb radiation from the body and warm up. The body is losing energy by emitting radiations at a rate. P1 = eAT4 and is receiving energy by absorbing radiations at a rate P2 = aAT04 Here ‘a’ is a pure number between 0 and 1 indicating the relative ability of the surface to absorb radiation from its surroundings. Note that this ‘a’ is different from the absorptive power ‘a’. In thermal equilibrium, both the body and the surrounding have the same temperature (say Tc) and, P1 = P2 or eATc4 = aATc4 or e=a Thus, when T > T0, the net rate of heat from the body to the surroundings is, dQ eA (T4 – T04) dt dT 4 4 or mc eA (T – T0 ) dt Rate of cooling
dT dt
or
eA 4 (T T0 4 ) mc
dT (T4 – T04) dt
NEWTON’S LAW OF COOLING According to this law, if the temperature T of the body is not very different from that dT of the surroundings T0, then rate of cooling – is proportional to the temperature dt difference between them. T0 prove it let us assume that T = T0 + T 4
T So that T = (T0 + T) = T 1 T0 4T (from binomial expansion) T04 1 T0 4
4
4 0
(T4 – T04 ) = 4 T03 (T)
or (T4 – T04 ) T (as T0 = constant) Now, we have already shown that rate of cooling dT 4 4 dt (T T0 ) and here we have shown that (T 4 T04 ) T , if the temperature difference is small. Thus, rate of cooling d dT dt T or dt as dT = d or T = Variation of temperature of a body according to Newton’s law Suppose a body has a temperature i at time t = 0. It is placed in an atmosphere whose temperature is 0. We are interested in finding the temperature of the body at time t, assuming Newton’s law of cooling to hold good or by assuming that the temperature difference is small. As per this law, rate of cooling temperature difference d eA 3 or dt mc (40 ) 0 d or dt 0 4eA30 Here = is a constant mc t d i 0 0 dt
0
0 i 0 et
cons tan t
i
t0
0 cons tan t
tt
FROM THIS EXPRESSION WE SEE THAT = I AT T = 0 AND = 0 AT T = , I.E. TEMPERATURE OF THE BODY VARIES EXPONENTIALLY WITH TIME FROM I TO 0 (<I). THE TEMPERATURE VERSUS TIME GRAPH IS AS SHOWN IN FIG.
i
0
t
NOTE: IF THE BODY COOLS BY RADIATION FROM 1 TO 2 IN TIME T, THEN TAKING THE APPROXIMATION d 1 2 and = av = 1 2 t dt 2 d The equation 0 becomes dt 1 2 1 2 0 t 2 This form of the low helps in solving numerical problems related to Newton’s law of cooling.
WEIN’S DISPLACEMENT LAW At ordinary temperatures (below about 600C) the thermal radiation emitted by a body is not visible, most of it is concentrated in wavelengths much longer than those of visible light. Figure shows how the energy of a black body radiation varies with temperature and wavelength. As the temperature of the black body increases, two distinct behaviours are observed. The first effect is that the peak of the distribution shifts to shorter wavelengths. This shift if found to obey the following relationship called Wein’s displacement law maxT = b Here b is a constant called Wein’s e constant. The value of this constant in 4000 K SI unit is 2.898 10–3 m–K. Thus, 3000 K 1 max T 2000 K Here max is the wavelength corresponding to the maximum spectral emissive power e. 0 3 1 2 4 The second effect is that the total Wavelength ( m) amount of energy the black body emits per unit area per unit time (= T4) increases with fourth power of absolute temperature T. This is also known as the emissive power. We know e =
0
ed = Area under e–
graph = T4
Area T4
or
e
A1
T m
A2
2T m 2
A2 = (2)4A1 = 16A1 Thus, if the temperature of the black body is made two fold, max remains half while the area becomes 16 times. Example-7. Estimate the surface temperature of sun. Given for solar radiations, m = 4753 A° Solution: From Wien’s displacement law m T = b T=
b (2.89 10 3 m k ) = 6097°K. m ( 4753 10 10 m)
Example-8. Experimental investigations show that the intensity of the solar radiation is maximum for wavelength m 4753 A o in the visible region. Estimate the surface temperature of the sun. Assume the sun to be a black body. Wien’s constant (b) = 2.892 10 3 mk . Solution : Wien’s law states that m T = constant = b T
b 2.892 10 3 mk m 4753 10 3 m
= 6060 K Example-9: A body cools from 62°C to 50°C in 10 minutes and to 42°C in the next 10 minutes. Find the temperature of surroundings. Solution: For the first ten minutes, 62o 50o dT = –1.2°C/min and dt 10 62 50 T = T0 (56 T0 )o C 10
1.2°C/min = KA (56T0)°C Similarly for the next ten minutes
…(1)
dT 42o 50o = 0.8 °C/min and dt 10 42 50 T = T0 ( 46 T0 )o C 2
0.8°C/min = KA (46To)°C Dividing (1) by (2) 3 56 T0 2 46 T0
T0 = 26°C.
…(2)
measurment OF SPECIFIC HEAT CAPACITY & error analysis As shown in the figure Regnault’s apparatus to determine the specific heat capacity of a solid heavier than water, and insoluble in it. A wooden partitions P separates a steam chamber O and A calorimeter C. The steam chamber O is a double walled cylindrical vessel. Steam can be passed in the space between the two walls through an inlet A and it can escape out through an outlet B. The upper part of the vessel is closed by a cork. The given solid may be suspended the vessel is closed by a cork. The given solid may be suspended in the vessel by a thread passing through the cork. A thermometer T1 is also inserted into the vessel to record the temperature of the solid. The stem chamber is kept on a wooden platform with a removable wooden disc D closing the bottom hole of the chamber. To start with, the experimental solid (in the form of a ball or a block) is weighed and then suspended in the steam chamber. Steam is prepared by boiling water in a separate boiler and is passed through the steam chamber. A calorimeter with a stirrer is weighed and sufficient amount of water is kept in it so that the solid may be completely immersed in it. The calorimeter is again weighed with water to get the mass of the water. The initial temperature of the water is noted. When the temperature of the solid becomes constant (say for 15 minutes), the partition P is removed. The calorimeter is taken below the steam chamber, the wooden disc D is removed and the thread is cut to drop the solid in the calorimeter. The calorimeter is taken to its original place and is stirred. The maximum temperature of the mixture is noted. Calculation: Let the mass of the solid =m1 mass of the calorimeter and the stirrer =m2 mass of the water =m3 specific heat capacity of the solid =s1 specific heat capacity of the material of the calorimeter(and stirrer) =s 2 specific heat capacity of water =s3 initial temperature of the solid =1 initial temperature of the calorimeter, stirrer and water =2 final temperature of the mixture = We have heat lost by the solid =m1s1(1-) heat gained by the calorimeter (and the stirrer) =m2s2(-2) Assuming no loss of heat to the surrounding the heat lost by the solid goes into the calorimeter stirrer and water. Thus m1s1(1 – ) = m2s2 (–2)+ m2s3( – 2) ……(1) (m2s2 m3 s3 )( 2 ) or s1 = m1(1 ) Knowing the specific heat capacity of water (s3 = 4186 J/kg-K) and that of the material of the calorimeter and the stirrer (s2=389 J/kg-K if the material be copper), one can calculate s1.Specific heat capacity of a liquid can also be measured with the Regnault apparatus. Here a solid of known specific heat capacity s1 is used and the
experimental liquid is taken in the calorimeter in place of water. The solid should be denser than the liquid. Using the same procedure and with the same symbols we get an equation identical to equation (1) above, that is, m1s1(1– ) = m2s2 (–2) + m3s3( – 2) in which s3 is the specific heat capacity of the liquid. We get m s (1 )m2s2 S3 = 1 1 m3 ( 2 )m3 Error analysis After correcting for systematic errors, equation (1) is used to estimate the remaining errors.
OBJECTIVE 1.
A piece of metal floats on mercury. The coefficients of volume expansion of the metal and mercury are 1 and 2 respectively. If their temperature is increased by T, the fraction of the volume of metal submerged in mercury changes by a 1 2 T 1 2 T (A) (B) 1 1T 1 1T
1 2 T 2 (D) 1 1 1T Solution: The correct choice is (A). It follows from the fact that the volumes of metal and mercury increase to V0 (1 + 1 T) and (1+ 2 T) respectively; Vo being the initial volume. (C)
2.
The coefficient of expansion of a crystal in one direction (x–axis) is 2.0 10–6 K–1 and that in the other two perpendicular (y–and z–axes) direction is 1.6 10–6 K–1. What is the coefficient of cubical expansion of the crystal? (A) 1.6 10–6 K–1 (B) 1.8 10–6 K–1 (C) 2.0 10–6 K–1 (D) 5.2 10–6 K–1 Solution: (D) Coefficient of cubical expansion is = x + y + z = x + 2 y ( y = z) –6 = 2.0 10 + 2 1.6 10–6 = 5.2 10–6 K–1 Hence the correct choice is (D). 3.
uniform metal rod of length L and mass M is rotating with angular speed about an axis passing through one of the ends and perpendicular to the rod. If the temperature increases by t C, then the change in its angular speed is proportional to (A) (B) 1 2 (C) (D) Solution: (B) At tC, the length of the rod becomes L’= L (1 + t), where is the coefficient of linear expansion. From the law of conservation of angular momentum, we have
or
I = I’’ 1 1 ML2 = ML’2 ’ 3 3 2
1 ' L or = L' (1 t)2 Now, for a given value of t, (1+ at)–2 is constant, say k. ' =k ' or k 1 or ’ – = (k – 1) i.e. (’ – ) . Hence the correct choice is (B).
4.
Solution:
o cylindrical rods or lengths l1 and l2, radii r1 and r2 have thermal conductivities k1 and k2 respectively. The ends of the rods are maintained at the same temperature difference. If l1 = 2l2 and r1 = r2/2, the rates of heat flow in them will be the same if k1/k2 is: (A) 1 (B) 2 (C) 4 (D) 8 (D) The rate of heat flow in rods A and B are Q1 k1r12 Q2 k 2 r22 and t l1 t l2 Q1 = Q2, if k1 l1 r22 k1r12 k 2r22 or 2 = 2 (2)2 = 8 k 2 l2 r1 l1 l2 Hence the correct choice is (D).
5: The temperature of the two outer surfaces of a x 4x composite slab, consisting of two materials having coefficients of thermal conductivity K and 2K and thickness x and 4x, respectively, are T2 and T1 (T2 > T1). T 2K The rate of heat transfer through the slab, in a steady 2 K A(T2 T1 )K state is f, with f equal to (see Fig.) x (A) 1 (B) 1/2 (C) 2/3 (D) 1/3 Solution: (C) Let A be the area of each slab. In the steady state, the rate of heat flow through the composite slab is given by A T2 T1 T2 T1 Q … (1) l1 l2 l1 l2 t K1A K 2 A K1 K 2 Given l1 = x, l2 = 4x, K1 = K and K2 = 4K. Using these values in (1) we get Q A T2 T1 A T2 T1 K 2 x 4x t x 3 K 2K
T1
Comparing this with the given rate of heat transfer, we get f =
2 . Hence the 3
correct choice is (C). 6.
Three identical rods A,B and C of equal lengths and equal diameters are joined is series as shown in following fig. Their thermal conductivities are 2K,K and K/2 respectively. Calculate the temperature at two junction points.
T1 100C
Solution:
7:
Solution:
or
T2
C 0C A B 0.5K 2K 85.7, 57.1C 77.33, 48.3C
(A) (B) 80.85, 50.3C (C) (D) 75.8,49.3C (A) ith 1 = ith 2 = ith 3 (100 T1 )2KA (T1 T2 )KA (T2 0)KA l l 2l 3T2 T (100 – T1) = (T1 – T2) = 2 T1 = 2 2 T (100 – T1) = 1 3 2T1 2 85.7 600 = 85.7C and T2 = = 57.1C r T1 = 7 3 3 Three rods made of the same material ane having the same cross-section have been joined as shown in. Each rod is of the same length. The left and right ends are kept at 0C and 90C respectively. The temperature of the junction of the three rods (A) 45C (B) 60C (C) 30C (D) 20C I=I1+I2 kA(90 T) (90 t)kA (t 0)kA I l l l 0 TC 3T = 180C or T = 60C
90C
0C
90C
90C
I1
I2 90C
8:
A solid copper sphere (density p and specific heat C) of radius r at an initial temperature 200 K is suspended inside a chamber whose walls are at almost 0 K. What is the required for temperature of the sphere to drop to 100 K? 7rpc ´ 10- 6 7rpc ´ 10- 8 (A) (B) 72es 72es - 10 7rpc ´ 10 7rpc ´ 10+6 (C) (D) 72es 72es
Solution:
(A) According to Stefan’s law P = eAT4
dt eAt 4 dQ mcdT eAT 4 or dr mc dt dt 2 4 t rpc dT dT e4r T or dt 3 p4rc 3e T 4 o dt 3
or
rpc 1 9e T3
100
200
7rpc 106 seconds 72e
A double pane window used for insulating a room thermally from outside consists of two glass sheets each pf area 1 m2 and thickness 0.01 m separated by a 0.05 m thick stagnant air space. In the steady state the room glass interface and glass outdoor interface are at 27C and 0C respectively. Calculate the rate of flow of heat through the windowpane. Also find the flow of heat through the windowpane. Also find the temperature of other interface if, conductivities of glass and air are 0.8 and 0.08 Wm-1 K-1 respectively. (A) 0.72C (B) 0.52C (C) 0.192C (D) 0.32C dQ KA dt L R L 0.05 1 0.01 = Req = 2 KA A 0.8 0.08 1 5 26 A = 1 m2 Req = 40 8 40 dQ (27 0) 40 = 41.5 W dt R 26 27 1 or 1 = 26.48C 41.5 = 0.8 12 0.01 0.8 12 (2 0) 41.5 = or 2 = 0.52C 0.01
(B)
Room
27C
0.01
Ce Outside
Air K 0.08
Kg 0.8
Solution:
Kg 0.8
9:
t=
0.05m
1
2
0.01
10:
A space object has the space of a sphere of radius R. Heat sources ensure that the heat evolution at a constant rate is distributed uniformly over its volume. The amount of heat liberated by a thermodynamic temperature. In what proportion would the temperature of the object change if its radius is decreased to half? (A) 2.19 (B) 5.19 (C) 1.19 (D) 4.19
Solution:
(C)
Heat liberated
T4 R T14 R Thus = 4 T2 R/2 Or
dQ dQ 2 4 R3 and RT dt dt
0C
4
T1 1 Or T2 = T1 or T2 = 1.19 2 That is , temperature decreases by a factor of 1.19.
11:
Solution:
12:
Solution:
13.
14.
The room temperature is +20C when outside temperature is 20C and room temperature room temperature is + 10C when outside temperature is -40C. Find the temperature of the radiator heating the room. (A) 30C (B) 60C (C) 90C (D) 45C (B) Applying Newton’s law In case (1) K1(T-Tr1)= K2(Tr2-Tout 1) And in case (2) K1(T-Tr2)= K2(Tr2-Tout 2) Dividing these equations T -Tr1 Tr2 -Tout1 T 20 20 ( 20) = T -Tr2 Tr2 -Tout2 T 10 10 ( 40) or T=60C Some water at 0C is placed in a large insulated enclosure (vessel). The water vapour formed is pumped out continuously. What fraction of the water will ultimately freeze, if the latent heat vaporization is seven times the latent heat of fusion? (A) 7/8 (B) 8/7 (C) 3/8 (D) 5/8 (A) m = mass of water, f = fraction which freezes L1 = latent heat of vaporization L2 = latent heat of fusion L1 = 7L2 Mass of water frozen = mf Heat lost by freezing water = m fL2 Mass of vapour formed = m(1 - f) Heat gained by vapour = m (1 – f) L1 mfL2 = m (1 - f) x 7L2 f = 7 – 7f or f = 7/8 A substance is in the solid form at 0C. The amount of heat added to this substance and its temperature are plotted in the following graph. If the relative specific heat capacity of the solid substance is 0.5, find from the graph, the mass of the substance is (Specific heat capacity of water = 1000 cal kg-1 K-1 ) (A) 0.02 kg (B) 2 kg (C) 0.04 kg (D) 0.05 kg
t
120 100 80 60 40 20
C A
B 2160
0
1000 Q
A substance is in the solid form at 0C. The amount of heat added to this substance and its temperature are plotted in the following graph. If the relative specific heat capacity of the solid substance is 0.5, find from the graph, the specific latent heat of the melting process is (Specific heat capacity of water = 1000 cal kg-1 K-1 )
2000
(calories)
(A) 6000 cal kg-1 (C) 1000 cal kg-1
4000 cal kg-1 2000 cal kg-1
(B) (D)
A substance is in the solid form at 0C. The amount of heat added to this substance and its temperature are plotted in the following graph. If the relative specific heat capacity of the solid substance is 0.5, find from the graph, the specific heat of the substance in the liquid state is (Specific heat capacity of water = 1000 cal kg-1 K-1 ) (A) 300 cal kg –1 K-1 (B) 500 cal kg –1 K-1 –1 -1 (C) 700 cal kg K (D) 100 cal kg –1 K-1
15.
Solution 13: (A) 800 calories of heat raise the temperature of the substance from 0C to 80C. 800 = m (1000 0.5 ) 80 ( specific heat = relative sp. heat x sp. heat of water) or m = 0.02 kg Solution 14: (B) Latent heat = 200 4 = 800 cal ( = 0.02 L L= 4000 cal kg-1
1 div reads 200 cal )
Solution 15: (C) In the liquid state temperature rises from 80C to 120C, that is, by 40C after absorbing (2160 – 1600) cal.
0.02 s 40 = 2160 – 1600 or s = 700 cal kg –1 K-1
16:
An earthenware vessel loses 1g of water per second due to evaporation. The water equivalent of the vessel is 0.5 kg and the vessel contain 9.5 kg of water. Find the time required for the water in the vessel to cool to 28C from 30C. Neglect radiation losses. Latent heat of vaporization of water in this range of temperature is 540 cal g-1 . (A) 2 min 5 s (B) 3 min 5 s (C) 5 min 3 s (D) 6 min 2 s
Solution:
(B) Here water at the surface is evaporated at the cost of the water in the vessel losing heat. Heat lost by the water in the vessel = (9.5 + 0.5) 1000 (30 – 20) = 105 cal Let t the required time in seconds. Heat gained by the water at the surface = (t 10-3) 540 103 ( L = 540 cal g-1 = 540 103 cal kg-1) 105 = 540 t or t = 185 s = 3 min 5 s
17:
In an industrial process 10 kg of water per hour is to be heated from 20C to 80C. To do this, steam at 150C is passed from a
boiler into a copper coil immersed in water. The steam condenses in the coil and is returned to the boiler as water at 90C. How many kg of steam are required per hour? Specific heat of steam = 1 kilo cal kg–1C–1. Latent heat of steam = steam = 540 kilo cal kg–1. (A) 1 kg (B) 2 kg (C) 3 kg (D) 4 kg Solution:
(A) Let the mass of steam required per hour be m kg. Heat gained by water in boiler per hour is = 10 kg 1 kilo cal kg–1 C–1 (80 – 20)C = 600 kilo cal … (1) Heat lost by steam per hour is = heat needed to cool m kg of steam from 150C to 100C + heat needed to convert m kg of steam at 100C into water at 100C + heat needed to cool m kg of steam at 100C to 90C = m 1 (150 – 100) + m 540 + m 1 (100 – 90) = 50 m + 540 m + 10 m = 600 m kilo cal … (2) Heat lost = heat gained. Equating (1) and (2) we have 600 m = 600 or m = 1 kg
18:
A closed cubical box made of a perfectly insulating material has walls of thickness 8 cm and the only way for the heat to enter or leave the box is through two solid cylindrical metal plugs, each of cross–sectional area 12 cm2 and length 8 cm fixed in the opposite walls of the box. The outer surface A of one plug is kept at a temperature of 100C while the outer surface of the other plug is maintained at a temperature of 4C. The thermal conductivity of the material of the plug is 0.5 cal cm–1 s–1 (C)–1. A source of energy generating 36 cal s–1 is enclosed inside the box. Find the equilibrium temperature of the inner surface of the box assuming that it is the same at all points on the inner surface. (A) 70C (B) 45C (C) 65C (D) 76C Insulator
(D) At equilibrium, the total energy generated by the source per second must Metal plug equal the heat leaving per second through the two metal plugs (Fig.). Let TC be the A equilibrium temperature. Then heat leaving atSurface 100 C the box per second through surface A k(T 100) 12 = cal s–1 8 and heat leaving the box per second through the surface B k(T 4) 12 cals1 8 12k Hence (T – 100 + T –4) = 36 8
Solution:
Metal plug Energy source Surface B at 4 C
or 2T – 104 =
36 8 36 8 = = 48 or T = 76C 12k 12 0.5
19:
Find the time during which a layer of ice of thickness 2.0 cm on the surface of a pond will have its thickness increased by 2 mm when the temperature conductivity of ice = 5 10-3 cal cm-1 s-1 (C)-1, density of ice at 0C = 0.91 g cm-3 and latent heat of fusion = 80 cal g-1 (A) 6 min 5 s (B) 2 min 6 s (C) 5 min 6 s (D) 3 min 5 s
Solution:
(C) substituting the given quantities in the expression pL t= ( x 22 - x12 ) 2kT we have 0.91 80 t= [(2.2)2-2.0)2] = 306 s = 5 min 6 s. 3 2 5 10 20
20:
Water is being boiled in a flat-bottomed kettle placed on a stove. The area of the bottom is 300 cm2 and the thickness is 2 mm. If the amount of steam produced is 1 g/min, calculate the difference of temperature between the inner and other outer surfaces of the bottom. The thermal conductivity of the material of kettle = 0.5 cal cm-1 s-1 (C)-1 and the latent heat of stem = 540 cal g-1. (A) 0.012 C (B) 0.0.04 C (C) 0.02 C (D) 0.0.08 C
Solution:
(A) If T is the temperature difference between the inner ad outer surface of the bottom of the kettle, then the amount of heat flowing through the bottom per second is Q kAT 0.5 300T = 750 cal per second … (1) t d 0.2 Q mL 1g 540calg1 But … (2) t t 60s Equating (1) and (2), we have 9 750 T = 9 or T = = 0.012 C 750
21:
If an anisotropic solid has coefficients of linear expansion x, y and z for three mutually perpendicular directions in the solid, what is the coefficient of volume expansion for the solid? (A) x + y + z (B) 2x + y + z 2 2 2 (C) x+ y+ z (D) 3x + 3y + 3z
Solution:
(A) Consider a cube, with edges parallel to X, Y, Z of dimension L 0 at T = 0. After a change in temperature T = (T –0), the dimensions change to
given by
Lx = L0 (1 + xT) Ly = L0(1 + yT) Lz = L0(1 + zT) and the volume of the parallelepiped is V = V0 (1 + xT)(1 + yT)(1 + zT) V0 [1 + (x + y + z)T] Where V0 = L30 . Therefore, the coefficient of volume expansion is x + y + z.
22.
In aluminum sheet there is a hole of diameter 2m and is horizontally mounted on a stand. Onto this hole an iron sphere of radius 2.004 m is resting. Initial temperature of this system is 25° C. Find at what temperature, the iron sphere will fall down through the hole in sheet. The coefficients of linear expansion for aluminum and iron are 2.4 10–4and 8.6 10 –5 respectively. (A) 82C (B) 43C (C) 45C (D) 15C
Solution:
(B) As value of coefficient of linear expansion for aluminum is more than that for ion, it expends faster then iron. So at some higher temperature when diameter of hole will exactly become equal to that of iron sphere, the sphere will pass through the hole. Let it happen at some higher temperature T. Thus we have at this temperature T, (diameter of hole)Al = (diameter of sphere)iron 2[1 + Al (T – 25)] = 2.004[1 + iron(T – 25)] 2Al (T – 25) = 0.004 + 2.004 iron (T – 25) 0.004 25 C or T 2 AL 2.004iron 0.004 or T 25 2 2.4 104 2.004 8.6 105 or T = 43C
23.
An iron ball has a diameter of 6 cm and is 0.010 mm too large to pass through a hole in a brass plate when the ball and plate are at a temperature of 30C. At what temperature, the same for ball and plate, will the ball just pass through the hole? (A) 23.8C (B) 13.8C (C) 53.8C (D) 83.8C (C) We let I stand for the iron ball and B stand for the brass plate. LI = 6 cm and LI – LB = 0.001 cm at t = 30C. Since the brass plate expands uniformly, the hole must expand in the same proportion. Then heating both the ball and the plate leads to increases in the diameters of the ball and the hole, with the hole increasing more, since B > I. We require LB – LI = 0.001 cm. LB = BLB t; LI = ILI t. We can approximate LB in this formula by 6 cm = LI. Then LB –LI = (B – I)LI t = 0.001 cm solving, t = 23.8C, and finally t = 30C + 23.8C = 53.8C
Solution:
24.
It is desired to put an iron rim on a wooden wheel. The diameter of the wheel is 1.1000m and the inside diameter of the rim is
1.0980 m. If the rim is at 20C initially, to what temperature must it be heated to just fit onto the wheel? (A) 52C (B) 72C (C) 102C (D) 152C (D) for iron is found to be 1.2 10–5 C–1. L = 1.1000 – 1.0980 = 0.0020 m = L t 0.0020 = (1.2 10–5)(1.098) t t + 20 = 172C t = 152C
Solution:
25.
Find the coefficient of volume expansion for an ideal gas at constant pressure. 1 (A) (B) g =T T 1 1 (C) (D) g= 3 g= 2 T T (A) For an idea gas PV= nRT As P is constant, we have P.dV = nRdT dV nR dT P 1 dV nR nR 1 V dT PV nRT T 1 T
Solution:
26.
What should be the lengths of steel and copper rod so that the length of steel rod is 5cm longer then the copper rod at all the temperatures. Coefficients of linear expansion for copper and steel are 1.7 and 1.1 (A) 2.17, 14.17 cm (B) 9.17, 14.17 cm (C) 9.17, 18.17 cm (D) 3.17, 5.17 cm (B) It is given that the difference in length of the two rods is always 5 cm. Thus the expansion in both the rods must be same for all temperatures. Thus we can say that at all temperature differences, we have LCu Lsteel or Cu l1t st l 2t [If l1 and l2 are the initial lengths of Cu and steel rods] or Cu l1 st l 2 or 1.7 l1 1.1 l 2 …..(1) It is given that l2 – l1 = 5cm ….(2) 1.7 1.1 1 l1 5cm 5 1.1 9.17cm or l1 0.6
Solution:
Now from equation (2) l2 = 14.17 cm A steel wire of cross-sectional area 0.5 mm2 is held between two rigid clamps so that it is just taut at 20 o C . Find the tension in the wire at 0 o C . Given that Young’s modulus of steel is Yst = 2.1 1012 dynes / cm2 and coefficient of linear expansion of steel is st 1.1 105 C .
27.
(A) (C)
2.31 10–4 9.31 10–6
(B) (D)
2.31 10–2 2.31 10–6
Solution:
(D) We know that due to drop in temperature, then tension increment in a clamped wire is T = YA T = 2.1 1012 0.5 10–2 1.1 10–5 20 = 2.31 10–6
28.
Two bodies have the same heat capacity. If they are combined to form a single composite body, show that the equivalent specific heat of this composite body is independent of the masses of the individual bodies. 2s1s2 s1s2 (A) (B) s2 - s1 s2 + s1 s2 2s1s2 (C) (D) s2 + s1 s2 + s1
Solution:
(C) Let the two bodies have masses m 1, m2 and specific heats s1 and s2 then m1s1 = m2s2 or m1/m2 = s2/s1 Let s = specific heat of the composite body. Then (m1 + m2) s = m1s1 + m2s2 = 2m1s1 2m1s1 2m1s1 2s1s2 s= m1 m2 m1 m1(s1 / s2 ) s2 s1
29.
20 gm steam at 100C is let into a closed calorimeter of water equivalent 10 gm containing 100 gm ice at – 10C. Find the final temperature of the calorimeter and its contents. Latent heat of steam is 540 cal/gm, latent heat of fusion of ice=80 cal/gm, specific heat of ice = 0.5 cal/C gm. (A) 13C (B) 63C (C) 93C (D) 33C
Solution: (D) Heat lost by steam = mL + ms (100 - ) where, is the equilibrium temperature Heat lost by steam = 20 540 + 20 1 (100 - ) = 10800 + 2000 - 20 Heat gained by (ice + calorimeter) = 100 80 + 100 0.5 10 + 100 Now Heat lost = Heat gained or
10800 + 2000 – 20 = 8000 + 500 + 110 130 = 4300
or
=
4300 = 33C. 130
30.
Victoria Falls in Africa is 122 m in height. Calculate the rise in temperature of the water if all the potential energy lost in the fall is converted to heat. (A) 0.29 K (B) 29 K (C) 0.69 K (D) 0.99 K
Solution:
(A) Consider mass m of water falling. mgy = mc t gy = c t We express both sides in joules by noting c = 1 kcal/kg K = 4184 J/kg K Then 9.8(122) = 4184 t and t = 0.29 K
31.
An electric heater supplies 1.8 kW of power in the form of heat to a tank of water. How long will it take to heat the 200 kg of water in the tank from 10 to 70C? Assume heat losses to the surroundings to be negligible. (A) 1.75 h (B) 7.75 h (C) 4.75 h (D) 5.75 h
Solution: (B) The heat added is (1.8 kJ/s)t and the heat absorbed is cm T = (4.184 kJ/kg K) (200 kg)(60 K) = 5.0 104 kJ Equating heats, t = 2.78 104 s = 7.75 h. 32.
What will be the final temperature if 50 g of water at 0C is added to 250 g of water at 90C? (A) 15C (B) 30C (C) 45C (D) 75C
Solution: (D) Heat gained = heat lost. (We assume no hat transfer to or from container.) (50 g)(1.00 cal/g C)(t – 0C) = (250 g)(1.00 cal/g C)(90C – t) where t is the final equilibrium temperature. 50t = 22500 – 250t or 300t = 22500 t = 75C 33.
A 500 g piece of iron at 400C is dropped into 800 g of oil at 20C. If c = 0.40 cal/gC for the oil, what will be the final temperature of the system. Assume no loss to the surroundings. (A) 15.7C (B) 30.7C (C) 45.7C (D) 75.7C
Solution:
(D) Heat lost = heat gained is written as 0.11(500)(400 – t) = 0.40(800)(t – 20) from which t = 75.7C
34.
A copper rod 2 m long has a circular cross section of radius 1 cm. One end is kept at 100C and the other at 0C, and the surface is insulated so that negligible heat is lost through the surface. Thermal conductivity of copper is 401 W/m–K. Find the thermal resistance of the bar. (A) 15.9 K/W (B) 20.9 K/W (C) 40.9 K/W (D) 50.9 K/W
35.
A copper rod 2 m long has a circular cross section of radius 1 cm. One end is kept at 100C and the other at 0C, and the surface is insulated so that negligible heat is lost through the surface. Thermal conductivity of copper is 401 W/m–K. Find the thermal current H. (A) 1.3 watt (B) 6.3 watt (C) 12.3 watt (D) 18.3 watt
36 .
A copper rod 2 m long has a circular cross section of radius 1 cm. One end is kept at 100C and the other at 0C, and the surface is insulated so that negligible heat is lost through the surface. Thermal conductivity of copper is 401 W/m–K. Find the dT temperature gradient . dx (A) –10C/m (B) –40C/m (C) –50C/m (D) –90C/m
37 .
A copper rod 2 m long has a circular cross section of radius 1 cm. One end is kept at 100C and the other at 0C, and the surface is insulated so that negligible heat is lost through the surface. Thermal conductivity of copper is 401 W/m–K. Find the temperature 25 cm from the hot end. (A) 20.5C (B) 40.5C (C) 77.5C (D) 87.5C
Solution 34 (A) Thermal resistance R = or R = Solution 35 (B) or
2
401 10 2
2
l l kA k ( r 2 )
= 15.9 K/W
Thermal current, H =
T 100 R R 15.9
H = 6.3 watt
0 100 = – 50 K/m = –50C/m 2 Solution 37 (D) Let 0c be the temperature at 25 cm from the hot 100 C end, then ( – 100) = (temperature gradient) (distance) or – 100 = (– 50) (0.25) or = 87.5C
Solution 36 (C) Temperature gradient =
38.
C
0.25 m 2.0 m
Two metal cubes with 3 cm–edges of copper and aluminium are arranged as shown in figure. Thermal conductivity of copper is
0 C
401 W/m–K and that of aluminium is the total thermal current from one other. (A) 0.14 K/W (B) (C) 0.74 K/W (D) 39.
237 W/m–K. Find reservoir to the 100 C
0.34 K/W 0.94 K/W
Al Cu
wo metal cubes with 3 cm–edges of copper and aluminium are arranged as shown in figure. Thermal conductivity of copper is 401 W/m–K and that of aluminium is 237 W/m–K. Find the ratio of the thermal current carried by the copper cube to that carried by the aluminium cube. (A) 0.75 K/W (B) 0.05 K/W (C) 0.15 K/W (D) 0.35 K/W
Solution: 38 (A) Thermal resistance of aluminium cube R1 = or R1 =
(3.0 10 2 )
(237) 3.0 10
2
2
= 0.14 K/W
Solution: 39 (B) Thermal resistance of copper cube R2 = or R2 =
(3.0 10 2 )
(401) 3.0 10 2
2
l kA
l kA
= 0.08 K/W
As these two resistances are in parallel, their equivalent resistance will be
R1R2 R1 R2 (0.14)(0.08) = = 0.05 K/W (0.14) (0.08)
R=
40.
What temperature gradient must exist in an aluminum rod in order to transmit 8 cal/s per square centimeter of cross section down the rod? k for aluminum is 0.50 cal/s cm C. (A) 36C/cm (B) 46C/cm (C) 86C/cm (D) 16C/cm
Solution:
(D) H = kA
T x
or 8 cal/s = (0.50 cal/s cm C)(1 cm2)
T x
T = magnitude of temperature gradient = 16C/cm. x The actual sign of the gradient depends on whether the heat flow is in the positive or negative x direction, being negative or positive for the two cases, respectively.
41.
What is heat conduction. (A) (A T)/d (C) (kA T)/d
(B) (D)
(k T)/d (T)/d
20 C
Solution: (C) The conduction equation, H Q/t = (kA T)/d can be reexpressed as H = T/R, where R d/(kA) is called the thermal resistance of the slab. It has units of K/W in SI. 42.
Consider the two insulating sheets with resistances R1 and R2 shown in Fig. What is the value of T' ? R T + R2T2 R T R1T2 (A) (B) T ¢= 1 1 T 2 1 R1 + R2 R1 R2 R T + R1T2 R T + R1T2 (C) (D) T ¢= 2 1 T ¢= 2 21 R1 - R2 R1 + R22
Solution:
(B) For the two sheets, H1 = (T1 – T)/R1, H2 = (T – T2)/R2. Noting that H1 = H2 = H, we have (T1 – T)/R1 = (T – T2)/R2. Cross multiplying we get R2(T1 – T) = R1(T – T2), or rearranging terms, R2T1 + R1T2 = (R1 + R2)T, and T’ = (R2T1 + R1T2)/(R1 + R2).
T2 R2
R1
H
43.
A spherical blackbody of 5 cm radius is maintained at a temperature of 327C. What is the power radiated? (A) 231 W (B) 431 W (C) 531 W (D) 631 W
Solution:
(A) The surface area of a sphere is 4r2. In this case, then, the area is 4(25 10–4) = 0.01m2. The power radiated is given by Stefan’s law: P= T4A = (5.67 10–8 W/m2.K4)(600 K)4(0.01 m2) = 231 W.
44..
A spherical blackbody of 5 cm radius is maintained at a temperature of 327C. What wavelength is the maximum wavelength radiated. (A) 2.82 m (B) 4.82 m (C) 3.82 m (D) 5.82 m
Solution:
(B) The surface area of a sphere is 4r2. In this case, then, the area is 4(25 10–4) = 0.01m2. The power radiated is given by Stefan’s law: P= T4A = (5.67 10–8 W/m2.K4)(600 K)4(0.01 m2) = 231 W By Wien’s law, m(600 K) = 2898 m.K and m = 4.82 m.
45.
A sphere of 3 radius acts like a blackbody. It is equilibrium with its surroundings and absorbs 30 kW of power radiated to it from the surroundings. What is the temperature of the sphere? (A) 2300 K (B) 2800 K (C) 3000 K (D) 2600 K
Solution:
(D) The power absorbed by a blackbody is Pa = A Ta4 , or (30 103 W)= (5.67 10–8 W/m2.K4) 4(0.03 m)2 Ta4 .
T T1
Ta4 = 4.68 1013; Ta = temperature of surroundings = 2600 K. Since the body is in equilibrium with its surroundings, it is at the same temperature, 2600 K.
46 .
A blackbody is at a temperature of 527C. To radiate twice as much energy per second, what temperature must be increased ? (A) 951 K (B) 451 K (C) 551 K (D) 651 K
Solution:
(A) Since P T4, the temperature must be increased to 21/4(800 K) = 951 K.
47.
Use Stefan’s law to calculate the total power radiated per square meter by a filament at 1727C having an absorption factor of 0.4. (A) 0.16 MW/m2 (B) 0.26 MW/m2 (C) 0.36 MW/m2 (D) 0.46 MW/m2
Solution:
(C) Stefan’s law gives R = T4 = 0.4(5.67 10–8)(2000)4 = 0.36 MW/m2
48.
A blackbody is at a temperature of 527C. To radiate twice as much energy per second, how many times of increase in radiated power when the temperature of a blackbody is increased from 7 to 287C. (A) 5 times (B) 16 times (C) 6 times (D) 20 times
Solution:
(B) Since P T4, the temperature must be increased to 21/4(800 K) = 951 K. 4
P(560 K) 560 And 16 times P(280 K) 280
49.
The initial and final temperature of water as recorded by an observer are (40.6 0.2) °C and (78.3 0.3) °C. Calculate the rise in temperature with proper error limit. (A) (27.7 0.5)°C (B) (17.7 0.5)°C (C) (37.7 0.9)°C (D) (37.7 0.5)°C
Solution:
(D) Let 1 = 40.6°C, 1 = 0.2 °C 2 = 78.3°C, 2 = 0.3 °C = 2 – 1 = 78.3 – 40.6 = 37.7°C & = (1 + 2) = (0.2 + 0.3) = 0.5°C Hence rise in temperature = (37.7 0.5)°C
50.
A cylinder of radius R made of a material of thermal conductivity k1 is surrounded by a cylindrical sheet of inner radius R and outer radius 2R made of material of thermal conductivity k 2 . The two ends of the combined system are maintained at two different
temperatures. There is no loss of surface and the system is in steady thermal conductivity of the system. (A) 4K = 2K1 + 3K2 (C) 4K = 6K1 + 4K2 Solution :
heat across the cylindrical state. Calculate the effective (B) (D)
4K = 5K1 + 3K2 4K = K1 + 3K2
(D) Two cylinders are in parallel, therefore equivalent thermal resistance R is given by 1 1 1 R R1 R 2
kA kA k 1A 1 k 2 A 2 1 2
R
But
Here A 2 2R R2 3R2
1 2 , A1 R2
2
and
A 2R 4R 2
K 4R 2 K 1R 2 K 2 3R 2
i.e.
4K = K1 + 3K2
2
51.
A lake is covered with ice 2 cm thick. The temperature of ambient air is 15oC. Find the rate of thickening of ice. For ice k = 4 104k-cal-m1s1(C)1. Density =9103 kg/m3and latent heat L = 80 Kilo Cal/Kg. (A) 2.5 cm/ hour (B) 1.5 cm/ hour (C) 3.5 cm/ hour (D) 4.5 cm/ hour
Solution :
(B) Heat energy flowing per sec is given by H
Q KA x t
. . . (I)
if dm mass of ice is increased in time dt, then dm A dx dx A. . dt dt dt
Since,
dm H L dt dx L . . . (ii) H A dt
From eq. (I) and (II) A L
dx KA dt x
Rate of thickening of ice = dx/dt 0( 15) 410 4 dx KA = K = 3 dt AL x L x 0.910 80 210 2
= 4.166 106 m/s = 1.5 cm/ hour.
52.
A body cools from 60C to 50C in 10 minutes. If the room temperature is 25C and assuming Newton’s law of cooling to hold good, find the temperature of the body at the end of the next 10 minutes. (A) 42.85C (B) 12.85C (C) 20.85C (D) 52.85C
Solution :
(A) According to Newton’s law of cooling rate of loss of heat (T To) where T = mean temperature of the body
60 50 10
Also,
60 50 25 2
. . . (1)
50 T 50 T 25 10
2
. . . (2)
where T is the required temperature Solving both equation, we get, T = 42.85C 53.
Two bodies A and B have thermal emissivities of 0.01 and 0.81 respectively. The outer surface areas of the two bodies are same. The two bodies emit total radiant power at the same rate. The wavelength B corresponding to maximum spectral radiancy in the radiation from B is shifted from the wavelength corresponding to maximum spectral radiancy in the radiation from A by 1.00 m. If the temperature of A is 5802 K, calculate (a) the temperature of B and (b) wavelength B. (A) 2000K, 1.6 m (B) 3000K, 1.8 m (C) 1934K, 1.5 m (D) 4000K, 1.9 m
Solution :
(C) (a) According to stefan’s law the power radiated by a body is give by P eAT4
According to the given problem, PA PB , with A A A B So that e A TA 4 eB TB 4 i.e, 0.01 5802 4 0.81TB 4 1 3
or TB 5802 1934K (b)
According to Wien’s displacement law. A TA B TB ; i.e, B 5802 / 1934 A i.e, B 3 A and also B - A = 1m(given) 1 3
so B B 1m i.e. B 1.5m
A body which has a surface area 2.00 cm2 and a temperature of 727oC radiates energy at the rate of 5 W. What is its emissivity? (Stefan boltzman constant 5.67 10 8 W / m2k 4 ). (A) 0.14 (B) 0.44 (C) 0.34 (D) 0.24
54
(B) P e AT4
solution:
5 e 5.67 10 8 2 10 4 1000 5 e 8 5.67 10 2 10 4 1012 4
5 0.44 11.34
55. Two rods of equal cross-sectional area A and length L L L are joined and what is the temperature of junction as shown A A k 2k in the figure. The temperature of one end of the composite o o 100 C 0 C o o rod is 0 C and the other end of the rod is at 100 C . Draw a graph showing the variation of the temperature with the distance x from the end maintained at 0 o C . Assume that the flow is longitudinal and there is no loss of heat from the lateral surface. 100 (x + L ) 100x L (A) (B) T T= 3L 4L 100 (x + L ) 100 (x + L ) (D) (C) T= T= L 5L L L 3L kA 2kA 2kA 1002kA 200kA H 3L 3L 200 x T 200kA T 3L x / kA 3L
solution:
(A) R
L o
0C
L 2k
k
o
100 C
when x L
x L x L L kA 2kA 2kA T2kA 200kA x L 3L 100x L T 3L R
56.
The graph is as shown
T o
100 C
O
L
2L
x
Two spherical soap bubbles coalesce to form a bigger bubble without any leakage of air. If V is the total change in volume of the contained air and S the total change in surface area, what is the value of 3PV + 4ST, where T is the surface tension of the soap solution and P is the atmospheric pressure. (A) 0 (B) 1 (C) 2 (D) 3
solution:
(A) Let r1 and r2 be the initial radii of two bubbles and r is the final radius of the bubble after they coalesce. Then 4T 4 3 rT 4 4T 4 3 P r1 P r23 P r r1 3 r2 3 r 3 P r13 r23 r 3 4T r12 r22 r 2 0 …(1) 4 also V r13 r23 r 3 = change in total volume 3 and S 4 r12 r22 r 2 = change in surface area
…(2) …(3)
from (1), (2) & (3) we get 3PV 4ST 0
57.
The graphs gives the variation of temperature of two bodies having the same surface area with time where Ax and Ay represent absorptivity and x and y represent emissivity then (A) x > y and Ax < Ay (B) x < y and Ax > Ay (C) x > y and Ax > Ay (D) x < y and Ax < Ay
T
x y t
solution: (D)According to Kirchoff’s law good emitter is a good absorber Ey > Ex and Ay > Ax The temperature gradient in a rod of 0.5 m length is 80 oC/m. If the temperature of hotter end of the rod is 30oC, then the temperature of the cooler end is (A) 40oC (B) –10oC o (C) 10 C (D) 0oC. 30 Solution: (B) 80 0.5
58.
59. A body at 300oC radiates 10 J cm–2 s–1. If Sun radiates 105 J cm–2 s–1, then its temperature is (A) 3000oC (B) 5457oC 4o (C) 300 × 10 C (D) 5730oC Solution : (B) (573)4 105 T4 10 The ratio of energy of radiation emitted by a black body at 27oC and 927oC is (A) 1:4 (B) 1 : 16 (C) 1 : 64 (D) 1 : 256. 4 Solution:(B) E=T
60.
13-Electrostatics
ELECTROSTATICS ELECTRIC CHARGE Charge is the physical property of certain fundamental particles (like electron, proton). Charges are divided into two parts (i) positive (ii) negative. Like charges repel and unlike charges attract. SI unit of charge is coulomb and CGS unit is esu. 1C = 3 109esu. Magnitude of the smallest known charge is e = 1.6 10-19C (charge of one electron or proton).
PROPERTIES OF ELECTRIC CHARGE QUANTIZATION OF CHARGE charge on any body is the integral multiple of the charge on an electron Q = ne, where n = 0, 1, 2, ................. Distribution of Charges
The concentration of the charges is maximum on a surface with greater curvature.
COULOMB’S LAW The force between two point charges is directly proportional to the product of their charges and inversely proportional to the square of the distance between them, In free space F=
1 q1 q 2 4 o r2
- permittivity of the medium
In material medium F=
1 q1 q 2 4 r 2
Where = 0 in vector form F
r
q1 q 2 1 r 4 0 r3
0 – permittivity of free space (Vacuum) r = relative permittivity of the medium 1 4 0
9 10 9
Nm 2 coul 2
0 = 8.85410-12 1. 2. 3. 4.
coul 2 Nm 2
This is a fundamental law and is based on physical observation The force is an action - reaction pair. The direction of force is always along the line joining the two charges. Electrical force between two point charges is independent of presence or absence of other charges.
Example 1.
A particle ‘A’ having a charge of 2 10 C and a mass of 100g is fixed at the bottom of a smooth inclined plane of inclination 30 . Where should another particle B, having same charge and mass be placed on the incline so that it may remain in equilibrium ?
Solution :
First of all draw the F.B.D. of the masses. For equilibrium F = 0 N = mg cos30 . . . (1) Fe = mg sin30 . . . (2)
-6
From (2)
x=
Electric Field Intensity F q0 q q 1 r E 4 0 r 3 q 1 E 4 0 r 2
kq 2 mg x2 2
N
Fe B
x mg sin30 mg cos30
2kq 2 = 27cm. mg
A 30
It is the force experienced by a unit positive charge placed at a point in an electric field.
E lim
[In vacuum or free space] N/coul or Volt/m
Example 2.:
Two particles A and B having charges 8 x10 C and –2 x10 C respectively are held fixed with a separation of 20 cm. Where should a third charged particle be placed so that it does not experience a net electric force ?
Solution :
As the net electric force on C should be equal to zero, the force due to A and B must be opposite in direction. Hence, the particle should be placed on the line AB. As A and B have charges of opposite signs, C cannot be between A and B
-6
-6
Also A has larger magnitude of charge than B. Hence, C should be placed closer to B than A. The situation is shown in figure. Suppose BC=x and the charge on C is Q y 1 (8.0 106 ) Q ˆ i 2 4π 0 (0.2 x ) A B 6 20 cm 1 (2.0 10 )Q ˆ and FCB i 2 4 0 x 1 (8.0 10 6 ) Q (2.0 10 6 ) Q FC FCA FCB i 4 0 (0.2 x )2 x2 But | FC | 0
FCA
Hence
FCB
C
x
1 (8.0 10 6 ) Q (2.0 10 6 ) Q 0 4 0 (0.2 x )2 x2
Which gives x = 0.2 m Superposition principle Force experienced by a given charge in the field of a number of point charges is the vector sum of all the forces. Fnet F1 F2 F3 F4
Similarly for electric field
Enet E1 E 2 E 3 ...........
Lines of Force Lines of force originate from positive charges and terminate on negative charges Lines of force originate or terminate perpendicular to the surface Tangent to the lines of force at any point gives the direction of the electric field. Lines of force do not intersect.
1.
Electric Field Intensity
Due to a point charge E=
q 1 4 0 r 2
Due to linear distribution of charge = linear charge density of rod
FCA x
(i)
, L
At a point on its axis.
a
dx
1 1 E= 4 0 a a L
x
= linear charge density (ii)
At a point on the line perpendicular to one end 1 1 Ex = 4 0 y y 2 L2 1 L Ey = 2 4 0 y [L y 2 ]1/ 2
Due to ring of uniform charge distribution dq
At a point on its axis
(R2 + x2)1/2 a
dE cos
P
qx 1 Ex 4 0 [a 2 x 2 ] 3 / 2
x
dE dE sin
Due to uniformly charged disc. At a point on its axis. Ex 2 0
x 1 [ x 2 a 2 ]1/ 2
= surface charge density
dr
r x
P
Thin spherical shell E
q 1 4 0 r 2
2.
Non conducting solid sphere having uniform volume distribution of charge
(i)
Outside point E=
1 4 0
q r2
P
(ii)
Inside point E
q = total charge
qr 1 4 0 a 3
= volume density of charge
Cylindrical Conductor of Infinite length Outside point E
2 0 r
= linear density of charge (as charges reside only on the surface)
Inside point E=0
Non-conducting cylinder having uniform volume density of charge 2 0 r r E 2 0 E
Outside point Inside point
Infinite plane sheet of charge E=
2 0
= surface charge density
Two oppositely charged sheets (Infinite) (i) Electric field at points outside the charged sheets EP = E R = 0 (ii) Electric field at point in between the charged sheets EQ =
0
Example 3.
Find electric field intensity due to long uniformly charged wire. (charge per unit length is )
Solution:
Elemental charge dq = d. Field at point P is dE = K .
R 4 0
Also
R
dq R 2
dq R 2 cos d Ex 3
dE x
2
dE y
2
d
X
2
2 2
dq sin R 2 2
Y
4 0R
R
dEX P dEY
dE
3.
On integration EY
4.
4 0 R
E E 2X E 2Y
tan
Example 4.
2 2 0 R
EX EY
45
What is the electric field at any point on the axis of a charged rod of length `L' and linear charge density`'? The point is separated from the nearer end by a. a
dx
P
x
Solution :
Consider an element, dx at a distance x from the point, P, where we seek to find the electric field. The elemental charge, dq =dx Then, dE k.
dx x2 aL
or
E k
a
Thus , E
Example 5.
1 dx k 2 x
aL
1 x a
1 1 k a L a
1 1 4 o a L a
[ k=
A ring shaped conductor with radius R carries a
y
total charge q uniformly distributed around it as
dq
shown in the figure. Find the electric field at a
R
point P that lies on the axis of the ring at a
1 ] 4 0
O
(R2 + x2)1/2
dE cos
P
x
x
distance x from its centre. dE dE sin
Solution:
Consider a differential element of the ring of length q ds . 2πR This element sets up a differential electric field d E at point P. The resultant field E at P is found by integrating the effects of all
ds. Charge on this element is
dq =
the elements that make up the ring. From symmetry this resultant field must lie along the right axis. Thus, only the component of d E parallel to this axis contributes to the final result. E = dE E = dE cos
1 dq 1 4 0 r 2 4 0 x cos = 2 (R x 2 )1 / 2
dE =
1 q ds . 2 2 2 R R x
To find the total x-component Ex of the field at P, we integrate this expression over all segment of the ring. Ex =
dE cos =
1 qx 4 0 2R. R2 x 2
ds 3/2
The integral is simply the circumference of the ring = 2R E=
1 qx 2 4 0 R x 2
3/2
As q is positive charge, field is directed away from the centre of the ring, along its axis.
5.
DIPOLE
Two equal and opposite charges separated by a small distance constitute an electric dipole P q 2a
P is directed from –ve to the +ve charge
(i)
Electric field at an Axial Point 1 2Px 4 0 [ x 2 a 2 ] 2
E=
x +q
A
-q
For x >> a E
(ii)
1 2P . 4 0 x 3
At a point on its perpendicular bisector E=
1 P 2 4 0 [a y 2 ] 3 / 2
For y >> a E
(iii)
A
1 4 0
P
y
+q
-q
y3
At any point r >> a
2P cos P sin 1 1 and E Er 3 4 0 4 0 r r3 1 P E 3 cos 2 1 3 4 0 r
Angle = tan-1 (1/2 tan )
A r
+q
-q
6.
GAUSS THEOREM
The flux of an electric field E through an arbitrary closed surface S is equal to the algebraic sum of the charge enclosed by this surface divided by 0.
q E. ds in 0
Example 6.
Figure shows a section of an infinite rod of charge having linear charge density which is constant for all points on the line. Find electric field E at a distance r from the line.
r E + + + ++ + + + + + + + + + + + + + + + +
Solution : From symmetry, E due to a uniform linear charge can only be radially directed. As a Gaussian surface, we can choose a circular cylinder of radius r and length l, closed at each end by plane caps normal to the axis. o E.ds qin o E.ds E.ds = qin
Cylindrical
Plane Surface
o E (2rl) + E.ds. cos 90 o l λl λ ε o 2πrl 2πε 0 r The direction of E is radially outward for a line of positive charge.
E=
Example 7. Figure shows a sphericaly symmetric distribution of charge of radius R. Find electric field E for points A and B which are lying outside and inside the charge distribution respectively. Solution:
The spherically symmetric distribution of charge means that the charge density at any point depends only on the distance of the point from the centre and not on the direction. Secondly, the object can not be a conductor, or else the excess charge will reside on its surface.
R
r
B
A
Now, apply Gauss’s law to a spherical Gaussian surface of radius r ( r > R for point A) o E .ds qin E=
ε oE( 4πr 2 ) = q
1 q where q is the total charge 4 0 r 2
For point B (r
3
r q 1 R 1 qr E= 40 40 R3 r2
7.
Electric Potential
Amount of work done in moving unit positive charge from infinity upto the point under consideration, against the field of a given charge q. In other words, it is negative of the work done by the internal forces. r
V = E. d r
=
1 q unit-volt 4 0 r
It is a scalar quantity (i)
Potential at a point due to several charges V = V1 + V2 + V3 + …
(ii)
Potential of a point due to an electric dipole V= For r >> a V=
P cos 1 4 0 (r 2 a 2 cos 2 )
P cos 1 4 0 r2
8.
Potential due to a uniformly charged disc
(i)
At a point on its axis
V= (ii)
(R
2
x 2 )1 / 2 x
= surface charge density
on disc At the centre of the disc V=
9.
2 0
R 2 0
Potential at the edge of a uniformly charged disc V=
R 0
Potential at the apex of a cone having charge Q distributed uniformly on its curved surface and having slanting length L V=
Q 2 0 L
Potential of a conducting sphere of radius R at a distance r from the centre (i)
r>R
(ii)
r=R
(iii)
r
Kq r Kq V= R
V=
V
q = charge on the sphere K=
1 4 0
kq R
10. Potential of a dielectric sphere (i)
q – total charge in the dielectric sphere
r>R V=
Kq r
(ii)
r=R
V=
Kq R
(iii)
r
V=
Kq r2 3 2R R 2
Torque on a dipole in presence of external electric field Torque = qE. 2a sin = PE sin PE
Potential Energy U = P . E
Example 8.
Two conducting spheres having radii a and b charged to q1 & q2 respectively. Find the potential difference between 1 & 2 ?
q2 q1 2
1
a b
Solution :
The potential on the surface of the sphere 1 is given by v1 = The
1 q1 1 q2 40 a 40 b
. . . . (A)
potential on the surface of the sphere 2 is given by, v2 =
1 q1 1 q2 , 4 o b 4 o b
V1 V2
1 q1 1 q1 4 o a 4 o b
V1 V2
q1 4 o
1 1 a b
11.
Electric Potential Energy The electric potential energy of a system of point charges is the amount of work done in bringing the charges from infinity in order to form the system. For point charges q1 and q2 separated by a distance r12, Electric potential energy of the system q1 and q2 is given by U=
1 q1 q 2 4 0 r12
For three particle system q1, q2 and q3 U=
1 4 0
q1 q 2 q1 q 3 q 2 q 3 r r13 r23 12
We can define electric potential (VP) at any point P in a electric field as , VP = UP ; where Up, is the change in electric potential energy in bringing the test q
charge q0 from infinity to point P. Example 9.
Determine the interaction energy of the point charges of the following setup
Solution :
U = U12 + U13 + U14 + U23 + U24 + U34
-q
a
+q 1
kq 2 kq 2 2 a 2a
= 3
kq 2 a2
2
kq 2 kq 2 kq 2 2 2 a a 2a
2
kq 2 2 a
2
a
a 4
(Here k = 1/ 40)
3
-q
a
+q
12. CAPACITORS 13.
Capacitance If Q is the charge given to a conductor and V is the potential to which it is raised by this amount of charge, then it is found that Q V. Or Q = CV, where C is a constant called capacitance of the conductor. C=
Q V
For parallel plate capacitor C =
0 r A d
i.e. the capacitance depends only on geometrical factors namely, the plate area and plate separation.
14. Spherical Capacitor C=
4 0 R1 R 2 (R 2 R1 )
15. Isolated Capacitor An isolated sphere can be thought of as a capacitor where other plate is at infinity. R1 = R and R2 =
C=
4 0 R1 4 0 R R1 1 R2
16. Cylindrical Capacitor C=
17.
2 0 R n 2 R1
R2 R1
18.
Combination of Capacitors Series Combination: In series combination, each capacitors has equal charge for any value of capacitance. Equivalent capacitance C is given by 1 C
C1 q
-q
1 1 1 C1 C2 C3
C2
C3
q -q
q -q
V
Parallel Combination: In parallel combination the potential differences of the capacitor connected in parallel are equal for any of capacitor. Equivalent capacitance is given by C = C1 + C2 + C3
q1
C1
q2
C2
q3
C3
V
Example 10.
In the above circuit, find the potential difference across AB.
8f
8f P
A
1
3 8f
10 V
2 Q
Solution:
8f
4 B
Let us mark the capacitors as 1, 2, 3 and 4 for identifications. As is clear, 3 and 4 are in series, and they are in parallel with 2. Then 2, 3, 4 combine is in series with 1. C34
C3 .C4 4f , C2,34 8 4 12f C3 C 4
Ceq.
8x12 4.8f , q Ceq. V 4.8 x10 48C 8 12
The `q’ on 1 is 48 C, thus V1=q/c=6v
[v1 =
48c 6v ] 8F
VPQ = 10 6 = 4V
By symmetry of 3 and 4, we say, VAB = 2V. Dielectric Substances having polar atoms/molecules intrinsically or being polarized are called dielectrics.
19.
Polar Dielectric
Substances which have polar atoms/molecules intrinsically but are randomly arranged. On application of external electric field they get polarized parallel to the external electric field. Net electric field inside dielectric. E E 0 Ei 1 E i E 0 1 K
E i = electric field due to induced charges
– surface charge density of capacitor plates
1 qi q 1 K
i – induced charge density
1 i 1 K
20. Capacitance of a parallel plate capacitor having dielectric slabs in series. C=
A 0 t1 t 2 ... K1 K 2
21. Energy stored in a Capacitor UC
Q2 1 1 CV 2 Q . V 2C 2 2
Energy supplied by Source US = Q. V. Energy lost in form of heat =
1 Q. V 2
22. Energy density of the electric field u
1 in free space 0 E 2 2 1 u K 0 E 2 in presence of dielectric medium ; K - dielectric 2
const.
23. Force on a Dielectric in a Capacitor F
Q 2 dC 2C 2 dx
when source is connected
Q C 1 2 dc V F dx 2
V=
Example 11. A 4f capacitor is charged to 150 V and another 6f capacitor is charged to 200 V. Then they are connected across each other. Find the potential difference across them. Calculate the heat produced. Solution :
4f charged to 150 V would have q1 = C1V1 = 600C
C1
q1
6f charged to 200 V would have q2 = C2V2 = 1200C After connecting them across each other, they will have a common potential difference V. Charges will readjust as q1’ and q2’ q 1 ' q 2 ' q 1 ' q 2 ' q q2 1800C 1 C 1 C 2 C 1 C 2 C 1 C 2 ( 4 6)f
V
V q2
C2
[conservation of ch arg e]
V 180 volt. Initial energy Ui
1 1 1 1 2 2 C 1 V1 C 2 V2 ( 4f )(150 V ) 2 (6f ).(200 V ) 2 0.165J 2 2 2 2
Final energy Uf
1 C1 C 2 .V 2 2
1 ( 4f 6f ).(180) 2 2
= 0.161J
Heat produced = |Uf Ui| = 0.003 J OBJECTIVE 1:
A long string with a charge of per unit length passes through an imaginary cube of edge a. The maximum flux of the electric field through the cube will be (A)
a o
(B)
2 a o
(C)
6 a 2 o
(D)
3a o
Solution :
The maximum length of the string which can fit into the cube is 3 a, equal to its body diagonal. The total charge inside the cube is
3 a , and hence the total flux through the cube is
3 a . o
(D) 2
2:
Potential in the x-y plane is given as V = 5(x + xy) volts. The electric field at the point (1, 2) will be (A) 3J V/m (A) 5 J V/m (C) 5J V/m (D) 3J V/m
Solution :
Ex =
V = (10 x + 5y) = 10 + 10 = 0 x V = 5x = 5 Ey = x E 5ˆj V / m .
(B) 3:
In the circuit shown, the equivalent capacitance between the points A and B is (A) C/5 (B) C/3 (C) C/2 (D) C
C C
A
C
B
C C
Solution:
Rearranging the circuit, the points E and D are at the same potentital (by symmetry). Then the capacity between E and D can be removed. 1 1 1 C C C
C = C/2 C C and are in parallel. 2 2
Hence Ceq =
(D)
C C =C 2 2
C
E C
A A C
C D
B 6C
4:
Electric field in a region is given by E (2iˆ 3jˆ 4kˆ) V/m. The potential difference between points (0, 0, 0) and (1,2,3) will be (A) 2 V (B) 5 V (C) 4V (D) 6 V
Solution:
P.d. across the points = E. r V2 – V1 = (2 ˆi 3ˆj 4kˆ ).(ˆi 2ˆj 3kˆ ) = 2 6 + 12 = 4 volts. (C)
5:
What is the mechanical work done in pulling the slab out of the capacitor after disconnecting it from the battery (A)
1 2 E C(r 1) 2
(C) E C (r 1) 2
Solution:
2
(B) E C (D) none of these
Work done = change in potential energy = U2 – U1 v1 = (1/2) E2C v2 =
1 (EC )2 1 E2C2 2 r = (1/2) E Cr 2 C' 2 C
Work done = (1/2) E2C (r – 1). (A). 6:
Solution:
Two point charges each of charge +q are fixed at (+a, 0) and (-a, 0). Another positive point charge q placed at the origin is free to move along x- axis. The charge q at origin in equilibrium will have (A) maximum force and minimum potential energy. (B) minimum force & maximum potential energy. (C) maximum force & maximum potential energy. (D) minimum force & minimum potential energy.
Y q (-a, 0)
The net force on q at origin is F F1 F2
=
1 q2 ˆ 1 q2 ˆ . 2 i . ( i ) 0ˆ 40 r 40 r 2
The P.E. of the charge q in between the extreme charges at a distance x from the origin along +ve x axis is U = =
1 q2 1 q2 . . 2 40 x 40 (a x ) 1 1 1 .q2 . 40 a x a x
dU q2 dx 40
1 1 2 (a x )2 (a x )
q q
X (+a, 0)
For U to be minimum dU 0, dx
d2U dx 2
0,
(a-x)2 = (a + x)2 a + x = (a – x) x = 0, because other solution is irrelevent. Thus the charged particle at the origin will have minimum force and minimum P.E. (D). 7:
In a parallel plate capacitor of capacitance C, a metal sheet is inserted between the plates, parallel to them. The thickness of the sheet is half of the separation between the plates. The capacitance now becomes (A) 4C (B) 2C (C) C/2 (D) C/4
Solution :
Before the metal sheet is inserted, C = o A / d After the sheet is inserted, the system is equivalent to two capacitors in series, each of capacitance C
o A 4C d/ 4
The equivalent capacity is now 2C. (B) 8:
The value of q if it floats in air is mgε r (A)
(C)
Solution :
2mgε r
(B)
q
m
mgε r
r
(D) none of these
qE = q(/r)
The magnitude of electric field due to the charged conducting plate is E =
σ
r
As the charged particle is floating in air (neglecting the buoyant force due to air we obtain)
q
mg
mg = qE mg E mg r q=
q=
(A) 9:
The capacitance of an filled parallel plate capacitor is 20 F. The separation between the plates is doubled and the space between the plates
is then filled with wax giving the capacitance a new value of 40 10 farads. The dielectric constant of wax is (A) 12.0 (B) 10.0 (C) 8.0 (D) 4.2
Solution :
101012 = K=8 (C)
10 :
o A K o A . 401012 = d 2d
The ratio of the time periods of small oscillation of the insulated spring and mass system before and after charging the masses is (A) 1 (C) 1
Solution :
q
k
+ m
(B) > 1 (D) = 1
In equilibrium of the charged small bodies
1 q2 = kx0 . 4 ( 0 x 0 )2
where x0 is the elongration in the spring in equilibrium. Let a further small elongation of x is made in the spring. Then net restoring force on any of the charged particle is given by,
F = k( x0 x )
1 q2 2 4 ( 0 x0 x )
= kx. a=
Since x < < x0 from (1) 2k x m
As F = a where = Hence =
2k m
mm a = 2 x, mm
T=2
m 2k
In absence of charge is T0 = 2 Therefore (D)
12
T 1 T0
m . 2k
q + m
11:
The charge flowing across the cell on closing the key k is equal to (A) CV (B) CV/2 (C) 2CV (D) zero
C
C 2
1
V
Solution:
When the key is kept open, the charge drawn from the source is Q = CeqV =
C V 2
When the key is closed the capacitor 2 gets short circuited And Ceq = C Q = CV charge flown through cell Q Q =
C V 2
(B) is correct choice. 12:
The conducting spherical shells shown in the figure are connected by a conductor. The capacitance of the system is (A) 40
ab
(B) 40 a
ba
(C) 40 b
Solution:
b
(D) 40
a
a2 ba
Hence, the capacitance of the system is the capacitance due to outer sphere of radius b. C = 4ob (C) is correct choice.
13:
A point charge q moves from point P to S along the path PQRS in a uniform electric field E pointing parallel to the positive direction of the x-axis. The coordinate of the point P, Q, R
y
P(a, b, 0)
and S are (a, b, 0), (2a, 0, 0), (a, b, 0) and (0, 0, 0) respectively. The work done by the field in the above process is given by the expression (A) qaE (C) q
a
2
b
2
E
(B) qaE (D) 3qE
a
Q(2a, 0, 0) x 100V
S(0, 0, 0) R(a, b, 0)
2
b2
E
Solution:
The work done is independent of the path followed and is equal to
qE . r ,
where r = displacement from P to S
Here, r = a ˆi bˆj , while E = E ˆi
work = (qE ˆi ). a ˆi bˆj = qaE Hence, B is correct choice. 14:
Charge Q is given to the upper plate of an isolated parallel plate capacitor of capacitance C. The potential difference between the plates (A)
(C)
Solution:
Q
(B)
C Q 2
Q
C 2
(D) zero
C
In general, for charge Q1 and Q2on upper and lower plate respectively the charge distributions on outer and inner part of the plates are shown in figure. Here Q1 =Q, Q2 = 0 Q1 Charge on inner side of Q1+Q2 plate
Q 2
are
and
respectively. Hence V =
Q 2
Q 2
2 Q1Q2 2 Q2
C
(Q1Q2) 2 Q1+Q2 2
Hence (C) is correct choice. 15:
A particle A of mass m and charge Q moves directly towards a fixed particle B. Which has charge Q. The speed of A is v when it is far away from B. The minimum separation between the particle is proportional to 2 (A) v (B) v 1 2 (D) v (C) v
Solution:
From Conservation of energy (KE+EPE)minimum separation = (KE+EPE)far away 1 Q2 1 mv 2 0 4 o rmin 2 1 rmin 2 v
0+
Hence, (D) is correct choice.
Q
A dipole of dipole moment p is kept along an electric field E such that E and p are in the same direction. Find the work done in rotating the
16:
dipole by an angle .
(B) W = –2Ep (D) W = –Ep
(A) W = 2Ep (C) W = Ep Ans. (A) Solution :
W = U = U2 – U1 Now U2 = (Ep cos ) = Ep U1 = (Ep cos 0) = Ep W = 2Ep.
17:
A simple pendulum of mass m and length carries a charge q. Find its time period when it is suspended in a uniform electric field region as shown in figure.
(A) T = 4
g 2 ( Eq / m) 2
(C) T = 2 2
g2 (Eq / m)2
(B) T = 2 (D) T = 2
E
g2 (Eq / m)2
2 g 2 ( Eq / m) 2
Ans. (B) Solution:
Time period of the pendulum = 2
geff
Here, geff = =
(mg )2 (Eq )2 m
=
g2 Eq / m
T=2
18:
Tension in the string in equibliriu m position mass of bob
2
g2 (Eq / m)2
.
A particle of mass 100 gm and charge 2 C is released from a distance of 50 cm from a fixed charge of 5C. Find the speed of the particle when its distance from the fixed charge becomes 3 m. Neglect any other force.
(A) –1.73 m/s
(B) 2.73 m/s
(C) –2.73 m/s Ans. (D) Solution:
Loss of potential energy = gain in kinetic energy U1 – U2 = K. 1
KQq r1
1 2 = ½ mv – 0 r2
2kQq m
v
r2 r1 r1 r2
2 9 109 5 10 6 2 10 6 2.5 = 1.73 m/s. 0.1 3 0.5
=
19:
(D) 1.73 m/s.
If a point charge q is placed at the centre of a cube what is the flux linked with the cube? (A)
(C)
1 q o
(B)
6 q o
(D)
3 q o
1 q 6 o
Ans. (A) Solution :
1 o
From gauss’s law, flux linked with a closed body is
times the
charge enclosed. The cube encloses a charge q so flux linked with cube, 20:
1 q o
If a point charge q is placed at the centre of a cube what is the flux linked with each face of the cube?
1 q o
(A)
(C)
6 q o
(B)
3 q o
(D)
1 q 6 o
Ans. (B) Now as cube is a symmetrical body with a faces and the point charge is at its centre, so electric flux linked with each face will be F
1 q 6 6 o
21:
2
Supposing that the earth has a surface charge density of 1 electron/m ; calculate earth's potential (A) 0.115Volt (C) 0.105Volt Ans. (A)
Solution:
0.110 Volt 0.112 Volt
(B) (D)
If R be the radius and the surface charge density of earth, then the total charge q on its surface is given by Q = 4R2 (i) The potential V at a point on earth's surface is same as if the entire charge q were concentrated as its centre. Thus V=
1 4R 2 R 1 q = 4 0 R o 4 0 R
Substituting the given value: V=
6.4 10 6 1.6 10 19 8.9 10 12
= 0.115 N-m/C = 0.115 J/C = 0.115Volt. 2
22:
Supposing that the earth has a surface charge density of 1 electron/m ; calculate electric field just outside earth's surface. 19 6 The electronic charge is 1.610 C and earth's radius is 6.410 m. 2 12 2 (0 = 8.910 C /N-m ) 8 (A) 1.810 N/C (B) + 1.8108 N/C 9 (C) 1.810 N/C (D) + 1.8109 N/C Ans. (A)
Solution:
(ii) Again, the electric field E just outside the earth's surface is same is if the entire charge q were concentrated at this cetnre. Thus E=
1 4R 2 1 q = 2 2 4 0 R 4 0 R o
Substituting the given value:
1.6 10 C / m 19
E=
2
8.9 10 12 C 2 / N m 2
= 1.8108 N/C The minus sign indicates that E is radially inward.
23 :
A circular wire of radius R carries a total charge Q distributed uniformly over its circumference. A small length of the wire subtending angle at the centre is cut off. Find the electric field at the centre due to the remaining portion .
(A)
(C)
Q
sin
4 0 R Q sin 2 2 4 0 R 4 2
2
(B)
(D)
sin 4 0 R 2 Q sin 2 2 4 0 R 8
Q
2
2
Ans. (B)
Solution :
Electric field due to an arc at its centre is k 1 , 2 sin , Where k = R 40 2
= angle subtended by the wire at the centre, = Linear density of charge. Let E be the electric field due to remaining portion. Since intensity at the centre due to the circular wire is zero. Applying principle of superposition.
k 2 sin nˆ E 0 R 2
E
Q .2 sin 4 0 R 2R 2 1
.
sin 4 0 R 2 Q
2
2
24: An electric dipole, made up of a positive and a negative charge, each of 1 C and 5 placed at a distance 2 cm apart, is placed in an electric field 10 N/C. Compute the maximum torque which the field can exert on the dipole, and the work that must be done to turn the dipole from a position = 0 to = 180
(A) 4106N-m or Joule (C) 410+6N-m or Joule Ans. (D)
Solution :
(B) (D)
4109N-m or Joule 4103N-m or Joule
The torque exerted by an electric field E on a dipole of moment p is given by = pE sin, Where is the angle which the dipole is making with the field.
is a maximum, when = 90. That is max = pE Here p = q(2) = 1106 0.02C/m and E = 105N/C max = 1106 0.02 105 = 2103N-m The work done in rotating the dipole from an angle o to is given by W=
o
pE sin d = pE (coso cos)
Here o = 0 and = 180 W = pE (cos0 cos180) = 2pE = 4103N-m or Joule 3
25:
A parallel-plate air capacitor has a plate area of 100cm and separation 5mm. A potential difference of 300V is established between its plates by a battery. After disconnecting a battery, the space between the plates is filled by ebonite (K = 2.6). Find out new potential difference between plates of the capacitor. (A) 115V (B) 120V (C) 110V (D) 125V Ans. (A)
26:
A parallel-plate air capacitor has a plate area of 100cm and separation 5mm. A potential difference of 300V is established between its plates by a battery. After disconnecting a battery, the space between the plates is filled by ebonite (K = 2.6). Find out initial and final capacitances of the capacitor.
3
(A) (C)
2.771011F, 4.61011F 1.771011F, 6.61011F
(B) (D)
1.771011F, 4.61011F 1.771013F, 4.61011F
Ans. (B) 27:
3
A parallel-plate air capacitor has a plate area of 100cm and separation 5mm. A potential difference of 300V is established between its plates by a battery. After disconnecting a battery, the space between the plates is filled by ebonite (K = 2.6). Find out initial and final surface-density of charge on the plates. +7
2
(A) 5.3110 C/m 7 2 (C) 5.3110 C/m Ans. (C)
Solution:
(B) 6.31107C/m2 (D) 9.31107C/m2
Capacity of the parallel plate air capacitor =
o A 8.86 10 12 100 10 4 = 1.771011F 3 d 5 10
Final capacity of the capacitor with dielectric between the plates is C = KC = 2.61.771011, C = 4.61011F
Initial charge on the capacitor 1.771011 300 = 5.31109C Since, the battery has been disconnected, the charge remains the same, therefore the new potential difference is V =
q 5.31 10 9 = 115V C 4.6 10 11
The surface density of charge remains the same in both the cases, i.e., = 28:
q 5.31 10 9 = = 5.31107C/m2 4 A 100 10
A spherical condenser has 10cm and 12 cm as the radii of inner and outer spheres. The space between the two is filled with a dielectric of dielectric constant 3. Find the capacity when the outer sphere is earthed.
(A) 41010F (C) 21010F Ans. (C)
q2 q1
R
r
(B) 11010F (D) 61010F
3 0.1 0.12 ab = 21010F = 9 0 . 12 0 . 1 b a 9 10
Solution:
C = 4k0
29:
A spherical condenser has 10cm and 12 cm as the radii of inner and outer spheres. The space between the two is filled with a dielectric of dielectric constant 3. Find the capacity when the inner sphere is earthed.
15 10 10 F 32 32 (C) 10 9 F 15 Ans. (D)
(A)
(B)
32 10 10 F 15
(D)
32 10 10 F 15
q2 q1
R
r
By choice of reference potential at infinity and the potential of earthed conductor zero, the given system can be visualised as combination of two spherical capacitors, both being at same potential difference. Connection wise these may be considered to be in parallel connection. 32 10 10 F C = 4ob + 4k0 ab = b a
15
30:
A charge Q is distributed over two concentric hollow spheres of radii r and R (R >r) such that the surface densities are equal. Find the potential at the common centre.
(A)
1 Q R r 4 0 R 2 r 2
Q R r (C) 4 0 R 2 r 2 Ans. (A) 1
Solution:
q1 + q2 = Q =
q1 4r
2
(B)
Q R r 4 0 R 2 r 2
(D)
Q R r 4 0 R 2 r 2
1
1
2
. . . (i) q2
. . . (ii)
4R 2
from (i) and(ii) q1 =
Qr 2 r 2 R 2
q2 =
Vcentre = V1 + V2 = = 31:
1 Q R r 4 0 R 2 r 2
1 4 0
QR 2 r 2 R 2
q1 q 2 R r
An electric dipole of dipole moment P is placed in a uniform electric field E is stable equilibrium position. Its moment of inertia about the centroidal axis is I. If it is displaced slightly from its mean position find the period of small oscillation.
T 2
PE I
(B)
(C) T 4
I PE
(D) T 4 2
(A)
I PE
T 2
I PE
Ans. (B)
Solution:
When displaced at an angle , from its mean position the mean position the magnitude of restoring torque is
q
PE
For small angular displacement sin PE
The angular acceleration is,
pE 2 cos I I
E P
+q
+q q
PE I I T 2 PE
Where 2
32:
Figure shows two conducting thin concentric shells of
q
radii r and 3r. The outer shell carries charge q. Inner
R2
shell is neutral. Find the charge that will flow from inner shell to earth after the switch S is closed.
r s
3r
q 3 q (C) 3 Ans. (D)
(A)
Solution:
(B)
q 2
(D)
q 3
Let q be the charge on inner shell when it is earthed.
Vinner = 0
1 q q 0 4 o r 3r
q q / 3 i.e. 33:
q charge will flow froms inner shell to earth. 3
A capacitor stores 10C charge when connected across a battery. When the gap between the plates is filled with a dielectric, a charge of 20 C flows through the battery. Find the dielectric constant of the dielectric.
(A) k = 2 (C) k = 3 Ans. (C) Solution:
In absence of dielectric Q = CV = 10C With dielectric Q = kCV = 30C From (i) and (2)
(B) k = 4 (D) k = 1
. . . (1)
. . . (2)
k=3 34:
Two infinitely large sheets S1 and S2 having surface charge densities 1 and 2 (1 > 2) respectively are placed at a distance d. Find the work done by the electric field when a charge particle ‘ Q’ is displaced by a 0 distance ‘a’ (a < d) at an angle 45 with the normal of the sheets. It is assume that the charge does not affect the surface charge densities of the two plates.
(A)
WPQ =
(C)
WPQ =
q (s 2 - s 1 ) 2 2 q (s 1 - s 2 ) 2 2
d
(B)
WPQ =
d
(D)
WPQ =
q (s 1 - s 2 ) -2 2 q (s 2 - s 1 ) -2 2
d d
Ans. (C) Ans.
Electric field at any point between the plate
æs s ö =ç 1 + 2 ÷ è2 Î o 2 Î o ø P WPQ q E.dr
Q
æs s ö = - qòç 1 - 2 ÷dr cos 45o 2Îo 2Îoø Qè P
0 æs s ö = - qòç 1 - 2 ÷cos 45o dr 2Îo 2Îoø dè
1 1 o q (s 1 - s 2 ) r 2Îo 2 d q (s 1 - s 2 ) WPQ = d 2 2
= -
35.:
A positive charge (+q) is located at the centre of a circle as shown in figure. W1 is the work done in taking a unit positive charge from A to B and W2 is the work done in taking the same charge from A to C. Then. (A) W1 > W2 (B) W1 < W2 (C) W1 = W2 (D) W1 = W2 = 0
Solution:
Point A, B and C are at the same distance from charge +q; hence electrical potential is the same at these points, i.e. There is no potential difference between A, B and C. Hence W 1 = W 2 = 0.
36.
A point Charge q is placed inside the cavity of a metallic shell.
Which one of the diagram correctly represents the electric lines of force.
(A)
(B)
(C)
(D)
Solution:
(C) Because electric field inside the conductor is zero and electric field lines are perpendicular to Gaussian surface.
37.:
A soap bubble has radius R, charge Q, surface tension T. Find the excess pressure in it. 32π 2 R 2 ε0T - q 2 64π 2 R 3ε0T - q 2 (A) (A) 32π 2 R 4 ε0 32π 2 R 4 ε0 (C)
Solution:
128π 2 R 3ε0T - q 2 32π 2 R 4 ε0
none of these
(C) Pexces
38..
(D)
q2 4T R 32 2R 4 0
A charge of 2C is brought from B to C along the path as shown by arrow in the figure. The work done is (A) 0.75 J (B) 0.6J (C) 0.06J (D) 0.075J
Solution:
(D)
w
B
q1q2 4 0
1 1 rf ri
1 1 10 2 1012 9 109 0.6 0.8 = 0.075 J 39.
Solution:
40.
2 F
80 cm
The electric field intensity at a point is 20 i + 15 j N/C . Considering potential at origin to be zero, the potential at P (2, 2) is (B) (A) 40 i + 60 j V (C) –100V (D)
60 cm
10 i + 15 j V 20V
x y (C) V = Ex dx Ey dy Ex x Ey y 100V 0 0
The equivalent capacitance between A and B is (each of the capacitors obtained is of capacitance equal to C) B’
B A’
A
1 C 2 5 (C) C 3 Ans. (C)
3 C 5 2 (D) C 5
(B)
(A)
1 2
B
A
4
D
5 A
D A=E
1 2 3 2 4 3
5 4
3
B
E
B
C
41.
CAB
2C C 5C C 3C 3
28. Page 426 Four large metal plates are located a small distance apart from one another as shown in the Fig. 24. The extreme plates are connected by means of a conductor, while a potential difference V is applied to the internal plates. Find the electric fields between the neighbouring plates
(A) E23
V V , E34 d 2d
V V , E23 E34 2d 2 (D) None of these
(B) E12
(C) zero Ans.
(A) We may take the plates to be connected as shown in the Fig. 56. Let E23 be the electric field between plates 2 and 3. Then from V the fact that electric field = rate of change of potential, E22 . d 1
2
4
4
– + + –– + + – – + + –– + + – – + + –– +
A
B
Applying `loop rule' to the loop `A – 1 – 2 – 3 – 4 – B – A' V V' – V – + V' V' = 2 where V' is the numerical value of the potential difference between plates 1 and 2 or 3 and 4 V E12 or E34 2d E 0 1 (density of charge on the plate 1) V 0 E12 0 4 (numerically) 2d The surface density of charge of the right surface of plate 2 is ' 0 ,E23 0 Vd
2 (surface density of charge on plate 2)
0 V 0 V 3 0 V 2d d 2 d
Thus V V E12 or E34 d 2d V 3 V (ii) 1 4 0 2 0 0 . 2d 2 d
(i) E23
42.
A simple pendulum consists of a small sphere of mass m suspended by a thread of length l. The sphere carries a positive charge q. The pendulum is placed in a uniform electric field of strength E directed vertically upwards. With what period will the pendulum oscillate if the electrostatic force acting on the sphere is less that the gravitational force? Assume the oscillations to be small 1/2
1/2
l (A) T = 2π g
l (C) T = 2π g - qE m Solution:
ml (B) T = 2π qE 1/2
1/2
Let x be the small displacement given to the pendulum such that the angle is small. The forces acting at A are (i) tension T along the thread (ii) weight mg acting vertically downwards. qE θ (iii) electrical force qE vertically T l upwards. The resultant force vertically down A wards is (mg – qE). Therefore θ B qE Net acceleration g ' g m mg mg qE cosθ mg qE sinθ Time period l l 2 T 2 qE g' g m
43.
l (D) T = 2π g + qE m
1/ 2
A ring has charge Q and radius R. If a charge q is placed at its centre then the increase in tension in the ring is Qq (A) (B) zero 4πε0 R 2
(C)
Solution:
44.
Solution:
45.
Qq 4π 2 ε0 R 2
(D)
T cos
T cos
Consider a small element AB, is very small. Then AB = R(2) Q Q Change on AB is dQ = 2R 2 R dQ q Qqθ 2T sinθ = = 2 4π 0 R 4π 2 0 R 2 Qq Qq 2Tθ = or T= 2 2 2 4π 0 R 8π 0 R 2
T
A
B
q T sin T cos
A charge Q is uniformly distributed over a large plastic plate. The electric field at point P, close to the centre of the plate is 10Vm-1 . If the plastic plate is replaced by a copper plate of the same dimensions and carrying same charge Q then the electric field at the point P will be (A) 5Vm-1 (B) zero -1 (C) 10Vm (D) 20Vm-1 (C) because E = 0
A hemisphere of radius r is placed in a uniform electric field of strength E. The electric flux through the hemisphere is (A (C)
2Er2 -2Er2
(B) (D) x
A
B
E
Solution:
Qq 8π 2 ε0 R 2
E.dS
B
E.dS
A
X B
E.dS
E r 2 E r 2 E r 2 E r 2
A X
E.dS
–Er2 zero
T
x
A
dS
B
E
short cut AXB is symmetrical surface Electric flux due to this part is zero. However, electric flux due to AB part is –ER2. 46.
1. What is the potential difference at the centre C +q (A) Zero (B) Kq a 2 a Kq (D) none of these (C) 2 a -q Solution: (A) We know that potential is a scalar quantity 1 q V= 40 r Where r is the distance between charge and point where potential has to be found. total potential at center C 1 1 (q1 + q2 + q3 + q4). = r 40 1 = (+ q – q + q – q) 40r =0
-q
C
47.
MeV is (A) 1.6 1019 J (B) 1.6 1022 J (D) 1.6 1013 J (C) 1.6 1013 J Solution: (C) 1 MeV = 1.6 10–16 106 Joules = 1.6 10–13 Joules 48.
Solution: field 49.
What is the angle between electric dipole moment and the electric field strength due to it on the axial line (A) 0 (B) 90 (C) 180 (D) none of these (D) Direction of dipole movement will be opposite to the electric
The total electric flux, leaving spherical surface of radius 1 cm, and surrounding an electric dipole is
+q
(A) q (C) 2q
0 0
(B) zero (D) 8 r 2
q
0
+q 4pe 0 -q Electric flux passes due to – q charge of dipole 4pe 0
Solution: Electric flux passes as due to q charge of dipole
\ Net flux passes due to both the charges =
50.
Solution:
q q 0 4pe 0 4pe 0
An isolated sphere of radius R contains a uniform Electric field magnitude volume distribution of positive charge. Which of the A curve on the graph below correctly illustrates the B dependence of the magnitude of the electric field of A,B,C,D C the sphere as a function of the distance r from its D r centre O R (A) A (B) B (C) C (D) D KQTOTAL (C) E r for inside point and E for outside point r2 3 0
and it will be
represented by C.
51.
The variation of potential with distance R from a fixed point is as shown in figure 4 (A) the electric field at R =5m is V m 1 5 æ4 ö (B) the electric field at R =5m is ç ÷V - m +1 , è5 ø (C) the potential at R = 5m is + 2.5 V- m– 1 is continuous and a decreasing function of distance. (D) the segment AB corresponds to an equipotent surface.
Solution:
(D) (The segment AB corresponding to an equipotential surface)
52.
An electric dipole is placed inside a conducting shell. Mark the correct statement(s) (A) the flux of the electric field through the shell is zero (B) the electric field is zero at every point on the shell (C) the electric field is not zero anywhere any where on the shell
Solution:
53.
(D) the electric field is zero on a circle on the shell. (A) Inside the surface the total charge is zero. So flux must be zero.
A network of six identical capacitors, each of value C is made as shown in the figure. Equivalent capacitance between points A and B is (A) C/4 (B) 3C/4 (C) 4C/3 (D) 3C A
B Solution:
(C) The network is equivalent to – Therefore equivalent capacitance = [2C series C] // [C series 2C] 2C×C 4C = 2 = 2C +C 3
54.
The charge flowing across the cell on closing the key k is equal to (A) CV (B) CV/2 (C) 2CV (D) zero
C
C 2
1
V
Solution:
When the key is kept open, the charge drawn from the source is Q = CeqV =
C V 2
When the key is closed the capacitor 2 gets short circuited And Ceq = C Q = CV charge flown through cell Q Q =
C V 2
(B) is correct choice.
55.
The capacitance of the system of parallel plate capacitor shown in the figure is (A) (C)
2 0 A1 A2 ( A1 A2 )d
0 A1 d
(B) (D)
2 0 A1 A2 ( A2 A1 )d
0 A2 d
A1
d A2
Solution :
Since the electric field between the parallel charge plates is uniform and independent of the distance, neglecting the edge effect, the effective area of the plate of area A2 is A1. Thus the capacitance between the plates is A C= 0 1 d
+
A1
0
d
E -
A2
(C) 56.
Solution:
57.
The three capacitors in figure, store a total energy in J of (A) 12 (B) 36 (C) 48 (D) 80
6μF
Total capacitance 6 F Applied voltage 4v 1 1 Stored energy U CV 2 6 106 (4)2 48F 2 2 In the circuit shown in figure C = 6 F. The charge stored in capacitor of capacity C is
6μF
3μF
(C)
C
(A) (C)
2C
zero 40 C
4V
10V
(B) (D)
90 C 60 C
Solution: (C) Both the capacitors are in series. Therefore charge stored on them will be same. (C)(2C) 2 2 Net capacity C 6F 4F C 2C 3 3 Potential difference 10V q CV 40C 58.
Figure shows two parallel plates, R and S joined to a battery of voltage V and with charges +Q and –Q. 1. the energy stored is QV. 2. the electric field strength between the plates increases uniformly from S to R. 3. the electric potential between the plates decreases uniformly from R to S. Which of the statement is/are correct (A) 3 only (C) 2 and 3 only
(B) 1 and 2 only (D) 1 only
Solution: uniformly
(A)
The electric potential between the plates decreases from R to S.
59.
In the circuit shown in figure potential difference between A and B E = 190V C
3C
A
Solution:
B
C
is (A) 30 V (C) 10 V
3C
(B) 60 V (D) 90 V
(C) Potential difference is divided among two capacitance C1 and C2 in the inverse ratio of their capacities when it is final in series. Therefore E = 190V C
3C
A
C
B 3C
Voltage across P and Q is C =190 C CapacitanceacrossPandQ =190
C
P
3C
Q
C + C
3C
equivalent capacitance between P and Q C =190 40 volt 15C C 4 Potential difference between P and Q (i.e. 40 volt) Will divide between two capacitor C and 3C which is in series. Therefore voltage across 3C capacitance. C 10Volt volt 40 C 3C Hence (C) is correct choice
60.
A parallel plate capacitor has two layers of dielectrics as shown in figure. This capacitor is connected across a battery, then the ratio of potential difference across the dielectric layers is k 1= 2
k2 = 6
d
2d
(A) 4/3 (C) 1/3 Solution:
(D)
(B) 1/2 (D) 3/2
Capacitance (for k 2) kA C1 0 d C1 2C
Where C
0 A d
Capacitance (for k 6) 6 A C2 0 2d C2 3C Therefore ratio of potential difference across the dielectric layer is 3/2
CURRENT ELECTRICITY CURRENT ELECTRICITY Flow of electric charge constitutes electric current. For a given conductor, if 'Q' charge flows through a cross-section of area A in time 't', then the average electric current through the conductor is given as Q dQ I= and its instantaneous value is . dt t
Crosssection
MECHANISM OF CURRENT FLOW IN METALLIC CONDUCTOR When an external potential difference is applied across a metallic conductor then an electric field is set up within the conductor. Applied electric field Force on electrons drift of electrons Due to the externally applied electric field electrons drift with an average velocity called drift velocity. This causes an electric current Total charge crossing a cross-section in one second is equal to I = neAvd. Here Avd is the volume of a cylinder of cross-section A length vd and ne is charge density of charge carriers (e.g. electrons).
vd I S
The current density is defined by J = /A Example1 : A steady current passes through a cylindrical conductor. Is there an electric field inside the conductor ? Solution : Yes; No doubt under steady state conditions in electrostatics when a conductor is charged, electric field inside it is zero as metal is an equipotential surface. However when a potential difference is applied across a conductor and a steady current flows though it, the condition no longer remains static and there exists an electric field inside the conductor. OHM’S LAW It states that "the potential difference across a conductor is directly proportional to the current flowing through it at a given temperature". At constant temperature
V constant(R) I
the constant 'R' is called resistance of the conductor. Resistivity () and conductivity ():
The resistance R of a given conductor is directly proptinal to length () and inversitional proptional cross-sectional area (A) such that R =
, where = A
resistivity of the material of the given conductor. Its S.I. unit is m. Reciprocal of resistivity is called the electrical conductivity () of the material, thus =
1 = whereas reciprocal of resistance is called conductance of the RA
given conductor. S.I. unit of conductivity is ( - m)-1 and is usually written as mho/m. Temperature Dependence of Resistivity: The conductivity of a metal decreases as its temperature is increased. Thus resistivity increases with the rise in temperature. If T and 0 represent the resistivities at temperatures T and T0 respectively, then for small temperature variations, T= 0 [1 + (T T0)] Where is called the temperature coefficient of resistivity. The resistivity varies over a very wide range. For metals (good conductor) 10-8 -m and for insulators 1017 -m Semiconductors (silicon, germanium ) have intermediate value much smaller than insulator but much larger than metals. Temperature coefficient of resistivity is negative for semiconductors and positive for the metals. For superconductors resistivity is zero. Thermistor: A thermistor is a semiconductor electronic device in which the resistance decreases as its temperature increases. This is used as a thermometer. The temperature coefficient of resistivity is negative for semiconductors, hence thermistors are usually prepared from oxides of various metals such as nickel, iron, cobalt and copper etc. A thermistor is used to detect small changes in temperature of the order of even 10-3 0C. Colour code for carbon Resistors: tolerance
The four bands indicate digit -1, digit-2, multiplier and tolerance respectively and the values of different colours are given in the following table.
digit 1 digit 2
Resistance code (in )
multiplier
Colour
Digit
Multiplier
Black
0
1
Brown
1
10
Red
2
102
Orange
3
103
Yellow
4
104
Green
5
105
Blue
6
106
Violet
7
107
Gray
8
108
White
9
109
Tolerance
Gold
0.1
5%
Silver
0.01
10%
Sometimes the carbon resistor indicates only three bands and the tolerance is missing from the colour code. This means tolerance has to be taken as 20%. Example2: Find the resistance of a carbon resistor if the colour code from left to right indicates brown, yellow, green and gold. Solution:
Use diagram 5 1 4 10 5% 5 R = (14 10 5%) = (1.4 106 + 0.07 106 ) = (1.4 0.07)M
KIRCHHOFF’S LAWS Junction Rule: It is based on the law of conservation of charge. At a junction in a circuit the sum of incoming currents is always equal to the sum of outgoing currents. In otherwords the algebraic sum of the currents at a junction is zero. Loop rule The algebraic sum of the changes in potential around any closed path is zero. It is based on the law of conservation of energy. In case of a resistor of resistance 'R' potential will decrease in the direction of current. Hence, for the shown conductor Va – Vb = IR
For an emf source, the potential changes will be obtained as illustrated below, Emf = , internal resistance = r i
a
b
I
a
b
R
Emf = , internal resistance = r i
a
b
Va – Vb = + ir Va – Vb = + ir Students can use any sign convention which they find easy. Example3: In the series circuit shown, E,F,G,H are cells of emf 2V,1V,3V and 1V respectively, and their internal resistance are 2, 1, 3 and 1 respectively. Calculate (i) the potential difference between B and D and (ii) the potential difference across the terminals of each of the cells G and H. Solution: Let us redraw the circuit. I At junction D, we have applied the A I1 junction rule, whereby we get current in DB as shown. + 1V F Loop BADB 1 2I1 - 2 + 1 + I1 + 2 (I1 I2) = 0 D 5I1 2I2 = 1 Loop DCBD 3+3I2+I2+12(I1I2)=0 6I2 – 2I1 = 2
E A
+ G
D
2V +
B I2
2 2
I1- I2
1V
+
1
3
3V
G
2 5 V 1 13 13
VBD = 2(I1) 2 + 1 + I1= 3 I1 1 = 3
(ii)
Terminal voltage of G = |3 + 3.I2| = 3 3
21 6 V = 13 13
GROUPING OF RESISTANCES
C
E
(i)
6 19 (I) 1 V . 13 13
H
+
2
5 6 A, I2 A 13 13
Terminal voltage of H =
B
+ F
1
I1
+
C
H
Resistance in series Let the equivalent resistance between A & B equals Req , by definition. Req =
V I
A
R1
R2
R3
B
I V
. . . (1)
Using Kirchoff's 2nd rule for the loop shown in figure, V = IR1 + IR2 + IR3 . . . (2) From (1) and (2) Req = R1 + R2 + R3 Resistance in parallel Here again, Req =
R1 i1
V I
. . . (1)
R2
A
I = i1 + i2 + i3 =
V V V R1 R2 R3
. . . (2)
I
i2
B
R3
i3
I
From (1) and (2) V
1 1 1 1 Req R1 R2 R3
Example 4: Find the equivalent resistance between A and B in the circuit shown here. Every resistance shown here has a magnitude of 2 .
C 2
2 O
A 2
B 2
D
Solution:
Points C, O & D are at the same potential. Therefore, resistances AO, AC and AD are in parallel . Similarly BC, BO and BD are in parallel. Similarly BC, BO and BD are in parallel.
=
RAB =
1 1 (2) + (2) 3 3
4 = 1.33 . 3
ENERGY, POWER AND HEATING EFFECT When a current I flows for time t from a source of emf E, then the amount of charge that flows in time t is Q = It. Electrical energy delivered W = Q.V = VIt Thus, Power given to the circuit, = W/t = VI or V2/R or I2R E r In the circuit 2 2 E. I = I R + I r, where I EI is the rate at which chemical energy is converted to 2 R electrical energy, I R is power supplied to the external 2 resistance R and I r is the power dissipated in the internal resistance of the battery.
An electrical current flowing through conductor produces heat in it. This is known as Joule's effect. The heat developed is given by H = I2.R.t joule, where I = current in ampere , R = resistance in , t = time in second. Maximum Power Theorem In a circuit, for what value of the external resistance the maximum power be drawn from a battery? For the shown network power developed in resistance R equals P
2
r
I R
E .R E ( I = and P = I2R ) 2 (R r ) Rr
Now, for dP/dR = 0 (for P to be maximum
E
dP 0) dR
(R r )2 2(R)(R r ) 0 E2. (R r ) 4
R + r = 2R R = r The power output is maximum, when the external resistance equals the internal resistance. R=r Example5 : A copper wire having a cross-sectional area of 0.5 mm2 and a length of 0.1 m is initially at 25oC and is thermally insulated from the surroundings. If a current of 10 A is set up in this wire (a) Find the time in which the wire starts melting. The change of resistance of the wire with temperature may be neglected. (b) What will this time be, if the length of the wire is doubled? Density of Cu = 9 103 Kg m-3 specific heat of Cu = 9 10-2 Cal Kg-1 oC-1, M.P. (Cu) 1075 oC and specific resistance = 1.6 10-8m. Solution : (a) Mass of Cu = Volume density = 0.5 10-6 0.1 9 103 = 45 10-5 Kg. Rise in temperature = = 1075-25 = 1050 oC. Specific heat = 9 10-2 Kg-1 oC 4.2 J
I2Rt = mS
but
R
t
mS I2 .R
L 1.610 8 0.1 3.2 10 3 A 0.5 x10 6
t
454.210 5 10500.09 558 s 10103.210 3
(b) When the length of wire is doubled, R is doubled, but correspondingly mass is also doubled. Therefore, wire will start melting in the same time. WHEATSTONE BRIDGE
For a certain adjustment of Q, VBD = 0, then no current flows through the galvanometer. VB = VD or VAB= VAD I1.P = I2.R Likewise, VBC = VDC I1.Q = I2.S
BI
1
Q
P
A
G
C
I2
R
P R Dividing, we get, Q S
S D
I2
GROUPING OF IDENTICAL CELLS 1. Series Grouping emf of the cell is and internal resistance is r. R is the external resistance, I is the current passing through the circuit and n is the total number of cells. Applying Kirchhoff’s law ir + ir + ........ (to n times ) iR = 0 i
n R nr
r
I R
2. Parallel grouping Applying Kirchhoff’s law I r IR 0 n n I r nR
r
r
r
R
3. Mixed grouping Number of rows is m and number of cells in each row is n Apllying kirchhoff’s law I r IR 0 m mn I= mR nr n n
For current through R to be maximum, mR = nr R=
I/m I/m I/m
R
nr = one line total internal m
resistance/No. of Line RC-CIRCUIT Charging: Let us assume that the capacitor in the shown network is uncharged for t < 0. The switch is connected to position 1 at t = 0. Now, 'C' is getting charged. If the charge on capacitor at time 't' is q.
r
writing the loop rule, q + IR E = 0 C
1
S
R
2
I
I
q dq R E c dt dq 1 dt EC q RC
C
dq RC EC q dt I
Integrating
q
o
dq 1 t dt EC q RC o
1 .t RC
ln | EC q |oq EC q t EC RC
ln
q EC 1 e t / RC At t = 0, q = 0
and at t = , q = E C (the maximum charge.) = qmax Thus,
qmax = c
q qmax 1 e
t RC
q
dq qmax t / RC E t / RC e e dt RC R E i imax e t / RC where imax R
t
i
i imax = /R
t
Discharging Flip the switch to 2 Consider the same arrangement as we had in the previous case with one difference that the capacitor has charge qo for t<0 and the switch is connected to position 2 at t = 0. If the charge on capacitor is q at any later moment t then the loop equation is given as
q IR 0 c dq q R dt c
Integrating, at t = t, q = q
1 S
2
C
R
dq 1 dt q RC
t = 0, q = q0
I
q
q0
i
dq 1 t dt q RC o
ln
q t or qo RC
q q0 .e t / RC
q0 t / RC e RC
EC t / RC e RC i i0e t / RC i
'-ve' sign indicates that the discharging current flows in a direction opposite to the charging current. Discharging
O i
q
t
-imax = /R
O t
Example6: Calculate the steady-state current in the 2 resistor shown. The internal resistance of the battery is negligible and the capacitance of the capacitor is 0.2 F.
2 B
A 3 0.2F
6V
4
2.8
Solution : The resistance of the parallel combination of 2 and 3 resistors is given by 1 1 1 5 R 2 3 6
R 1.2
This resistance is in series with 2.8 giving a total effective resistance = 1.2 + 2.8 = 4 . In the steady state, charge on the capacitor C has stablised and hence no current passes through 4 resistor which is in series with the capacitor. Thus the current through the circuit = 6/4 = 1.5 A, VAB = 1.5 1.2 = 1.8 V, I through 2 resistor = 1.8/2 = 0.9 A. AMMETER An Ammeter is an instrument used for measuring current in electrical circuits. A galvanometer is a low resistance instrument. A large current passing through it may damage the instrument Changing the range of an ammeter Suppose the ammeter gives full scale deflection when a current Ig flows through it. Now if we want to convert the reading of the ammeter in such a
manner that it gives full scale deflection for a higher current I in the branch of the circuit, we connect a small resistance S in parallel to the coil of the galvanometer, which has a resistance G. The resistance value is so chosen that out of the total current I only Ig flows through the coil and the remaining current flows through S. As potential difference across S = potential difference across G.
(I – Ig)S = IgG
G
Ig
I
(I-Ig)
IgG I Ig
S
S=
So, effectively this ammeter will measure current up to I ampere and its effective resistance GS . S G
=
In practice G is large as compared to S. Therefore the effective resistance of the ammeter equals RA S, which is small. An ideal ammeter has zero resistance. VOLTMETER A voltmeter is an instrument used for measuring potential difference across the two ends of a current carrying conductor. It is connected in parallel with the conductor across which the potential difference is to be measured. The current through the conductor should not change on connecting the voltmeter, and so the voltmeter should draw a very small current, i.e. its resistance has to be high. When a galvanometer is used to measure potential difference across the ends of a current carrying conductor, a high resistance R is connected in series with the galvanometer. Consider the diagram VOLTMETER shown. Suppose the galvanometer gives full scale G Ig deflection when a current Ig passes through its coil. If G is resistance of the galvanometer coil then:
V
R
Potential difference to be measured = V = Ig (G R) So, effectively the voltmeter has resistance = Rv = (G + R)
R=
V -G Ig
I
In practice Rv is very large compared to G. An ideal voltmeter should possess infinite resistance.
POTENTIOMETER A potentiometer is used to compare electromotive forces of two cells or to measure the internal resistance of a cell. Principle: The potentiometer is based upon the principle that when a constant current is passed through a wire of uniform cross sectional area, the potential drop across any portion of the wire is directly proportional to its length. R Consider the network shown in the E diagram : I
E R1 R R R 2 1
Here, VA VB
I A
I
I
R1
B
R2 (Fig. i)
Now, let us consider that an EMF source having EMF same as (V A – VB) calculated above and internal resistance ‘r’ connected in parallel to R1. Suppose that a current ‘I2’ passes through the EMF source R
E
A
I1
B
R1
I2
R2
E', r
(I1+I2)
C
(Fig. ii)
We have, applying Kirchoff’s rules, the following equations R2 (I1+I2) + R1 I1 + R (I1+ I2) = E R1 I1 = I2 r + E
. . . (i) . . . (ii)
Solving (i) & (ii) we get I2 =
ER1 E' (R1 R2 R) R1R R2 r R1 R2 R
E R1 we get I2 = 0 R1 R2 R
For the chosen value of E' that is
Thus, taking a length of uniform resistance wire between A and C instead of the two resistors R1 & R2, it is possible to get zero (null) deflection in the galvanometer in the circuit shown below, provided that. E1 < IRAC
E R
A
B
1 P
I
C (I1+I2)
G
R2
E1
R2 (Fig. iii)
Let 1 be the length AB of the wire, corresponding to zero deflection in galvanometer, and let B be the point where the movable pointer ‘P’ (also called Jockey) makes contact with wire AC. If the same experiment is repeated but with another cell of EMF E2, we will be getting a different length ‘2’ at the instant galvanometer shows null deflection. If in both the cases the rheostat ‘R’ is kept unchanged, E1 1 . E2 2
The arrangement shown in figure (iii) is called Potentiometer and we use it to measure EMF using the above relationship. Application : A potentiometer can be used to compare emfs of two cells or to measure internal resistance of a cell. The method to compare the emfs of two cells has already been explained. When we want to measure the internal resistance of a cell, consider the following formula : E 1S 1 2 S , V 2
r=
where r = internal resistance of the cell, E = EMF, V = potential difference across the cell during current flow, S = resistance of resistance box, 1 , 2 are two balancing lengths. Example7:
A potentiometer wire of length 100 cm has a total resistance of 10. It is connected in series with a resistance R and a cell of emf 2 volts and of negligible internal resistance. A cell of emf 10 mV is balanced against a length of 40cm of potentiometer wire. What is the value of the external resistance R ?
Solution :
As shown in the figure, if R is the unknown resistance, the current in the circuit V 2 I= (r R ) (10 R )
10 mV G
A
B 40 cm 100 cm
2V
C
R
Now as the 100 cm wire has a resistance of 10 , the resistance of 40 cm of wire will be 40 (10/100) = 4 ohm.
Potential drop across 40 cm wire will be V = I 4 but here V = 10 mv (given) 10 10 3
Hence,
2 4 (10 R )
=
i.e. R = 790 . OBJECTIVE 1:
Solution:
A copper wire of diameter 1.02 mm carries a current of 1.7 amp. Find the drift velocity (vd) of electrons in the wire. Given n, number density of electrons in copper = 8.5 1028 /m3. (A) 1.75 mm/sec (B) 1.25 mm/sec (C) 2.5 mm/sec (D) 1.5 mm/sec Ans. (d) I = 1.7 A J = current density =
I r
2
1.7 (0.51 10 3 )2
= nevd = 8.5 1028 (1.6 10-19 ) vd
vd =
1.7 (0.51 10
3 2
) 8.5 10 28 1.6 10 19
= 1.5 10-3 m/sec. = 1.5 mm/sec. 2:
A cylindrical conductor of length and inner radius R1 and outer radius R2 has specific resistance . A cell of emf is connected across the two lateral faces of the conductor. Find the current drown from the cell.
(A) I =
(C) I =
2l R ln 2 R1
2l
R ln 2 R1 Ans. (a)
2l R ln 1 R2 2 (D) I = R ln 2 R1 (B) I =
2
R2 x
R1
dx
Solution: figure.
Consider the differential element of the cylinder as shown in the
R2
R
dx
l ( R ) a
2xl
R1
R2 ln 2l R1 I= R 2l I= R ln 2 R1
R=
3:
A copper wire of resistance 4 is melted and redrawn to thrice its original length. Find the resistance of stretched wire. (A) 40 (B) 60 (C) 45 (D) 36 Ans. (d)
Solution :
Since volume of the wire does not change. 1A1 = 2 A2 where 1 and A1 are the initial length and crosssection of the wire, and 2 and A2 are the final length and crosssection.
4:
l1 A 2 l2 A1
R1 =
R1 l1 A 2 l1 l1 R2 l2 A1 l2 l2
R1 l12 R2 l22
R2 = 9R1 = (9 4) = 36
. . . (1)
l1 A1
and
as
R2 =
l2 A2
. . . (2) 2 = 31
An electric bulb rated 220 v and 60 W is connected in series with another electric bulb rated 220 v and 40 W. The combination is connected across 220 volt source of e.m.f. Which bulb will glow more? (A) P1 P2 (B) P1 P2 (C) P1 P2 (D) P1 / P2 Ans. (b)
Solution :
R
V2 P
resistance of first bulb is R1
V2 P1
and resistance of the second bulb is R2
V2 P2
In series same current will pass through each bulb Power developed across first is P1 I2 and that across second is P2 I2
V2 P2
P1' P2 P2' P1
as P2 P1
V2 P1
P1 1 P2
P2 1 P1
P1 P2
The bulb rated 220 V & 40 W will glow more. 5:
A battery of emf 1.4 V and internal resistance 2 is connected to a 100 resistor through an ammeter. The resistance of the ammeter is 4/3 . A voltmeter is also connected to find the potential difference across the resistor. The ammeter reads 0.02A. What is the resistance of the voltmeter? (B) 300 (A) 200 (D) 150 (C) 400 Ans. (a)
6:
A battery of emf 1.4 V and internal resistance 2 is connected to a 100 resistor through an ammeter. The resistance of the ammeter is 4/3 . A voltmeter is also connected to find the potential difference across the resistor. (iii) The voltmeter reads 1.10 V. What is the error in reading? (A) 0. 3 V (B) 0.43 V (C) 0.53 V (D) 0.23 V Ans. (d)
Solution:
5&6 E = 1.4V r =2
A
100
4/3
V
(ii)
100R v 4 Total resistance in the circuit = 2 + 3 100 R v Emf I Total resis tan ce 1 .4 V 0.02A 100R v 4 2 + 3 100 R v
(iii)
Rv = 200 Potential difference
across
the
voltmeter
100 x 200 0.02. 100 200
=1.33 V. Voltmeter reading = 1.10 V Error = 1.33 - 1.10 = 0.23 V.
7:
B
Five equal resistances each of R value R are connected to form a network as shown in figure. Calculate the equivalent A resistance of the network between the points B and D.
R
R
C
R
R D
(A) Re q BD
3 R 2
(C) Re qBD R 1 2
(B) Re q BD (D) Re q BD
5 R 2 7 R 2
Ans. (c) Solution :
The circuit can be redrawn as in the figure. P=R (a) for points B and D, resistance (P+R), G an (Q+S) are in A parallel. With P = Q = R = S = G = R 1 1 1 1 i.e. Re qBD 2R 2R R Re qBD 1 R 2
B Q=R G=R
R S=R
So,
D
C
=
8:
Five equal resistances each of R value R are connected to form a network as shown in figure. Calculate the equivalent A resistance of the network between the points A and C, and A R
B R
R
C
R D
(A) R (C) 2 R Ans. (a) Solution :
(B) R/2 (D) 5 R
for points A and C, the given network is balanced wheatstone bridge as
P R . So excluding G, (P+Q) is in parallel with Q S
(R+S) i.e.
1 1 1 Re qAC 2R 2R
i.e. (Req)AC = R 9:
Five equal resistances each of R value R are connected to form a network as shown in figure. Calculate the equivalent A resistance of the network between the points A and C
B R
R
C
R
R D
(A)
5 R 8
(C)
7 R 8
6 R 8 9 (D) R 8
(B)
Ans. (a) Solution :
for points A and B, starting from opposite side of AB, (Q+S) in parallel with G gives RBCD =
2R R 2 R 2R R 3
RBCD in series with R gives RBDO =
2 5 RR R 3 3
And RBDO in parallel with P gives the equivalent resistance between AB, i.e. (Req)AB =
5 / 3R R i.e. (Req) = AB 5 / 3R R
5 R 8
10:
E Figure shows a potentiometer circuit for determining the internal resistance of a cell. When switch S is open, Athe E G balance point is found to be at 76.3 cm of the wire. When switch S is closed and the value of R is R S 4.0 , the balance point shifts to 60.0 cm. Find the internal resistance of cell E. (A) 1 (B) 1.1 (C) 2.1 (D) 5.1 Ans. (b)
Solution :
r
J J B
EV I E r 1R , V
But
E / , hence V
76.3 60.0 r R 4 .0 1.1 60.0
11:
In the circuit shown in figure each cell has emf 5V and has an internal resistance of 0.2 ohm. What is the reading of ideal voltmeter V. (A) 0V (C) 2V Ans. (a)
Solution:
+ V
(B) 1V (D) 5V
As internal resistance of an ideal voltmeter is infinite, the resistance of the battery across which it is connected will not change by its presence as 1 1 1 r1 r
r'
=
r
Now as the given 8 batteries are discharging in series i.e. Eeq = 8 5 = 40 V and req = 8 0.2 = 1.6 so current in the circuit I = Hence
= 25 A potential difference across the v = E – Ir = 5 – 25 0.2 = 0V.
Eeq req
required
40 1.6
battery
12:
When two resistances X and Y are put in the left hand and right hand gaps in a wheatstone meter bridge, the null point is at 60cm. If X is shunted by a resistance equal to half of itself then find the shift in the null point. (A) 26.7 cm (B) 36.7 cm (C) 46.7 cm (D) 96.7 cm Ans. (a)
Solution:
Arrangement is shown in the figure. X 60 3 Y 40 2
X
Y
A
. . . . (1)
When X is shunted then resistance in the left gap becomes X 2 X X' X 3 X 2
X/2
X.
Now
13 :
. . . (2)
40 cm
X 1 3 l l 3 l = 33.3cm Y (100 l ) 3 2 (100 l )
Shift = 60 – 33.3 = 26.7 cm.
Each resistor in the network of the given figure has a resistance of 10. The resistance between points A and B is (A) 10 (C) 30
Solution:
60 cm
A
C
D
E
F
B
(B) 20 (D) 40
Resistors in arms CD and EF are in series which add up to 10+10=20 . Resistors in arms CE and EF are also in series which add upto 20 Hence resistance between points C and F, is given by 1 1 1 or R 20 20
R = 10
Hence, resistance between A and B = 10 + 10 + 10 = 30 (C)
14:
In a gas discharge tube if 3 1018 electrons are flowing per sec from left to right and 2 1018 protons are flowing per second from right to left through a given cross section. The magnitude and direction of current through the cross section (A) 0.48, left to right (B) 0.48 A, right to left (C) 0.80A, left to right (D) 0.80 A, right to left
Solution :
As current is rate of flow of charge in the direction in which positive charge will move, the current due to electron will be ie =
neqe = 3 1018 1.6 1019 t
= 0.48 A (Opposite to the motion of electrons, i.e. right to left) Current due to protons ip =
np qp t
= 2 1018 1.6 1019
= 0.32 A (Right to left) so total I = ie + ip = 0.48 + 0.32 = 0.80 A (Right to left) Hence correct answer is (D) 15 :
The current in a wire varies with time according to the equation I = 4 + 2t, where I is in ampere and t is in sec. The quantity of charge which has passed through a cross-section of the wire during the time t = 2 sec to t = 6 sec will be (A) 60 coulomb (B) 24coulomb (C) 48 coulomb (D) 30 coulomb
Solution :
Let dq be the charge which has passed in a small interval of time dt, then dq = idt = (4 + 2t)dt Hence total charge passed between interval t = 2 sec and t = 6 sec 6
q = ( 4 2t )dt = 48 coulomb 2
(C) 16 :
A uniform copper wire of length 1 m and cross-sectional area 5 10 7 m 2 carries a current of 1 A. Assuming that there are 8 10 28 free electrons per m3 in copper, how long will an electron take to drift from one end of the wire to the other? (A) 0.8 10 3 sec (B) 1.6 10 3 sec (C) 3.2 10 3 sec (D) 6.4 10 3 sec Solution : The drift velocity of electrons is given by Vd
I enA
If is the length of the wire, the time taken is t
e n A 1 1.6 10 19 8 10 28 5 10 7 Vd I 1
= 6.4 10 3 sec (D) 17 :
A electric current of 16 A exists in a metal wire of cross section 106 m2 and length 1m. Assuming one free electrons per atom, the drift speed of the free electrons in the wire will be (Density of metal =5 103 kg/m3, atomic weight = 60) (A) 5 103 m/s (B) 2 103 m/s (C) 4 103 m/s (D) 7.5 103 m/s
Solution :
According to Avogadro's hypothesis N m NA M N m N so n = NA A v VM M
Hence total number of atoms n =
6 10 23 5 10 3 60 10 3
= 5 10 /m As I = ne eA vd 28
3
Hence drift velocity vd = vd =
I neeA
16 5 10
= 2 10 (B)
28
3
1.6 10 19 10 6
m/s
18:
A copper wire is stretched to make it 0.1 % longer. The percentage change in its resistance is (A) 0.2 % increase (B) 0.2% decrease (C) 0.1 % increase (D) 0.1 % decrease
Solution :
For a given wire, R =
L s
with L s = volume = V = constant so that R =
L2 V
R L =2 = 2 (0.1 %) L R
= 0.2 % (increase) (A) 19 :
The driver cell of a potentiometer has an emf of 2V and negligible internal resistance. The potentiometer wire has a resistance of 5 and is 1m long. The resistance which
must be connected in series with the wire so as to have a potential difference of 5mV across the whole wire is (A) 1985 (B) 1990 (C) 1995 (D) 2000 Solution :
In order to have a potential drop of 5 mV = 5 10 3 V across a wire of resistance 5 , the current flowing in the wire should be I
5 10 3 1 10 3 A 5
If R is the resistance to be connected in series with the wire, then 2 1 10 3 R5
which gives R = 1995 (C) 20 :
A battery of 10 volt is connected to a resistance of 20 ohm through a variable resistance R. The amount of charge which has passed in the circuit in 4 minutes, if the variable resistance R is increased at the rate of 5 ohm/min. (A) 120 coulomb (B) 120 loge2 coulomb (C)
Solution :
120 loge 2
coulomb
(D)
60 loge 2
coulomb
dq V dt R dq dR V . dR dt R dR dq = 12 V R 40 dR q = 12 V = 12 V (loge 40 – loge20) R 20
I=
= 12 10 loge2 (B)
21:
In the given circuit, R1 = 10 , R2 =6 and E = 10 V. Then effective resistance (A) Effective resistance of the circuit is 20 (B) Effective resistance of the circuit is 30 (C) Effective resistance of the circuit is 40 (D) Effective resistance of the circuit is
R1
A2
R1
R1
R2 R1
A3 R1
A1
R2 R1
R1 R2 R1
50 Solution:
Potential difference across R2 resistances is zero, therefore current in three branches is zero, therefore current in two branch containing R1 will be same, simplified circuit will be Effective resistance of the circuit Reff = current through the circuit I =
10
A2
10
10
10
10
A1
10
10
10
10 V
40 40 = 20 40 40
10 = (1/2) amp 20
Hence reading of A1 = ½ amp. Hence reading of A2 =1/4 amp. (A)
22:
Solution:
In the given circuit, R1 = 10 , R2 =6 and E = 10 V. Then reading of A1 (A) Reading of A1 is 1 amp. (B) Reading of A1 is 1/2 amp. (C) Reading of A1 is 2 amp. (D) Reading of A1 is 3 amp. Potential difference across R2 resistances is zero, therefore current in three branches is zero, therefore current in two branch containing R1 will be same, simplified circuit will be
R1
R1
R2 R1
R1 R2
R2
A3
R1
R1
R1
A1
10
A2
10
10
10
10
A1
Effective resistance of the circuit Reff = current through the circuit I =
R1
A2
10
10
10
10 V
40 40 = 20 40 40
10 = (1/2) amp 20
Hence reading of A1 = ½ amp. Hence reading of A2 =1/4 amp. (B)
23:
In the given circuit, R1 = 10 , R2 =6 and E = 10 V. Then reading of A2 (A) Reading of A2 is 1 amp (B) Reading of A2 is 2 amp (C) Reading of A2 is 1/4 amp. (D) Reading of A2 is 3 amp
R1
A2
R1
R1
R2 R1
A3 R1
A1
R2 R1
R1 R2 R1
Solution:
Potential difference across R2 resistances is zero, therefore current in three branches is zero, therefore current in two branch containing R1 will be same, simplified circuit will be
10
A2
10
A1
Effective resistance of the circuit Reff = current through the circuit I =
10
10
10
10
10 V
40 40 = 20 40 40
10 = (1/2) amp 20
Hence reading of A1 = ½ amp. Hence reading of A2 =1/4 amp. (c) 24:
Solution :
One billion electrons pass from A to B in 1 ms. What is the direction and magnitude of current? (A) 1.6 A (B) 0.8 A (C) 0.16 A (D) 1.6 A i
Ne (109 )(1.6 x10 19 C) 1.6 x10 7 A 3 t 10
i = 0.16 A. The current flows from B to A. 25:
Solution:
26:
Solution:
27:
(C)
An electron gun in a TV set shoots out a beam of electrons. The beam current is 10A. How many electrons strike the TV screen each second? (A) 2.78 1014 (B) 6.3 1013 (C) 6.78 104 (D) electron will not reach (B) The number of electrons per second N = I/e = (1.0 10–5 C/s)/(1.6 10–19 C) = 6.3 1013 electrons per second. In above question, how much charge strikes the screen in a minute? (A) –600C (B) 600C (C) 1300C (D) –1300C (A) The charge Q striking the screen obeys |Q| = IT = (10 C/s)(60 s) = 600 C. Since the charges are electrons, the actual charge is Q = – 600 C. A copper bus bar carrying 1200 A has a potential drop of 1.2 mV along 24 cm of its length. What is the resistance per m of the car?
10
10
Solution:
28:
Solution:
29:
(A) 5.2 (B) 3.2 (C) 8 (D) 4.2 (D) From Ohm’s law, applied to 24 cm of the bar, V24 = IR24, or (1.2 10–3V) = (1200 A) R, and R24 = 1.0 . By proportion, R100 = (100/24) R24 = 4.2 A 20–cm–long copper tube has an inner diameter of 0.85 cm and an outer diameter of 1.10 cm. Find its electric resistance when used lengthwise.( = 1.7 10–8) (B) 89 (A) 51.2 (C) 80 (D) 42 (B) R= (L / A). The cross–sectional area is [(1.102 – 0.852)]/(4 104)] = 3.83 10–5 m2; then with L = 0.20 m and = 1.7 10–8, then we get R = 89. At what value of resistance Rx in the circuit in the figure will the total resistance between points A and B is R? 2R
A
R
R B
(A) (C) Solution: If
Rx =
R R
3 1
3 1
2
R
(B) (D)
Rx
R
R R
3 1 3 1
3 1
R x 2R R or Rx2 + 2RRx – 2R2 = 0 On solving and rejecting the negative root of the quadratic equation, we have Rx = R 3 1
A
2R
A Rx
R B
–0.444 0.222
3
3
(B) (D)
0A –0.222
The loop equations are: –6 + 6 – 10I4 = 0, so I4 = 0. (This result tells us that I1 and I2 flow through their respective 3 ) +6 –3I1 + 12I3 = 0 –6 –12I3 –3I2 = 0
Rx
B
What will be the I1 for the following circuit. (A) (C)
Solution:
2R
(A) R x 2R R
30:
2R
12 6V
I1
E
I4
10
I3
6V
I2
The point equation is I2 = I1 + I3. Solve simultaneously to find I1, I2, I3 and I4, 0.222, –0.222, –0.444 and 0 A. Hence the correct choice is (C) 31:
Solution:
What will be the I2 for the following circuit.
3 3 (A) –0.222 (B) –0.212 12 (C) –0.222 (D) 0.222 6V 6V I The loop equations are: –6 + 6 – 10I4 = 0, so I I4 = 0. (This result tells us that I1 and I2 flow 10 E through their respective 3 ) +6 –3I1 + 12I3 = 0 I –6 –12I3 –3I2 = 0 The point equation is I2 = I1 + I3. Solve simultaneously to find I1, I2, I3 and I4, 0.222, –0.222, –0.444 and 0 A. Hence the correct choice is (C)
3
1
I2
4
32:
3 What will be the I3 for the following circuit. 3 12 (A) 0.222 (B) –0.444 6V 6V I (C) –0.444 (D) 0.444 I The loop equations are: –6 + 6 – 10I4 = 0, so I4 = 0. (This result tells us that I1 and I2 flow 10 E through their respective 3 ) I +6 –3I1 + 12I3 = 0 –6 –12I3 –3I2 = 0 The point equation is I2 = I1 + I3. Solve simultaneously to find I1, I2, I3 and I4, 0.222, –0.222, –0.444 and 0 A. Hence the correct choice is (C)
3
Solution:
1
I2
4
33:
Solution:
What will be the I4 for the following circuit. (A) 1A 3 (B) 0.222 (C) 0A 6V (D) 2A The loop equations are: –6 + 6 – 10I4 = 0, so I4 = 0. (This result tells us that I1 and I2 flow through their E respective 3 ) I +6 –3I1 + 12I3 = 0 –6 –12I3 –3I2 = 0 The point equation is I2 = I1 + I3. Solve simultaneously to find I1, I2, I3 and I4, 0.222, –0.222, –0.444 and 0 A. Hence the correct choice is (C)
3
12
I3
6V
I1
I2
10
4
34:
Three resistances of 4 each are connected as shown in figure. If the point D divides the resistance into two equal halves, the resistance between points B and C will be (A) 12 (B) 6
4
4 4
B
4
C
8 3 The circuit can be rearranged as shown in Fig. because it is Wheatstone’s Bridge circuit The resistance between points B and C is given by 1 1 1 R 4 4 2 2 8 or R = . Hence the correct choice is (d) 3
3
(C)
Solution:
(D)
4
2
C
4
A
2
D
B
R
35:
Solution:
What will be the equivalent resistance of the network shown in the figure between the R terminals 1 and 2. R 3 2 8 (A) R (B) R 5 5 R 7 R R (D) none (C) 1 5 (C) This is the familiar problem, which can be solved using Kirchhoff’s rule. Here, the problem is going to be simplified using Delta–star Transformation. Before going through the solution, let us understand this Delta–Star Transformation. This figure (a) and (b) show the Delta and Star respectively. a
a
Ra
R1
R2 Rc
b
R3
c
b
Rb
c
(a) (b) For obtaining the expression for equivalent we first derive expression for resistance between points of terminals and then equating the expression for resistance of the corresponding terminals of two circuits. Now for Delta Configuration the equivalent resistance between a and c, R1R2 Ra = R1 R2 R3 R1R3 Rb = R1 R2 R3 R2R3 Rc = R1 R2 R3 The given network can be reduced in the form as shown using series and parallel grouping.
2
R 4 R
2
2 R
2R R
3 R
R
2R 4
3
2R
4
R/4
R/2
1
R/2 1
(c) (d) Using Delta–Star Transformation, the figure (c) can be reduced to figure (d). Now the reduced network is very simple and using series and parallel grouping, the equivalent resistance between 7 terminal 1 and 2 becomes R. 5 CMP I:
A cell having a steady emf of 2volt is connected across the potentiometer wire of length 10m. The potentiometer wire is of manganine and having of 11.5 /m. A new resistance of 5 with negligible length is put in series with the potentiometer wire.
36:
What is total resistance of wire? (B) 115 (A) 11.5 (C) 1150m (D) 120 Solution: Resistance per unit length of wire = 11.5/m Length of wire = 10 m Resistance of wire = 11.5 10 = 115 Total resistance = 115 + 5 = 120 Hence correct choice is (D) 37:
Solution:
What is potential gradient when 5 resistance is not connected (A) 0.2 V/m (B) 0.4 V/m (C) 0 V/m (D) 1.38 V/m appliedvoltage Fall in potential per meter = = 2/10 = lengthof wireof potential
0.2 v/m potential gradient = 0.2 V/m Hence correct choice is (A) 38:
Solution:
What is potential gradient after connect 5 resistance. (A) 0.2 V/m (B) 0.15 V/m (C) 0.104 V/m (D) 2.0 V/m
5
115
wire length 10 m
2V
2 115 230 23 V 115 5 220 22 voltage across potentiometer Potential gradient = lengthof wireof potential 23 / 22 0.104 V / m = 10 Hence correct choice is (C)
Voltage across 115 resistance =
CMP II:
An ammeter and a voltmeter are connected in series to a battery with an emf E = 6 volt when a certain resistance is connected in parallel with voltmeter, the reading of latter decreases two times, where as the reading of the ammeter increasing the same number of times.
39:
What is ratio of resistance of voltmeter to resistance of ammeter (A) 2 (B) 1/2 (C) 1/3 (D) 3
40:
What will be voltmeter reading before the connecting of the resistance. (A) 1V (B) 2V (C) 3V (D) 4V
41:
What will be voltmeter reading after the connecting of the resistance. (A) 1 V (B) 2V (C) 4V (D) 3V
Solution:
Suppose RA = Resistance of ammeter, Rv = resistance of Voltmeter In the first case current is the circuit 6 i ….(1) ( RA RV ) And voltage across voltmeter V = 6 – Voltage across ammeter V = 6 – iRA 6 V 6 RA ….(2) ( RA RV ) In the second case reading of ammeter becomes two times i.e. the total resistance become half while the resistance of ammeter remains unchanged. Hence 6 12 i =i ( RA RV ) / 2 ( RA RV )
and voltage across voltmeter V ’ = 6 – i RA 12 V' 6 …..(3) ( RA RV ) It is given that V V' … (4) 2 6 6 R R 12RA A V [From (2) and (3)] 6 ( RA RV ) 2 R RA 1 V 2 (Ans. of Question 39) RA RA RV 3 Substituting this value into equation (3) V’ 6 12 1/3 V’ 2 (Ans. of Question 40) From Eq, (4) V 4 (Ans. of Question 41)
42:
Solution:
43:
Solution:
In the Bohr model, the electron of a hydrogen atom moves in a circular orbit of radius 5.3 10–11 m with a speed of 2.2 106 m/s. Determine its frequency f and the current I in the orbit. (A) 1.06 mA (B) 2.06 mA (C) 0.06 mA (D) 3.06 mA 6 2.2 10 m / s = 6.6 1015 rev/s f= 11 2 r 2(5.3 10 m) Each time the electron goes around the orbit, it carries a charge e around the loop. The charge passing a point on the loop each second is current = I = ef = (1.6 10–19C) (6.6 1015s–1) = 1.06 mA. Note that the current flows in the opposite direction to the electron, which is negatively charged. (A)
A wire carries a current of 2.0 A. What is the charge that has flowed through its cross–section in 1.0 s. How many electrons does this correspond to? (A) 3.0 C, 1.25 1019 (B) 2.0 C, 1.25 1019 (C) 4.0 C, 1.25 1019 (D) 2.0 C, 5.25 1019 q i= t q = it = (2.0 a) (1.0s) = 2.0 C q = ne q 2.0 n= = 1.25 1019 19 e 1.6 10
(B) 44:
Solution:
45:
A current of 7.5 A is maintained in a wire for 45s. In this time (a) how much charge and (b) how many electrons flow through the wire? (A) 400.5C, 2.1 1021 (B) 300.5C, 2.1 1021 (C) 337.5C, 2.1 1021 (D) 700.5C, 2.1 1021 (a) q = It = (7.5A) (45 s) = 337.5C (b) The number of electrons N is given by 337.5 C q N= = 2.1 1021 19 e 1.6 10 C where e = 1.6 10–19 C is the charge of an electron. (C)
What will be the current if charge q revolves with the frequency f?
(A) f (C) q Solution:i = qf (D)
(B) (D)
q/f qf
46:
Calculate the mean free time between collision in copper at room temperature? and electron average collision per second. (A) 2.4 10–14 s (B) 4 10–14 s (C) 5.4 10–14 s (D) .4 10–14 s Solution: n = 8.5 1028 m–3 = 1.72 10–8 e = 1.60 10–19 c and m = 9.11 10–31 kg m = 2.4 10–14 s 2 ne Taking the reciprocal of this time, we find that each electron average about 4 1013 collision every second. (A)
T=
47:
Solution:
48:
A coil of wire has a resistance of 25.00 at 20C and a resistance of resistance of 25.17 at 35C. What is its temperature coefficient of resistance? (A) 4.5 10–4C–1 (B) 5 10–4C–1 (C) 0.5 10–4C–1 (D) 4.0 10–4C–1. R = Ro[1+ (T–T0)], or = R/(R0T), with R = R–R0 = 0.17 and T = T – T0 = 15C. Then = (0.17)/(25.00 15) = 4.5 10–4C–1. (A) The three resistors in Fig. are R1 = 25 , R2 = 50 , and R3 = 100 . What is the total resistance of the circuit? (A) 50.3 (B) 60.3
Solution:
(C) 58.3 (D) 80.3 The sum of R2 and R3 in parallel is E 12V
I R1
R2
I1 R3 I2
1 1 1 1 1 3 R R 2 R3 50 100 100
or R = 33.3
Since R is in series with R1, the total resistance of the circuit is R = R + R1 = 33.3 + 25 = 58.3 (C) 49:
Solution :
The three resistors in Fig. are R1 = 25 , R2 = 50 , and R3 = 100 . What are the currents l1, l2 and l3 for a 12–V battery? E 12V I= = 0.206 A E 12V R 58.3 I (A) 0.206 A, 0.137, 0.0685 A R (B) 0.506 A, 0.137, 0.0685 A R (C) 0.606 A, 0.137, 0.0685 A I (D) 0.706 A, 0.137, 0.0685 A 1
2
1
R3 I and R3 is The potential V across R2 V = E – R1I = 12 V –(25)(0.206 A) = 6.85 V. Therefore, V 6.85V V 6.85V = 0.137 A I3 = = 0.0685 A I2 = R2 50 R3 100 (A) 2
50:
What is total resistance across AB in the following network. C
8
4
A
B
4
8
2
D
(A) 6.4 (C) 7.4 Solution:
(B) 2.4 (D) 5.4
We can’t apply Wheatstone’s bridge condition because 4 2 8 8
Break delta ACD into star Y connection, For that assign 1,2 and 3 like following fig. 2C 8
4 1 A
B
4
8
2 3
D
R12 R13 R12 R13 R23 42 8 = = 0.8 4 2 4 10 R12 R23 44 16 R2 = = = 1.6 4 2 4 10 R12 R23 R13 R23 R13 42 8 = = 0.8 R3 = 4 2 4 10 R12 R23 R13 Connect R1, R2 and R3 like following fig. R1 =
2
C
4 R
2
1 .6
8
1
A
R1 0.8
4
B
R 3 0.8
8
2
3
D
remove delta circuit from above fig., we get 1.6
A
8
0.8
B
0.8
8
resistance across A and B is R = 0.8 + [(8 + 1.6) || (0.8 + 8)] 9.6 8.8 = 0.8 = 5.4 9.6 8.8 51:
(D)
Find the effective resistance between points A and B of the network shown if Fig.
3
E
D
3
C
6
6
3
6
F
3
3
A
3
(A) 3 (C) 5
B
(B) 4 (D) 7
Resistors AF and FE of 3 each are in series with each other. Therefore, the network AEF is a parallel combination of two 6 resistors. Thus the resistance between points A and E is given by
Solution:
3
E
D
3
C
6
6
3
6
F
3
3
A
3
B
1 1 1 R AE 6 6
giving RAE = 3 . The network reduces to that shown in Fig. (a). Similarly the resistances between points A (A) 52:
Kirchhoff’s current law obeys conservation of (A) charge (B) momentum (C) energy (D) none of these. Solution: (A)
53:
Solution:
The slide wire Wheatstone bridge shown in Fig. is balanced when the uniform slide wire AB is divided as shown. Find the value of the resistance X. (A) 3 (B) 4 (C) 2 (D) 7
N 3
X
G A
L 40 cm D
M
B
60 cm
X L 40cm or 3 M 60cm
54:
X = 2
(C)
A dry cell delivering 2 A has terminal voltage 1.14V. What is the internal resistance of the cell if its open–circuit voltage is 1.59 V? (A) 5.09 (B) 6.09 (C) 7.09 (D) 0.09
Solution:
55.
Solution :
56.
Solution :
57.
Solution : 58.
Solution :
The open–circuit voltage is simply the emf of the cell, so V = E – ir with V = 1.41 V, i = 2 A, E = 1.59 V. 1.41 = 1.159 – 2r, and r = 0.09 The sensitivity of a galvanometer of resistance 406 ohm is increased by 30 times. The shunt used is (A) 88 (B) 14 (C) 6 (D) 16 . Gi g 226 i 14 . (B) = 30. The shunt S = i ig 29 ig The sensitivity of a galvanometer of resistance 8722 is decreased by 90 times. The shunt used is (A) 88 (B) 90 (C) 94 (D) 98 . i (D) Here g (90)–1 . i Gi g Gi g G (90)–1 S 98 89 i ig 1 ig / i 90 A wire l = 8.00 m long, of uniform cross-sectional area A = 8.00 mm2, has a conductance of G = 2.45 –1. What is the resistivity of the material of the wire ? (A) 2.1 × 10–7 S (B) 3.1 × 10–7 S –7 (C) 4.1 × 10 S (D) 5.1 × 10–7 S. R A 8.00 106 4.1 107 S (C) A Gl 2.45 8.00 A wire 250 cm long and 1 mm2 in cross-section carries a current of 4 A when connected to a 2 V battery. The resistivity of the wire is (A) 0.2 × 10–6 m (B) 2 × 10–7 m (C) 5 × 10–6 m (D) 4 × 10–6 m l (A) R = A
1 106 A 2 P 100S l 250
59.
Solution :
An electrical cable of copper has just one wire of radius 9 mm. Its resistance is 5 . This single wire of cable is replaced by 6 different well-insulated copper wires each of radius 3 mm. The total resistance of the cable will now be equal to? (A) 7.5 (B) 45 (C) 90 (D) 270 Sl 5 11 81 or S (A) R l Sl 5 l 81 45 R1 l 9 1 6 45 or R 7.5 . R 45 5
60.
When cells are arranged in parallel (A) the current capacity decreases (B) the current capacity increases (C) the e.m.f. increases (D) the e.m.f. decreases.
Solution :
(B) When cells are connected in parallel, the current capacity increases.
THERMAL & CHEMICAL &THERMOELECTRICITY
EFFECTS
OF
CURRENT
HEATING EFFECT OF CURRENT, JOULE'S LAW (i) (ii)
(iii)
Whenever work is transformed into heat or heat into work, the quantity of work done is mechanically equivalent to the quantity of heat. Thus Work Heat. or Work = a constant × heat or W = JH Where J is known as Joule's constant or mechanical equivalent of heat. It is defined as that mechanical work which produces unit calorie of heat. Expression for heat produced in a conductor due to current flow through itA
B
i R
(a) (b)
(c)
Let the potential difference between the points A and B of a conductor is V, on account of which a current i flows through it. As potential difference is the work done per unit test charge W = QV But Q = it W = Vit W Vit i2RT But heat developed H = J J J Where R is the resistance of wire. If i i is in ampere, V is in volt and R is in ohm, then W = Vit Joule = Vit × 107 ergs. Vit Heat developed H = calories J Vit 0.24 Vit cals. 4.2 = 0.24 i2Rt cals.
JOULE'S LAWS OF HEATING EFFECTS OF CURRENT (i) (ii) (iii)
(iv)
The heat developed in a conductor is given by H = i2Rt Joule = 0.24 i2Rt cals The amount of heat developed in a conductor, in a given time, is directly proportional to the square of the current. i.e. H i2 , when R and t are constants. The amount of heat developed in a conductor by a given current in a given time is directly proportional to the resistance of the conductor. i.e. H R , when i and t are constants. The amount of heat developed in a given conductor due to a given current is proportional to the time of flow of the current. i.e. H t , when i and R are constants.
ELECTRICAL ENERGY OR WORK If Q units of charge be carried between two points differing in potential by V, then electrical work done is W = Q × V Joule = Q × V × 107 ergs
POWER (i)
(ii)
(iii) (iv)
The rate at which work is done is defined as power Q V Power Vi t = i2R V2 R Thus electric power = potential difference × current i.e. 1 watt = 1 Volt × 1 Ampere = 1 Joule/sec = 107 ergs/sec The practical unit of power is kilowatt 1 kilowatt = 1000 watt Horse-power = 550 ft-lbs per sec. = 550 × 12 × 2.54 ×453.6 gm-cm/sec = 746 × 107 ergs/sec = 746 watt.
Illustration 1:
Solution:
A coil of resistance 2 is immersed in 1 kgm of water and is connected to the terminals of the battery of internal resistance 4 and emf 6 volt for 3 minutes. The increase in temperature of water is : (A) 0.93C (B) 0.085C (C) 1.92C (D) 4.31C Heat developed in coil = Heat absorbed by water Rt i2 = ms Dt J 2 i2Rt æ E ö Rt \ Dt = =ç ÷ Jms è R + r ø Jms 2 ´ 180 æ 6 ö =ç ´ » 0.085°C ÷ è 4 + 2 ø 4.2 ´ 1 ´ 103 2
Illustration 2:
Solution:
A Daniel cell has an emf of 1.08 volt and internal resistance 0.5 . It is successively connected to two wires whose resistances are 2 and 3. The ratio of the amounts of heat developed in two wires will approximately be : (A) 4:3 (B) 3:4 (C) 2:3 (D) 3:2 Heat developed in a wire H = i2Rt E2Rt or (R + r)2 R \ Hµ (R + r)2 2
\
H1 æR1 ö æR2 + r ö = ´ H2 çè R2 ÷ ø çè R1 + r ÷ ø 2
2 æ3.5 ö 2 49 = ´ ç = ´ » 4:3 ÷ 3 è 2.5 ø 3 25
Illustration 3:
Solution:
Illustration 4:
Solution:
Illustration 5:
Solution:
(v)
For tungsten wire, the average temperature coefficient above 20 C is 5.1 10–3C–1. The filament of an electric lamp has a cold resistance of 9.7 at 20C. When glowing it has a resistance of 121. Its temperature while glowing is : (A) 2250C (B) 2190C (C) 2270C (D) 2430C R1 = R20 (1 + t) or 121 = 9.7 (1 + 5 10–3 t) \ t = 2250°C above 20C the temperature of glowing = 2250 + 20 = 2270C If a cell has an e.m.f. of 1.08 volt and internal resistance 0.5 . Its terminals are connected to two wires of 1 and 2 in parallel. The current in each wire will be : (in ampere) (A) 0.617, 0.308 (B) 0.412, 0.206 (C) 0.824, 0.412 (D) 2.618, 1.309 Total current E 1.08 1.08 ´ 6 i= = = R1R2 1 2 7 + +r 2 3 R1 + R2 1 iµ R 2 2 1.08 ´ 6 current in 1 resistance = i = ´ = 0.617 amp. 3 3 7 i 1.08 ´ 6 current in 2 resistance = = = 0.308 amp. 3 7´ 3 In the above problem the ratio heats generated in the two wires will be : (A) 1:2 (B) 2:1 (C) 1:3 (D) 3:1 2 V Heat generated H = t R In parallel combination V = constant H1 R2 2 \ = = = 2 :1 H2 R1 1
Kilo-watt-hour or Board of Trade (B.O.T.) unit(a) Energy consumed in a given time is the product of power and time (b) When power of one watt is consumed for an interval of one hour, then energy consumed = 1 watt × 1 hour = 1 watt – hour = 0.001 kilo-watt-hour (c) Board of Trade (B.O.T.) unit of electric power is kilo-watt-hour by which the consumption of electric energy is measured and charged by power supply authorities. 1 kilo watt-hour = 103 × 60 × 60 watt-sec = 36 × 105 Joule
(d)
Rule for calculation of cost(i) Number of B.O.T. units = Sum of wattages of all systems × time (in hour) 1000 (ii) Total cost = Number of B.O.T. units × rate of charge per unit (iii) Total cost = V2 ( P)or ( Vi) or ( i2R)or time in hour rate per unit R 1000
Illustration 6:
Solution:
Illustration 7:
Solution:
Illustration 8:
Solution:
A dwelling house is installed with 15 lamps, each of resistance 10 3 and 4ceiling fans each driven by 1/8th horse-power motor. If the lamps and fans are run on an average for 6 hours daily, then the number of B.O.T. units consumed by lamps in a month of 31 days will be: (A) 135 (B) 150 (C) 165 (D) 180 Number of B.O.T. units consumed by 15 lamps V 2 ´ 15 ´ 6 ´ 31 = R ´ 1000 220 ´ 220 ´ 15 ´ 6 ´ 31 = » 135 103 ´ 1000 In the above problem the number of B.O.T. units consumed by the fans in the month will be : (A) 193 (B) 173 (C) 143 (D) 113 No. of B.O.T. unit consumed by 4 fans 1 ´ 746 ´ 6 ´ 31 ´ 4 = » 143 3 ´ 1000 In Illustration 6, the total cost of electric power consumed in the month, at the rate of Rs. 1.40 per B.O.T. unit, will be : (A) Rs. 279/(B) Rs. 389/(C) Rs. 421/(D) Rs. 538/Total number of B.O.T. units consumed = 135 + 143 = 278 cost of electric power consumed = 278 1.40 = Rs. 389.20 or 389/-
CHEMICAL EFFECTS OF CURRENT Definitions of various terms (i) Electrolysis: The process of splitting up or decomposing a liquid by passing an electric current through it, is defined as electrolysis. (ii) Electrolyte: The compound, whether fused or in solution, which undergoes decomposition by an electric current, is defined as electrolyte. (iii) Anions and Kations: The decomposed substance appears in the form of ions. The ions appearing at the anode are known as anions and those at the cathode are known as cations.
(iv) (v) (vi)
(vii) (viii)
(ix)
(x) (xi) (xii)
Ionisation: The phenomenon of separation of a molecule into oppositely charged ions is known as ionisation. Equivalent weight- The ratio of the atomic weight of an element to its valency is defined as its equivalent weight. Atomic weight: The ratio of the average mass of an atom of an element of the mass of an atom of hydrogen, taken as 1.008, is defined as the atomic weight of that element. Valency: The valency of an element is the number of atoms of hydrogen or chlorine which combines with or is displaced by one atom of the element. Molecular weight: The molecular weight of a substance is the ratio of the mass of one molecule of the substance to the mass of an atom of oxygen which is taken as 16. Gram-equivalent: It is the weight in grams of an element which will combine with or replace 1 gm of hydrogen. Gram - atom Gram-equivalent = valency Gram-molecule: The molecular weight of any substance. expressed in grams is defined as gram-molecule of that substance. Avogadro number: The number of molecules of a substance in its one-grammolecule is known as Avogadro number. Normal solution: A solution containing one gram-equivalent of the solute per litre is called a normal solution.
FARADAY'S LAWS OF ELECTROLYSIS (i)
(ii)
(iii)
First law: The towal mass of ions liberated at an electrode, during electrolysis, is proportional to the quantity of electricity which passes through the electrolyte. But Q = it i.e. m Q m it Hence the first law may also be stated as follows The mass of ions liberated at an electrode during electrolysis is proportional to (a) the current following through the electrolyte, and (b) the time for which the current flows Second law: If same quantity of electricity is passed through different electrolytes, the masses of the substances (ions) deposited at the respective cathodes are directly proportional to their chemical equivalents (equivalent weights). i.e. m E (chemical equivalent) Electro-chemical equivalent (E.C.E.): (a) The electro-chemical equivalent of an element is its mass in grams deposited on the electrode by the passage of 1 coulomb of charge through it i.e. by the passage of 1 ampere current for 1 second. (b) According to Faraday's first lawm i.t or m = Z i.t Where Z is the constant of proportionality known as the electro- chemical equivalent of the substance. It is numerically equal to the mass in grams of the element deposited when a unit current flows in unit time. (c) According to Faraday's first law m = ZQ If same charge is passed two electrolytes, the m1 Z1 m2 Z2
(iv)
According to Faraday's second law m1 E1 m2 E2 m Z 1 1 m2 Z2 E or Z2 = Z1 2 E1 (d) E.C.E. of any substance = E.C.E. of hydrogen × chemical equivalent of the substance. (e) If silver is taken as the standard substance for which E.C.E. is 0.001118 gm per coulomb, then E.C.E. of any substance = Chemical equivalent of subs tance 0.001118 chemical equivalent of silver Faraday: (a) The quantity of electricity (i.e. charge) required to liberate a gram equivalent of a substance during electrolysis. (b) As derived above E1 Z1 E or cons tant F E2 Z2 Z The constant F is known as one Faraday. For example for copper E = 31.5 gm and Z = 0.000329 gmC–1 31.5 F 96500 coulomb 0.000329 (c) According to Faraday's first law m = Z.i.t. = ZQ If p is the valency of the element, then electrons has to flow through the solution to deposit one atom. charge required to deposit 1 mol of the substance = Npe, where N = Avogadro number and m = M M = ZNpe M 1 E E Z p Ne Ne F Where F = Ne = Faraday constant or F = 6.0229 × 1023 × 1.602 × 10–19 = 96487C 96500C
Illustration 9:
Solution:
In producing chlorine through electrolysis 100 KW power at 125 volt is being consumed. If the E.C.E. of chlorine is 0.367 10–6 kg/coul, then the mass of chlorine liberated per minute will be: (A) 16.3 10–4 kg (B) 17.61 10–3 kg (C) 18.2 10–3 kg (D) 10–4 kg P Mass of chlorine liberated = Zit = Z t V -6 5 0.367 ´ 10 ´ 10 ´ 60 = = 17.61 ´ 10- 3 125
Illustration 10: The chemical equivalents of copper and silver are 32 and 108 respectively. When copper and silver voltammeters are connected in series and electric current is passed through them for sometime, 1.6 gm of copper is deposited. The mass of silver deposited will be :
Solution:
(A) 3.5 gm (B) 2.8 gm (C) 5.4 gm (D) none of these E1 m1 = E2 m2 E 108 \ m2 = 2 ´ m1 = ´ 1.6 = 5.4 gm E1 32
Illustration 11: A silver and a copper voltmeters are connected across a 6 volt battery of negligible resistance. In half an hour, 1 gm of copper and 2 gm of silver are deposited. The rate at which energy is supplied by the battery will approximately be : (Given E.C.E. of copper = 3.294 10–4 g/C & E.C.E. of silver = 1.118 10–3 g/C) (A) 64 watt (B) 32 watt (C) 96 watt (D) 16 watt m Solution: m = Zit or i = Zt For silver voltmeter m 2 i1 = 1 = = 0.994 amp Z1t 1.118 ´ 10- 3 ´ 1800 For copper voltmeter m 1 i2 = 2 = = 1.687 amp. Z2 t 3.294 ´ 10- 4 ´ 1800 power of circuit = V(i1 + i2) = 6 (0.994 + 1.687) = 6 2.681 16 W.
SOME IMPORTANT POINTS (i) (ii) (iii)
Capacity of a cell- It is expressed in kilowatt hours which involves both current and the voltage. Energy capacity- It is expressed in kilowatt hours which involves both current and the voltage. Efficiency- (a) It is the ratio of discharging capacity to the charging capacity. (b) If discharge takes place slowly, the maximum available energy is 80% of that spent in charging. (c) If the cell is short circuited, the discharge takes place suddenly and the cell is spoiled.
APPLICATIONS OF ELECTROLYSIS The phenomenon of electrolysis has been used as a valuable tool in (i) identifying the components of certain liquids (ii) making accurate measurement of current (iii) calibrating an ammeter (iv) determining the electro-chemical equivalents of elements (v) electroplating (vi) producing pure metals (vii) electrolying
THERMOELECTRICITY-SEEBECK EFFECT
(i) (ii)
Thermocoule: If two wires of different metals are joined at their ends so as to from two junctions, then the resulting arrangement is called a thermocouple. Seebeck effect: (a) When the two junctions of a thermocouple are maintained at different temperatures, then a current starts flowing through the wires. This is called Seebeck effect and the e.m.f. developed in the circuit is called thermo-emf. (b) The Seebeck effect is perfectly reversible, i.e. if the hot and the cold junctions are interchanged, the direction of current is reversed. (c) The thermo-emf is of the order of a few micro-volts per degree temperature difference.
THERMOELECTRIC SERIES (i)
(ii) (iii)
Seebeck arranged a number of metals in the form of a series called thermoelectic series. The arrangement of some of the metals forming the series areSb, Fe, Zn, Ag, Au, Mo, Cr, Sn, Pb, Hg, Mn, Cu, Co, Ni, Bi. In the above series, current flows through the cold junction from the metal which appear earlier to the metal which appears later. Greater the separation of the two metals in the series, greater is the thermo emf generated.
VARIATION OF DIFFERENCE
WITH
TEMPERATURE-
If the cold junction is kept at 0o and the temperature of hot junction (t) is gradually increased, it is found that the thermo emf e first increases, attains a maximum value and then decreases to become zero again (Fig.) Thermo emf V
(i)
THERMO-EMF
tn 0oC
(ii) (iii) (iv)
(v)
ti
Temperature of hot junction t
If the temperature is further increased, the emf changes direction. Netural temperature (tn)- The temperature at which thermo-emf is maximum is defined as neutral temperature. tn is fixed for a given thermo-couple. Temperature of inversion (ti)- The temperature at which thermo-emf changes its direction is called the temperature of inversion. ti is an much above tn as the tenoeratyre if cikd hybctuib us bekiw tn. Mathematically 1 e t t 2 2 where and are constants for a given thermo- couple.
Illustration 12: The e.m.f. of a thermocouple, one junction of which is kept at 0 C, is given by e = at + bt2 The neutral temperature will be: a a (A) (B) b b
a a (D) 2b 2b At neutral temperature e = maximum de \ =0 dt or a + 2btn = 0 (C)
Solution:
\
tn = -
a . 2b
Illustration 13: In the above problem, the temperature of inversion will be : a a (A) (B) b b a a (D) (C) 2b 2b t +t Solution: tn = o i 2 \ ti = 2tn - to or ti = 2 ´ -
a a - 0=2b b
Illustration 14: In Illustration 12, the Peltier co-efficient will be : (A) (t – 273) (a + 2bt) (B) (t + 273) (a – 2bt) (C) (t – 273) (a – 2bt) (D) (t – 273) (a – 2bt) de Solution: Peltier coeff. p = T dT tC = T – 273 \ e = a(T - 273) + b(T - 273)2 d.w.r.t. T de = a + 2b(T - 273) dT de \ p=T = T[a + 2b(T - 273)] dT or p = (t + 273) (a + 2bt) . Illustration 15: In Illustration 12, the Thomson co-efficient will be : (A) 2a(t + 273) (B) 2a(t – 273) (C) 2b(t + 273) (D) 2b(t – 273) 2 de Solution: Thomson co-efficient s = T 2 dT de = a + 2b(T - 273) dT d.w.r.t. T d2e = 2b dT 2 d2e \ s = T 2 = 2b(t + 273) dT
SEEBECK COEFFICIENT The rate of change of thermo emf with temperature is called thermo-electric power or Seebeck coefficient (S). de X t dt When t = tn, e is maximum de 0 dt tn 0 or tn
PELTIER EFFECT (i) (ii) (iii)
(iv)
This effect is the converse of Seebeck effect. If a current is passed through a junction of two dissimilar metals, heat is eigher absorbed or evolved at the junction. On reversing the direction of current, the heating effect is also reversed. If the seebeck current is in a certain direction through the hot junction, then an external current sent in the same direction through this junction produces a cooling at this junction and a heating at the other junction. Peltier coefficient ()- It is the amount of heat absorbed or evolved at a junction per second when a current of one ampere flows through it. de T dT
THOMSON EFFECT (i) (ii)
(iii)
An emf is developed between two parts of a single metal if they are at different temperatures. This is called Thomson effect. Thomson coefficient- If de is the potential difference between two points in a metal which have a temperature difference dt, then the ratio de d2e T 2 dT dT is defined as the thomson coefficient. If one part of a metal is at a higher temperature than the other, the free electrons at hot part will have more kinetic energy and as a result these electrons will diffuse faster towards colder part than the electrons from cold to the hot part. This would result in the net transfer of electrons setting an electric current in the metal.
APPLICATIONS OF THERMOELECTRIC EFFECTS (i) (ii) (iii) (iv)
Measurement of temperature Detection of heat radiations Refrigeration Power generation.
OBJECTIVE ASSIGNMENT 1.
Sol: 2.
Sol:
3.
Sol:
4.
Sol:
5.
Sol: 6.
Consider the two statements A and B (i) the neutral temperature does not depend on temperature of cold junction (ii) the inversion temperature does not depend on temperature of cold junction (A) both A and B are correct (B) A is correct but B is wrong (C) A is wrong but B is correct (D) both A and B are wrong. (B) The neutral temperature of a thermocouple with cold junction at 20 oC is 220oC. Its temperature of inversion is (A) 420oC (B) 120oC (C) 110oC (D) 440oC (A) We know tn - t c = ti - tn tc = neutral temperature ti = inversion temperature \ ti = 2tn - tc = 440° - 20° = 420° C For a thermocouple if the cold junction is maintained at 0oC the inversion temperature is 680oC. Its Neutral temperature is (A) 1360oC (B) 680oC (C) 340oC (D) 170oC (C) We know inversion temperature = 2(neutral temperature) 680 Neutral temperature = = 340° C 2 Consider the following two statements A and B and identify the correct choice in the given answers. (i) Thermo e.m.f. is minimum at neutral temperature of a themocouple. (ii) When two junctions made of two different metallic wires are maintained at different temperatures, an electric current is generated in the circuit (A) A is false and B is true (B) A is true and B is false (C) Both A and B are false (D) Both A and B are true (A) At neutral temperature of a thermocouple, thermo emf should be maximum, but not minimum A is false and B is true. According to the Faraday's Law of electrolysis, the mass deposited or liberated at an electrode is proportional to : m Q2 (B) (A) mQ (C) (D) m does not depend on Q m l2 (A) Two resistors having equal resistances are joined in series and a current is passed through the combination. Neglect any variation in resistance as the temperature changes. In a given time interval. (A) equal amounts of thermal energy must be produced in the resistors (B) unequal amounts of tgernak ebergt nat be oridyced (C) the temperature must rise equally in the resistors
Sol: 7.
Sol:
(D) (A)
the temperature must rise unequally in the resistors.
A constant voltage is applied between the two ends of a uniform metallic wire. Some heat is developed in it. The heat developed is doubled if (A) both the length and radius of the wire are halved (B) both the length and radius of the wire are doubled (C) the radius of the wire is doubled (D) the length of the wire is doubled Where the time remains the same in both the cases. V 2 V 2 pr 2 r 2 (B) H = = µ rL R L 2
H ær1 ö æL 2 ö = ´ H' çè r2 ÷ ø çè L1 ÷ ø H 1 2 1 = ´ = H' 4 1 2 Þ H' = 2H Þ
8.
Which of the following plots may represent the thermal energy produced in a resistor in a given time as a function of the electric current? (A) a (B) b (C) c (D) d
d
c b a
i
Sol:
(A) H = I 2 Rt = U
Þ U µ I2 It represents a parabola. 9.
Sol:
Two heater wires, made of the same material and having the same length and the same radius, are first connected in series and then in parallel to a constant potential difference. If the rates of heat produced in the two cases are H s and Hp respectively, H then s will be : Hp 1 1 (A) (B) 2 (C) (D) 4 2 4 (C) Since l1 = l2 , r1 = r2 & r 1 = r 2 So resistance R1 = R2 = R Hs =
V2 V2 = Req 2R
V2 V2 = Hp = R Req 2
æV 2 ö çè 2R ÷ ø
Hs 1 = = Hp æ ö 4 2 çV ÷ çR ÷ çè ÷ 2 ø
10.
Sol:
11.
Sol:
12.
Sol:
13.
Sol:
A hot electric iron has a resistance of 80 and is used on a 200 V source. The electrical energy spent, if it is used for 2hr. will be (A) 800 Wh (B) 1000 Wh (C) 2000 Wh (D) 8000 Wh 2 2 V (200) (B) (P) = = = 500W R 80 Therefore energy spent = power time = 500 2 = 1000 Wh. Antimony and Bismuth are usually used in a thermocouple, because (A) higher thermo e.m.f. is produced (B) lower thermo e.m.f. is produced (C) constant thermo e.m.f. is produced (D) negative thermo e.m.f. is produced Since antimony and bismuth form the pair of metals producing higher thermo emf in comparison with other pairs of thermocouple. Therefore pair is usually used in thermocouple. At 40C resistance of platinum is 3.14 and at 100C its resistance is 3.76. Its temperature coefficient is (A) 3.75 10–1/C (B) 3.75 × 10–2/C (C) 0.00329/C (D) None of these R2 - R1 (C) a = R1 ´ (t 2 - t1 ) 3.76 - 3.14 Þ a = = 0.00329 3.14 ´ (100 - 40) Two resistance filaments of same length are connected first in series and then in parallel. Fine the ratio of power dissipated in both cases assuming that equal current flows in the main circuit. (A) 1:4 (B) 4:1 (C) 1:2 (D) 2:1 (D) Required ratio i/2
R
i i/2 R Parallel Combination
R R Series Combination
i2R + i2R 2
2
æi ö æi ö R+ç ÷ R çè ÷ è2 ø 2ø
14.
Sol: 15.
Sol:
16.
Sol:
17.
Sol:
18.
Sol:
19.
= 2 :1
Thermocouple thermometer is based on (A) Seeback effect (B) (C) Peltier effect (D) (A)
Crompton effect Photoelectric effect.
A wire has a resistance of 3.1 at 30C and a resistance 4.5 at 100oC. The temperature coefficient of resistance of the wire (A) 0.0064oC–1 (B) 0.0034oC–1 o –1 (C) 0.0025 C (D) 0.0012oC–1 R - R30 4.5 - 3.1 (A) a = 100 = R30 ´ Dt 3.1 ´ (100 - 30) 1.4 1.4 = = = 0.0064°C- 1 3.1 ´ 70 217 Faraday constant (F), chemical equivalent (E) and electrochemical equivalent (Z) are related as : E E Z F F 2 (C) (D) (A) F = EZ (B) F= Z Z E (C) We know frmo the Faraday's second law of electrolysis that electrochemical equivalent 1 M 1 E (Z) = ´ = ´ E or F = F P F Z Two heater coils separately take 10 minutes to boil a certain amount of water. If both coils are connected in series, time taken to boil the same amount of water will be : (A) 15 min. (B) 20 min. (C) 7.5 min. (D) 2.5 min. 2 V t Þ tµR (B) Q = R t R \ 1 = 1 t 2 R2 10 R Þ = Þ t 2 = 20 min t 2 2R Resistance of a copper coil is 4.64 at 40oC and 5.6 at 100oC. Then its resistance at 0oC is (in ): (A) 4 (B) 0.96 (C) 5.12 (D) 4.2 R - R0 5.6 - R0 4.64 - R0 (A) a = t = = R0 ´ 4t R0 ´ 100 R0 ´ 40 Þ R0 = 4W Resistance of a coil at 100oC is 4.2 and the temperature coefficient of resistance of its material is 0.004oC–1. Then its resistance at 0oC is () : (A) 5 (B) 3 (C) 4 (D) 3.5
Sol:
(B) Rt = R0 (1 + a t) Þ 4.2 = R0 (1 + 0.004 ´ 100)
Þ R0 = 20.
Sol:
4.2 = 3W 1.4
An electric kettle has two heating elements. One brings it to boil in ten minutes and the other in fifteen minutes. If the two heating filaments are connected in parallel, the water in the kettle will boil in (A) 5 minutes (B) 25 minutes (C) 8 minutes (D) 6 minutes. V2 V2 (D) Q = ´ 10 = ´ 15 R1 R2
Þ
R1 2 = R2 3
V2 V2 ´ 10 = Q= t R1 æ R1 R2 ö çè R + R ÷ 1 2 ø
æ R ö 5 Þ 10 = ç1 + 1 ÷t = t è R2 ø 3 Þ 21.
Sol:
22.
Sol: 23.
Sol: 24.
t = 6 min
The resistance of a conductor is 5 ohm at 50oC and 6 ohm at 100oC. Its resistance at 0oC is (A) 2 ohm (B) 1 ohm (C) 4 ohm (D) 3 ohm. R - R0 (C) a = 1 R0 t 5 - R0 6 - R0 Þ a = = Þ 10 - 2R0 = 6 - R0 50R0 100R0 R0 = 4W The mass liberated at any electrode when electricity is passed through it is directly proportional to 1 (A) q2 (B) q (C) (D) q3 q (B) m = ZIt = Zq Þ m µ q Which one of the following causes production of heat when current is set up in a wire? (A) fall of electrons from higher orbits to lower orbits (B) interatomic collisions (C) interelectron collisions (D) collisions of conduction electrons with atoms. (D) In the Seeback series bismuth occurs first followed by Cu and Fe among other. The Sb is the last in the series. If V1 be the thermo-emf at the given temperature difference for Bi-Sb themocouple and V2 be hat for Cu-Fe thermocouple, then
Sol: 25.
Sol: 26.
Sol:
27.
Sol:
28.
Sol: 29. Sol: 30.
Sol:
(A) (C) (D)
V1 = V 2 V1 > V 2
(B) (D)
V1 < V 2 data insufficient.
Which of the following is a characteristic temperature for the themocouple ? (A) temperature of cold junction (B) temperature of hot junction (C) temperature of inversion (D) neutral temperature. (D) When a copper voltameter is connected with a battery of electromagnetic field 12V, 2gm of copper is deposited in 30 minutes. If the same voltmeter is connected across a 6V battery, the mass of copper deposited in 45 minutes would be (A) 1 gm (B) 1.5 gm (C) 2 gm (D) 2.5 gm (B) m µ zIt, m µ zvt m2 v 2 t 2 = m1 v1t1 v t 6 45 Þ m2 = 2 2 ´ m1 = ´ ´ 2 = 1.5 gm v1t1 12 30 The electro-magnetic field of a Cu-Fe thermo couple varies with temperature of the hot junction (cold junction at 0oC) as E = 14 – 0.022. The neutral temperature is given by (A) 350oC (B) 400oC (C) 450oC (D) 500oC. dE (A) E = 14q - 0.02q2, =0 dq Þ q = 350° C Same current is being passed through a copper voltmeter and a silver voltmeter. The rate of increase in weights of the cathodes in the two voltameters will proportional to: (A) atomic masses (B) atomic numbers (C) relative densities (D) none of these. (A) Which of the following acts a depolariser in Leclanche cell? (A) ZnCl2 (B) Mn2O3 (C) MnO2 (D) NH4Cl (C) In the circuit shown the 5 resistor develops 20W power due to current flowing through it, then power dissipated in 4 resistance is (A) 4W (B) 6W (C) 10 W (D) 20 W (A) If the current in 5 branch is i, then current in 4 branch will be i/2. P = i2R = i2 ´ 5 = 20 \ i = 2A
P' = i2R' = (1)2 ´ 4 = 4W 4Rt = R0 (1+ µ t) 1. 3R0 = R0 (1 + 4 ´ 10- 3 t)
1 = 500° C (B) 2 ´ 10- 3 H = Pt = 210 ´ 300 J 210 ´ 300 = cal = 15000 cal 4.2 Þ t=
2.
3.
Q = It = 1.6 ´ 60 = 96C 96 No. of Cu++ions = = 3 ´ 1020 3.2 ´ 10- 19
For Test 1.
Sol: 2.
Sol: 3.
Sol: 4.
Sol:
5.
Sol: 6.
Sol: 7.
At what temperature will the resistance of a copper wire become three times its value at 0oC? [Temperature coefficient of resistance for copper = 4 × 10–3 per oC] (A) 400oC (B) 450oC (C) 500oC (D) 550oC. (C) How much heat is developed in a 210 watt electric bulb in 5 minutes ? [Mechanical equivalent of heat = 4.2 joule/calorie] (A) 7500 calories (B) 15000 calories (B) 22500 calories (D) 30000 calories. (B) A current of 1.6 A is passed through a solution of CuSO 4. How many Cu++ ions are liberated in one minute ? [Electronic charge = 1.6 × 10–19 coulomb]. (A) 3 × 1020 (B) 3 × 1019 (C) 6 × 1020 (D) 6 × 1019 (A) The smallest temperature difference that can be measurred with a combination of a thermocouple of thermo e.f.f. 30V per degree and a galvanometer of 50 ohm resistance, capable of measuring a minimum current of 3 × 10–7 amp, is : (A) 0.5 degree (B) 1.0 degree (C) 1.5 degree (D) 2.0 degree. (A) Minimum voltage that can be measured by the galvanometer V = ig G = 3 ´ 10- 7 ´ 50 = 1.5 ´ 10- 5 volt Smallest temperature difference that can be mesured 1.5 ´ 10- 5 Dt = = 0.5°C 30 ´ 10- 6 If 2.2 kilowatt power is transmitted through a 10 ohm line at 22000 volt, the power loss in the form of heat will be(A) 0.1 watt (B) 1 watt (C) 10 watt (D) 100 watt. 2 P R 2200 ´ 2200 ´ 10 = 0.1 W (A) H = I 2R = 2 = V 22000 ´ 22000 For electroplating a spoon, it is placed in the voltameter at (A) the position of anode (B) the position of cathode (C) exactly in the middle of anode and the cathode (D) anywhere in the clectrolyte (B) Positive ios get deposited on cathode. The quantity of electricity needed to liberate one gram equivalent of an element is(A) 1 ampere (B) 96500 amperes
Sol:
8.
Sol: 9.
Sol:
10.
Sol:
11.
Sol:
(C) 96500 coulomb (D) 96500 faradays. (C) The amount of electricity required to liberate 1 gm equivalent is 1 Faraday or 96500 coulomb. When current is passed through a junction of two dissimilar metals, heat is evolved or absorbed at the junction. This process is called(A) Seebeck effect (B) Joule effect (C) Peltier effect (D) Thomson effect. (C) Peltier effect A 5oC rese in temperature is observed in a conductor by passing a current. When the current is doubled the rise in temperature will be approximately(A) 10oC (B) 16oC (C) 20oC (D) 12oC. (C) As H µ I2 , on doubling the current, heat produced and hence rise in temperature becomes fuor times. Rise in temperature 4 5 = 20C Three equal resistors connected across a source of e.m.f. together dissipate 10 watt of power. What will be the power dissipated in watts if the same resistors are connected in parallel across the same source of e.m.f.? 10 (A) 10 (B) (C) 30 (D) 90. 3 V2 (D) In series Power = = 10W \ V 2 = 30R 3R 2 V 30R ´ 3 = = 90W In parallel P' = R æR ö çè ÷ 3ø If nearly 105 coulombs liberte 1 gm-equivalent of aluminium, hen the amount of aluminum, (equivalent weight 9) deposited through electrolysis in 20 minutes by a current of 50 amp. will be(A) 0.6 gm (B) 0.09 gm (C) 5.4 gm (D) 10.8 gm. (C) m = zit or m µ it
m1 i1 t1 = m2 i2 t 2 or m2 = 5.4 gm \
12.
Sol:
13.
or
9 105 = m2 50 ´ 20 ´ 60
Two filaments of same length are connected first in series and then in parallel. For the same amount of main current flowing, the ratio of the heat produced is : (A) 2:1 (B) 1:2 (C) 4:1 (D) 1:4 2 (C) Q = I R or Q µ R Qs Rs 2R \ = = = 4 :1 Qp Rp æR ö çè ÷ 2ø Constantan wire is used in making standard resistances, because its (A) specific resistance is low (B) density is high (C) temperature coeff. of resistance is negligible
Sol:
14.
Sol:
(D) melting point is high. (C) The temperature coefficient of resistance of constantan is quite negligible, hence its resistance almost remains constant. As the temperature of hot junction increases, the thermo cmf : (A) always increases (B) always decreases (C) may increase of decrease (D) always remains constant (C) Thermo e.m.f.
0o C
Temp. of hot junction
t
The variation of thermo-e.m.f. with temperature of hot junction is as shown in the figure. 15.
Sol:
16.
Sol:
17.
A certain charge liberates 0.8 gm of Oxygen. The same charge will liberate how many gms of Silver? 108 gm . (A) 108 gm (B) 10.8 gm (C) 0.8 gm (D) 0.8 MAg EAg (B) = M0 E0 108 ´ 0.8 or mAg = = 10.8 g 8 An electric kettle has two heating coils. When one coil is used, water in the kettle boils in 5 minutes, while when second coil is used, same water boils in 10 minutes. If the two coils, connected in parallel are used simultaenously, the same water will boil in time. (A) 3 min. 20 sec. (B) 5 min. (C) 7 min. 30 sec. (D) 2 min. 30 sec. 2 V t Þ tµR (A) Q = R t 5 R1 \ 1 = = Þ R2 = 2R1 t 2 10 R2 In second case R1 R 5 = = 1 R1R2 2 t R R1 + R2 3 1 10 Þ t= min = 3 min 20 sec. 3 An external resistance R is connected to a battery of e.m.f. V and internal resistance r. The joule heat produced in resistor R is maximum when R is equal to
r (C) 2r (D) infinitely large. 2 (A) According to maximum power transfer theroem, transfer of power from the surce to the load will be maximum when load resistance is equal to the internal resistance of the source. R=r
(A)
Sol:
18.
Sol:
19.
Sol:
20.
Sol:
21.
Sol: 22.
r
(B)
If Tc, Tn and Ti denote the temperatures of cold junction, neutral temperature and inversion temperature of a thermo-couple respectively, then : (A) Tc + Ti = Tn (B) Ti + Tc = 2Tn (C) Tc + Ti = 2Tn (D) Tc – Ti = 2Tn T + Ti (A) Tn = c 2 Þ Tc + Ti = 2Tn The electrochemical equivalent of a material in an electrolyte depends on (A) The nature of the material (B) The current through the electrolyte (C) The amount of charge passes through electrolyte (D) The amount of material present in electrolyte. (A) m = Zit = ZQ m \ Z= Q It depend on the nature of material. A wire of length I and 3 identical cells of negligible internal resistances are connected in series. Due to the current, the temperature of the wire is raised by T in a time t. A number of N similar cells is now connected in series with a wire of the same material and cross section but the length 2L. The temperature of the wire is raised by the same amount T in the same time t. The value of N is(A) 4 (B) 6 (C) 8 (D) 9. 2 V t (B) The heat generated in the wire is givein by H = R 2 (3E) t In the first case, we have mCp DT = . . . (1) R in the second case, we have (NE)2 t (2m) Cp Dt = (\ R µ L) . . . (2) 2R Dividing Equation (2) by Equation, (1) we get N2 / 2 2= Þ N2 = 36 Þ N = 6 9 Two wires having resistance of 2 and 4 are connected to same voltage. Ratio of heat dissipated in the resistances is(A) 1:2 (B) 4:3 (C) 2:1 (D) 5 : 2. H1 R2 2 (C) As heat dissipated = V 2 /R , so it is inversely proportional to 1/ R \ = = H2 R1 1 If thermo e.m.f. is given by E = at + bt2, then neutral temperature is-
a 2b (B) 2b a (A) At neutral temperature dE/dt = 0 or a + 2bTN = 0 a TN = 2b
(A) Sol:
23.
Sol:
24.
Sol:
(C)
a b
(D)
b a
Two parallel and straight conductors are at a distance d apart, each have current i following through them. The force per unit length is m0i2 m0i (A) (B) 2p d 4pd 2 m0i m0i (C) (D) 4pd d2 (B) Force due to magnetic field induction = B12, where m 2I B1 is the magnetic field due to first = 0 × 1 4p d m0 i2 \ F= × 2p d Thomson coefficient of a conductor is 10V/K. The two ends of it are kept at 50C and 60C respectively. Amout of heat abosrbed by the conductor when a charge of 10 C flows through it is : (A) 1000 J (B) 100 J (C) 100 mJ (D) 1 mJ (D) Amount of heat absorbed by the conductor H = Q dt H = (10 10–6) (10) (10) = 10–3 J = 1 mJ
25_57. In an electroplating experiment 4A current for 2 minutes deposits m gram of silver. When 6A current flown for 40 seconds then the deposited silver will be (A) 4m (B) m/2 (C) m/4 (D) 2m Sol: (B) m i t m1 i1 t1 m 4 ´ 120 \ = = or m2 i2 t 2 m2 6 ´ 40 m or m2 = . 2 26_58. The temperature coefficient of a resistance wire is 0.00125 per degree. The resistance at 300K is 1 ohm. At what temperature its resistance will be 2 ohm. (A) at 1154 K (B) at 1100 K (C) at 600 K (D) at 1400 K Sol: (B) Rt2 = Rt1 [1+ a (t 2 - t1)] 2 = 1[1 + 0.00125(t 2 - 300)]
or or
1 = 800 0.00125 t 2 = 1100 K t 2 - 300 =
27_59. Two electric heater wires are made from the same material and have the same dimensions. First these are connectted in series and then these are connected in parallel (to a 220 V AC source) then the ratio of heat generated will be (A) 2:1 (B) 1:2 (C) 4:1 (D) 1:4 2 V 1 Sol: (D) H = µ R R R H1 R2 \ = = 2 = 1: 4 H2 R1 2R 28_60. The mass of a product liberated on anode in an electrochemical cell depends on (A) ( t)1/2 (B) t (C) /t (D) 2 t (Where t is the time period for which the current is passed). Sol: (B) m Q But Q = t m t or m = Z t 29_61. If i is the inversion temperature, n is the neutral temperature, c is the temperature of the cold junction, then (A) i + c = n (B) i – c = 2n qi + qc (D) c – i = 2n (C) = qn 2 Sol: (C) Neutral temperature is the mean of temperature of inversion and temperature of q + qc cold junction, i.e., qn = i . 2 30_75. The thermo e.m.f. of a thermo-couple is 25V/C at room temperature. A galvanometer of 40 ohm resistance, capable of detecting current as low as 10 –5 A, is connected with the thermocouple. The smallest temperature difference that can be detected by this system is (A) 16C (B) 12C (C) 8C (D) 20C Sol: (A) Minimum p.d. to be measured V = iG = 10 - 5 ´ 40 = 4 ´ 10 - 4 V 4 ´ 10- 4 = 16°C . Minimum temp. diff. detected = 25 ´ 10- 6
1.
VARIOUS DEFINITIONS & FEATURES RELATED TO THEM
1.1
BAR MAGNET
A system composed of two poles, equal in magnitude but opposite in polarity, placed at a small displacement apart is known as a bar magnet. It is also known as a magnetic dipole. –mp
+mp N
S
N
S (a)
(b)
1.2
POLE STRENGTH (MP)
(Ia)
The strength of a magnetic pole to attract magnetic materials towards it, is known as pole strength. Greater the number of unit poles in a magnetic pole, greater will be its strength. F Magnetic force mp B Magneticinduction Its unit is ampere-meter.
(b) (c) (d)
Illustration 1:
Solution :
The magnetic flux density of magnetic field is 1.5 Wb/m2. A proton enters this field with a velocity of 2 107 m/s in a direction making an angle of 30 with the field. The force acting on the proton will be : (A) 2.4 10–12 Newton (B) 0.24 10–12 Newton (C) 24 10–12 Newton (D) 0.024 10–12 Newton (A) F = qvBsin q = 1.6 ´ 10- 19 ´ 2 ´ 107 ´ 1.5 ´ 0.5 = 2.4 ´ 10- 12
1.3
MAGNETIC LINES OF FORCE
(a)
The imaginary lines which represent the direction of magnetic field, are known as magnetic lines of force. The imaginary path traced by an isolated (imaginary) unit north pole is defined as a line of force. Magnetic lines of force are closed curves. Outside the magnet their direction is from north pole to south pole and inside the magnet these are from south to north pole.
(b) (c)
N
(d) (e) (f) (g)
S
They neither have origin nor end. These lines do not intersect, because if they do so then it would mean two value of magnetic field at a single point, which is not possible. The tangent drawn at any point to the line of force indicates the direction of magnetic field at that point. At the poles of the magnet the magnetic field is stronger because the lines of force there are crowded together and away from the poles the magnetic field is week. i.e. magnetic field intensity number of lines of force.
(h)
(i)
The number of magnetic lines of force passing through unit normal area is defined as magnetic induction (B) whereas the number of lines of force passing through any area is known as magnetic flux. The lines of force can emerge out of the north pole of magnet at any angle and these can merge into the south pole at any angle.
1.4
MAGNETIC FIELD
(a)
The space around a magnet in which a torque acts on a magnetic needle is known as magnetic field. The space around a magnet in which a net force acts on a magnetic test pole is known as magnetic field. The space around a magnet in which its effect is experienced is known as magnetic field. There are four types of magnetic field : (i) Uniform magnetic field: (a) The magnetic field, in which the intensity of magnetic field is same at all points, is known as uniform magnetic field. (b) In such a magnetic field the magnetic lines of force are parallel and equidistant. e.g. the magnetic dines of force of earth's magnetic field. (ii) Non-uniform magnetic field: (a) The magnetic field, in which the intensity of magnetic field at different points is different, is known as non-uniform magnetic field. (b) It is represented by non-parallel lines of force. (iii) Varying magnetic field: (a) The magnetic field, which keeps on changing with respect to time is known as a variable magnetic field. (b) Example :– B = B0 sin t or B = B0 cos t (iv) Non-varying magnetic field: (a) The magnetic field which does not change with time is known as a constant magnetic field. (b) The direction of magnetic field is that in which a force acts on a unit test pole. (c) It can be produced by moving charges, current carrying loops, and variations in electric currents. F F N S
(b) (c) (d)
B
B
1.5
MAGNETIC DIPOLE
(a)
The structures, which tend to align along the direction of magnetic field, are known as magnetic dipoles. Bar magnet, current carrying solenoid, current carrying coil, current carrying coil, current loop, magnetic needle etc. are the examples of magnetic dipole. It is not possible to separate out the two poles of a magnet. Magnetic dipole is not a system composed of two poles because the existence of monopoles is not possible. A loop of single turn is also a magnetic dipole. One face of the loop behaves as north pole and the other face behaves as south pole. The face of the coil, in which current is anticlockwise, behaves as north pole and the face in which current is clockwise, behaves as south pole.
(b) (c) (d)
1.6
MAGNETIC FLUX
(a) (b) (c)
The number of lines of force passing through a given area is defined as magnetic flux. The magnetic flux passing through unit normal area is defined as magnetic induction (B). When the magnetic field is normal to the plane then = BA, when A = 1m2 then = B. B nˆ Normal to surface
A
Vibrative lines Of force Surface
(d)
When magnetic field makes an angle with the normal to the plane : (i) Magnetic flux linked with the plane = Area of the plane (A) × Component of magnetic field normal to the plane (B cos ) i.e. = AB cos If the number of turns in the coil is N. then = NAB cos os Bc
B
Bcos
A
B
nˆ
or A
(ii)
= magnetic field normal to the plane (B) × component of A in the direction of magnetic field (A cos ) i.e. = BA cos
A
os Ac
B
In both cases q is the angle between B and nˆ .
(e)
(iii)
When B and A are mutually parallel then = 0o and = BA.
(iv)
When B and A are mutually perpendicular, = 90o and = 0.
(v) When B and A are antiparallel, then = 180o and = BA. When the angle between B and the plane of coil is then = BA sin . If the number of turns in the coil is N then = NBA sin . B O
B sin nˆ
180o A
B
B cos
(i) (ii)
When the plane of coil is parallel to B then = 0o and = 0. When B and the plane of coil are mutually perpendicular i.e. = 90, then = BA.
(f)
(iii) When B and the plane of coil are mutually antiparallel i.e. = 180o, then = 0. Magnetic flux linked with a small surface element dA d B.d A B dA cos where d A = area of small element. nˆ = unit normal vector. B
DA
nˆ
(g)
The flux linked with total area of the surface A d B.dA B dA cos
(h)
Positive magnetic flux: When the magnetic induction B and the unit normal vector nˆ are in the same direction then is called the positive magnetic flux. B A = BA Negative magnetic flux: When the magnetic induction B and unit normal vector are mutually in opposite directions then is called negative magnetic flux = – BA The flux emanating out of a surface is positive and the flux entering the surface is negative. is a scalar quantity. The net magnetic flux coming out of a closed surface is always zero, i.e. B.dA 0
(i)
(j) (k) (l)
A
or
.B 0
1.7
MAGNETIC FLUX DENSITY OR MAGNETIC INDUCTION
(a)
The magnetic lines of force passing through unit normal area in a magnetic field is defined as magnetic induction. B when A 1m.2 then B A
(b)
F B
i
i
(c)
(d) (e)
L
Direction of magnetic induction: The direction in the magnetic field in which if a current carrying conductor is placed then no force acts on it, is known as the direction of magnetic induction. Magnetic induction is a vector quantity. The magnetic induction due to a bar magnet 2KM (i) In axial position B 3 r KM (ii) In equatorial position B 3 r m Here K = 0 = 10- 7 Weber / A.m 4p
1.8
MAGNETISING FIELD OR INTENSITY OF MAGNETIC FIELD H
(a)
The ratio of magnetic induction produced in vacum (B o) and magnetic permeability of Bo vacuum is defined as magnetising field (H), i.e. H o The intensity of magnetic field due to a pole of strength m p at a distance r from it is m H p2 r M Due to a small magnet H = 3 1 3 cos2 r
(b)
(c)
Illustration 2:
Solution: Illustration 3:
1000 turns per meter are wound over a Rowland ring of ferromagnetic material. On passing a current of 2 ampere in the coil, a magnetic field of 10 Wb/m2 is produced in it. The magnetising force generated in the material will be : (A) 1.2 10–3 A/m (B) 2.6 10–3 A/m –4 (C) 2.6 10 A/m (D) 2 103 A/m Ni = ni = 103 ´ 2 = 2 ´ 103 A / m (D) H = 2p r In the above problem, the value of intensity of magnetisation in A/m will be : (A) 7.96 106 (B) 7.96 10–6 3 (C) 3.98 10 (D) zero
Solution:
(A) B = m0 (I + H)
\ I= Illustration 4:
Solution:
B 10 - H= - 2 ´ 103 = 7.96 ´ 106 A / m m0 4p ´ 10- 7
In the above question 115 the relative permeability of the material will be : (A) 4.98 103 (B) 4.98 10–3 (C) 2.98 10–3 (D) 3.98 103 7.96 ´ 106 I (D) mr = 1 + X = 1 + = 1 + = 3.98 ´ 103 3 H 2 ´ 10
1.9
MAGNETIC MOMENT (M)
(a)
If a magnet of length l and magnetic moment M is bent in the form of a semicircular 2M are then its new magnetic moment will be M' = mp
mp
N
S
l = r
l l N
S
N
2r
S
M = mpI
(b)
The magnetic moment of an electron due to its orbital motion is 1B whereas that due to its spin motion it is B . 2 eh eh i.e. Morbital = l 4ml 2ml
eh eh B Mspin = s s sB 2 4ms 2ms Here B = Bohr magneton eh (i) The value of Bohr magneton B = 4 m (ii) B = 0.93 × 10–23 Amp-m2 and
(c)
Other formulae of M: (i) M = nir2 eVr er 2 er 2 2f er 2 M (ii) 2 2 2 T e (iii) 2mJ (iv) M = nB
(d)
Resultant magnetic moment : (i) When two bar magnets are lying mutually perpendicular to each other, then
M M12 M22 2mpl
–mp S l mp N S –mp
l
N mp
M1
M 2
M 1
M 2
2
M2
(ii)
When two coils, each of radius r and carrying current i, are lying concentrically with their planes at right angles to each other, then
M M12 M22 2ir 2 if M1 = M2
Illustration 5: Solution:
A square loop OABCO of side carries a current i. It is placed as shown in figure. Find the magnetic moment of the loop. Magnetic moment of the loop can be written as,
M i BC CO
Here, BC k
CO cos 60 i sin 60 j
2
i
3 j 2
3 j M i ( k) i 2 2 i 2 or M j 3 i Ans. 2
Illustration 6:
Find the magnitude of magnetic moment of the current carrying loop ABCDEFA. Each side of the loop is 10 cm long and current in the loop is i 2.0 A.
C D
i B
A
E
F
Solution :
By assuming two equal and opposite currents in BE, two current carrying loop (ABEFA and BCDEB) are formed. Their magnetic moments are equal in magnitude but perpendicular to each other. Hence
Mnet M2 M2 2 M
Where M ia (2.0)(0.1)(0.1) 0.02 A-m2 Mnet ( 2 ) (0.02) A – m2 0.028 A-m2 C D B
A
E
F
Illustration 7:
Solution:
Two identical bar magnets each of length L and pole–strength m are placed at right angles to each other with the north pole of one touching the south pole of the other. Evaluate the magnetic moment of the system. As magnetic moment is a vector M2 sin and as here M1 = MR (M12 M22 2M1M2 cos )1/ 2 with tan M1 M2 cos M2 = mL and = 90°, so MR = (M2 + M2 + 2MM cos 90)1/2 = ( 2)mL Msin90 tan 1, i.e., = tan–1(1) = 45° And, M Mcos90 S
M2
M1
M1 N
N
MR
M2 (A)
(B)
1.10 INTENSITY OF MAGNETISATION (I) (a)
(b) (c)
(d) (e) (f) (g) (h)
The magnetic moment per unit volume of a material is defined as intensity of M magnetisation (I). i.e. I = , when V V = 1m3 then I = M. The unit of intensity of magnetisation is ampere/meter and its dimensions are M 0L– 1 0 1 TA The pole strength per unit area of cross-section is defined as intensity of m magnetisation. i.e. I p when A = 1m2 then A=1m2 then I = mp A It is a vector quantity whose direction is along the magnetic field. In para- and ferro-magnetic materials its direction is in the direction of H and in diamagnetic materials it is opposite to that of H. I is produced in materials due to spin motion of electrons. The value of I and its direction in a material depend on the nature of that material. I - M curve
(i)
For para magnetic materials
(ii)
I
For diamagnetic materials I
Para
dia M
M
(iii)
For ferromagnetic materials I
Ferro
(i)
I - H curve Ferromagnetic I
H Diamagnetic
(j) (k) (l) (m) (n)
Its value depends on temperature. It is produced on account of induction in a material. For low magnetising field I H. i.e. I H or I = Ht is a dimensionless constant i.e. it carries no unit. Other types of intensity of magnetisation: (i) Mass intensity of magnetisation (Imass)Intensity of magnetisation 1 Imass orIm density of magnetic material (ii) Molar intensity of magnetisationImolar = IMW = Mo. Wt.× mass intensity of magnetisation = MW × Im (iii) Molecular intensity of magnetisationMolar int ensity of magnetisation IMW Imolecular IM Avogadro No. N
1.11 MAGNETIC SUSCEPTIBILITY OR ( OR K) (a)
(b) (c) (d) (e)
(f) (g) (h)
(i) (j)
The ratio intensity of magnetisation (I) in a material and the magnetising field (H) is defined as magnetic susceptibility (). I i.e. H When H = 1 Oersted then = I. The intensity of magnetisation induced in a material by unit magnetising field is defined as magnetic susceptibility. It has no unit and no dimensions. It is a measure of ease with which a material can be magnetised by a magnetised by a magnetising field (H). Magnetic susceptibility of various materials(i) For diamagnetic materials – = low and negative (ii) For paramagnetic materials – = low but positive (iii) For ferromagnetic materials – = high and positive I For paramagnetic substances it is inversely proportional to temperature i.e. T For low magnetising field the value of is constant. Different types of magnetic susceptibility(i) Volume susceptibility I V X H (ii) Mass or specific susceptibility m Volume susceptibility m v density of meterial (iii) Molar susceptibility MW MW = m × MW = Specific susceptibility × Mol. Wt. (iv) Molecular susceptibility m m = Atomic wt. × specific susceptibility AI = A m H –T curve 1 – : T
(a)
T
For paramagnetic substances Illustration 8:
(b)
T–1
For paramagnetic substances
The magnetic induction along the axis of an air solenoid is 0.03 Tesla. On placing an iron core inside the solenoid the magnetic induction becomes 15 Tesla. The relative permeability of iron core will be : (A) 300 (B) 500
(C) Solution:
700
(D) 900 15 m Bm (B) Bm = mH, B = m0H \ mr = = = = 500 B 0.03 m0
1.12 ABSOLUTE MAGNETIC PERMEABILITY () (a) (b) (c) (d) (e) (f) (h) (i) (j)
(k)
The ratio of magnetic induction (B) to magnetising field (H) is defined as magnetic permeability (). The extent to which magnetic permeability of that medium. It is the characteristic property of a magnetic material because it represents the amplification of magnetising field in that material. Its value is always positive and is different for different materials. For materials its value can be greater or less than 0. Its value depends on H and T. = 0 [1 + ] = 0r (i) For feeromagnetic materials = high (ii) For paramagnetic materials = low (iii) For diamagnetic materials = very low The unit of magnetic permeability is Weber per ammere-meter or Henry per meter and its dimensions are M1L0T–2A–2.
1.13 RELATIVE PERMEABILITY (R) (a)
(b) (c) (d) (e) (f) (g)
The ratio of magnetic permeability of medium () to the magnetic permeabiliy of free space (0) is defined as relative permeability (r). i.e. r 0 Number of lines of force passing throughunit area in medium r Number of lines of force passing throughunit area in vacuum magnetic flux density in medium r Magnetic flux density in vacuum The limit unto which a magnetic field penetrates matter, is known as relative permeability of that material. It has no unit and no dimensions. r = 1 + Relative permeability of various substances(i) For diamagnetic substances the value of r is slightly less than one i.e. r < 1. (ii) For paramagnetic substances the value of r is slightly greater than one i.e. r > 1. (iii) For ferromagnetic substance the value of r is much greater than one i.e. r >> 1.
2.
MAGNETIC DIPOLE IN A UNIFORM MAGNETIC FIELD
(i)
When a dipole is placed (or suspended) in a uniform magnetic field, then no net force acts on it i.e. the resultant force acting on it is zero. When a dipole is placed in a uniform magnetic field then a torque acts on it which is given by M B Newton-meter The magnitude of torque is given by MB sin 2 ml B sin Where m is the pole strength and 2l is the length of dipole.
(ii)
(iii)
(iv) (v)
B mB q
N
2l S mB
The units of M are ampere-meter2 or Joule/Tesla and its dimensions are M0L2A1 When a magnetic dipole is placed in a non-uniform magnetic field, then a torque as well as a force both act on it. The force acting on the dipole is given by
(vi)
(vii)
dB F M Newton dr (a) The work done in rotating a dipole in a magnetic field from an angle 1 to angle 2 with the field direction is given byW = MB (cos 1 – cos 2) Joule (b) If initially the dipole is lying along the field direction, then 1 = 0 W = MB (1 – cos ) Potential energy (U): (a) The potential energy of a dipole in a magnetic field is given by U = – M.B Joule (b) Special cases: When = 0, then U = – MB When = 90o, U = 0 When = 180o, U = – MB (–1) = MB
Illustration 9:
Solution:
A Magnet is suspended in the magnetic meridian with a untwisted wire. The upper end of the wire is rotated through 180 to deflect the magnet by 30 from magnetic meridian. Now this magnet is replaced by another magnet. Now the upper end of the wire is rotated through 270 to deflect the magnet by 30 from magnetic meridian. Compare the magnetic moments of magnets. If be the twist of the wire, then C , where C being restoring couple per unit twist of wire. Here 1 180 – 30 150 (150 /150) radian 2 (270 – 30 ) 240 (240 /180) radian If M be the magnetic moment and H, the horizontal component of earth’s field, then MH sin C MH sin If M1 and M2 be the magnetic moments of the two magnets respectively, then C 1 M1H sin for first magnet
C 2 M2H sin
for second magnet
M 150 /180 5 1 M1 or 1 M2 240 /180 8 2 M2
M1 : M2 5 : 8
3.
CLASSIFICATION OF MATERIALS
(i) (ii)
The root cause of magnetism in matter is the motion of electric charges. The motion of electrons and protons in atoms is responsible for their magnetic properties. The variation in the number of fundamental charged particles and variation in their arrangement in different materials are responsible for differences in their magnetic properties. On the basis of mutual interactions or behaviour of various materials in an external magnetic field, the materials are divided in three main categories. (1) Diamagnetic substances (2) Paramagnetic substances (3) Ferromagnetic substances
(iii)
(iv)
Comparative study of these materials: Diamagnetic Paramagnetic Property substances substances Cause of Orbital motion Spin motion of electrons magnetism substances Explanation On the basis of On the basis of spin and of magnetism orbital motion of orbital motion of electrons electrons Principle Electron principle Electron principle Behaviour in These are repelled These are feebly a non-uniform in an external attracted in an external magnetic field magnetic field i.e. magnetic field i.e. have a have a tendency to tendency to move from move from high to low to high field region. low field region. State of These are weakly These get weekly magnetisation magnetised in a magnetised in the direction opposite direction of applied to that of applied magnetic field magnetic field When a rod The materials align The materials align of the themselves at right themselves along the material is angles to the direction of magnetic suspended direction of field. between the magnetic field. pole pieces of a magnet. N
S (a)
N
S (b)
Ferromagnetic substances Ferromagnetic substances On the basis of domains formed. Domain principle These are strongly attracted in an external magnetic field. i.e. have an easy tendency to move from low to high field region. These get strongly magnetised in the direction of applied magnetic field. The materials easily align themselves in the direction of magnetic field.
N
S (a)
N
S (b)
Liquid or power in a watch glass when placed between the pole pieces (a) when poles are far apart. (b) When poles are close to each other. When the material in the form of liquid is filled in the U-tube and placed between pole pieces.
(a) The liquid gets bulged in the middle N
S
(b) The liquid gets depressed in the middle. N
(a) The liquid gets depressed in the middle N
S
(b) The liquid gets bulged in the middle. N
S
(a) The liquid is very much depressed in the middle. N
(b) The liquid gets very much bulged in the middle.
S
Liquid level in that limb gets depressed
S
N
Liquid level in that limb rises up.
S
Liquid level in that limb rises up very much.
N S
N S
N S
liquid
liquid
liquid
On placing the gaseous materials between pole pieces. The value of magnetic induction B Magnetic susceptibility Dependence of on temperature
The gas expands at right angles to the magnetic field.
The gas expands in the direction of magnetic field.
The gas rapidly expands in the direction of magnetic field.
B < B0
B > B0 Here B0 = magnetic induction in vacuum Low but positive 1
B > > B0
1 C or c = T - TC T - TC This is called Cure-Weiss law. TC = Curie temperature.
Dependence of on H
does not depend
Inversely proportional to temperature 1 C cµ or c = . This T T is called Curie law where C = Curie constant. does not depend
Low and negative 1 Does not depend on temperature (except Bi at low temperature)
Positive and high 102.
cµ
does not depend
Relative permeability (r) Intensity of magnetisation ()
r < 1
r > 1
r > > 1 r 102
is a direction opposite to that of H and its value is very low.
is in the direction of H but value is low.
is in the direction of H and value is very high.
-H Curves I
I
I
H
H
Magnetic moment (M)
Transition of materials (at Curie temperature) –T Curve
The value of M is very low ( 0 and is in opposite direction to H.) These do not change
c
The value of M is very low and is in the direction of H.
The value of M is very high and is in the direction of H.
On cooling, these get converted to ferromagnetic materials at Curie temperature.
These get converted into paramagnetic materials above Curie temperature.
c
T
The property of magnetism
Examples
Nature of effect
Diamagnetism is found in those materials in the atoms of which the number electrons is even. Cu, Ag, Au, Zn, Bi, Sb, NaCl, H2O air and diamond etc. Distortion effect
H
c
T
TC
T
Paramagnetism is found in those materials in the atoms of which the majority of electron spins are in the same direction. Al, Mn, Pt, Na, CuCl2, O2 and crown glass.
Ferro-magnetism is found in those materials which when placed in an external magnetic field are strongly magnetised.
Orientation effect
Hysteresis effect
Fe, Co, Ni, Gd, Fe3O4 etc.
4.
CURIE LAW AND CURIE TEMPERATURE
(a)
Curie law: (i) The magnetic susceptibility of paramagnetic substances in inversely 1 proportional to its absolute temperature i.e. c µ T C (ii) c µ Where C = Curie constant, T = absolute temperature T (iii) On increasing temperature, the magnetic susceptibility of paramagnetic materials decreases and vice versa.
(iv)
(b)
The magnetic susceptibility of ferromagnetic substances does not change according to curie law. Curie temperature (TC): (i) The temperature above which a ferromagnetic material behaves like a paramagnetic material is defined as curie temperature (TC). (ii) The minimum temperature at which a ferromagnetic substance is converted into paramagnetic substance is defined as curie temperature. (iii) For various ferromagnetic materials its values are different. e.g. for Ni TCNi = 358°C for Fe TCFe = 770°C for (iv)
5.
CO
TCCO = 1120°C
At this temperature the ferromagnetism of the substances suddenly vanishes.
CURIE-WEISS LAW
–T Cure c
T
TC
6.
OTHER IMPORTANT FEATURES AND FORMULAE
(A)
Angle of declination and Geographical meridian: Geographical meridian
BH
Angle of declinetion
f q
Angle of dip
B
BV Magnetic meridian
(i) (ii)
The horizontal component of earth's magnetic field is from S to N. If at any place the angle of dip is and magnetic latitude is then, tan = 2 tan The total intensity of earth's magnetic field I = I 0 1 + 3 sin2 l
M R3 Here M and R are the magnetic moment of bar magnet of earth and radius of earth respectively. At magnetic equator of earth = 0 and at poles = 90. Ipole = 2I equator here
(iii)
I0 =
(B)
(iv)
At the poles and equator of earth, the values of total intensity are 0.66 and 0.33 oersted respectively.
(i) (ii) (iii)
In vacuum :– B0 = 0H In medium :– B = H Resultant magnetic field due to and H :– (a) B = B1 + BH = 0 + 0H B = 0 ( + H) B = 0H (1 + /H) or B = 0 (1 + )H + = H (a) = 0 [1 + ] (b) = 0r B (c) m= H m (a) mr = m0 (b) mr = 1 + c (a) c = mr - 1
(iv)
(v)
(vi)
I H (vii) The magnetic potential due to a small magnet at a point distant r is given by :– Mcos q V= r2 (viii) The mutual interaction force between two small magnets of moments M 1 and M2 is given by 6M1M2 F=K r4 Magnetic torque (a) = MB sin (b) = BiNA sin t =M´ B (c) (b)
(C)
(D)
Magnetic potential energy (UB) (a) UB = M ×B (b)
7.
c=
UB = M ×B (1- cos q)
ASSIGNMENT
1.
The magnetism of atomic magnet is due to (A) only spin motion of electrons (B) only orbital motion of electrons (C) both spin and orbital motion of electrons (D) the motion of protons. Solution: (C) The magnetism of atomic magnet is due to both spin an orbital motions of electrons. 2.
The earth's magnetic field inside an iron box as compared to that outside the box, is-
(A) Solution:
less
(B)
more
(C)
zero
(D)
same.
(A) The earths magnetic field inside an iron box is less than that outside the box.
The magnetic susceptibility of a paramagnetic materials varies with absolute temperature T as cons tan t . T (A) (B) (C) (D) eT T 1 Solution: (C) The magnetic susceptibility of a paramagnetic substance is inversely 1 proportional to the absolute temperature i.e. T 3.
4.
There are two points A and B on the extended axis of a 2 cm long bar magnet. Their distances from the centre of the magnet are x and 2x respectively. The ratio of magnetic fields at points A and B will be(A) 8 : 1 approximately (B) 4 : 1 (approximately) (C) 4:1 (D) 8 : 1.
Solution:
(A) The intensity of magnetic field on the axis 2m B 0 3 4 r 3
B r 2x 1 2 8 :1 B2 r1 x 5.
3
The resultant magnetic moment of neon atom will be(A)
Solution:
B . 2 (B) Neon atom is diamagnetic, hence its net magnetic moment is zero.
infinity
(B)
zero
(C)
mB
(D)
A magnetic material of volume 30 cm3 is placed in a magnetic field of intensity 5 oersted. The magnetic moment produced due to it is 6 amp-m2. The value of magnetic induction will be(A) 0.2517 Tesla (B) 0.025 Tesla (C) 0.0025 Tesla (D) 25 Tesla. Solution: (A) B = m0 (I + H) M 6 2 155 amp / m I= 6 V 30 10 5 amp / m H = 5 oersted = 4 10 3 –7 m0 Wb/amp-m 5 B 4 10 7 2 105 3 4 10 = 0.2517 Tesla 6.
The mass of an iron rod is 80 gm and its magnetic moment is 10 A–m2. If the density of iron is 8gm/C.C. hen the value of intensity of magnetisation will be(A) 106 A/m (B) 104 A/m (C) 102 A/m (D) 10 A/m M Md Magnetic moment density I Solution: (A) I V m mass
7.
10 8 103 80 103 = 106 A/m
8.
The main difference between electric lines of force and magnetic lines of force is(A) Electric lines of force are closed curves whereas magnetic lines of force are open curves. (B) Electric lines of force are open curves whereas magnetic lines of force are closed curves. (C) Magnetic lines of force cut each other whereas electric lines of force do not cut. (D) Electric lines of force cut each other whereas magnetic lines of force do not cut. Solution: (B) The magnetic lines of force are in the form of closed curves whereas electric lines of force are open curves.
9.
The mass of a specimen of a ferromagnetic material is 0.6 kg. and its density is 7.8×103 kg/m3. If the area of hysteresis loop of alternating magnetising field of frequency 50 Hz is 0.722 MKS units then the hysteresis loss per second will be(A) 277.7×10–5 Joule (B) 277.7×10–6 Joule (C) 277.7×10–4 Joule (D) 277.7×10–4 Joule. Solution: (A) W H = VAft m Aft d 0.6 0.722 50 or WH 7.8 103 = 277.7 × 10–5 Joule
The horizontal component of flux density of earth's magnetic field is 1.7×10 –5 tesla. The value of horizontal component of intensity of earth's magnetic field will be? (A) 24.5 A/m (B) 13/5A/m (C) 0.135 A/m (D) 1.35 A/m. 5 2 B 1.7 10 Wb / m = 13.5 A/m Solution: (B) H 0 4 10 7 Wb / A m 10.
A magnetising field of 2×103 an iron rod. The relative permeability of the rod will be(A) 102 (B) 10o (C) 104 B Solution: (C) r 0 H0 8 r 104 or 3 2 10 4 10 7
11.
(D)
101
In a hydrogen atom the electron is revolving in a circular path of radius 5.1×10 –11m with a frequency 6.8×1015 Hz. The equivalent magnetic moment will be(A) 8 × 10–24 A–m2 (B) 8 × 10–22 A–m2 –20 2 (C) 8.9 × 10 A–m (D) 8 × 10–18 A–m2 2 Solution: (A) –19 or M = 1.6×10 ×6.8×1015×3.14×(5.1×10–11)2 = 8.9 ×10–24 A–m2
12.
13.
A cube of side l is placed in a magnetic field of intensity B. The magnetic flux emanating out of it will be(A) zero (B) Bl2 (C) 2Bl2 (D) 6Bl2
Solution: (A) The net flux coming out of a closed surface is zero. Hence the flux coming out of the cube will be zero. 14.
A circular disc of area (4iˆ 5ˆj) 10 3 m2 is placed in a uniform magnetic field of intensity (0.2iˆ 0.3ˆj) Tesla. The flux crossing the disc will be-
(A) (C) Solution:
23 Weber 23×10–3 Weber (C) B.A (0.2iˆ 0.3ˆj).[0.004iˆ 0.005ˆj]
(B) (D)
23×10–2 Weber 23×10–4 Weber
= 0.0008 + 0.0015 = 0.0023 Weber = 23×10–4 Weber 15.
The total magnetic flux in a material, which produces a pole of strength mp when a magnetic material of cross-sectional area A is placed in a magnetic field of strength H, will be(A) 0 (AH + mp) (B) 0 AH (C) 0 mp (D) 0 [mp AH + A] Solution: (A) = BA ....(A) ....(B) B 0 (I H) From eqs. (A) and (B) 0 (I H)A m I p .....(C) A From eqs. (B) and (C) m 0 A p H 0 [mp AH] A
The relation between and H for a specimen of iron is as follows 0.4 12 10 –4 Henry / meter H The value of H which produces flux density of 1 Tesla will be(A) 250 A/m (B) 500 A/m (C) 750 A/m (D) 103 A/m Solution: (B) B = H 0.4 12 1004 H or B H –4 or 1 = 0.4 + 12×10 H H = 500 A/m
16.
17.
The correct curve between X and
1 for paramagnetic materials isT
(A)
(B)
1/T
(C)
1/T
(D)
1/T
Solution:
(A)
X
1/T
1 T
18.
The SI unit of magnetic flux is(A) Weber (B) Maxwel (C) 2 m Solution: (A) BA Weber 2 Weber m
Tesla
(D)
Gauss
19.
The intensity of magnetic field due to an isolated pole of strength m p at a point distant r from it will bemp mp r2 2 (A) (B) m r (C) (D) p mp r2 r Solution: (A) The magnetic intensity due to an isolated pole of strength mp at a distance r = mp r2
20.
A uniform magnetic field is directed from left towards right in the plane of paper. When a piece of soft iron is placed parallel to the field. The magnetic lines of force passing through it will be(A)
(B)
(C)
(D)
Solution: (C) The magnetic lines of force in a ferrmagnetic material are crowded together. 21.
A bar magnet of magnetic moment M is cut into two equal parts. The magnetic moment of either of the parts will beM (A) (B) 2M (C) 2M (D) I H2 2 l M Solution: (A) M ml M' m 2 2 22.
A loop of area 0.5 m2 is placed in a magnetic field of strength 2 Tesla in direction making an angle of 60o with the field. The magnetic flux linked with the loop will be1 3 3 Weber Weber (A) (B) 2 Weber (D) Weber (C) 2 2 1 1 Solution: (A) BA cos , 2 cos 60o 2 2 23.
The force experienced by a pole of strength 100 A-m at a distance of 0.2m from a short magnet of length 5 cm and pole strength of 200A-m on its axial line will be (A) 2.5×10–2 N (B) 2.5×10–3N (C) 5.0×10–2N (D) 5.0×10–3N. 2m'l Solution: (A) F mB 0 3 m 4 x 7 10 2 200 0.05 100 8 103 –2 = 2.5×10 N
The magnetic susceptibility of a paramagnetic substance is 3×10 –4. It is placed in a magnetising field of 4×104 amp/m. The intensity of magnetisation will be(A) 3 108 A/M (B) 12 108 A/M (C) 12 A/M (D) 24 A/M –4 3 Solution: (C) I = XH = 3×10 ×4×10 = 12 A/m
24.
25.
Volt-second is the unit of(A) B (B) d or et volt sec Solution: (B) e dt
(C)
I
(D)
x
The volume susceptibility of a magnetic material is 30×10–4. Its relative permeability will be(A) 31 10–4 (B) 1.003 (C) 1.0003 (D) 29 10–4 1 X 1 30 10 4 1.003 Solution: (B) r 0
26.
27.
A current of 2 ampere is passed in a coil of radius 0.5 m and number of turns 20. The magnetic moment of the coil is(A) 0.314 A–m2 (B) 3.14 A–m2 (C) 314 A–m2 (D) 31.4 A–m2 2 Solution: (D) N = 2×3.14×0.25×20 = 31.4 A–m2
28.
The value of magnetic susceptibility for super-conductors is(A) zero (B) infinity (C) +1 (D) Solution: (D) For superconductor r = 1 + X = 0 X = –1
–1
29.
A current of 1 ampere is flowing in a coil of 10 turns and with radius 10 cm. Its magnetic moment will be(A) 0.314 A-m2 (B) 3140 A-m2 (C) 100 A-m2 (D) 0 A-m2 Solution: (A) M = iA = R2 Ni = 3.14×0.01×10×1=0.314 Am2
The magnetic moment of a magnet of mass 75 gm is 9×10 –7 A-m2. If the density of the material of magnet is 7.5×103 kg/m3 then intensity of magnetisation will be(A) 0.9 A/m (B) 0.09 A/m (C) 9 A/m (D) 90 A/m –7 3 M Md 9 10 7.5 10 Solution: (B) I 0.09 A / m V m 75 10 –3 30.
For test 1.
The area of hysteresis loop of a material is equivalent to 250 Joule. When 10 kg material is magnetised by an alternating field of 50Hz then energy lost in one hour will be if the density of material is 7.5 gm/cm3. (A) 6×104 Joule (B) 6×104 Erg (C) 3×102 Joule (D) 3×102 Erg. m Solution: (A) E nA Vt nA t d 50 250 10 3600 6 10 4 Joule 3 7.5 10
2.
The corecivity of a bar magnet is 100 A/m. It is to be demagnetised by placing it inside a solenoid of length 100 cm and number of turns 50. The current flowing the solenoid will be(A) 4A (B) 2A (C) 1A (D) zero. H 100 i 2A Solution: (B) H = ni n 50
3.
A current i is flowing in a conductor of length l. When it is bent in the form of a loop then its magnetic moment will be4 l2i l2 (A) (B) (C) (D) 4l2i. l2i 4 4 l2 il2 Solution: (A) M = iA = i r2 But 2r = l M i 2 4 4
4.
A rod of ferromagnetic material with dimensions 10 cm×0.5 cm×0.2 cm is placed in a magnetic field of strength 0.5×104 amp/m as a result of which a magnetic moment of 5 amp-m2 is produced in the rod. The value of magnetic induction will be(A) 0.54 Tesla (B) 0.358 Tesla (C) 2.519 Tesla (D) 6.28 Tesla. M Solution: (D) B 0 (I H) 0 H V 5 4 3.14 10 7 6 5000 6.28 Tesla . 10
5.
The ratio of intensities of magnetic field in the axial and equatorial positions of a magnet will be(A) 1:4 (B) 4:1 (C) 1:2 (D) 2:1 Solution: (D) 0 2M M2 Ba 4 3 2 :1 0 M Bc 4 r 3 r O
r
M
6.
The resultant magnetic moment due to two current (i) carrying concentric coils of radius r, mutually perpendicular to each other will be(B) (C) (D) (A) 2ir 2 2r 2 2ir 2 2ir Solution: (A) M' = 2M = 2r 2i The magnetic susceptibility of a paramagnetic material at –73oC is 0.0075 then its value at –173oC will be(A) 0.0045 (B) 0.0030 (C) 0.015 (D) 0.0075. 1 Solution: (C) For paramagnetic materials X T X T 0.0075 200 2 1 or X2 0.015 X1 T2 100 7.
8.
The area of cross-section of three magnets of same length are A, 2A and 6A respectively. The ratio of their magnetic moments will be(A) 6:2:1 (B) 1:2:6 (C) 1 : 4 : 36 (D) 36 : 4 : 1 Solution: (B) M µ m µ A (Area of cross-section) M1 : M2 : M3 = 1: 2 : 6 9.
If the radius of a circular coil is doubled and the current flowing in it is halved then the new magnetic moment will be if its initial magnetic moment is 4 units(A) 8 units (B) 4 units (C) 2 units (D) zero. 2 M = p R Ni Solution: (A) 2
M1 æR1 ö = M2 çè R2 ÷ ø
æi1 ö 4 1 çè i ø ÷ or M = 4 ´ 2 \ M2 = 8 units 2 2
10.
Newton/Weber is the unit of
(A) Solution:
(B) M F i f = =H (A) F = Bil = il or A f l H
(C)
B
(D)
11.
The inner and the outer radii of a toroid are 9cm and 11cm respectively and the number of turns in it is 3140. A magnetic field of 2.5 Tesla is produced in it when a current of 0.5 ampere is passed in it. The permeability of core material is (in Henry/meter)(A) 10–1 (B) 10–2 (C) 10–3 (D) 10–4 B B 2.5 ´ 2 ´ 3.14 ´ 0.1 Solution: (C) m= = = = 10- 3 Henry / meter Ni H 3140 ´ 0.5 2p r 12.
In the above problem the relative permeability of the core will be(A) 684.5 (B) 864.7 (C) 369.4 (D) -3 m 10 = = 796.2 Solution: (D) mr = m0 12.57 ´ 10- 7
796.2
A magnetising field of 5000 A/m produces a magnetic flux of 5×10 –5 Weber in an iron rod. If the area of cross-section of the rod is 0.5 cm2, then the permeability of the rod will be (in henry/m)(A) 1 10–3 (B) 2 10–4 (C) 3 10–5 (D) 4 10–6 f B = = mH A Solution: (B) 5 ´ 10- 5 f \ m= = = 2 ´ 10- 4 Henry / m 3 -4 AH 0.5 ´ 10 ´ 5 ´ 10 13.
14.
In the above problem the magnetic susceptibility of the rod will be(A) 158.2 (B) 198.0 (C) 295.3 (D) 343.6 Solution: (A) m mr = = 1+ X m0 \
X=
m 2 ´ 10- 4 - 1= - 1 = 159.2 - 1 = 158.2 m0 12.56 ´ 10- 7
15.
In the above problem the intensity of magnetisation will be (in A/m)(A) 7.9×102 (B) 7.9×102 (C) 7.9×102 (D) 7.9×105 Solution: (D) I = XH = 158.2 ´ 5000 = 7.9 ´ 106 Amp / m
16.
At any place on earth, the horizontal component of earth's magnetic field is the vertical component. The angle of dip at that place will be(A) 60o (B) 45o (C) 90o (D) 30o
3 times
Solution: =
(D) tan q = 1 3
= tan30° \
BV BV = BH 3 BV
BH q
q = 30°
BV
The horizontal component of earth's magnetic field at any place is 0.36×10 –4 Weber/m2. If the angle of dip at that place is 60o then the value of vertical component of earth's magnetic field will be-(in Wb/m2)(A) 0.12 10–4 (B) 0.24 10–4 (C) 0.40 10–4 (D) 0.62 10–4 Solution: (D) BV = BH tan q = 0.36 ´ 10 - 4 tan60° 17.
= 0.36 ´ 10- 4 ´
18.
3 = 0.62 ´ 10 - 4 Wb / m2
The value of angle of dip at a place on earth is 45 o. If the horizontal component of earth's magnetic field is 5×10–5 Telsa then the total magnetic field of earth will be(A) (B) 5 2 105 Tesla 10 2 105 Tesla (D) zero. (C) 15 2 105 Tesla
Solution:
(A) B = B2V + BH2 = BH2 tan2 45° + BH2 = BH 2
= 5 2 ´ 10- 5 Wb / m2
19.
The ratio of intensities of magnetic field, at distances x and 2x from the centre of a magnet of length 2cm on its axis, will be(A) 4:1 (B) 4 : 1 approx (C) 8:1 (D) 8 : 1 approx. m m 2M 2M » 0 3 Solution: (D) B = 0 2 2 3/2 4 p (l + x ) 4p x 3
\
B1 æx 2 ö = = 8 : 1 approximately B2 çè x1 ÷ ø
The length of a bar magnet is 10 cm and its pole strength is 10 –3 Weber. It is placed in a magnetic field of induction 4 10–3 Tesla in a direction making an angle of 30o with the field direction. The value of orque acting on the magnet will be(A) 2 10–7 N-m (B) 2 10–5 N-m (C) 0.5 102 N-m (D) 0.5 N-m. Solution: (A) t = MBsin q = m/Bsin q -3 = 10 ´ 0.1 ´ 4p ´ 10- 3 ´ 0.5 = 2p ´ 10- 7 N - m
20.
A magnetic needle of magnetic moment 60 amp-m2 experiences a torque of 1.210–3 N-m directed in geographical north. If the horizontal intensity of earth's magnetic field 2 , then the angle of declination will be(A) 30 (B) 45 (C) 60 (D) 90 Solution: (A) t = MBsin q 1.2 ´ 10- 3 1 t \ sin q = = = \ q = 30° -6 MB 60 ´ 40 ´ 10 2
21.
22.
The ratio of total intensities of magnetic field at the equator and the poles will be-
(A) 1:1 (B) 1:2 (C) 2:1 (D) Solution: (A) Total intensity of magnetic field remains constant Be \ = 1: 1 Bp
1:4
The magnetic flux density of magnetic field is 1.5 Wb/m 2. A proton enters this field with a velocity of 2×107 m/s in a direction making an angle of 30o with the field. The force acting on the proton will be(A) 2.4 10–12 Newton (B) 0.24 10–12 Newton (C) 24 10–12 Newton (D) 0.24 10–12 Newton Solution: (A) F = qvBsin q = 1.6 ´ 10- 19 ´ 2 ´ 107 ´ 1.5 ´ 0.5 = 2.4 ´ 10 - 12
23.
24.
The intensity of magnetic field at a point X on the axis of a small magnet is equal to the field intensity at another point Y on its equatorial axis. The ratio of distances of X and Y from the centre of the magnet will be(A) (B)–3 (B) (B)–3 (C) 23 (D) 21/3 m 2M m M dx = 21/ 3 Solution: (D) 0 3 = 0 3 \ 4p d x 4p d y dy
25.
Two magnets A and B are equal in length, breadth and mass, but their magnetic moments are different. If the time period of B in a vibration magnetometer is twice that of A, then the ratio of magnetic moments will be1 (B) 2 (C) 4 (D) 12 (A) 2 1 I Solution: (C) T = 2p \ Mµ 2 MB T 2
\ 26.
MA æTB ö æ2 ö = = ç ÷= 4 MB çè TA ÷ ø è1ø
The period of oscillation of a freely suspended bar magnet is 4 second. If it is cut into two equal parts lengthwise then the time period of each part will be(A) 4 sec. (B) 2 sec. (C) 0.5 sec. (D) 0.25 sec.
Solution:
I ml2 = 2p = 4 sec (A) T = 2p MB 12 ´ m / B
27.
The time period of a small magnet in a horizontal plane is T. Another magnet B oscillates at the same place in a similar manner. The size of two magnets is the same but the magnetic moment of B is four times that of A. The time period of B will beT T (A) (B) (C) 2T (D) 4T 4 2 MA TB MA T Solution: (B) = or TB = T ´ = MB TA 4MA 2
28.
The value of current enclosed by a circular path of radius 0.30 cm is 9.42 ampere. The value of magnetic field along the path will be(A) 500 amp/m (B) 1000 Amp/m
(C) Solution:
5 104 Amp/m (D) Zero i 9.42 (A) H = = = 500 amp / m 2pr 2 ´ 3.14 ´ 3 ´ 10- 3
A magnetic wire is bent at its midpoint at an angle of 60o. If the length of the wire is l and its magnetic moment is M then the magnetic moment of new shape of wire will beM M (A) 2M (B) M (C) (D) 2 2 1 M Solution: (C) M' = m × = 2 2 l/2 l/2 29.
l/2
30.
A current of 2 ampere is flowing in a coil of radius 50 cm and number of turns 20. The magnetic moment of the coil will be(A) 3.14 amp-m2 (B) 31.4 amp-m2 (C) 314 amp-m2 (D) 0.314amp-m2 Solution: (B) M = p R2 Ni = 3.14 ´ 0.25 ´ 20 ´ 2 = 3.14 amp / m2
1.
1000 turns per meter are wound over a Rowland ring of ferromagnetic material. On passing a current of 2 ampere in the coil, a magnetic field of 10 Wb/m2 is produced in it. The magnetising force generated in the material will be(A) 1.2 10–3 A/m (B) 2.6 10–3 A/m –4 (C) 0.6 10 A/m (D) 2 10–3 A/m Ni = ni = 103 ´ 2 = 2 ´ 103 amp / m Solution: (D) H = 2p r
2.
A magnet makes 10 oscillation per minute at a place where the horizontal component of earth's magnetic filed (H) is 0.33 oersted. The time period of the magnet at a place where the value of H is 0.62 oersted will be(A) 4.38 S (B) 0.38 S (C) 2.38 S (D) 8.38 S H 1 0.33 Solution: (A) T µ \ T2 = T1 1 = 6 ´ = 4.38 s H2 0.62 H
The magnetic induction inside a solenoid is 6.5×10 –4T. When it is filled with iron medium then the induction becomes 1.4T. The relative permeability of iron will be(A) 1578 (B) 2355 (C) 1836 (D) 2154. B 1.4 Solution: (D) mr = = = 2154 B0 6.5 ´ 10- 4 3.
1.
FARADAY’S LAWS OF ELECTROMAGNETIC INDUCTION
According to Faraday’s Law, whenever the magnetic flux through a circuit changes, an emf is induced in the circuit. The magnitude of the induced emf is equal to the rate at which flux changes with time. d Magnitude of the induced emf = dt If represents the flux through a single turn, and the loop has N turns, then d Induced emf = N dt This induced emf creates an induced current in the circuit whose magnitude is given as 1 d induced emf i= = R dt net resis tance of circuit
2.
LENZ’S LAW
According to Lenz’s law, induced emf in a circuit opposes the cause due to which it was induced. Consider the following examples. (a) Suppose that the north-pole of a bar magnet is moved N towards a conducting wire loop as shown in the figure. s Due to a change in the magnetic flux associated with the loop, a current is induced. Due to induced current, a magnetic field is induced and this magnetic field opposes the motion of bar magnet. The direction of the induced current can be deduced by the following argument: the north pole is moving towards the loop; therefore to oppose the motion of the bar magnet only a north pole will be induced on that face of the loop which faces the magnet. (b) A rectangular loop ABCD is being pulled out of the x x x x x x OBx x B x x x x x xA x x magnetic field which is directed into the plane of the x x x x x x x x paper. Perpendicular to the plane of the paper. As the xx xx xx xx xx xx xx xx loop is dragged out of the field, the flux associated with x x x x x xD x x C the loop decreases. The induced current flows in the loop in a sense so as to oppose the decrease in this flux. For this to happen the magnetic field due to the induced current in the loop must be directed into the plane of the paper. Thus the current in the loop must flow be clockwise.
Illustration 1:
Solution:
Magnetic field is increasing into the page with time when a conducting loop of definite radius is placed on the plane of the paper. The find the direction of current in the loop.
x
x
x
x x
x
x B x
x
x
x
x
x
x x
As the flux is increasing inside then the current in the loop will be such that it will be opposing the increase in magnetic field, i.e., the induced current in the loop will create such a magnetic field which is directed out ward.
Thus the direction of current will be anticlockwise. Illustration 2:
Solution:
A square loop ACDE of area 20cm2 and X XB X resistance 5 is rotated in a magnetic field B = A C 2T through 180 X X X (a) in 0.01 s and E (b) in 0.02 s D X X X Find the magnitude of e, i and q in both the cases. Let us take the area vector S perpendicular to plane of loop inwards. So initially, S B and when it is rotated by 180, S B Hence, initial flux passing through the loop, i = BS cos 0 = (2) (20 10-4) (1) = 4.0 10-3 Wb Flux passing through the loop when it is rotated by 180, f = BS cos 180 = (2) (20 10-4) (-1) = - 4.0 10-3 Wb Therefore, change in flux, B = f - i = - 8.0 10-3 Wb (a) Given t = 0.01 s, R=5 8.0 103 e B 0.8 volt 0.01 t i
e
R and
(b)
0.8 0.16 A 5 q = it = 0.16 0.01 = 1.6 10-3 C
t = 0.02 s
e e
B 8.0 103 0.4 volt 0.02 t
0.4 0.08 A R 5 and q = it = (0.08) (0.02) = 1.6 10-3 C Note: Time interval t in part (b) is two limes the time interval in part (a), so e and i are half while q is same. i
Illustration 3: A conducting circular loop with variable radius r is placed in a uniform magnetic field B=0.020T with plane perpendicular to the field. While the radius of the loop is contracting at a constant rate of 1.0 mm/s; find the induced emf in the loop when the radius is 2 mm. Solution:
Radius (variable) = r B = 0.02 T Induced emf E = dr 1.0 10 3 mt dt
dr d dA = B 2r (BA ) B dt dt dt
r = 2 10-3 mt
E = 0.02 2 2 10-3 1 10-3
= 6 10-8 = 18.85 10-8 V. Illustration 4:
Solution:
A rectangular loop of N turns of area A and resistance R rotates at a uniform angular velocity about Y-axis. The loop lies in a uniform magnetic field B in the direction of X-axis. Assuming that at t = 0, the plane of the loop is normal to the lines of force, find an expression for the peak value of the emf and current Induced in the loop. What is the magnitude of torque required on the loop to keep it moving with constant ? As is maximum at t = 0, (t) = BA cos t Magnitude of induced emf = N d / dt
BAN Sin t Magnitude of induced current =
BAN Sin t R
(a) peak value of emf = BAN peak value of induced current = BAN/R (b) Power input = heat dissipation per sec I2R B2 A 22N2 sin2 t R
3.
MOTIONAL EMF
Induced emf due to motion of a conductor in magnetic field Suppose a conductor is moving with velocity v in a uniform magnetic field B . We take a small element PQ = d of the conductor, then induced e.m.f. in the element. B e (v B).d VQ VP P d v Q or, e = (vB sin). d cos Where is the angle between v and B and is the angle between element and direction of force Illustration 5:
Solution:
A metal rod 1.5 m long rotates about its one end in a vertical plane at right angles to the magnetic meridian. If the frequency of rotation is 20 rev/s. Find the emf induced between the ends of rod. BH = 0.32 10-4T: Horizontal component of earth’s magnetic field. Length of rod = 1.5 mt. Frequency of rotation = 20 rev/sec. BH = 0.32 10-4 T Angular velocity of rod = 20 2 rad/sec. emf induced
=B
dA = 0.32 10-4 20 (1.5)2 dt
= 4.5 10-3 V
4.
INDUCTANCE
4.1
SELF INDUCTION
We have already discussed capacitors – devices that store energy using electric fields. Like a capacitor, an inductor is also quite a commonly used element in electric circuits. It stores magnetic energy. As we know that when current flows through a conductor a magnetic field is set-up in surrounding of it, and hence it is associated with magnetic flux. If magnetic flux associated with a coil is and current flowing through it is I, then its inductance is given by the expression L
. I
The quantity 'L' is called self-inductance of
the coil. It does not depend on the current, but it depends on the permeability of the core and the dimensions of the coil. S.I. unit of inductance is Henry. Consider the circuit, in which a solenoid is connected across a cell through a resistor. When the switch is open, the current in the circuit is zero. When the switch is closed, a current flow in it. Since current in the circuit increases from zero to a certain value, N magnetic field associated with it changes that causes induction of an emf across the solenoid. Induction of an emf due to variation in current flowing through the coil itself is known as self induction. Since B = LI, and =
dB dt
L
dI . dt
Inductance of an ideal solenoid: Let a current I flow through a solenoid. The magnetic field due to the current flowing within the solenoid is, B = 0nI, where n is the number of turns per unit length. If area of cross section of the solenoid is A then flux associated with length is equal to = nBA. (Assuming that the solenoid is ideal and long) where is the length of the solenoid. Now B = nI (n )(0n A) = 0 n2 A Self Inductance of a Coil Consider a coil of N turns and area of crossection A carrying a current i. The length of the coil is ( A). iN N B A N 0 A 2 NA 0 i
Comparing with = L i, we get: L =
4.2
R - L CIRCUIT
GROWTH OF CURRENT
i
0N2 A
i
L
R
When switch is on at t = 0. Then current starts from 0 to I. And self inductor opposes the grow of current L
LdI IR 0 dt I t LdI L I 1 LdI IR dI t dt 0 dt R 0I IR 0 R
I
I
L I / R Rt In I t In L R R 0 / R I Rt Rt R L I = 1 e L e /R R Rt dI L e dt L Rt LdI potential difference across inductor = e L dt Rt I = I e L R At t = 0, I = 0 At t I R
Rt
V= e L At t = 0 potential difference across inductor = At t potential difference across inductor = 0. So inductor will behave like plane wire. Consider graph betn I and time. (Amp.)I
R t (sec.)
Consider graph betn potential difference across Inductor and time
R
(volt.)V
t (sec.)
Illustration 6:
A coil of inductance 1.0 H and resistance 100 is connected to a battery of emf 12 V. Find the energy stored in the magnetic field of the coil 10 ms after the circuit is switched on.
Solution:
L = 1.0 H E = 12 V
R = 100
Rt E i(t) = 1 e L R
Energy stored in the magnetic field is 1/2 L i2. (12)2 1 = 1.0 2 10 4
= Illustration 7: Solution:
100 10 10 3 1 e 1
2
2
1 1 28.8 mJ 4 e 2 10 144
A coil of metal wire is kept stationary in a non-uniform magnetic field. An e.m.f. is induced in the coil. For induced emf to develop in a coil the magnetic flux through it must change. But in this case the number of magnetic lines of force through the coil is not changing. Therefore the statement is false.
4.3
INDUCTANCE OF COMMON ELEMENTS
1.
Wire L= 0 8 2a
2.
3.
Hollow cylinder 2 L = 0 In 1 2 a >> a Parallel wires d L = 0 In a >> d, d >> a
2a
2a d
4.
Coaxial conductor b L = 0 In a a b
5.
o
6.
Circular loop 4 L = 0 In 2.45 2 d = –2 o,o >> d
d
Solenoid 0N2S L= a
>> a 7.
Torus (of circular cross section) L = 0N2[o –
02 a2 ] o
8.
Sheet L = o 2
2 In b t 0.5 t b
4.4
MUTUAL INDUCTION
Consider two coils C1 and C2 placed as shown. By varying current i1 in coil C1, we change the flux not only through C1 but also through coil C2. The change in flux 2 through C2 (due to change in current i1) induces an emf in the coil C2. This emf is known as mutually induced emf and the process is known as mutual induction. 2 i1 2 = M i1
C1 I1
primary
Where M is called as mutual inductance of the pair of coils. The coil C 1 in which i varies is often called primary coil and the coil C2 in which the emf is C1 2 induced is called secondary coil. I1 induced emf in coil C2 = E2 E2
d2 di M dt dt
a secondary
The mutual inductance is maximum when the coils arc wound up on the same axis. It is minimum when the axes of coils are normal to each other. Note down the following points regarding the mutual inductance:
1.
The SI unit of mutual inductance is henry (1H).
2.
M depends upon closeness of the two circuits, their orientations and sizes and the number of turns etc.
3.
Reciprocity theorem: M21 = M12 = M e2 = - M(di1/dt) and N2B2 M12 and i1
4.
e1 = - M(di2/dt) N1B1 M21 i2
A good approach for calculating the mutual inductance of two circuits consists of the following steps: (a) Assume any one of the circuits as primary (first) and the other as secondary (second). (b) Suppose a current i1 flows through the primary circuit. (c) Determine the magnetic field B produced by the current i1. (d) Obtain the magnetic flux B2 . (e)
With the flux known, the mutual inductance can be found from, N2B2 M i1
Illustration 8:
Solution:
A long solenoid of length 1 m, cross sectional area 10 cm2, having 1000 turns has wound about its centre a small coil of 20 turns. Compute the mutual inductance of the two circuits. What is the emf in the coil when the current in the solenoid changes at the rate of 10 Amp/s? Let N1 = number of turns in solenoid; N2 = number of turns in coil A1 and A2 be their respective areas of crossection. (A1 = A2 in this problem) Flux 2 through coil crated by current i1 in solenoid is 2 = N2(B1A2) iN NN A 2 N2 0 1 1 A 2 2 0 1 2 2 i2 Comparing with 2 = M i1; we get NN A Mutual inductance = M = 0 1 2 2 4 107 1000 20 10 104 2.51 105 H 1 di Magnitude of induced emf = E2 = M 1 dt M
Illustration 9:
Solution:
The coefficient of mutual induction between the primary and secondary of a transformer is 5 H calculate the induced emf in the secondary when 3 Amp current in the primary is cut-off in 2.5 10-4 sec. M = coefficient of mutual induction = 5 H.
3 Amp current is a cut off in 2.5 10-4 sec. E = M
4.5
di 3 = -6 104 V 5 4 dt 2.5 10
L–C CIRCUIT
L.C. OSCILLATIONS A capacitor is charged to a potential difference of V o by connecting it across a battery and then is allowed to discharge through an inductor of inductance L. Initial charge on the plates of the capacitor qo = CVo At any instant, let the charge flown through the circuit be q and the current in the circuit be i. Applying Kirchhoff’s law di q0 q L =0 C dt
qo-q -(qo-q)
C
Differentiating w.r.t. time we get dq d2 i LC 2 = 0 dt dt 2 1 di i = 2i, = 2 LC dt
I
f=
1 2 LC
The charge q on the plates of the capacitor and current I in the circuit vary sinusoidally as q = q0 sin (t + ) and I = q0 cos (t + ). where is the initial phase and it depends on initial conditions of the circuit and
=
1 LC
.
The total energy of the system remains conserved 1 1 1 1 CV 2 Li 2 = constant = CV02 Li 02 2 2 2 2
Illustration 10: A–2–F capacitor is initially charged to 20 V and then shorted across a 6–H inductor. What are the frequency of oscillation and the maximum value of the current? Solution: The frequency of oscillation is independent of the initial charge and depends only on the values of the capacitance and inductance. The frequency is 1 1 1 f= 2 2 LC 2 LC 1 = = 4.59 104 Hz 6 6 2 (6 10 )(2 10 ) According to equation the maximum value of the current is related to the maximum value of the charge by Q0 Im = Q0 = LC The initial charge on the capacitor is Q0 = CV0 = (2 F)(20 V) = 40 C Thus 40 C Im = = 11.5 A (6H)(2 F)
4.6
MASS–SPRING SYSTEM VS INDUCTOR–CAPACITOR CIRCUIT
A comparison of oscillations of a mass spring system and an L - C circuit. S. No.
Mass Spring System
1. 2.
Displacement (x) Velocity (v)
3.
Acceleration (a)
11.
d2 x k 2 x, where 2 dt m x = A sin (t ) or x = A cos (t ) dx v A 2 x2 dt dv a 2 x dt 1 Kinetic energy = mv2 2 1 Potential energy = kx 2 1 1 1 mv2 + kx2 = constant = kA2 = 2 2 2 1 2 mv max 2 vmax A
12.
amax 2 A
4. 5. 6. 7. 8. 9.
10.
13. 14.
5.
Inductor-Capacitor Circuit Charge (q) Current (i) di Rate of change of current dt 2 dq 1 2q, where 2 dt LC q = q0 sin (t ) or q = q0 cos (t ) dx i q02 q2 dt di 2 q Rate of change of current dt 1 Magnetic energy = Li 2 1 q2 Potential energy = 2 C 2 1 2 1 q 1 q02 Li + = constant = = 2 2 C 2 C 1 2 Limax 2 imax = q0
di 2q0 dt max
1 k m
C L
ALTERNATING CURRENTS AND CIRCUITS
The basic principle of the ac generator is a direct consequence of Faraday’s law of induction. When a conducting loop is rotated in a magnetic field at constant angular frequency a sinusoidal voltage (emf) is induced in the loop. This instantaneous voltage is, V = V0 sin t
(i)
The usual circuit diagram symbol for an ac source is shown in Figure.
In Equation (i) V0 is the maximum output voltage of the ac generator, or the voltage amplitude and is the angular frequency, equal to 2 times the frequency f. = 2f
The frequency of ac in India is 50 Hz, i.e., f = 50 Hz So, = 2f 314 rad/s The time of one cycle is known as time period T, the number of cycles per second the frequency f. 1 2 T or T f A sinusoidal current might be described as, i = i0 sin t If an alternating current is passed through an ordinary ammeter or voltmeter, it will record the mean value for the complete cycle, as the quantity to be measured varies with time. The average value of current for one cycle is, T
i
(iav )T
one cycle
0 T 0
Thus, i
idt
2 /
(i0 sin t)dt
0
dt
2 /
0
0
dt
0
one cycle
Similarly, the average value of the voltage (or emf) for one cycle is zero. V
one cycle
0
Since, these averages for the whole cycle are zero, the dc instrument will indicate zero deflection. In ac, the average value of current is defined as its average taken over half the cycle. Hence, T/2
i
Half cycle
(iav )T / 2
idt
0 T/2 0
/
0
(i0 sin t)dt
dt
/
0
dt
2 i0
This is sometimes simply written as, iav. Hence, 2 iav i Half cycle i0 0.637 i0 2 Vav V0 0.637V0 Similarly, A dc meter can be used in an ac circuit if it is connected in the full wave rectifier circuit. The average value of the rectified current is the same as the average current in any half cycle, 2 i.e., time the maximum current i0 . A more useful way to describe a quantity is the root mean square (rms) value. We square the instantaneous current, take the average (mean) value of i2 and finally take the square root of that average. This procedure defines the rootmean-square current denoted as irms. Even when i is negative, i2 is always positive so irms is never zero (unless i is zero at every instant). Hence, T
2
i
0
one cycle
T
0
irms
i2
i2dt
2 /
0
dt
one cycle
(i02 sin2 t)dt
2 /
0
i0 2
dt
0.707i0
i02 2
Thus, irms
i0
0.707i0 2 Similarly, we get V Vrms 0 0.707V0 2 The square root of the mean square value is called the virtual value and is the value give by ac instruments. Thus, when we speak of our house hold power supply as 220 volts ac, this means that the rms voltage is 220 volts and its voltage amplitude is, V0 = 2 Vrms = 311 volt
Form Factor
i0
V / 2 rms value 0 1.11 The ratio, average value 2V0 / 2 2 is known as or factor,
irms 0.707i0 iav 0.637i0 t
Note: 1. The average value of sin t, cos t, sin2t, cos2t, etc. is zero because it is positive half of the time and negative rest half of the time. Thus,
sin t cos t sin2 t cos2t 0 then 2.
i = i0 sin t i i0 sin t i0 sin t 0 1 2 1 sin2 t cos2 t 2 2 2 2 i i0 sin t
The average value of sin2 t and cos2 t is or If then
3.
If
i2 i02 sin2 t i02 sin2 t
i02 2
Like SHM, general expressions of current/voltage in an sinusoidal ac are,
or
i = i0 sin (t ), i = i0 cos (t ),
and
V = V0 sin (t ) V = V0 cos (t )
Illustration 11: If the current in an ac circuit is represented by the equation, i = 5 sin (300t - /4) Here, t is in second and i in ampere. Calculate, (a) peak and rms value of current. (b) frequency of ac (c) average current Solution: (a) An in case of ac,
i = i0 sin (t ) The peak value
(b)
5
3.535 A 2 2 Angular frequency = 300 rad/s 300 47.75 HZ f 2 2 2 2 iav i02 (5) 3.18 A and
irms
i02
i0 = 5A
(c)
Capacitor in an AC Circuit If a capacitor of capacitance C is connected across the alternating source, the instantaneous charge on the capacitor, q = CVC = CV0 sin t and the instantaneous current i passing through it, is given by: V V0 sin t dq i CV0 cos t dt V0 q -q sin(t / 2) 1/ C a b or
i
i = i0 sin (t + /2)
Here, V0
i0 C
1 is the effective ac resistance or the capacitive C reactance of the capacitor and is represented as XC. It has unit as ohm. Thus, 1 XC C i, V It is clear that the current leads the voltage by 90 or the potential i drop across the capacitor lags the current passing it by 90. V i Figure shows V and i as functions of time t.
This relation shows that the quantity
0
0
t
Vc
Inductor in an AC Circuit Consider a pure inductor of self inductance L and zero resistance connected to an alternation source. Again we assume that an instantaneous current i = i 0 sin t flows through the inductor. Although there is no resistance, there is a potential difference VL between the inductor terminals a and b because the current varies with time, giving rise to self induced emf. L di VL = Vab = - (induced emg) = L a b dt i di Li0 cos t Or VL = L dt
VL= V0 sin t (i) 2 Here V0 = i0(L) (ii) V Or i0 = 0 L V0 (iii) i sin(t) L Equation (iii) shows that effective ac resistance, i.e., inductive reactance of inductor is,
Or
XL = L And the maximum current, V i0 0 XL The unit of XL is also ohm. From Equations. (i) and (iii) we see that the voltage across the inductor leads the current passing through it by 90. Figure shows VL and i as functions of time. i, VL V0 i0 i
t
VL
Illustration 12: A 100 resistance is connected in series with a 4H inductor. The voltage across the resistor is, VR = (2.0V) sin (103 rad/s)t: (a) Find the expression of circuit current (b) Find the inductive reactance (c) Derive an expression for the voltage across the inductor. V (2.0V)sin(103 rad / s)t Solution: (a) i R R 100 -2 (2.0 10 A) sin (103 rad/s)t (b) XL = L = (103 rad/s) (4H) = 4.0 103 ohm (c) The amplitude of voltage across inductor, V0 = i0XL = (2.0 10-2 A) (4.0 103 ohm) = 80 volt In an ac voltage across the inductor leads the current by 90 or /2 rad. Hence, VL = V0 sin (t + /2)= (80 volt) sin (103 rad / s)t rad 2 Note: That the amplitude of voltage across the resistor (=2.0 volt) is not same as the amplitude of the voltage across the inductor (=80 volt), even though the amplitude of the current through both devices is the same.
5.1
SERIES L–R CIRCUIT
As we know potential difference across a resistance in ac is in phase with current and it leads in phase by 90 with current across the inductor.
VR
5.2
VL
SERIES C–R CIRCUIT
Potential difference across a capacitor in ac lags in phase by 90 with the current in the circuit. Suppose in phasor diagram current is taken along positive x-direction. Then VR is also along positive x-direction but VC is along negative y-direction. So we can write.
VC
VR
y VR
i
x
VC
V
V = VR – jVC = iR – j(iXC) i =iR – j =iZ C 1 Z = R - j C The medulus of impedance is,
Here, impedence is,
2
1 Z R C and the potential difference lags the current by an angle, V X 1/ C 1 tan1 or tan1 C tan1 C = tan1 VR R R CR 2
5.3
SERIES L–C–R CIRCUIT
Potential difference across an inductor leads the current by 90 in phase while that across a capacitor, it lags in phase by 90. VL
VC
VR
Suppose in a phasor diagram current is taken along positive x-direction. Then VR is along positive x-direction, Then VR is along positive x-direction, VL along positive y-direction and VC along negative y-direction.
V VR2 (VL VC )2
VL i
So, we can write, Here impedence is,
VL VC VR
y
x VR
VC
V = VR + jVL – jVC = iR + j(iXL) – j(iXC) = iR + j [ i (XL – XC)] = iZ 1 Z = R + j(XL – XC) = R + j L C 2
1 Z R2 L C And the potential difference leads the current by an angle, 1 L V V X X C C C or tan1 tan1 L tan1 L R VR R The modulus of impedence is,
Illustration 13: An alternating emf 200 virtual volts at 50 Hz is connected to a circuit of resistance 1 and inductance 0.01 H. What is the phase difference between the current and the emf in the circuit. Also find the virtual current in the circuit. Solution: In case of an ac, the voltage leads the current in phase by an angle, X tan1 L R Here, XL = L = (2fL) = (2) (50) (0.01) = and R = 1 = tan-1 () 72.3 V Vrms Further, irms = rms Z R2 XL2 Substituting the values we have, 200 irms = 60.67 amp. (1)2 ( )2 Illustration 14: Find the voltage across the various elements, i.e., resistance, capacitance and inductance which are in series and having values 1000, 1F nd 2. henry respectively. Given emf as, V = 100 2 sin 1000 t volt Solution: The rms value of voltage across the source, 100 2 Vrms 100 volt 2 = 1000 rad/s irms =
Vrms Vrms 2 Z R (XL XC )2
Vrms
1 R L C 2
2
100
= 0.0707 amp 2 1 (1000)2 1000 2 1000 1 106 The current will be same every where in the circuit, therefore, P.D across resistor VR = irmsR = 0.0707 1000 = 70.7 volt P.D across inductor VL = irmsXL = 0.0707 1000 2 = 141.4 volt and 1 P.D across capacitor VC = irmsXC = 0.0707 = 70.7 volt 1 1000 10 6 Note: (i) The rms voltages do not add directly as, VR + VL + VC = 282.8 volt Which is not the source voltage 100 volts. The reason is that these voltages are not in phase and can be added by vector or by phasor algebra.
5.4
CIRCUIT ELEMENTS WITH AC
Circuit elements
Amplitude relation
Circuit quantity
Phase of V
Resistor
V0 = i0R
R
In phase with i
Capacitor
V0 = i0XC
XC
1 C
Lags i by 90
Inductor
V0 = i0XL
XL = L
Leads i by 90
6.
ASSIGNMENT
1.
In a series resonant circuit the ac voltages across resistance R, inductance L and capacitance C are 5V, 10V and 10V respectively. The ac voltage applied to the circuit will be (A) 25 V (B) 20 V (C) 10 V (D) 5 V. Solution: (D) V2 V = [ R + (VL – VC)2]1/2. 2.
There is a current of 20 milliampere through an ideal choke coil of 1.5 H when connected to 220 volt, 50 hertz ac supply. The power consumed is: (A) Zero (B) 220 20 10–3 watt (C) 220 50 watt (D) 220 20 1.5 watt. Solution: (A) Ideal choke is a wattles device. Because the phase angle between the current and p p efm is . This gives power factor cos = cos = 0. 2 2
3.
The ratio of induced emf in a coil of 50 turns and area A oscillating at frequency 50 Hz to that in a coil of 100 turns and same area oscillating at frequency 100 Hz is : (A) 0.25 (B) 0.50 (C) 0.75 (D) 1.00 Solution: (A)
E nAB. Therefore
E1 n1w1 50 ´ 50 . = = E2 n2w 2 100 ´ 100
4.
In the above questions what is the potential drop across the capacitor? (A) 110 V (B) 110 2 V (C) 220 V (D) 220 2 V. Solution: (A) AT resonance, the potential drop across the capacitance is equal and opposite to that inductance. The rms current in an a.c. circuit is 2 A. If the wattles current be 3 A, what is the power factor? 1 1 1 1 (B) (C) (D) . (A) 2 3 3 2 Solution: (C) Iwattless =Irms sin 3 Hence sin = 2 = 60 1 Therefore cos = 2 . 6. A circular wire loop of radius r can withstand a radial force T before breaking. A particle of mass m and charge q(q > 0) is V sliding over the wire. A magnetic field B is applied normal to the q plane of the wire. The maximum speed Vmax the particle can have m before the loop breaks is B rT (A) Zero (B) m r é qB q2B2 4T ù ré 4T ù 2 2 + + + + qB q B (C) (D) ê ú ê ú 2 êë m m2 mr ú 2ë mr û û Solution: (D) The following forces act on the particle. Force T acting radially inwards. mv 2 Centrifugal force acting radially outwards. r Magnetic force qvB acting radially inwards. mv 2 T + qvB = r 2 mv - qvB - T = 0 r Tr æqBr ö =0 v2 = ç v÷ è m ø m
5.
v=
1 é qBr q2B2r 2 4Tr ù + + ê ú 2 ëê m m2 m ú û
v=
r é qB q2m2 4T ù + + ê ú 2 êë m m2 mr ú û
7.
A rectangular loop of sides a and b, has a resistance R and lies at a distance c from an infinite straight wire carrying current I0. The current decreases to zero in time (t - t) ,0
a
b c I0
dx
x
c
I0
Consider an infinitesimal element of length b. thickness dx at a distance x from the wire. Since d = BdA mI df = 0 (bdx0 2p x c +a m0Ib dx f = òdf = 2p ò x c
m0bI æc + a ö loge ç è c ÷ ø 2p mb æc + a ö é æ t öùì æ t öü f = 0 loge ç I0 ç1 - ÷úí I = I0 ç1 - ÷ý ÷ ê è c ø ë è t øûî è t øþ 2p
f =
Since
df dt m bI æc + a ö e = 0 0 loge ç è c ÷ ø 2pt dq m0bI0 æc + a ö = loge ç è c ÷ ø dt 2pt R
m bI æc + a ö q = 0 0 loge ç dt è c ÷ øò 2pt R 0
e=-
m0bI0 æc + a öæ 1 ö - ÷ loge ç è c ÷ç øè t ø 2p m bI e æc + a ö I = = 0 0 loge ç è c ÷ ø R 2pt R m bI æc + a ö dq = 0 0 loge ç dt è c ÷ ø 2pt R e =-
t
8.
q=
m0bI0 æc + a ö loge ç è c ÷ ø 2pt R
A network of inductances, each of value 1H, is shown in figure. The equivalent inductance of the circuit between points A and B is
A
B
(A) 6.218 H (B) 0.268 H (C) 8.162 H (D) Solution: (D) 1 and 2 in series in parallel with 3. The result of this is in series with 4, which is in parallel with 5, again in series with A 6, which again is in parallel with 7, in series with 8 and in parallel with 9. Finally result of above steps, 10 & 11 are in B series to get 2.618 H.
2.618 H 10
6
4 7
9 11
5 3
8
2
1
A uniformly wound solenoid coil of self inductance 1.8 10–4 henry and resistance 6 ohm is broken up into two identical coils. These identical coils are then connected in parallel across a 15-voll battery of negligible resistance. The time constant for the current in the circuit and the steady state current through the battery is (A) 0.3 × 10-4S, 8A (B) 1.3 × 10-4S, 8A -4 (C) 0.3 × 10 S, 5A (D) none of these Solution: (A) (i) L, R The coil is broken into two identical coils. L/2´ L/2 L Leq = = = 0.45 ´ 10- 4 H , L/2 +L/2 4 R/2´ R/2 R = = 1.5W Req = R/2 +R/2 4 Leq. 0.45 ´ 10- 4 Time constant = = = 0.3 ´ 10- 4 s Req. 1.5 E 12 Steady current I = = = 8 A. R 1.5
9.
10.
If 0 and 0 are, respectively, the electric permittivity and magnetic permeability of free space, and the corresponding quantities in a medium, the index of refraction of the medium in terms of the above parameters is me me (A) 2 (B) m0e 0 m0e 0 (C)
Solution:
1 2
me m0e 0
(D)
none of these
(B) We know that the velocity of light in vaccum c =
And the velocity of light in a medium n =
1 m0e 0
1 me
Also the refractive index n=
11.
Velocity of light in medium n 1/ m0e 0 = = = Velocity of light in vaccum c 1/ me
me m0e 0
The network shown in figure is part of a complete circuit. If at a certain instant the current (I) is 5A, and is decreasing at a rate of 103 A/s then VB – VA =
I A
(A) Solution:
15V (A)
(B)
A
1 20V
15 V
1
15 V
5mH
(C)
B
25V
5mH
VB + e - 15 + I ´ 1 = VA
Here, I = 5A,
e=
(D)
30V
B
VB – VA = 15 – e – I
df é - dI ù =Lê = - 5 ´ 10- 3 ´ 103 = - 5V ú dt ë dt û
VB – VA = 15V 12. What is the nature of RCL series circuit for frequency less than the resonant frequency? (A) Resistive (B) Inductive (C) Capacitive (D) None of the above Solution: (C) For < 0, XL < XC. So, the circuit is capacitive. 1 1 13. If the power factor changes from to then what is the increase in impedance in A.C.? 2 4 (A) 20 % (B) 50 % (C) 25 % (D) 100 % Solution: (D) 1 cos Z Z1 cosq1 1/ 4 1 = = = Z2 cosq 2 1/ 2 2 Z2 = 2Z1 2Z1 - Z1 Percentage change = 100 = 100 %. Z1 14.
What is the impedance of a purely antiresonant circuit at resonance? (A) Zero (B) R (C) 2R (D) . Solution: (D) L For antiresonant circuit Z = . For purely antiresonant circuit R = 0. Hence Z = . CR
15.
A small dc motors operating at 200 V draws a current of 3 A at its full speed of 2500 r.p.m. If the resistance of the armature is 7, the back e.m.f. of the motor is (A) 149 V (B) 159 V (C) 169 V (D) 179 V Solution : (D) E = V-I R = 200-3x7 = 179 V. 16.
A motor (A) converts electrical energy into mechanical energy (B) converts mechanical energy into electrical energy (C) converts electrical energy into magnetic energy (D) produces mechanical energy. Solution : (A)
17.
For maximum output power in D.C. motor, the induced back emf (E) should be (A) applied voltage (B) double applied voltage (C) half of applied voltage (D) one third of applied voltage
Solution (C) For maximum output, E = V / 2 18.
In ac motor, a capacitor is used to (A) reduce ripple
(B)
reduce A.C.
(C)
(D)
increase A.C.
introduce phase difference of 2
Solution : (C) The capacitor introduces a phase difference of /2 which is necessary for motors. 19.
A generator develops an e.m.f. of 130 V and has a terminal potential difference of 125 V, when the armature current is 25 A. The resistance of the armature is (A) 0.5 (B) 0.2 (C) 1.5 (D) 2.4 Solution : (B) E V 130 125 0.2 I R 25 20.
A coil is wound of a frame of rectangular cross-section. If the linear dimensions of the frame are doubled and the number of turns per unit length of the coil remains the same, then the self inductance increases by a factor of (A) 6 (B) 12 (C) 8 (D) 16. Solution : (C) The number of turns has become twice and area has gone up four times. Therefore self inductance is 4x2 = 8 times. XL = L = 2 vL with iron core XL increases. Since E 1 , I decreases. XL
21.
A Circuit contains two inductors of self inductance L 1 and L2 in series. If M is the mutual inductance, then the effective inductance of the circuit shown will be (A) L1+L2 (B) L1+L2 -2M (C) L1+L2 +M (D) L1+L2 +2M Solution : (D) when inductances are connected like this in series, then L = L1+L2+2M. 22.
A circuit contains two inductors of self inductance L 1 and L2 in series as shown in fig. If M is the mutual inductance then the effective inductance of the circuit is (A) L1+L2 (B) L1+L2 -2M (C) L1+L2 +M (D) L1+L2 +2M
Solution : (B) in this type of combination, L = L1+L2-2M
L2
L1
L2
L1
23.
Two coils of self inductance L1 and L2 are in parallels as shown in fig. If their Mutual inductance is M, then the effective inductance will be given by L1L 2M2 L1L2M (A) (B) L1 L 2 2M L1 L2 2M
L1
L2
L1L2 M L1L2 M (D) L1 L 2 2M L1 L2 2M Solution : (D) When two inductances are in parallels then L1L2 M2 L= L1 L 2 2M 2
(C)
24.
Air cored chokes are used for reducing high frequency ac because (A) air is free (B) air core has high permeability (C) with air core self inductance increases (D) with air core self inductance is not much. Solution : (D) XL = L = 2 vL, since v is large therefore L should not be large. Hence air cored chokes are used for high frequency ac. 25.
In an ac circuit V and I are given by V = 150 sin (150 t) V and I = 150 sin 150 t A 3 The power dissipated in the circuit is (A) 5625 W (B) 4825 W (C) 7450 W Solution : (A) E0 = 150 V and I0 = 150 x 10–3 mA and = 600 P
E0 2
x
I0 2
cos
(D)
3425 W
150x150 1 x 5625 watt 2 2
26.
The induced e.m.f. in a coil rotating in a uniform magnetic field depends upon (A) only on total number of turns in the coil (B) only on magnetic field (C) area of coil and speed of rotation (D) all of the above Solution : (D) E = n AB , so it depends on all these. 27.
A resistance and an inductor are connected in series to a 220 V A.C. supply. When measured with an A.C. voltmeter, the P.D. across resistor is 140 V. The P.D. across the terminals of the inductor will be nearly (A) 150 V (B) 160 V (C) 170 V (D) 180 V
solution (C) 220 = VL =
1402 VL 2 or 48400 = 19600 + VL2
28800 = 170 V
28.
If q0 is the maximum charge then while charging, the charge on a capacitor at time t is given by q0 (1 - et / RC ) (D) q0 (1 - e - t / RC ) (A) q0 t (B) q0 (1 + e - t / RC ) (C) Solution : (D) q = q0(1 e t / RC ) where RC is the time constant. 29.
In the case of charging of a capacitor, at t = RC, the charge on the capacitor (q0 is the maximum charge) is (A) 0.432 q0 (B) 0.532 q0 (C) 0.632 q0 (D) 0.732 q0 Solution : (C) When t = RC, q = q0 (1-e–1) = 0.6321 q0
30.
In the case of charging or discharging of a capacitor RC has the units of (A) Ohm (B) A-F (C) Ohm - m (D) sec. solution : (D) The quantity RC has units of volt current sec ond x sec ond current volt
For test 1.
The capacitor gets almost fully charged after time t equal to (A) RC (B) 3 RC (C) 4 RC (D) Solution : (D) After t = 5 RC, the capacitor is about 99.3% charged.
5 RC
The time constant for the given RC circuit where R = 1 K and C = 103 pF will be (A) 2 m second (B) 1 second (C) 1.5 m second (D) 0.5 second. Solution : (B) t = RC = 103 x 103 x 10–12 = 10–6 s = 1s
2.
3.
A capacitor of 4F is connected to a 15 V supply through 1 mega ohm resistance. The time taken by the capacitor to charge upto 63.2 % of its final charge will be (A) 2s (B) 3s (C) 4s (D) 5 s. Solution : (A) 63.2 % = RC = 106 x 4 x 10–6 = 4 s If is a real number, cos + j is a complex number where j is equal to 1 (C) –1 (D) (A) 1 (B) 1 Solution : (D) i = 1
4.
The Exponential function e j is equal to (A) cos (B) (C) cos - j sin (D) Solution : (D) e j = cos + j sin
5.
cos + sin cos + j sin
6.
If LCR are connected is series, then the total effective resistance called impedance in terms of complex number is giver by 1 1 jR L R jL (A) (B) C C 1 1 (D) (C) RL R jL jC jC Solution : (D) 1 Z = R jL jC
7.
The magnetic flux through a coil perpendicular to its plane and directed into paper is varying according to relation = 5t2 + 10t + 5 milli weber. The e.m.f. induced in the loop after 5 sec 15 (A) 0.03 volt (B) 0.06 volt (C) 0.08 volt (D) 0.02 volt Solution : (B) = (5t2 + 10t + 5) x 10–3 Wb d as e = (in magnitude) dt d (5t2 + 10t + 5) x 10–3 Wb sec–1 dt = (10t + 10) 10–3 volt e = (10 x 5 + 10) x 10-3 = 0.06 volt
A train is moving with a speed of 30 m s–1 NS on the rails separated by 2 m. If the vertical component of earth's field is 8 x 10–5 T, the e.m.f. is (A) 0.0048 V (B) 0.048 V (C) 0.48 V (D) 4.8 V Solution : (A) e = Blv = 30 x 2 x 8 x 10-5 = 0.0048 V 8.
An aircraft with a wingspan of 40 m flies with a speed of 1080 km h –1 in the eastward direction at a constant altitude in the northern hemisphere, where the vertical component of earth's magnetic field is 1.75 x 10 –5 T. Then the e.m.f. that develops between the tips of the wings is (A) 0.5 V (B) 0.34 V (C) 0.21 V (D) 2.1 V Solution : (C) L = 40 mv = 1080 km h-1 = 300 m sec-1 B = 1.75 x 10-5 T e = Blv = 1.75 x 10-5 x 40 x300 = 0.21 V
9.
A Uniform flux density of 0.1 Wb m2 extends over a plane circuit of area 1 m 2, and is normal to it. How quickly must the field be reduced to zero if an emf of 100 V is to be produced? 1 sec (A) 10–2 Sec (B) 10–3 sec (C) (D) 1 micro sec. 10 Solution : (B) 1 = BA = 0.1 x 1 = 0.1 Wb; 2 = 0 10.
e=-
1 d 2 dt t
0 0.1 100 = , t = 10-3 s t
11.
Current in a 10 milli henry coil increases uniformly form zero to one to one ampere in 0.01 second, then the value of induced e.m.f. is (A) +1V (B) 2V (C) –1 V (D) –2 V Solution : (C) L = 10 x 10-3 = 10-2 henry change in current = 1-0 = 1 amp. dt = 0.01 sec dl 1 e== -10-2 x = -1 volt dt 0.01
12.
A step up transformer is used on 120 V line to provide a P.D. of 2400 V. If the number of turns in primary is 75, then the number of turns in the secondary shall be (A) 25 (B) 150 (C) 1500 (D) 500 Solution : (C) Here EP = 120 V, ES = 24000 V, np = 75 and ns = ? E ns 2400 ns As S = np 120 75 EP 2400 ns x75 1500 120 13.
The current drawn from the primary of a transformer which steps down 220 V to 22 V at 0.1 amp. shall be (A) 1 amp. (B) 0.1 amp (C) 0.2 amp. (D) 0.01 amp Solution : (D) Here np = 220 V, ns = 22 V, Is = 0.1 amp. and Ip = ? E I As s P EP IS E 22 IP S xIS x0.1 0.01amp EP 220
14.
What is the r.m.s. value of current ? I = I 2 sin t A 3 (A)
2A
(B)
Solution : (B) Peak value of current, I0 = 15.
1A
(C)
1 2
A
(D)
2 A.
2 A
A hot wire instrument reads 15 A. The peak value of the current is 15 15 2 A (A) (B) (B) 15 A (D) A 2
15 A
Solution : (A) The hot wire instruments measure virtual values only. Irms 15A Ipeak 2 I r.m.s. 2x15 15 2A
16.
The exponential function ex is equal to (A)
x+x2+x3+....
x x2 (C) 1 x .... 2! 3! x 2 x3 Solution : (B) ex = 1 x .... 2! 3!
(B) (D)
x 2 x3 .... 2! 3! x 2 x3 1 x .... 2! 3! 1 x
17.
An A.C. voltmeter reads 20 V across a resister, 30 V across an inductor and 15 V across a capacitor in LCR series circuit. In this case the supply voltage will be (A) 100 V (B) 60 V (C) 50 V (D) 25 V Solution : (D) e= =
VR2 (VL VC )2 400 225 25 V
18.
An dc voltage, E (in volt) = 200 2 sin 100 t is connected through an dc ammeter. The reading of the ammeter shall be (A) 10 mA (B) 20 mA (C) 40 mA (D) 80 mA Solution : (B) 1 1 E V 200V, XC 10 4 6 C 100x10 200 I 20 mA 10 4 19.
An LCR series circuit its connected to a source of A.C. at resonance, the applied voltage and the current flowing through the circuit will have a phase difference of (A) (B) (C) (D) 0. / 2 / 4 Solution : (D) 1 At resonance L C Z R or cos 1or 0 20.
'
The Capacitor shown in the Fig. is initially charged to a voltage V and the circuit is then closed. The charge on C is found to oscillate with frequency w. The frequency will be doubled if 1 1 2 2 LC LC 22 L (A) the voltage V is doubled C (B) both L and C are halved (C) both L and C are doubled
(D) both L and C are creased by a factor Solution : (B) 1 1 4 ' 2 2 . LC L / 2.C / 2 LC 21.
2
In a parallel LCR circuit shown in the Fig. at resonance (A) the source current is maximum (B) the impedance of the circuit is minimum and is equal to R (C)
E
R
C
L
the resonance frequency will be the same as for a series resonance circuit with the same value of L, C and R the voltage across L and C are in phase.
(D) Solution : (C) In a parallel resonance circuit, at resonance
L / C RL 2 2 LC L / C RC for a good qual;ity capacitor Rc = 0 and we put RL = R. Then
1
1 R2 LC L2 If resistance is neglected then condition of resonance in both series and parallel circuit become the same, namely (LC)1/ 2
22.
In a series LCR ac circuit, off resonance, the current is (A) always in phase with the generator voltage. (B) always lags the generator voltage (C) always leads the generator voltage (D) may lead or lag behind the generator voltage. Solution : (D) In a series LCR circuit, off resonance, the power factor is R R cos 2 Z 1 2 R L C 1 R So, tan L C 1 Now depending on whether L is greater than or less, the lead or lag will C occur. 23.
Who discovered laws of EMI ? (A) Faraday (C) P.A.M. Dirac
(B) (D)
Einstein R.P. Feynman.
(B)
MLT-2A-1
Solution : (A) 24.
The Dimensions of flux is (A) MLT-2 A
(C) ML-1T-3A-1 Solution : (D) F BA cos V A cos q
25.
(D) (
ML2T-2A-1 F = Bqv)
(MLT 2 )(L2 ) ML2T 2 A 1 1 (AT)(LT )
When a current of 4 A through primary gives rise to a flux of magnitude 1.35 Wb through secondary, what is the coefficient of mutual induction (M)? (A) 0.43 H (B) 0.38 H (B) 0.34 H (D) None of these.
Solution : (C) MI M M
I
1.35 0.34 henry 4
26.
When current changes from 13 A to 7 A in 0.5 second through a coil, the e.m.f. induced is 3 x 10-4 V. What is the coefficient of self induction? (A) 25 x 10-3 H (B) 25 x 10-4 H (C) 25 x 10-5 H (D) 25 x 10-6 H Solution : (D) e = 300 x 10–6 V dl 13 7 6A dt = 0.5 s dI e=L in magnitude dt dt 300x106 x0.5 L e = 25 x 10-6 H dI 6
27.
For two coils inductively coupled, the mutual inductance is 1.9 H. The current through primary changes by 6 A in 1 ms. The induced e.m.f. is 1.4 V (A) 11.4 mV (B) 11.4 V (B) 11.4 kV (D) Solution : (A) M = 1.9 H, dI = 6 A, dt = 10-3 sec dI 6 e m 1.9x 3 dt 10 = 11.40 x 10-3 V 28.
In the middle of a long solenoid there is a coaxial ring of square cross-section, made of conductivity material of resistivity p. The thickness of the ring is equal to h, its inside and outside radii are equal to 'a' and 'b' respectively. What is the current induced in a radial width (dr), where the magnetic field varies with time as B t ? hdr hrdr hrdr hrdr (A) (B) (C) (D) 4p p 2p 2p
Solution : (D) According to the problem, the e.m.f. induced in an elementary ring of radius r 1 h dr and width dr is r 2 . The conductance of this ring is d R p 2r thus, the current induced is dI = hrdr / 2p . Upon integration we get the total current
I
b
a
hrdr h(b2 a2 ) 2p 4p
29.
In the Q.95, what is the total current ? h(b2 a2 ) h(b2 a2 ) h(b2 a2 ) h(b2 a2 ) (A) (B) (C) (D) 4p 8p 16p 2p Solution : (A) According to the problem, the e.m.f. induced in an elementary ring of radius r 1 h dr and width dr is r 2 . The conductance of this ring is d R p 2r thus, the current induced is dI = hrdr / 2p . Upon integration we get the total current
I
b
a
hrdr h(b2 a2 ) 2p 4p
30.
Suppose p is the mass density, po is the resistivity, l 0 is the length and S is the cross-section of a wire. The mass of the wire is m. What is the active resistance of the wire ? l2 2l2 (B) (D) (A) l2 / m l2 / 2m (C) 4m m Solution: (A) The time constant is the ratio L/R, where L is the inductance of the solenoid and R is the active resistance. The latter is given by R = 0l0/S where 0 is the resistivity, l0 is the length and S is the cross-section of the wire. Also, if is the mass density of the material of the wire, mass of the wire m = l0S. Upon eliminating S one can write R = 0 l0 / S The time constant () is therefore given by L mL t = = 2 ær r 0l0 ö r r 0l02 çè m ÷ ø The length of the wire l0, in terms of the length of the solenoid l, is given by 1/ 2
æ4p lL ö l0 = ç è m0 ÷ ø
Thus, the time constant is given by m æm ö mL t = = 0 ç ÷ 4 p lL 4 p è r r 0l ø rr0 ´ m0
1.
SOME IMPORTANT DEFINITIONS USED IN OPTICS
We define a varied collection of useful terms. Some of these definitions will be revisited when the appropriate topics are taken up. (i)
Ray: The path of light, as determined within the approximations of geometric optics, is a ray. In a homogeneous medium, it is a straight line.
(ii)
Beam: A collection of rays, usually referred to as a bundle of rays, forms a beam. One may have a convergent, divergent or, a parallel beam. A convergent beam may converge to a point or, a line, a divergent beam may diverge from a point or, a line. A parallel beam consists of parallel rays.
(iii)
Collimation: A process whereby a divergent (or, convergent) beam is rendered parallel usually through the use of lenses and/or mirrors.
(iv)
Object & Image: The term object is used to refer to any object (being photographed or observed) that is a source of light, and its likeness (usually two dimensional) formed or observed by an optical system is the image.
(v)
Optical System: An optical system consists of elements like lenses, mirrors, prisms, etc.
(vi)
Axis: The axis of an optical system is frequently an axis of symmetry such that a ray directed along the axis continues in the same direction or, returns backwards (if reflected within the system)
(vii)
Centre of Curvature: Most lenses and curved mirrors being manufactured spherical (i.e., their surfaces are spherical), the centre of curvature of the curved mirror or, the curved surface of a lens is important and is frequently denoted by the letter C. The radius of curvature is also equally important in the analysis.
(viii) Pole: The pole of a spherical surface (refracting or, reflecting) is the central point of the surface involved in the formation of the image. It is denoted by O or, P. The axis, for a single spherical surface, is the join of the pole with the centre of curvature; it is known as the principal axis. (ix)
Optical Centre: The optical centre of a thin lens is a point on the axis of the lens such at a ray directed towards that point emerges parallel to itself after passing through the lens.
(x)
Paraxial Rays: It is observed that the formation of clear images by spherical surfaces takes place only with rays which are close to the principal axis and make very small angles with it.
(xi)
Wave speed: It is the distance travelled by the wave disturbance in a unit time. It is denoted by the letter .
(xii)
Frequency: It is the number of vibrations made by the medium particle in 1 s. In other words, it is number of waves passing through a point of the medium in 1 second. It is generally represented by the letter n or . Its unit is hertz and it is represented as Hz or s–1.
(xiii) Wavelength: It is the distance travelled by the wave in one complete period of a medium particle. In other words, distance between two consecutive crest or trough. Relationship between the wavelength, wave speed and frequency: If the wave speed is and period is T, then by definition
Wavelength T or But T 1/ x1/ Or
Distance travelled in one period Distance travelled in T s x T
i.e., Wave speed Frequency X Wavelength.
2.
REFLECTION AND REFRACTION AT PLANE AND SPHERICAL SURFACES
2.1
REFLECTION AT PLANE SURFACE
(i) Reflection of Light When light rays strike the boundary of two media such as air and glass, a part of light is turned back into the same medium. This is called Reflection of Light. The wavelength and the velocity of the light wave remains the same. Incident
N
Reflected
N N
i r
ir i r
Laws of reflection are obeyed at every reflecting surface, i.e. (ii) Laws of Reflection (a) The incident ray (AB), the reflected ray (BC) and normal (BN) to the surface (SS') of reflection at the point of incidence (B) lie in the same plane. This plane is called the plane of incidence (or plane of reflection). (b) The angle of incidence (the angle between normal and between the reflected ray and the normal are equal, i.e. i r A
N
C
i r S
S B
(iii) Types of reflection (a) Regular Reflection (a)
(b)
Diffused Reflection or Scattering or Diffusion
Regular Reflection: When the reflection takes place from a perfect plane surface it is called Regular Reflection.
(a) Regular reflection
(b)
Diffused Reflection or Scattering or Diffusion: When the surface is rough light is reflected from the surface from bits of its plane surfaces in irregular directions. This is called diffusion. This process enables us to see an object from any position.
(b) Irregular reflection
Characteristics of Reflection by a Plane Mirror (i) Distance of object from mirror = Distance of image from the mirror. (ii) The line joining the object point with its image is normal to the reflecting surface. (iii) The image is laterally inverted (left right inversion). (iv) The size of the image is the same as that of the object. (v) For a real object the image is virtual and for a virtual object the image is real. (vi) The minimum size of a plane mirror, required to see the full self image, is half the size of that person. (vii) For a light ray incident at an angle 'i' after reflection angle of deviation d = p - 2i (viii) (ix) (x) (xi) (xii)
(xiii)
(xiv)
If i = 0 then r = 0, this implies that a ray of light incident normally on a mirror retraces its path. The eye always observes an object in the direction in which the rays enter the eye The laws of reflection holds good for all kinds of reflection. Image of an object is the point at which rays after reflection (or reflection) actually converge or appear to diverge from that point. If the direction of the incident ray is kept constant and the mirror is rotated through an angle about an axis in the plane mirror then the reflected ray rotates through an angle 2. If an object moves towards (or away from) a plane mirror at a speed v, the image will also approach (or recede) at the same speed. Further the relative speed of image to the object will be v – (–v) = 2v. If two plane mirror are inclined to each other at 90, the emergent ray is always antiparallel to incident ray, if reflected from each mirror, irrespective of angle of incident.
M1
M2 (xv)
When two plane mirror, inclined to each other at an angle , the number of images formed can be determined as follows: 360 (a) If is an even integer, say ‘p’, then number of image formed say q = p – 1, for all position.
360 (b) If is an odd integer say ‘q’ then number of image formed say p = q, if the object is not on the bisector of the angle between mirrors. Also, p = q – 1, if the object is on the bisector. 360 (c) If is a fraction, then the number of image formed will be equal to its integral part. (xvi) The images are laterally inverted. (xvii) The linear magnification is unity. ∧
∧
∧
Illustration 1:
A ray of light on a plane mirror along a vector i + j - k . The normal on
Solution:
incident point is along i + j . Find a unit vector along the reflected ray. Reflection of a ray of light is just like an elastic collision of a ball with a horizontal ground. Component of incident ray along the inside normal gets reversed while the component perpendicular to it remains unchanged.
∧
∧
Thus the component of incident ray vector A i j k parallel to normal,
i.e., i j gets reversed while perpendicular to it, i.e., k remains unchanged. Thus, the reflected ray can be written as
R i j k A unit vector along the reflected ray will be
R i j k = r R 3
2.2
REFLECTION-AT SPHERICAL MIRRORS
A spherical mirror is a reflecting surface which forms a part of a sphere (as shown in following a and b diagram). When the reflection takes place from the inner surface and outer surface is polished or silvered the mirror is known as concave mirror. Vice- versa, it is convex.
(a) concave mirror
C
Concave Mirror
F
(b) convex mirror
P
P
F
C
Convex Mirror
Ray diagram for concave and convex mirror
2.3
CHARACTERISTICS OF REFLECTION BY A REFLECTING SURFACES (FOR SMALL APERTURE)
(i)
Pole (P) is generally taken as the mid point of reflecting surface.
SPHERICAL
(ii) (iii) (iv) (v)
(vi) (vii) (viii)
Centre of curvature (C) is the centre of the sphere of which the mirror is a part. Radius of curvature is the radius of the sphere of which the mirror is a part. Distance between P and C. Principal Axis is the straight line connecting pole P and centre of curvature C. Principal focus (F) is the point of intersection of all the reflected rays which strike the mirror (with small aperture) parallel to the principal axis. In concave mirror it is real and in the convex mirror it is virtual. Focal length (f) is the distance from pole to focus. Aperture is the diameter of the mirror. If the incident ray is parallel to the principal axis, the reflected ray passes through the focus. (Fig (a)) C
FP
(a)
(ix)
If the incident ray passes through the focus, then the reflected ray is parallel to the principal axis (Fig.(b)) C
F
P
(b)
(x)
Incident ray passing through centre of curvature will be reflected back through the centre of curvature. (Fig.(c)) C F P (c)
(xi) (xii)
(xiii)
R where f = focal length R = Radius of curvature. 2 Sign convention and magnification. We follow Cartesian coordinate system convention according to which: (a) The pole of the mirror is the origin (b) The direction of the incident rays is considered as positive x-axis. (c) Vertically up is positive y-axis Note: Radius of Curvature and Focal Length of: (a) Concave mirror is positive (b) Convex mirror is positive (a) Linear Magnification or lateral magnification or transverse magnification h v =M= 2 = h1 u h2 = y coordinate of image; h1 = y coordinate of the object (both perpendicular to the principle axis of mirror) v 2 - v1 v2 (b) Longitudinal Magnification = (for any size of the object) = - 2 (for u u2 - u1 short object) f=
Tracing for spherical mirror Concave
Convex
(1) A ray going through centre of curvature is reflected back along the same direction.
(2) A ray parallel to principal axis is reflected through the focus, and vice-versa. Also, mutually parallel rays after reflection intersect on the focal plane.
(3) A ray going to the pole and the reflected ray from it make equal angles with the principal axis.
2.4
C
C
F
F
F
i i
P F
i i
REAL AND VIRTUAL SPACES
A mirror, plane or spherical divides the space into two: (i) A side, where the reflected rays exist. This is real space. (ii) The other side where the reflected rays do not exist. This is virtual space Real Space
virtual Space
Real Space
virtual Real Space Space
virtual Space
Object Object is decided by incident rays only. The point object for mirror is that point. (a) From which the rays actually diverge to be incident on the mirror (Real object). Or (b) Towards which the incident rays appear to converge (virtual object). Point Object Real
Point Point Object Object Virtual (Real)
Image Image is decided by reflected or refracted rays only. The point image for a mirror is that point. (a) Towards which the rays reflected from the mirror, actually converge (real image). Or (b) From which the reflected rays appear to diverge (virtual image). NEW CARTESIAN SIGN CONVENTION – FOR MIRRORS AND LENSES There are several sign-conventions. Students should try to follow any one of these. In this package, the new Cartesian sign convention has been used wherever required.
Incident Ray +ve
-ve
P
Mirror / Lens
(i) (ii) (iii) (iv) (v) (vi)
All distance are measured from the pole. Distances in the direction of incident rays are taken as positive. Distances in the direction of incident rays are taken as negative. Distances above the principal axis are taken as positive. Distances below the principal axis are taken as negative. Angles measured from the normal, in anti-clockwise direction are positive, while in clockwise direction are negative.
2.5
MIRROR FORMULA
Consider the shown figure where O is a point object and I is corresponding image. CB is normal to the mirror at B. B q a O
C
q g
b I
P’
P
By laws of reflection, OBC = PBC = + = , + = + = 2 For small aperture of the mirror, , , 0 tan , tan , tan , P’ P tan + tan = 2 tan BP' BP' BP' Þ + =2 OP' IP' CP' Applying sign convention, u = -OP, v = -IP, R = -CP 1 æ 1ö 2 Þ - + ç- ÷ = u è vø R 1 2 If u = , , but by definition, if u = , v = f. v R R 1 1 1 f and Hence, 2 u f For convex mirrors, an exactly similar formula emerges. Illustration 2: An object is placed 21cm infront of a concave mirror of radius of curvature 10cm. A glass slab of thickness 3cm and refractive index 1.5 is then placed closed to the mirror in the space between the object and mirror. Find the position of the final image formed. (You may take the distance of the near surface of the slab from the mirror to be 1cm) Solution: Given that radius of curvature = 10 cm focal length f = 5 cm 1 1 Here u x t 1 21 3 20cm. 3 Where 1.5 = refractive index of slab Using the formula for concave mirror
1 1 1 f u v 1 1 1 5 20 v 20 cm. 3 Applying the correction due to the slab, we have final image distance = |v| + 1 20 23 1 cm. 3 3
Solving we get v
2.6
REFRACTION AT A PLANE SURFACE
Laws of Refraction (Snell’s Law) (i) The incident ray, the refracted ray and the normal to the refracting surface at the point of incidence all lie in the same plane. (ii) The ratio of the sines of the angle of incidence (i) and of the angle of refraction (r) is a constant quantity for two given media, which is called the refractive index of the second medium with respect to the first. sin i cons tan t sin r When light propagates through a series of layers of different medium as shown in the figure, then the Snell’s law may be written as 1 sin 1 = 2 sin 2 = 3 sin 3 = 4 sin 4 = constant In general, sin = constant 1
2
3 4
1 1
2
2
1
2
water
3
4
Fig. A light ray passing from air to water Fig. A series of transparent layers of bends toward the normal different refractive indices When light passes from rarer to denser medium it bends toward the normal as shown in the fig. According to Snell’s law 1 sin 1 = 2 sin 2 When a light ray passes from denser to rarer medium it bends away from the normal as shown in the fig. 1
1
2
2
air water
Fig. A light ray passing from water to air bends away from the normal. For a given point object, the image formed by refraction at plane surface is illustrated by the following diagrams.
Case - I
Case - II O I1
I2
I
O
Denser medium
I2
Rarer medium
I1
I
R2
R1
R1
R2
Determination of the image formed for the cases given becomes much simpler when we restrict ourselves to nearly normal incident rays. CASE – I (Object in the denser medium) Rarer medium m2 A B I
q2 q1
q1
q2
Denser medium m1
O
For nearly normally incident rays 1 and 2 will be very small. AB tan 1 sin 1 Object dis tance from therefractingsurface Similarly, AB sin 2 Image dis tance from therefractingsurface Imagesdist ance from therefractingsurface Object dis tance from therefractingsurface sin 1 1 2 2 sin 2 1 The same result is obtained for the other case also. The image distance from the refracting surface is also known as Apparent depth or height. Apparent Shift Apparent shift = Object distance from refracting surface – image distance from refracting surface. 1 y (apparent shift) t 1 where t is the object distance and 1 2 If there are a number of slabs with different refractive indices placed between the observer and the object. Total apparent shift = yi Illustration 3:
A person looking through a telescope T just sees the point A on the rim at the bottom of a cylindrical vessel when the vessel is empty. When the vessel is completely filled with a liquid ( = 1.5), he observes a mark at the centre, B, of the vessel. What is the height of the vessel if the diameter of its cross-section is 10cm?
Solution:
It is mentioned in the problem that on filling the vessel with the liquid, point B is observed for the same setting; this means that the images of point B, is observed at A, because of refraction of the ray at C. N
T r
C Liquid m=1.5
i h
A
D 5 cm
For refraction at C, sinr a l 1.5 sini AD 10 Now sin r AC 102 h2 Where h is the height of vessel. BD 5 sini = = BC 5 2 + h2 \
100 + h2
.
25 + h2 = 1.5 5
\
25 + h2 9 = 2 100 + h 16 2 100 + 16h = 900 + 9h2
\
7h2 = 500
\
2.7
10
\ h = 8.45 cm
REFRACTION AT SPHERICAL SURFACE
When light strikes a spherical (curved) surface, the position of object (O), the image (I) and the radius of curvature (R) are related as: i
P r
O
1
2
I
Q
2 1 2 1 ; here 2 > 1 v u R Consider another case, a spherical surface of radius R separating two media with refractive indices m1 & m2 . The radius of curative R will be taken positive when incident rays strike the convex side of the surface. If u is the object distance and v is the image distance,
i r
O
C
R
u
μ1
μ2
I
v
For light rays going from medium 1 (1) to medium 2 (2): m1 m2 m2 - m1 + = u v R The corresponding equation for refraction at a plane surface will be: m1 m2 (taking R = ) + =0 u v Illustration 4:
Solution:
A transparent rod 40 cm long is cut flat at one end and rounded to a hemispherical surface of 12 cm radius at the other end. A small object is embedded with in the rod along its axis and half way between its ends. When viewed from the flat end of the rod, the object appears 12.5 cm deep. What is its apparent depth when viewed from the curved end? For the flat surface:
I1
O
Real depth of the object = 20 cm Apparent depth = 12.5 cm real depth Using m= apparent depth 20 m= = 1.6 12.5 For the curved surface:
I2
O
m1 m2 m2 - m1 + = u v R u = 20 cm, R = –12 cm, 1.6 1 1 - 1.6 Þ + = - 12 20 V 1 1 1.6 Þ = V 20 20 Þ V = - 33.3 cm. Hence the object appears 33.3 cm deep from the curved side. We will use
3.
CRITICAL ANGLE AND TOTAL INTERNAL REFLECTION
Consider a ray of light that travels from a denser medium to a rarer medium. The angle of incidence for which the angle of refraction becomes 90 is called critical angle. This is used for constructing totally reflecting prisms. m1
m2
c
m sin c =2 m1 = 2 sin 90° m1 1 Þ sin c = 1 m2 and if i > c When the angle of incidence of a ray travelling from denser to rarer medium is greater then the critical angle, no refraction occurs. The incident ray is totally reflected back into the same medium. Here the laws of reflection hold good. Some light is also reflected before the critical angle is achieved, but not totally. Illustration 5:
Solution:
A point source of light a placed at the bottom of a tank containing a liquid of refractive index The level of the liquid is at a height h above the bottom of the tank. A bright circular spot is seen on the surface of the liquid when viewed from above. Find the radius of the spot. The rays of light emerging from the source, which are incident at an angle q > qc . (where q c is the critical angle) will be reflected. 1 ...(i) sin qc = m R = tan qc Further, h sin qc = 1 - sin2 qc 1 = m2 - 1 h ...(ii) or, R = m2 - 1 R air
h
N
4.
THIN LENS
Consider the thin lens shown here with the two refracting surfaces having radii of curvature equal to R1 and R2 respectively. The refractive indices of this surrounding medium and of the material of the lens are 1 and 2 respectively.
O
m1
m2 (1)
m1
I
(2)
Now using the result that we obtained for refraction at single spherical surface we get, m2 m1 m2 - m1 For first surface, ...(1) = v1 u R1 m m m - m2 For second surface, 1 - 1 = 1 ...(2) v v1 R2
é1 m2 m1 1ù = (m1 - m2 ) ê ú v u ëR1 R2 û öé 1 1 1 æm2 1ù - = ç - 1÷ê ú v u è m1 ø ëR1 R2 û
Adding (1) and (2),
Þ
When u = , v = f, Also,
1 1 1 - = v u f
öé 1 1 æm2 1ù = ç - 1÷ê ú f è m1 ø ëR1 R2 û
Lensmaker's Formula
Thin Lens Formula
EQUIVALENT FOCAL LENGTH OF TWO OR MORE THIN LENSES IN CONTACT
1 1 1 = + + ... F f1 f2 Special cases (a) If the lens is immersed in a liquid whose refractive index is greater than refractive index of material of the lens then 1 1 1 (I g 1) f(liquid) R1 R2 fliquid is the focal length of lens in the liquid Ig is the R.1. of glass w.r.t. liquid. The focal length of the lens in the liquid becomes negative. (b)
If the lens is immersed in a liquid whose R.1. is equal to the R.1. of the material of the lens then a b 1 I g a I
1 1 (1 1) 0 f(liquid) R1 R2 The focal length becomes infinite
(c)
1
If the lens be immersed in a liquid whose R.1. is less than the R.1. of the material of the lens 1 1 1 (I g 1) f(liquid) R1 R2
The focal length of the lens increases. In general, if decreases, f increases Note: A lens is converging if its focal length is positive and diverging if focal length is negative (in neo-Cartesian sign convention). From this we can conclude that a convex lens need not necessarily be a converging and a concave lens diverging. Ray tracing for lens Convex (i)
Concave
A ray passing through optic centre goes undeviated. O
(ii) When a ray is incident parallel to principal axis, the refracted ray passes or appears to pass through the focus.
Illustration 6:
F
F
A glass rod has ends as shown in figure, The refractive index of glass is . The object point O is at a distance 2R from the surface of larger radius of curvature. The distance between the apexes of the ends is 3R. Show that the image point of O is formed at a distance of
Solution:
O
(9 - 4μ )R (10μ - 9 )(μ - 2 )
From the right hand vertex. R Glass AIR
R/2
O
A
B
m AIR 2R
For the first surface m2 m1 m2 - m1 = v' u R1 Here 2 =, 1 =1, = –2R, R1 = +R u 1 m- 1 + = +R v ' 2R u m- 1 1 2m- 3 = = v' R 2R 2R 2mR \ v' = 2m- 3 For the second surface é æ 2mR öù u = - ê3R - ç ú è 2m- 3 ÷ øû ë
3R
Here
2 = 1, 1 = , R2 = +
R 2
m2 m1 m2 - m1 = v u R2 m 1 1- m + = v é 2mR ù R / 2 3R - ê ú ë2m- 3û m(2m- 3) 1 2 - 2m = v R (6mR - 9R - 2mR)
1 2 - 2m m(2m- 3) - 10m2 + 29m- 18 = = v R R [4m- 9] R [4m- 9] v=
5.
(9 -
4m)R (9 - 4m)R = 10m2 - 29m+ 18 (10m- 9)(m- 2)
MAGNIFICATION
(i) Magnification in Mirror In new Cartesian sign convention, we define magnification is such a way that a negative sign (of m) implies inverted image and vice-versa. A real image is always inverted one and a virtual one is always erect. Keeping these points in mind and that the real object and its real image would lie on the same sides in case of mirror and on opposite sides in case of lenses, we define m as in case of reflection by spherical mirror as: B A’ F A
C
m=(ii)
P
B’
v u
Lateral Magnification for Refracting Spherical Surface B
m1
m2 A’
A
P B’
M=
- (A 'B') . AB
As i, r ® 0, Þ
AB
Now,
Þ
PA = m2 A 'B' m1 PA '
sini m2 = sinr m1
tani m2 = tanr m1 Þ
- PA ' m2 A 'B' = PA AB m1
v
Hence,
(iii)
m=
u
m2 m1
Magnification of A Thin Lens v - (OA ') M= = u (AO) B
A’ A
B’
Illustration 7:
Solution:
An object is placed at a distance of 10 cm to the left on the axis of a convex lens A of focal length 20 cm. A second convex lens of focal length 10 cm is placed co-axially to the right of the lens A at a distance of 5 cm from A. Find the position of the final image and its magnification. Trace the path of the rays. Here for 1st lens, u1 = –10 cm f1 = 20 cm. 1 1 1 1 1 1 = Þ = v1 u1 f1 v1 20 10 Þ v1 = - 20 cm
Q’
L1
L2
O1
O2
f1
f1
Q
i.e. the image is virtual and hence lies on the same side of the object. This will behave as an object for the second lens. For 2nd lens 1 1 1 = v 2 u2 f2 Here
u2 = - (20 + 5)
f2 = 10 cm
1 1 1 + = v 2 25 10
50 2 = 16 cm 3 3 2 i.e. the final image is at a distance of 16 cm on the right of the second lens. 3 The magnification of the image is given by, v v 20 50 4 m= 1× 2 = × = = 1.33 u1 u2 10 3 ´ 25 3 Þ
v2 =
6.
POWER OF A LENS
1 f If two lens in are separated by a distance d then power of combination of lens is d P = P1 + P2 - PP m 1 2 Where P1 and P2 are optical powers of the two lenses, and is the refractive index of the medium in between them. If f is in metres then the power P of the lens in dioptres is given by, P =
Illustration 8:
A lens has a power of +5 diopters in air. What will be its power if completely immersed in water? and a μ g = 3/2. a μ w = 4/3
Solution:
Let fa and fw be the focal lengths of the lens in air and water respectively, then 1 1 Pa = or +5 = fa fa fa = 0.2 m = 20 cm é1 1 1ù Now ...(1) = a mg - 1 ê ú fa R R ë 1 2û
(
)
é1 1 1ù = w mg - 1 ê ú fw ëR1 R2 û Dividing equation (1) by equation (2), we get, fw é a mg - 1 ù =ê ú fa êë w mg - 1ú û m 3/2 9 a g = = Again, w mg = 4/3 8 w mg
(
and
\
)
...(2)
fw (3 / 2) - 1 (1/ 2) = = =4 fa (9 / 8) - 1 (1/ 8) fw = fa ´ 4 = 20 ´ 4 = 80 cm = 0.8 m
Pw =
7.
1 1 = = 1.25 diopter. fw 0.8
SILVERING AT ONE SURFACE OF LENS
When one surface of a thin lens is silvered, then the focal length F of the effective lensmirror combination is expressed as, 1/F = å 1/ f i , where fi is the focal length of the lens or mirror to be repeated as many times as the refraction or reflection respectively is repeated. Some Cases: (i) Focal length of plano–convex lens when silvered at its plane surface When an object is placed in front of such a lens. The ray first of all are refracted from the convex surface, then reflected from the polished plane surface and again refracted out from the convex surface. If f and fm be the focal lengths of lens and mirror.
R
R
Þ
Þ (ii)
1 1 1 1 1 1 2 2 é1ù = + + , Here, = ( m - 1) ê úand = = =0 F f fm f f fm Rm ¥ ëR û f 1 2 R = Þ F= = F f 2 2( m - 1)
Focal length of plano –convex lens when silvered at convex surface 1 1 2 é1ù = ( m - 1) ê úand = f fm R ëR û 1 2 2 1 R é( m - 1) ù 2 Þ = + Þ = 2ê + Þ F= ú F f R F 2m ë R û R
Þ
(iii)
1 1 1 1 = + + , F f fm f
Here,
Focal length of convex lens where convex surface of radius R2 is silvered
é1 1 1ù 1 2 = ( m - 1) ê úand = f fm R2 ëR1 R2 û é æ1 1 2 2 1 1 öù 2 Þ = + Þ = 2 ê( m - 1) ç ú+ ÷ F f R2 F R R è ø 1 2 ë û R2 Þ
1 1 1 1 = + + , F fl fm f
Illustration 9:
Here,
An object is 1 metre in front of the curved surface of a plano–convex lens whose flat surface is silvered. A real image is formed 120 cm in front of the lens. What is the focal length of the lens?
Solution: I Q 100cm 120cm
Here u = 100 cm v = 120 cm 1 1 1 1 1 11 F v u 120 100 600 1 1 1 1 Now F FL FM FL Where FL – focal length of the lens. FM – focal length of the mirror. 2 1 = + ( FM = ¥ ) FL FM
1 2 = F FL 11 2 \ = 600 FL 1200 FL = = 109.1cm 11
8.
ANGLE OF DEVIATION OF A RAY WHEN IT PASSES A LENS
O is the object and is the image is the angle of deviation. d = a + b » tan a + tan b
h h é 1 1ù + = hê - ú -u v ëv u û h d= f
= Þ
9.
DETERMINATION OF FOCAL LENGTH OF A CONCAVE MIRROR BY U–V METHOD
The relation between object distance u and the image v from the pole of the mirror is given by, 1 1 1 v u f Where f is the focal length of the mirror. The focal length of the concave mirror is obtained 1 1 either from versus graph or from u–v graph. v u 1 1 and graph u v When the image is real (of course only upon then it can be obtained on screen), the object lies between focus (F) and infinity. In such a situation u, v and f all are negative. Hence the mirror formula 1 1 1 v u f 1 1 1 Becomes, v u f 1 1 1 or again, v u f 1 1 1 or, v u f Comparing with y = mx + c, the desired graph will be a straight line with slope –1 and 1 intercept equal to . f
1 1 versus graph is as shown in figure. The intercepts on the v u 1 horizontal and vertical axes are equal. It is equal to . A straight line OC at an angle 45o f 1 1 with the horizontal axis intersects line AB at C. The coordinates of point C are , . 2f 2f The focal length of the mirror can be calculated by measuring the coordinates of either of the points A, B or C.
The corresponding
u–v graph The u–v graph comes out to be a hyperbola as shown in figure. A line drawn at angle 45o from the origin intersects the hyperbola at point C. The coordinates of point C are (2f, 2f). The focal length of mirror can be calculated by measuring the coordinates of point C.
Determination of focal length of convex lens using u–v method The relation between u, v and f for a convex lens is, 1 1 1 v u f 1 1 and graph u v Using the proper sign convention, u is negative, u negative, v and f are positive. So we have, 1 1 1 1 1 1 or v u f v u f 1 1 Comparing with y = mx + c, graph between and is negative with v u 1 slope –1 and intercept . The corresponding graph is as shown in figure. f Proceeding in the similar manner as discussed in case of a concave mirror the focal length of the lens can be calculated by measuring the coordinates of either of the points A, B and C.
u–v Graph The u versus v graph is a hyperbola as shown in figure. By measuring the coordinates of point C whose coordinates are (2f, 2f) we can calculate the focal length of the lens.
10. PRISM A prism has two plane surfaces AB and AC inclined to each other as shown in following figure. A is called the angle of prism or refracting angle. (i) Refraction Through A A X i
D r1
r2
Y r
emergent ray
incident ray
A light ray striking at one face of a triangular glass prism gets refracted twice and emerges out from the other face as shown above. The angle between the emergent and the incident rays is called the angle of deviation (D). The angle between the two refracting faces involved is called the refracting angle (A) of the prism. From AXY, we have: A + (90° – r1) + (90° – r2) = 180° (i) Þ r1 + r2 = A Deviation D = (i – r1) + (e – r2) D = (i + e) – (r1 + r2) D=i+e–A Þ A+D=i+e (ii) Þ We also have two equations from Snell's Law at X & Y. sinr2 1 sinr =m & = sinr1 sine m (i) Angle of Deviation It can be easily seen that if we reverse the emergent ray, it goes back along the same path. The angles of incidence and emergence get interchanged but the angle of deviation remains same. A
A
1
2
angle of incidence = 1 angle of emergence = 2 q1 + q2 = A + D
1
2
angle of incidence = 2 angle of emergence = 1 q2 + q1 = A + D
Hence the same angle of deviation D is possible for two different angles of incidence : 1 and 2, where 1 + 2 = A + D. (ii) Minimum Deviation The angle of deviation is minimum when the path of light ray through the prism is symmetrical. i.e., angle of incidence = angle of emergence A
r1=r2=r
i=e Þ r1 = r2 and hence A + Dmin = i + i = 2i also A = r1 + r2 = 2r from
sini = m, we have: sinr1 æA + Dmin ö sin ç ÷ è ø 2 =m A sin 2
(iii) Grazing Incidence When i = 90°, the incident ray grazes along the surface of the prism and the angle of refraction inside the prism becomes equal to the critical angle for glass - air. This is known as grazing incidence. A i 90
r
e
(iv) Grazing Emergence When e = 90°, the emergent ray grazes along the prism surface. This happens when the light ray strikes the second face of the prism at the critical angle for glass - air. This is known as grazing emergence. A i
e = 90
(v) Maximum Deviation The angle of deviation is same for both the above cases (grazing incidence & grazing emergence) and it is also the maximum possible deviation if the light ray is to emerge out from the other face without any total internal reflection. (vi) Dispersion of Light When a ray of light passes through a prism, it splits up into rays of constituent colours or wavelengths. This phenomenon is called dispersion of light. The refractive index of a medium is different for light rays of different wavelengths. Larger the wavelengths, the lesser is the refractive index. mred < mviolet because l red > l violet. i.e.,
A ray of white light passing through a prism gets splits into different colours because the deviation is more for violet and less for red. This phenomenon is known as dispersion. Dr
white
Dv red
angle of dispertion
violet
For a prism with very small refracting angle A, the deviation D is given by: d » (m- 1) A Hence deviation of violet = Dv = (v, – 1)A and for red = Dr = (r – 1)A The angle between the red and violet rays is known as the angle of dispersion. The angle of dispersion = f = Dv - Dr = (mv - mr ) A
Dv + Dr æmv + mr ö =ç - 1÷A è 2 ø 2 mv - mr f Dispersive power of glass = w = = Dm mv + mr -1 2
Angle of mean deviation = Dm =
This phenomenon arises due to the fact that refractive index varies with wavelength. It has been observed for a prism that decreases with the increase of wavelength, i.e. mblue > mred . Angular dispersion, q = dv - dr (vii)
Dispersive power d - dr where is deviation of mean ray (yellow) w= v d
As
dv = (mv - 1) A,
\
w=
dr = (mr - 1) A,
mv - mr m- 1
(viii) Deviation without Dispersion This mean an achromatic combination of two prisms in which net or resultant dispersion is zero and deviation is produced. For the two prisms,
White
White
Þ
(mv - mr ) A + (m'v - m'r ) A ' = 0 (m - mr ) A and wd + w ' d ' = 0 A' = - v
Spectrum
m'v - m'r
Where and ' are the dispersive powers of the two prisms and and ' their mean deviations.
(ix) Dispersion without deviation A combination of two prisms in which deviation produced for the mean ray by the first prism is equal and opposite to that produced by the second prism is called a direct vision prism. This combination produces dispersion without deviation. For deviation to be zero, ( + ') = 0 Þ (m- 1) A + (m'- 1) A ' = 0 Þ
A' = -
(m- 1) A (- ve sign (m'- 1)
Þ
prism A' has to be kept inverted)
Illustration 10: A prism of crown glass refracting angle of 50 and mean refractive index = 1.51 is combined with one flint glass prism of refractive index = 1.65 to produce no deviation. Find the angle of flint glass and net dispersion. Given: v = 1.523, R = 1.513 (for crown glass) 'v = 1.665, 'R = 1.645 (for flint glass) Solution: Let A' be the angle of flint glass prism. Here A = 5 and = 1.51 for crown glass prism. d = (m- 1) A = (1.51 - 1) ´ 5 = 2.55° Deviation produced by flint glass d = (m'- 1) A ' = (1.65 - 1) A ' = 0.65 A ' For no deviation ' = or 0.65 A' = 2.55 2.55 A' = = 3.92° 0.65 Net dispersion,
(mv - mR ) A - (m'v - m'R ) A '
= (1.523 - 1.513)5 - (1.655 - 1.645)3.92 = 0.05 - 0.0784 = - 0.0284
11. ASSIGNMENT 1.
Two plane mirrors A and B are aligned parallel to each other, as shown in the figure. A light ray is incident at an angle of 30 at a point just inside one end of A. The plane of incidence coincides with the plane of the figure. The maximum number of time the ray undergoes reflections (including the first one) before it emerges out is: 2 3m 0.3mm
B
30
A
(A) Solution:
28 (B)
No. of ref =
(B) 2 3 0.2 3
= 30
30
(C)
32
(D)
34
30 30
02 2 3
2.
Consider the metal coin in figure. The right half is polished, while left half is unpolished. Polished Unpolished
In a dark room, this coin is placed face up on the floor and illuminated with a thin-beamed flashlight, as shown in figure. A and B look at the coin from the position as indicated in figure B eye Flashlight beam
A eye
Floor Coin
Solution:
According to A, which half of the coin (if either) is brighter? (A) The polished half (B) The unpolished half (C) Both halves look equally bright (D) Both halves look completely dark (A) The light reflects of the coin and reaches A’s eye. B's eye
A's eye
As the polished half is smoother than the unpolished half, hence the polished half behaves more like a mirror. So, most of the light striking the polished half of the coin reflects instead of going diffusion reflection. Hence, A sees a high percentage of the light that hits the polished part of the coin. Light hitting the unpolished half undergoes more diffuse reflection, bouncing off the coin in all different directions. Hence, a smaller percentage of that light reaches A’s eyes. So, for A the polished half looks more brighter than the unpolished half.
3.
A glass prism of refractive index 1.5 is immersed in water (R.I. = 4/3). The beam of light incident normally on the face AB is totally reflected to reach the face BC, if (A) sin 8/9 (B) sin < 2/3 (C) 2/3 < sin < 8/9 (D) None of these Solution: (A) The light ray is totally reflected internally at the point P (say). Therefore the critical angle C . Also, sin c n w sin 90 ng
sin c =
4/3 8 3/2 9
B
A
C
B
A
p
sin 8/9
C
4.
A diminished image of an object is to be obtained on a screen 1.0 m away from it. This can be achieved by approximately placing: (A) a convex mirror of suitable focal length (B) a concave mirror of suitable focal length (C) a convex lens of focal length less than 0.25 m (D) a concave lens of suitable focal length. Solution: (C) mage can be formed on the screen if it is real. Real image of reduced size can be formed by a concave mirror or a convex lens. Let u = 2f + x, then 1 1 1 + = u v f 1 1 1 Þ + = 2f + x v f 1 1 1 f +x Þ = = v f 2f + x f (2f + x ) f (2f + x ) f +x It is given that u + v = 1m. Þ
v=
2f + x +
f (2f + x ) f ù é = (2f + x ) ê1 + < 1m f +x ë f + xú û
2
f (2f + x ) < 1m or, f +x or, (2f + x)2 < (f + x ) This will be true only when f < 0.25 m.
5.
A screen beaming a real image of magnification m 1 formed by a convex lens is moved through a distance x. The object is the moved until a new image of magnification m2 is formed on the screen. The focal length of the lens is: (A) (C)
x m2 m1 x
m1m2
(B)
x m1 m2
(D)
None of these
Solution:
(A) In first case, 1+m1 =
q f
In the second case And
q 1 1 1 and =m1 p p q f
q x m2 p
…(1) 1 1 1 q x p f q x m2 = f
…(2)
(1) and (2) m2 – m1 = x/f 6.
f=
x m2 m1
If incident ray MP and reflected ray QN are parallel to each other, find the angle between the mirrors. Q
P i2
i i1 1
i2
M
N
(A) 60 (B) 45 (C) 90 (D) 180 Solution: (C) The deviation produced by first mirror = 180 – 2i1 and the deviation produced by second mirror = 180 – 2i2. The total deviation is 180 as MP and QN are anti parallel So, 180 = 360 – 2(i1 + i2) i1 + i2 = = 90 7.
A 16 cm long image of an object is formed by a convex lens on a screen placed normal to its principal axis. On moving the lens towards the screen, without changing the positions of the object and the screen, a 9 cm long image is formed again on the screen. The size of the object is: (A) 9 cm (B) 11 cm (C) 12 cm (D) 13 cm. Solution: (C) y = y1 ´ y 2
= 16 ´ 9 = 4 ´ 3 = 12cm. 8.
A small mirror of area ‘A’ and mass ‘m’ is suspended in a vertical plane by means of a weightless string. A beam of light of intensity ‘I’ falls normally on the mirror and the string is deflected form the vertical through a very small angle ‘’. Assuming the mirror to be perfectly reflecting, obtain an expression for .
2I/ C
mg
(A) (C)
2IA c mg 3I c mg
(B)
I c mg
(D)
none of the above
Solution: (A) When the light is incident on the mirror is totally reflected and there is no loss of energy. Change in momentum = 2 incident momentum Energy Intensity = Time Area Intensity Momentum = Velocity of light Force Rate of change in momentum Pressure, P = = Area A 2IA = (where c = speed of light) c P A 2IA = = tan = mg c mg 9. If x and y be the distances of the object and image formed by a concave mirror from its focus and f be the focal length then (A) xf = y2 (B) xy = f2 (C) x/y = f (D) x/y = f2 2 Solution: (B) According to Newton's formula xy = f . Note that m = or 10.
f y x x
f f v f u f
xy = f2
An observer can see, through a pin-hole, the top end of a thin rod of height h, placed as shown in the figure. The beaker height is 3h and its radius is h. When the beaker is filled with a liquid up to a height 2h, he can see the lower end of the rod. Then the refractive index of the liquid is:
3h
h 2h
(A) 5/2 Solution: (B) sin i 1 = sin r m
(B)
5/2
(C)
3/2
(D)
3 / 2.
Eye r
h
x
2h-x 2h
x =
1 m
...(i)
=
x h
...(ii)
x 2 + 4h2 2h - x
(2h - x )2 + h2 2h - x
sin r =
2
(2h - x)
+ h2
5 2 11. The sun (diameter d) subtends an angle radians at the pole of a concave mirror of focal length f. What is the diameter of the image of the sun formed by the mirror? (A) f (B) f/2 (C) f2 (D) f2 B Solution: (A) Since the sun is at a very large distance, u is very large and so (I/u) is practically zero. Using (i) and (ii), m=
So,
1 1 0 v f
i.e., v = f
A
i.e., the image of sun will be formed at the focus and will be real, inverted and diminished. Now as the rays from the sun subtend an angle radians at the pole, then in accordance with figure. = i.e. 12.
A B d FP f
A
C F B
M
P M
[where d is the diameter of the image of the sun]
d = f
A ray of light is incident on the left vertical face of a glass cube of refractive index 2, as shown in figure. The plane of incident is the plane of the page, and the cube is surrounded by liquid (1). What is the largest angle of incidence 1 for which total internal reflection occurs at the top surface? 1
B A 1
2
c
c
2
2
(A)
sin1 =
μ2 -1 μ1
2
(B)
sin1 =
μ2 +1 μ1
2
(C) Solution:
μ sin1 = 1 +1 (D) sin1 = μ2 (A) Consider A Point Applying Snell’s law, 1sin1 = 2sin2 …(1) But 2 = 90 – c cos2 = sinc = 1 …(2) 2 Elimination of 2 between (1) and (2), we get
2
μ1 -1 μ2
2
sin1 = 2 1 1 13. A ray incident at a point as an angle of incidence of 60° enters a glass sphere of R.I. n = 3 and is reflected and refracted at the farther surface of the sphere. The angle between the reflected and refracted rays at this surface is: (A) 50° (B) 60° (C) 90° (D) 40° P Solution: (C) Refraction at P. 600 Sin 600 3 Sin r1
Sinr1
1 r1= 30° 2
r1
Since r2 = r1 r2 = 30° Refraction at Q
r2 r’2
Q
i2
Sin r2 1 Sin i2 3
Putting R2= 30° we obtain i2 = 60° Reflection at Q r2 r2 30o
180 0 (r2 i2 )
= 180°- (30°+60°) = 90° 14.
What is the number of images of an object formed by two mirrors if the angle between the mirrors is in between 2/4 and 2/5? (A) 10 (B) 8 (C) 6 (D) 4 2 2 Solution: (D) As we know, if angle between two mirrors is between and then n n 1 the no. of images formed is ‘n’ 2 2 So here in the question it is and 4 4 1 no. of images will be 4. If one face of a prism of prism angle 30 and = 2 is silvered, the incident ray retraces its initial path. What is the angle of incident? (A) 90 (B) 180 (C) 45 (D) 55 Solution: (C) As in AED 30 + 90 + D = 180
15.
A 30
i
D
E
r
2
C
B
D = 60 Now as by construction D + r = 90 r = 90 – 60 = 30 Applying Snell’s law at surface AC 1 1 sin i = ( 2 )sin30 = 2 = 2 2 1 i = sin–1 = 45 2
16.
Two plane mirrors are inclined to each other at angle . A ray of light is reflected first at one mirror and then at the other. Find the total deviation of the ray (A) 360 – 2 (B) 360 + 2 (C) 180 – 2 (D) 180 + 2
Solution (A) A ray AB is incident on mirror OM1 at angle and is reflected along BC suffering a deviation 1 = FBC A
M1 E d a
B F
d1
a
N1 N2 b
q O
b d2
C
M2 G
The ray BC falls on mirror OM2 at an angle of incidence and is reflected along CD suffering another deviation 2 = GCD The total deviation is = 1 + 2 It is clear from the diagram that 1 = 180 - 2 and 2 = 180 - 2 = 1 + 2 = 360 –2( + ) Now, in triangle OBC, OBC + BCO + BOC = 180 or (90 – ) + (90 – ) + = 180 or += Hence = 360 – 2 Which is independent of the angle of incidence at the first mirror. 17.
A layer of oil 3cm thick is floating on a layer of coloured water 5cm thick. Refractive index of coloured water is 5/3 and the apparent depth of the two liquids appears to be 36/7cm. Find the refractive index of oil. (A) 1.6 (B) 1.4 (C) 1.9 (D) 0.9
Solution (B)
24 cm
I1
A
I2
O
10 cm x
Apparent depth (A) =
\ \ \
B 6 cm
m
24 - x
t1 t 2 + m1 m2
36 5 3 = + 7 5 / 3 m2 3 36 15 = - 3= m2 7 7 7 m2 = = 1.4 5
18.
A glass sphere ( = 1.5) of radius 8 cm is placed in sunlight. Where is the image of the sun formed by the light passing through the sphere? When refraction by first surface of sphere (A) 24 cm (B) 30 cm (C) 34 cm (D) 50 cm
19.
A glass sphere ( = 1.5) of radius 8 cm is placed in sunlight. Where is the image of the sun formed by the light passing through the sphere? When refraction by second surface of sphere (A) (C)
4 cm 15 cm
(B) (D)
6 cm 50 cm
Solution (18(A); 19(A)) As the sun is very far away, the incident ray can be assumed as coming from infinity. For refraction at first surface: normal normal
I2
I1
m1 m2 m2 - m1 + = u v R 1 1.5 1.5 - 1 Þ + = v1 ¥ +8 Þ v1 = 24 cm. . . . [18 (A)] The rays try to converge at 1 and hence for the second surface it is a virtual object. For second surface: m1 m2 m2 - m1 + = u v R u = –(24 – 16) = –8 cm 1.5 1 1 - 1.5 Þ + = - 8 v2 -8
Using
[19 (A)] Þ v 2 = 4 cm. Hence the final image is formed at 2 which is 4 cm from the other surface of the sphere. 20.
If the critical angle is 4842’ when the media concerned are air and water and 3647’ when they are in air and glass, what is it when they are in water and glass. (A) 5252' (B) 6252' (C) 6262' (D) 7070' Solution (A) We know that 1 sinC = mw 1 1 mw = = sinC sin48°42' 1 Similarly mg = sin36°47' Now for glass and water media sin C = refractive index of water w.r.t. glass mw 1/ sin 48°42' sin36°47' = = or C = 52°52' mg sin36°47' sin 48°42' 21.
Find the size of the image formed in the situation shown in figure. m= 1 1 cm
m= 1.33
C
20 cm
O 40 cm
(A) (C)
0.9 cm 1.2 cm
(B) (D)
0.6 cm 1.4 cm
Solution (B) Here u = –40 cm, R = –20 cm = 1, 2 = 1.33 m= 1 1 cm
C
20 cm
O 40 cm
We have, m2 m1 m2 - m1 = v u R 1.33 1 0.33 =v 40 20 v = –32 cm. The magnification is m = h2 32 =Þ 1 1.33 (- 40) The image is erect.
h2 m1v = h1 m2u h2 = 0.6 cm
m= 1.33
22.
A convergent lens of 6 diopters is combined with a diverging lens of -2 diopters. Find the power of combination? (A) 4 diopter (B) 6 diopter (C) 8 diopter (D) 10 diopter
Solution (A) Here P1 = 6 diopters, P2 = -2 diopters Using the formula P = P1 + P2 = 6 – 2 = 4 diopters 23.
Select a graph between ‘v’ and ‘u’ for a concave mirror. v
v
(A)
(B) O
O
u
u
v
v
(C)
(D) O
O
u
u
Solution (A) For a concave mirror 1 1 1 u v f Curve would be a hyperbola for spherical mirror. v
O
24.
u
A prism is made up of flint glass whose dispersive power is 0.053. Find the angle of dispersion if the mean refractive index of flint glass is 1.68 and the refracting angle of prism is 3. (A) 20.08 (B) 10.08 (C) 0.208 (D) 0.108
Solution (D) As the refracting angle is small deviation = ( – 1) A angle of dispersion = Dv - Dr
= (mv - 1) A - (mr - 1) A = (mv - mr ) A æ m - mr ö =ç v A (mmean - 1) è mmean - 1÷ ø
= w A (mmean - 1) = (0.053) ´ 3 ´ (1.68 - 1) = 0.108°
CMP:
The following figure shows a simple version of a zoom f1 r0
Q
f2=-|f2| r'0
d f
I'
s'2
The converging lens has focal length f1’ and the diverging lens has focal length f2 = –|f2|. The two lenses are separated by a variable distance d that is always less than f1’ also the magnitude of the focal length of the diverging lens satisfies the inequality |f2| > (f1 –d). If the rays that emerge from the diverging lens and reach the final image point are extended backward to the left to the diverging lens, they will eventually expand to the original radius r0 at the same point Q. To determine the effective focal length of the combination lens, consider a bundle of parallel rays of radius r0 entering the converging lens. 25.
At the point where rays enters the diverging lens, the radius of the ray decreases to: f -d f -d (B) r0 = 1 (A) r’0 = 1 r0 r’0 f1 f1 f -f (C) r’0 = 1 2 r0 (D) r’0 = r0 f1 Solution: (A)
bundle
A r0
r'0 B
d f
D C E
From the similar triangles ABC and DEC r0 r' (f d)r0 0 so r’0 1 f1 f1 d f1 26.
To the right of the diverging lens the final image I’ is formed at a distance s’2. (f1 - f2 )d (f1 - d) | f2 | (A) (B) f2 - f1 + d | f2 | -f1 + d f1 - f2 + d (d - f1 )f2 (C) (D) f1 - f' 2 f1 - f2 - d Solution: (B) The image at focal point of the first lens, a distance f 1 to the right of first lens, serves as the object for the second lens. The image is at distance f 1–d to the second lens so s2 f2 (f d)f2 s2 = –(f1 –d) = d – f1 so s’2 = 1 s2 f2 | f2 | f1 d 27. Find the effective focal length? f1 | f2 | f1f2 (B) (A) | f2 | -f1 + d f1 - f2 - d
(C) Solution:
f1 f2 - f1 + d (A)
(D)
r'0 r'0
f2 f1 + f2 + d
I' s'2
From the similar triangles in the sketch r0 r '0 f1(f2 ) so f = | f2 | f1 d f s'2 If f1 = 12 cm, f2 = – 18 cm find the maximum focal length of the combination? (A) 36 cm (B) 34 cm (C) 32 cm (D) 30 cm Solution: (A) Put the numerical values in the expression in the earlier question and we get 36 cm.
28.
29.
In continuation of fourth questions determine the minimum focal length of the combination (A) 21.6 cm (B) 24 cm (C) 30 cm (D) 35 cm Solution: (A) Put the numerical values in the above expression which we found and we will get 21.6 cm. 30.
What value of d gives f = 30 cm? (A) 12 cm (B) 10 cm 216 Solution: (C) f = 30 cm then 30 = 6d 6 + d = 7.2 hence d = 1.2 cm.
(C)
1.2 cm
(D)
14 cm
For test 1.
An achromatic combination of two prisms - one of quartz and other of flint glass is to be formed. If the refracting angle of the quartz prism is 4, find the refracting angle of the flint glass prism. Also find the net mean deviation produced by the combination. Given that for the flint glass : v = 1.694, r = 1.65 and for the quartz glass : v = 1.557, r = 1.542. (A) –1.282 (B) +1.282 (C) –5.282 (D) –8.282
Solution (A) For an achromatic combination, net dispersion = 0 dispersion by one prism = dispersion by other prism. Þ Þ m1v - m1r A1 = (m2v - m2r ) A 2
(
)
Þ
(1.694 - 1.65) A1 = (1.557 - 1.542) ´
Þ
A1 = 1.364°
4
Dm1 - Dm2 = (m1m - 1) A1 - (m2m - 1) A2 æm + m1r ö æ1.557 + 1.542 ö = ç 1v - 1÷ ´ 1.364 - ç - 1÷ ´ 4 è ø è 2 ø 2
= - 1.641° ´ 2 = - 1.282 Net mean deviation = Dm1 - Dm2 = - 1.282 2.
Find the dispersion produced by a thin prism of 18 having refractive index for red light = 1.56 and for violet light = 1.68. (A) 4.16 (B) 6.16 (C) 2.16 (D) 8.16 Solution (C) We know that dispersion produced by a thin prism q = (mv - mR ) A Here mv = 1.68, mR = 1.56 and A = 18°. q = (1.68 - 1.56) ´ 18° = 2.16°.
3.
Calculate the dispersive power for crown glass μv = 1.523, and μR = 1.5145 (A) 0.1639 (B) 0.01639 (C) 1.639 (D) 2.639 Solution (B) Here mv = 1.523 and mR = 1.5145 Mean refractive index m=
from
the
given
data
1.523 - 1.5145 = 0.01639 (1.51875 - 1)
Dispersive power is given by, m - mR 1.523 - 1.5145 w= v = = 0.01639 (m- 1) (1.51875 - 1) 4.
Two plane mirrors are inclined to each other such that a ray of light incident on the first mirror and parallel to the second is reflected from the second mirror parallel to the first mirror. Determine the angle between the two mirrors. (A) 60 (B) 30 (C) 90 (D) 180 Solution: (A) Let be the angle between the two mirrors OM1 and OM2. The incident ray AB is parallel to mirror OM2 and strikes the mirror OM1 at an angle of incidence equal to . From figure we have M1BA = OBC = M1OM2 = M1 D
E
q B
A
a q
d
N1
a
N2 q
q
q M2
O
C
Similarly for reflection at mirror OM2, we have M2CD = BCO = M2OM1 = Now in triangle OBC, 3 = 180, therefore = 60
5.
A small filament is at the centre of a hollow glass sphere of inner and outer radii 8 cm and 9 cm respectively. The refractive index of glass is 1.50. Calculate the position of the image of the filament when viewed from outside the sphere. (A) 9 cm (B) – 9 cm (C) – 19 cm (D) +19 cm Solution: (B) For refraction at the first surface, u = –8 cm, R1 = –8 cm, 1 = 1, 2 = 1.5 m2 m1 m2 - m1 = v u R1 1.5 1 0.5 + = v' 8 - 8 v' = –8 cm OA = 8 cm OB = 9 cm
m= 1.5
Glass
O
A
B Air
It means due to the first surface the image is formed at the centre. For the second surface u = –9 cm, 1 = 1.5, R2 = –9 cm m2 m1 m2 - m1 = v u R2 1 1.9 1 - 1.5 + = -9 v 5 v = –9 cm Thus, the final image is formed at the centre of the sphere. 6.
A thin rod of length f/3 is placed along the optical axis of a concave mirror of focal length f. Such that its image which is real and elongated just touches the rod. Calculate the magnification. (A) 3/2 (B) 5/2 (C) 1/2 (D) 2/3 Solution: (A) Let be the length of the image. 1 mf Þ 1= Then, m = f 3 3
l
rod
image
f/3
u’ u v
Also image of one end coincides with the object, Þ u' = 2f f f 5f u' = u + Þ 2f - = 3 3 3
æ f mf ö v = - çu + + ÷ . Putting in mirror formula, è 3 3 ø 1 1 1 + = f mf u f + u+ 3 3
Þ
1 2 = m + 6 15
Þ
Þ
3 3 1 + = 5f + f + mf 5f f
m=
3 2
7.
An object of length 2.5 cm is placed at a 1.5 f from a concave mirror where f is the magnitude of the focal length of the mirror. The length of the object is perpendicular to the principal axis. Find the length of the image. Is the image erect or inverted? (A) – 8 cm (B) – 10 cm (C) – 5 cm (D) – 12 cm Solution: (C) f O
F 1.5 f
The focal length F = –f and u = –1.5 f, we have, 1 1 1 1 1 1 + = or + = u v F - 1.5f v f 1 1 1 1 = - =v 1.5f f 3f v = –3f 3f Now, m = - v / u = =-2 1.5f h2 or =-2 or h2 = - 2h1 = - 5 cm h1 The image is 5 cm long. The minus sign shows that it is inverted. 8.
Select a graph for concave mirror between 1/v and 1/u. 1/v
1/v
(A)
(B) O
O
1/u
1/v
1/u
1/v
(C)
(D) O
Solution: (C) For a mirror
1/u
O
1/u
1 1 1 u v f 1 1 let y = , x = v u 1 x+y= , f
y = – x + 1/f
This is the equation of a straight line with slope = (–1) and intercept =
1 f
Hence the graph would be 1/v
1/f
O
9.
1/f
1/u
Show with the help of ray diagrams that a right angled isosceles prism can produce a deviation of 90° If the refractive index is greater than 2 . 45 (A) (B) 45 45
45 45 45 45
45
(C)
(D)
None of these.
45
45
Solution: (A) m> 2 1 1 < m 2 Þ sinC < sin45° Þ C < 45° Hence a ray striking a glass-air surface from inside the prism will be reflected back if angle of incidence is not less than 45°. 90° Deviation is produced if the ray strikes on one of the sides normally as shown. Here, 1 = 45° and hence reflection takes place. 10.
Convex lens of 10 cm focal length is combined with a concave lens of 6 cm focal length. Find the focal length of the combination. (A) 15 cm (B) – 15 cm (C) 30 cm (D) – 30 cm Solution: (B) Here f1 = 10 cm, f2 = -6 cm, F = ?
Use the formula \
1 1 1 1 1 1 = + = - =F f1 f2 10 6 15
F = - 15 cm
11.
Focal length of a convex lens of refractive index 1.5 is 2 cm. Focal length of lens when immersed in a liquid of refractive index 1.25 will be (A) 5 cm (B) 7.5 cm (C) 10 cm (D) 2.5 cm Solution (A) ( g 1) fp 5 fl xfa fa g 2 1 l 5 x 2cm = 5 cm. 2
12.
A double convex lens of focal length 20 cm is made of glass of refractive index 3/2. When placed completely in water (a 4 / 3) , its focal length will be (A) 80 cm (B) 15 cm (C) 17.7 cm (D) 22.5 cm Solution (A) (g 1) fw f w 4 fa g fa 1 w fw = 4 x fa = 4 x 20 = 80 cms.
13.
Find the minimum angle of deviation for a prism with angle A = 60° and = 1.5. (A) 37 (B) 40 (C) 55 (D) 65 Solution: (A) Minimum Deviation The angle of minimum deviation occurs when i = e and r1 = r2 and is given by: æA + Dm ö sin ç è 2 ÷ ø m= A sin 2 Aö æ Dm = 2sin- 1 çmsin ÷ - A Þ è 2ø Substituting = 1.5 and A = 60°, Dm = 2sin- 1 (0.75) - 60° = 37°
14.
Find the maximum angle of deviation for a prism with angle A = 60° and = 1.5. (A) 50 (B) 58 (C) 64 (D) 60 Solution: (B) Maximum Deviation
i 90
r1 r2
e
The deviation is maximum when i = 90° or e = 90° that is at grazing incidence or grazing emergence. Let i = 90° Þ r1 = C = sin- 1 (1/ m) Þ
r1 = sin- 1 (2 / 3 ) = 42°
Þ
r2 = A - r1 = 60° - 42° = 18°
sinr2 1 = sin e m Sin e = u sin r2 = 1.5 sin 18° Þ sine = 0.463 Þ e = 28° Deviation = Dmax = i + e – A = 90° + 28° – 60° = 58° u sin g
CMP :
15.
Inside a substance such as glass or water, light travels more slowly than it does in a vacuum. If c denotes the speed of light in a vacuum and v c denotes its speed through some other substance, then v = n Where n is a constant called the index of refraction. To almost exact approximation, a substance’s index of refraction does not depend the wavelength of light. For instance, when red and blue light waves enter water, they both slow down by about the same amount. More precise measurements, however, reveal that n varies with wavelength. Table 1 presents some indices of refraction of Cutson glass, for different wavelengths of visible light. A nanometer (nm) is 10–9 meters. In a vacuum, light travesl at c = 3.0 108 m/s. Indices of refraction of Cutson glass (Table 1) approximate wavelength in vacuum (nm) n yellow 580 1.500 yellow orange 600 1.498 orange 620 1.496 orange red 640 1.494
Inside Cutson glass (A) Orange light travels faster than yellow light (B) Yellow light travels faster than orange light (C) Orange and yellow light ravels equally fast (D) We cannot determine which colour of light travels faster Solution: (A) The velocity of light is inversely proportional to the index of refraction c v = , So, the lower the n, the higher the v. According to table, Cutson glass has a n slightly lower n for orange light than it has for yellow light. Therefore, inside Cutson glass, orange light travels slightly faster.
16.
For blue green of wavelength 520 nm, the index of refraction of Cutson glass is probably closest to: (A) 1.49 (B) 1.50 (C) 1.51 (D) 1.52 Solution: (C) Table suggests a linear relationship between wavelength and index of refraction. Starting with yellow light, we can extrapolate with linear relationship to blue green light. As compared to yellow light waves, blue green light waves are 60 nm shorter (520 nm vs. 580 nm). By reading table from bottom to top, we see that each 20 nm decreases in wavelength corresponds to a 0.002 increase in n. So, since yellow and blue green differ in wavelength by 60 nm, the corresponding n’s should differ by 0.006. nblue green = nyellow + 0.006 = 1.500 + 0.006 = 1.506 which is closer to 1.51 than it is to 1.50. You can also address this question by extrapolating table. blue green 520 1.506 540 1.504 560 1.502 yellow 580 1.500 yellow orange 600 1.498 orange 620 1.496 orange red 640 1.494 17.
For visible light, which graph bes expresses the index of refraction of glass as a function of the frequency of the light?
frequency
frequency
frequency
Cutson
frequency
(A) (B) (C) (D) Solution: (A) As just discussed, index of refraction decreases linearly with wavelength Graph b show a linear relationship. But the problem wants the relationship between frequency and n, not between wavelength and n. Since higher wavelengths correspond to lower frequencies, and vica versa, n increasing with frequency. To clarify this reasoning, we should review the relationship between wavelength and frequency. For all waves, not just light waves, v = f, where v denotes velocity, denotes wavelength, and f denotes frequency. In a vacuum, all light waves travel at c speed c. So, frequency and wavelength are inversely proportional. f = . Higher wavelengths correspond to lower frequencies, and vice versa. Therefore, since n decreases at higher wavelengths, it increases at higher frequencies. c Again, extending table 1 can help us spot this relationship. Using f = , we can find the frequency of different colors of light. For instance c 3.0 108 m / s fyellow = = 5.17 1014 Hz 580 10 9 m yellow fyellow–orange =
c yellow orange
3.0 108 m / s = 5.0 1014 Hz 9 600 10 m
18.
Which of the following phenomena happens because n varies with wavelength? (A) A lens focuses light (B) A prism brakes sunlight into different colours (C) Total internal reflection ensures that light travels down a fiber optic cable (D) Light rays entering a pond change direction at the pond’s surface Solution: (B) 19.
The time required for light to pass through a glass slab of 2 mm thickness is ( glass 1.5)
10–8 s (B) 10–11 s (D) 3 d d d. 2x10 x1.5 Solution (C) t sec 1011 sec v c/ c 3x108 (A) (C)
10–6 s None of these
20.
A spherical surface of curvature R separates air (=1.5). The centre of curvature is in the glass. A point object P placed in air is found to have a real image Q in the glass. The line PQ cuts the surface at a point O and PO = OQ. The distance PO is equal to (A) 2R (B) 5R (C) 3R (D) 1.5R 2 1 1 Solution (B) As v u R 1.5 1 1.5 1 2.5R x 5R or x x R 0.5
21.
If the tube length of astronomical telescope is 105 cm and magnifying power is 20 for normal setting, calculate the focal length of the objective (A) 100 cm (B) 10 cm (C) 20 cm (D) 25 cm f Solution (A) L = f0 + fe = 105 cm and M = 0 20 fe f0 = 100 cm and fe = 5 cm. 22.
Time of exposure for a photographic print is 10 seconds, when a lamp of 50 cd is placed at 1 m from it. Then another lamp of luminous intensity I is used, and is kept at 2 m from it. If the time of exposure now is 20 s, the value of I (in cd) is (A) 25 (B) 100 (C) 200 (D) 20 8 c 3x10 3x108 Hz Solution (B) v 1 8 c 3x10 v' 3x107 Hz. ' 10
23.
How does refractive index () of a material vary with respect to wavelength () ? A and B are constants B A 2 (A) = A + B2 (B)
(C)
= A + B
Solution (B) Cauchy's formula is A
(D)
A
B
B 2
24.
Time required for making a print at a distance of 40 cm from a 60 watt lamp is 12.8 second. If the distance is decreased to 25 cm, then time required in making the similar print will be (A) 15 sec (B) 10 sec (C) 5 sec (D) remains some. Solution (C) If prints are similar, I I E1t1 E2 t 2 12 t1 22 xt 2 r1 r2 But I1 = I2 r22 t 2 2 xt1 t 2 5 sec. r1 25.
An achromatic convergent doublet has power of +2D. If power of convex lens is + 5D, then ratio of the dispersive powers of a convergent and divergent lenses will be (A) 3:5 (B) 2:5 (C) 2:3 (D) None of these Solution (A) P = +2D P1 5D P2 3D
As
P2 f1 w w 3 3 1 1 P1 f2 w2 5 w2 5
26.
If a convex lens having focal length f and refractive index 1.5 is immersed in water then new focal length will be (w = 4/3) (A) f (B) 1.5 f 4 f (D) 4f (C) 3 1 1 1 Solution (D) (g 1) . . . (1) f R1 R2 1 g w 1 1 . . . (2) fw w R1 R2 fw 4f. 27.
Radii of curvature of a convex lens are 20 cm and 30 cm and its focal length if 24 cm. Refractive index of the lens will be (A) 1.33 (B) 1.5 (C) 1.71 (D) None of these Solution (B) 1 1 1 As ( 1) f R1 R2
1 1 ( 1) 1 ( 1) 12 1.5 24 20 30
28.
Both radii of curvature of a convex lens are 20 cm and refractive index of the material of the lens is 1.5. Rays parallel to the axis of the lens will converge at (A) 10 cm (B) 20 cm (C) 30 cm (D) 40 cm Solution (B) 1 1 1 As ( 1) f R1 R2
1 1 0.5x2 1 1 (1.5 1) f 20 20 20 20 or f=20 cm i.e., rays will converge at 20 cm from the optical centre of the lens.
29.
Eye piece of an astronomical telescope has focal length of 5 cm. If angular magnification in normal adjustment is 10, then dustance between eye piece and objective should be (A) 15 cm (B) 35 cm (C) 55 cm (D) 75 cm Solution (C) In normal adjustment, L = f 0 + fe f0 f0 and M or 10 f0 50 cm fe 5 L = 5 cm + 50 cm = 55 cm. 30.
Power of a convex lens is + 5D (g= 1.5). When this lens is immersed in a liquid of refractive index , it acts like a divergent lens of focal length 100 cm. Then refractive index of the liquid will be 5 5 (B) (A) 3 4 6 (C) (D) none of these 5 Solution (A) g 1 Pa a 5 5 Pl g 100 /100 1 l 5 l 3
Unit 20 - Wave Optics 1.
WAVE NATURE OF LIGHT
Huygens’ Wave Theory (i)
Each point on a wavefront acts as a source of new disturbance and emits its own set of spherical waves called secondary wavelets. The secondary wavelets travel in all directions with the velocity of light so long as they move in the same medium.
The envelope or the locus of these wavelets in the forward direction gives the position of new wavefront at any subsequent time. A surface on which the wave disturbance is in the same phase at all points is called a wavefront. Wave optics involves effects that depend on the wave nature of light. In fact, it is the results of interference and diffraction that prove that light behaves as a wave rather than a stream of particles (as Newton believed). Like other waves, light waves are also associated with a disturbance, which one consists of oscillating electric and magnetic field. The electric field associated with a plane wave propagating along the x-direction can be expressed in the form: E = E o[sin(t - kx + o)] where , k and o bearing their usual meanings. (ii)
Points to remember regarding Interference When two waves with amplitude A1 and A2 superimpose at a point, the amplitude of resultant wave is given by A=
A12 A 22 2A1A 2 cos
Where is the phase difference between the two waves at that point. Intensity (I) =
1 E02 . C = speed of light, E0 = electric field amplitude 20C
Intensity (I) = I1 + I2 + 2 I1I2 cos. Hence for I to be constant, must be constant. When changes randomly with time, the intensity = I1 + I2. When does not change with time, we get an intensity pattern and the sources are said to be coherent. Coherent sources have a constant phase relationship i.e. one that does not change with time. The intensity at a point becomes a maximum when = 2n (n = 0, 1, 2, . .) and there is constructive interference. If = (2n 1) there is destructive interference. (Here n is a non-negative integer) Determination of Phase Difference The phase difference between two waves at a point will depend upon (a) the difference in path lengths of the two waves from their respective sources. (b) the refractive index of the medium (c) initial phase difference, between the source, if any. (d) Reflections, if any, in the path followed by waves.
In case of light waves, the phase difference on account of path difference Optical path difference μ(Geometri cal path difference ) 2 = 2 π λ
=
where is the wavelength in free space. In case of reflection, the reflected disturbance differs in phase by with respect to the incident one if the wave is incident on a denser medium from a rarer medium. No such change of phase occurs when the wave is reflected in going from a denser medium to a rarer medium.
2.
YOUNG’S DOUBLE SLIT EXPERIMENT
A train of plane light waves is incident on a P
barrier containing two narrow slits separated by S1
a distance 'd'. The widths of the slits are small
d
compared with wavelength of the light used, so
O
that interference occurs in the region where the
S2 Screen
light from S1 overlaps that from S2. A series of Double slit
alternately bright and dark bands can be
D
observed on a screen placed in this region of overlap. The variation in light intensity along the screen near the centre O shown in the figure Now consider a point P on the screen. The phase difference between the waves at P is , where 2 = Po
O a bright fringe.
a dark fringe.
fringe width
(where Po is optical path difference, Po=Pg; Pg being the geometrical path difference.)
2 S2P S1P
(here = 1 in air)
As P
As, D >> d, S2P - S1P d sin sin tan( = y/D). [for very small ] Thus, =
S1 d
2 dy D
S2
For constructive interference, = 2n (n = 0, 1, 2...)
2 dy 2n D
y
Similarly for destructive interference,
O dsin D
y=n
D d
y = (2n 1)
D 2d
(n = 1, 2, .........)
Fringe Width W It is the separation of two consecutive maxima or two consecutive minima. Near the centre O [where is very small], W = yn+1 - yn [yn gives the position of nth maxima on screen] =
D d
Intensity Variation on Screen. If A and Io represent amplitude of each wave and the associated intensity on screen, then, the resultant intensity at a point on the screen corresponding to the angular position as in above figure, is given by
Two coherent sources
Intensity
Two Incoherent sources
4I0 One source 2I0 I0 -5
I = Io + Io + 2 I o2 cos,
-5
-4
when =
-2
-
0
+
+2
+3
+4
+5
2d sin = 4Io cos2 2
Illustration 1: A beam of light consisting of two wavelengths 6500 oA and 5200 oA is used to obtain interference fringes in YDE. The distance between the slits is 2.0 mm and the distance between the plane of the slits and the screen is 120 cm. (a) Find the distance of the third bright fringe on the screen from the central maxima for the wavelength 6500 oA. (b) What is the least distance from the central maxima where the bright fringes due to both the wavelengths coincide? Solution:
(i)
y 3 n.
D 3x1.2mx 6500x10 10 m 0.12 cm d 2x10 3 m
Let nth maxima of light with wavelength 6500 Å coincides with that of mth maxima of 5200Å. (ii)
mx6500 A o xD nx5200 A o xD m 5200 4 d d n 6500 5 Least distance = y 4 =
4x6500 x10 10 x1.2 4.D(6500 A o ) 0.16cm = d 2x10 3 m
Illustration 2: The intensity of the light coming from one of the slits in a Young's double slit experiment is double the intensity from the other slit. Find the ratio of the maximum intensity to the minimum intensity in the interference fringe pattern observed. Solution :
Imax l1 l2 Imin l1 l2
2
2
As I1 2I2
2 1 I max 34 Imin 2 1
Displacement of Fringes When a film of thickness 't' and refractive index '' is introduced in the path of one of the sources, then fringe shift occurs as the optical path difference changes. Optical path difference at 1)
P = S2P [S1P+ t t] = S2P S1P ( 1)t = y.d/D ( nth fringe is shifted by
Δy
, t
P
S1
S2
D( 1)t w ( 1)t d
Illustration 3: Monochromatic light of wavelength of 600 nm is used in a YDSE. One of the slits is covered by a transparent sheet of thickness 1.8 x 10 -5 m made of a material of refractive index 1.6. How many fringes will shift due to the introduction of the sheet? Solution:
As derived earlier, the total fringe shift =
w ( 1)t .
As each fringe width = w, The number of fringes that will shift =
total fringe shift fringe width
w ( 1)t ( 1)t (1.6 1)x1.8x10 5 m 18 w 600x10 9 m
Illustration 4: In the YDSE conducted with white S1 light (4000Å-7000Å), consider two points P1 d=1cm and P2 on the screen at y1=0.2mm and S2 y2=1.6mm, respectively. Determine the wavelengths which form maxima at these points Solution: The optical path difference at P1 is p1 =
P2 y2 = 1.6mm P1
y1 = 0.2mm
4m
dy1 10 0.2 5 10 4 mm 5000 A D 4000
In the visible range 4000 - 7000Å n1 =
5000 1.25 4000
and
n2
5000 0.714 7000
The only integer between 0.714 and 1.25 is 1 The wavelength which forms maxima at P is For the point P2, Here n1 =
40000 10 4000
= 5000Å
dy 2 10 p2 = 1.6 4 10 3 mm 40000 A D 4000 40000 and n2 5.71 7000
The integers between 5.71 and 10 are 6, 7, 8, 9 and 10
The wavelengths which form maxima at P2 are 1 = 4000Å for n = 10 2 = 4444Å for n=9
3 = 5000Å for n=8 4 = 5714Å for n=7 5 = 6666Å for n=6 Illustration 5: A transparent paper ( = 1.45) of thickness 0.02 mm is pasted on one of the slits of a Young’s double slit experiment which uses monochromatic light of wavelength 620 nm. How many fringes will cross through the centre if the paper is removed? Solution: Due to pasting the fringes shift which will restore its position after removal. Path difference will be S2P S1P t + t
= t ( 1) +
yd D
for a bright fringe x = n t( 1) + yd/D = n. y = (n t( 1)).
D d
Again after removing y = n
D d
y y = t( 1)
D d
No of fringes shifted will be
y y D D t t( 1) ( D) d d N 3 0.02 10 (0.45 )
n=
=
3.
620 10 9
= 14.5
INTERFERENCE BY THIN FILM
A ray of light incident on a thin film of thickness ‘t’ gets partially reflected and refracted at A at surface I and thereafter it gets reflected and refracted at B of surface II. The rays after emerging in the first medium interfere. Now the inference will depend upon the path difference between AD and ABC, as beyond CD path difference is zero. x = (AB + BC) – AD = (AE + EB + BL) – AD
G
= AE–AD+(EB+BC) (I)
i
x = AE – AD + EF
C
i
In LBF = BC = BF
A r x
r
E
In ECF EF = CP cos r = 2 t cos r In ADC and AEC
Surface-II B
AE AD sini and sinr AC AC sini AD μ sinr AE
AE = AD (V)
Putting (IV) and (V) in (III) we obtain
Surface-I
r
r
F
x = 2 t cosr IAD as In reflected at a denser medium it suffers an additional path difference /2 Total path difference the taken place is 2 t cosr – λ / 2 For constructive interference 2 t cosr – / 2 = n. 2 t cosr = n + /2 = (2n + 1)/2 maxima For normal incidence r = 0
2 t = (2n + 1) / 2 = n
2 t cosr – / 2 = (2n – 1) / 2 For normal incidence
2 t cosr = n
2t=n
Illustration 6: White light is incident normally on a glass plate of thickness 0.50 -6 x 10 m and index of refraction 1.50. Which wavelengths in the visible region (400 nm - 700 nm) are strongly reflected by the plate? Solution :
The light of wavelength is strongly reflected if the light rays reflected are interfering constructively. As we know the condition for constructive interference
1 2
2t = n . Here 2t = 2 x 1.5 x (0.5 x 10-6)m = 1.5 x 10-6 m.
1 2
Putting = 400 nm, 1.5 x 10-6 = n 400 x 10-9 n = 3.25
Similarly, by putting = 700 nm.
1 2
1.5 x 10-6 = n (700 x 10-9)
n = 1.66 Thus, within 400 nm to 700 nm, for integral values of n = 2 and 3. Now,
4.
4t 600nm and 429 nm are strongly reflected. 2n 1
DIFFRACTION
5. Diffraction is the bending or spreading of waves that encounter an object ( a barrier or an opening) in their path. 6. In Fresnel class of diffraction, the source and/or screen are at a finite distance from the aperture. 7. In Fraunhoffer class of diffraction, the source and screen are at infinite distance from the diffracting aperture. Fraunhoffer is a special case of Fresnel diffraction. Single Slit Fraunhoffer Diffraction In order to find the intensity at point P on the screen as shown in the figure the slit of width 'a' is divided into N parallel strips of width x. Each strip then acts as a radiator of Huygen's wavelets and produces a characteristic wave disturbance at P, whose position on the screen for a particular arrangement of apparatus can be described by the angle .
The amplitudes Eo of the wave disturbances at P from the various strips may be taken as equal if is not too large. The intensity is proportional to the square of the amplitude. If Im represents the intensity at O, its value at P is
P
O
a
2
sin l = lm ;
where
A minimum occurs when, sin = 0 and 0, so = n, n = 1, 2, 3...
f
a sin 2
I
a sin n a sin n
Angular width of central maxima of diffraction pattern = 21 = 2 sin-1(/a) /a 2/a 0 3/a -3/a -2/a -/a [ 1 gives the angular position of first minima] The concept of diffraction is also useful in deciding the resolving power of optical instruments. Illustration 7: Light of wavelength 6 10-5cm falls on a screen at a distance of 100 cm from a narrow slit. Find the width of the slit if the first minima lies 1mm on either side of the central maximum. Solution:
Here n = 1, = 6 10-5 cm. Distance of screen from slit = 100 cm. Distance of first minimum from central maxima = 0.1 cm. sin = 1 =
Dis tan ceof 1st min ima from the centralmax ima Dis tan ceof the screen from the slit
0.1 1 100 1000
We know that asin = n a=
= 0.06 cm. 1
In YDSE if the source consists of two wavelengths 1 = 4000Å and 2 = 4002Å . Find the distance from the centre where the fringes disappear, if d=1cm ; D=1 m . The fringes disappear when the maxima of 1 fall over the minima of 2. That is
Illustration 8:
Solution:
p p 1 1 2 2
or
p=
Where p is the optical path difference at that point. 1 2 2 2 1
Here 1 = 4000Å, 2 = 4002Å
p = 0.04 cm
In YDSE,
p = dy/D
1000 0.4 40mm D y= p d 10
Illustration 9: A beam of light consisting of two wavelengths 6500A°and 5200A° is used to obtain interference fringes in a Young’s double slit experiment (i) Find the distance of the third fringe on the screen from the central maximum for the wavelength 6500A°. (ii) What is the least distance from the central maximum where the bright fringes due to both wavelength coincide? (iii)The distance between the slits is 2mm and the distance between the plane of the slits and screen is 120cm. What is the fringe width for = 6500A°? Solution:
(i)
The width of the fringe
D d
Then distance of the third fringe 3w =
3D 3 120 6500 10 8 d 0.2
= 0.117 cm (ii)
Let mth and nth bright fringe of the wavelength coincide. Now position mth bright fringe is ym = n1 and
D d
yn = n2
D d
m 1 5200 4 n 2 6500 5
Now (iii)
Fringe width w =
D 6500 10 10 1.2 d 2 10 3
= 3900 107m = 0.039 cm
Illustration 10: In YDSE light of two wavelengths of 700 nm and 500 nm. If D/d = 10 3 find the minimum distance from central maxima where the maxima of two wavelength coincide again. n1 1 n 2 2 Solution: n1 7 n2 5 D1 y n1 d y 7 10 3 500 10 9
35 10 4
y= 3.5 mm POLARISATION Polarisation of two interfering wave must be same state of polarisation or two source of light should be unpolarised.
8.
BREWSTER LAW
According to this law when unpolarised light is incident at polarising angle (i) on an interface separating a rarer medium from a denser medium, of refractive index as shown in Fig., below such that = tan i then light reflected in the rarer medium is completely polarised. Reflected and refractive rays are perpendicular to each other. en ay tr r
R e (P flec ol te ar d is ra ed y )
d ci In i
Rarer Medium
R ra ef ed ct ra y
9.
Denser medium of refractive index m
REDUCTION IN INTENSITY
Intensity of polarised light is 50% of that of the unpolarised light, i.e., Ip = Iu / 2 where I p = Intensity of polarised light and
Iu = Intensity of unpolarised light.
10. ASSIGNMENT 1.
In Young's double slit experiment the light emited from source has l = 6.5 × 10 –7 m and the distance between the two slits is 1 mm. Distance between the screen and slit is 1 metre. Distance between third dark and fifth birth fringe will be (a) 3.2 mm (b) 1.63 mm (c) 0.585 mm (d) 2.31 mm (b) D 5x6.5x107 x1 x5 n 32.5x104 m d 103 D 5x6.3x107 x1 x3 (2n 1) 16.25x104 m 3 2d 2x10 x5 - x3 1.63 mm.
Solution-
2.
Light is incident normally on a diffraction grating through which first order diffraction is seen at 32o. The second order diffraction will be seen at (a) 84o (b) 48o o (c) 64 (d) None of these
Solution(d) For second order diffraction,
2 2 sin 1 d = 2 sin 320 > 1 Which is not possible. Hence there is no second order diffraction. sin 2
3.
In Young's double slit experiment, if the widths of the slit are in the ratio 4:9, ratio of intensity of maxima to intensity of minima will be (a) 25:1 (b) 9:4 (c) 3:2 (d) 81:16
Solution(a) As ratio of slit widths = Ratio of intensities I 9 a12 9 a1 3 1 or 2 or I2 4 a2 4 a2 2 amax = a1 + a2 = 3 + 2 =5; amin. = 3 - 2 = 1 Imax. (a1 a2)2 (3 2)2 25 Imin. (a1 a2)2 (3 2)2 1
4.
A light source approaches the observer with velocity 0.5 cm. Doppler shift for light of wavelength 550 Å is (a) 616 Å (b) 1833 Å (c) 5500 Å (d) 6160 Å
Solution-
(b)
cv v' c v .v and c v c ' 0 c 5500xc 5500x2 3667 A or ' c v c 0.5c 3
In case of light, v '
0
0
(5500 3667) A 1833 A
5.
Angular width of a central max. is 30o when the slit is illuminated by light of wavelength 6000 Å. Then width of the slit will be approx. (a) 12 × 10–6 m (b) 12 × 10–7 m (c) 12 × 10–8 m (d) –9 12 × 10 m
Solution(b) Angular width = 2 = 300 sin 300 a 6000x1010 a 12000x1010 m 0 sin 30 1/ 2 –7 = 12 x 10 m 6.
Light of wavelength 6000 12 × 10–6 m is incident on a single slit. First minimum is obtained at a distance of 0.4 cm from the centre. If width of the slit is 0.3 mm, then distance between slit and screen will be (a) 1.0 m (b) 1.5 m (c) 2.0 m (d) 2.3 m
Solution-
(c)
Using a sin = n x D or a. n or a n. D x a.x or D n 4 3x10 x4x103 m 2.0m 1x6000x1010 7.
If velocity of a galaxy relative to earth is 1.2 × 106 ms –2 then % increase in wavelength of light from galaxy as compared to the similar source on earth will be (a) 0.3 % (b) 0.4 % (c) 0.5 % (d) 0.6 % (b) v Since c v x c 1.2x106 12 105 x 8 4x103 or 8 3x10 3 10 x100 0.4 i.e., 0.4%
Solution-
8.
Doppler shift for the light of wavelength 6000 Å emitted from the sun is 0.04 Å. If radius of the sun is 7 × 108 m then time period of rotation of the sun will be (a) 30 days (b) 365 days (c) 24 hours (d) 25 days.
Solution-
(d)
Doppler shift, d
v x R c c
2 or R d T c Substituting values for all parameter, we get Rx2 x or T d c T = 25 days
9.
On introducing a thin nuca sgeet if tgucjbess 2 × 10 –6 m and refractive index 1.5 in the path of one of the waves, central bright maxima shifts by n fringes. Wavelength of the wave used is 5000 Å, then n is (a) 1 (b) 2 (c) 5 (d) 10
Solution(d) shift in no. of fringes is given by n = (u-1) t (u 1)t (1.5 1)x2x106 n fringes 2 5000x1010
10.
Two beams of light having intensities I and 4I interfere to produce a fringe patternon the screen. Phase difference between the beams is at point a and at sities at A 2 and B is (a) 3 I (b) 4 I (c) 5 I (d) 6 I
Solution-
(b)
Given I1 = I, I2 = 4I; 1
and 2 2
IA = I1 + I2 + 2 I1I2 cos 1 I1 I2 5I 2 IB I1 I2 2 I1I2 cos 2 I1 I2 2 I1I2 cos
I1 I2 2 I1I2 cos I
IA IB 5I I 4I
11.
White light is used to illuminate the two slits in Young's double slit experiment, separation between the slits is b and the screen is at a distance d(>>b) from the slits. At a point on the screen, directly in front of the slits, certain wavelengths are missing. Some of these missing wavelengths are 2b2 2b2 b2 b2 (b) (c) (d) (a) 3d d 3d d
Solution-
(a,d)
b2 b2 Path difference = d d 2d 2d For missing wavelengths, b2 (2n 1) 2 2d b2 If n = 1, 2 d b2 b2 or If n = 2, 3 x 2 2d 3d 12.
A radar operates at wavelength 50.0 cm. If the beat freqency between the transmitted singal and the singal reflected from aircraft () is 1 kHz, then velocity of the aircraft will be (a) 800 km/hr (b) 900 km/hr (c) 1000 km/hr (d) 1032 km/hr
Solution(b) when source is fixed and observer is moving towards it ca ' . c when source is moving towards observer at rest
a 1 c c c a " v' . c a c a c a 1 c 1
a a 2a c 1 1 1 c c c 2a 2a ' c 0.5x1000 a 250 ms1 2 2 = 900 km/hr
13.
A plane electromagnetic wave of frequency wo falls normally on the surface of a mirror approaching with a relativisitic velocity . Then frequency of the reflected wave will be given c (1 )w 0 1 1 (a) (c) (d) (b) wo (1 )w 0 (1 ) 1 (1 ) (1 )w o
Solution-
(c)
c .w0 c Frequency of waves reflected from mirror and movi9ng towards source, c w" .w ' c c c c x .w0 or w " w0 c c c c 1 c w" w0 c 1 c 1 c wo 1 wo where c 1 1 c
Frequency of Em waves going towards the approaching mirror, w'=
14.
A spectral line of wavelength 0.59 mm is observed in the directions to the opposite edges of the solar disc along its equator. A difference in wavelength equal to () 8 picometer is observed. Period of Sun's revolution around its own axis will be about (Radius of sun = 6.95 × 108 m) (a) 30 days (b) 24 hours (c) 25 days (d) 365 days
Solution(c) or c c
Change in wavelength for two edges = Total change, 2 c 2 or c 2 2R 2 4R and T = 2Rx c c 4x3.14x6.95x108 x0.59x10 6 days 3x108 x8x1012 x86400 24.8 days 25 days
15.
If light with wavelength 0.50 mm falls on a slit of width 10 mm and at an angle o = 30o to its normal. Then angular position of first minima located on right sides of the central Fraunhoffer's diffraction will be at (a) 33.4o (b) 26.8o o (c) 39.8 (d) None of these
Solution(a) For first diffraction minima at angle d(sin - sin 0) = For right of C.M., sin 1 sin 0 d 0.5 = 0.55 i.e., 1 33.370 = 0.5 + 10 16.
Angular width of central maximum in the Fraunhoffer's diffraction pattern is measured. Slit is illuminated by the light of another wavelength, angular width decreases by 30%. Wavelength of light used is (a) 3500 Å (b) 4200 Å (c) 4700 Å (d) 6000 Å
Solution(b) For first diffraction min. d sin = and if angle is small, sin d i.e. Half angular width, d 2 Full angular width w = 2 = d 2 ' Also w' = d ' w' w' or ' w w 0
= 6000 x 0.7 = 4200 A 17.
A parallel beam of white light falls on a thin film whose refractive index is 1.33. If angle of incidence is 52o then thickness of the film for the reflected light to be coloured yellow ( = 6000 Å) most intensively must be (a) 14 (2n + 1) m (b) 1.4 (2n + 1) m
(c) 0.14 (2n + 1) m
(d) 142 (2n + 1) m
(c) sin i sin i 0.788 sin r 0.6 sin r 1.33
Solution-
cos r 1 sin2 r 1 (0.6)2 0.8 For constructive interference on reflection 2t cos r (2n 1) 2 (2n 1) (2n 1)x0.6 t 4 cos r 4x1.33x0.8 =0.14 (22+1) m
18.
A plane monochromatic light falls normally on a diaphragm with two narrow slits separated by a distance d = 2.5 mm. A fringe pattern is formed on the screen placed at D = 100 cm behind the diaphragm. If one of the slits is covered by a glass plate of thickness 10 m, then distance by which these fringes will be shifted will be (a) 2 mm (b) 3 mm (c) 4 mm (d) 5 mm
Solution(a) xd ( 1) tD ( 1)t x D d 3 (1.5 1)x10 x100 0.2 cm or x = 0.25 19.
A two slit Young's experiment is done with monochromatic light of wavelength 6000 Å. Slits are 2 mm apart and fringes are observed on a screen placed 10 cm away from the slits. If a transparent plate of thickness 0.5 mm is placed in front of one of the slit, interference pattern shifts by 5 mm. Then refractive index of transparent plate should be (a) 1.1 (b) 1.2 (c) 1.3 (d) 1.5 (b) xd xd ( 1)t or 1 D Dt 0.5x0.2 1 0.2 1.2 or 1 10x0.05
Solution-
20.
In Young's double slit experiment, using monochromatic light, fringe pattern shifts by a certain distance on the screen when a mica sheet of refractive index 1.6 and thichness 1.964 mm is introduced in the path of one of the two waves. If now mica sheet is removed and distance between slit and screen is doubled, distance between successive max. or min. remains unchanged. The wavelength of the monochromatic light used in the experiment is (a) 4000 Å (b) 5500 Å (c) 5892 Å (d) 6071 Å
Solution-
(c)
xd (u 1) tD (u 1) t or x D d
when distance between screen and slit is doubled, them fringe width As x 2D (u 1) tD d d ( 1) t (1.6 1)x1.964x106 or 2 2
2D d
0
= 0.5892 x 10–6 m = 5892 A
21.
In a two slit experiment with monochromatic light, fringes are obtained on a screen placed at some distance from the slits. If screen is moved by 5 × 10–2 m towards the slits, then change in fringe width is 3 × 10–5 m. If the distance between slits is 10–3 m then wavelength of the light used will be (a) 4000 Å (b) 6000 Å (c) 5890 Å (d) 8000 Å
Solution-
d d
(b)
1 2 or
22.
(D1 D2 ) d(1 2 ) or d D1 D2
0 3x105 x103 6000 A 5x102
White light is used to illuminate two slits in Young's double slit experiment. Separation between the slits is b and the screen is at a distance d (>>b) from the slits. Then wavelengths missing at a point on the screen directly in front of one of the slit are b2 b2 b2 b2 b2 b2 b2 b2 , , , , (a) (b) (c) (d) 2d 4d d 4d 2d 3d d bd (a) b 2 .b b2 d 2d
Solution-
For missing wavelength 1 b2 b2 or 1 2 2d 2d 2 3 2 b b2 or 2 2 2d 3d 2 53 b b2 or 3 2 2d 5d
1 32 53 7 4 , , , , etc. 2 2 2 2
23.
Interference fringes from sodium light (1 = 5890 Å) in a double slit experiment have an angular width 0.20o. To increase the fringe width by 10%, wavelength of light used should be (a) 5892 Å (b) 4000 Å (c) 8000 Å (d) 6479 Å (d) D and angular fringe width, D d d 1 1 / d, 2 2 / d
Solution-
1 1 or 2 1. 2 2 2 1 0 0.22 5890 x 6479 A 0.20
24.
In a Young's double slit experiment, angula width of a fringe formed on a distant screen is 0.1o. If wavelength of light used is 6000 Å , then distance between the slits will be (a) 0.241 mm (b) 0.344 mm (c) 0.519 mm (d) 0.413 mm (b) or d d 5 6x10 x180x7 0.344 mm or d 0.1x22
Solution-
25.
In a Young's double slit experiment two narrow slit 0.8 mm apart are illuminated by the same source of yellow light (l = 5893 Å). If distance between slits and screen is 2m then separation between adjacent bright lines will be (a) 14.73 mm (b) 14.73 cm (c) 1.473 mm (d) 147.3 mm
Solution(a) D 5893x108 x200 0.1473 cm d 0.08 26.
Monochromatic light of wavelength 5000 Å is incident on two slits separated by a distance of 5 × 10–4 m. Interference pattern is seen on the screen placed at a distance of 1 m from the slits. A thin glass plate of thickness 1.5 × 10 –6 m and refractive index 1.5 is laced between one of the slits and the screen. If intensity in the absence of plate was Io then new intensity at the centre of the screen will be I 3I (a) I0 (b) 2 I0(c) o (d) o 2 4 Solution(a) xd ( 1) t D
( 1) Dt (1.5 1)x1.5x106 x1 d 5x104 = 1.5 x 10–3 m
or x
27.
In Young's double slit experiment, separation between the slits is 2 × 10 –3 m and distance of the screen from the slit is 2.5 m. Light in the range of 2000–8000 Å is allowed o fall on the slits. Wavelength in the visible region that will be present on the screen at 10–3 m from the central maxima will be (a) 4000 Å (b) 5000 Å (c) 6000 Å (d) 8000 Å (a) 0 xd 103 x2x103 8000 A D 2.5 For bright line 8000 = n11 n22 n32
Solution-
0
If n1 2 1 4000 A i.e., second bright line for visible region is present. 28.
A double slit experiment is immersed in a liquid of refractive index 1.33. Separation between the slits is 1.0 mm and he distance between slit and screen is 1.33 m. If slits are illuminated by a parallel beam of light whose wavelength is 6300 Å, then fringe width will be (a) 6.3 mm (b) 63 mm (c) 0.63 mm (d) None of these (c) air t l D D D , in liquid 1 air ld d d
Solution-
6300x1010 x1.33 0.63 mm 1.33x103
29.
In a Young's interference experimental arrangement incident yellow light is composed of two wavelength 5890 Å and 5895 Å. between the slits is 1 mm and the screen is placed 1 m away. Order upto which fringes can be seen on the screen will be (a) 384 (b) 486 (c) 512 (d) 589
Solution-
(d)
1 n 1 2 2 n 2 n 1 or 2 n 1
1 2 1 5895 5890 5 2n 5890 5890 1 5890 n 589 10 or
30.
Ratio of intensities between a point A and that of central fringe is 0.853. Then path difference between two waves at point A will be (a) (c) (d) (b) 2 4 8
Solution(c) R2 = a2 + b2 + 2ab cos IR 0.853 Imax. IR 0.853 Imax. 0.853x4I IR = I + I0 + 2I cos = 2I (1 + cos) =0.853 x 4I 4 8
For Test 1.
When light is incident on a soap film of thickness 5 × 10–5 cm, wavelength reflected maximum in the visible region is 5320 Å. Refractive index of the film will be (a) 1.22 (b) 1.33 (c) 1.51 (d) 1.83.
Solution-
(b)
2 (2n 1) (2n 1)x5320x1010 or 2t cos r 2 2x5x105 x102 x1 1.33 2t cos r (2n 1)
2.
A ray of unpolarised light is incident on a glass plate of refractive index 1.54 at polarising angle, then angle of refraction is (a) 33o(b) 44o(c) 57o (d) 90o
Solution(a) tanip or tanip 1.54
i.e., ip = tan–1 1.54 = 570, But r + ip = 900 r = 900 - ip = 900 - 57 = 330.
3.
Two waves of same intensity produce interference. If intensity at maximum is 4I, then intensity at the minimum will be (a) 0 (b) 2I (c) 3I (d) 4I.
Solution - (a) I1 = I 2 = I = a 2 Imax. = (a + a)2 + (2a)2 = 4a2 = 4I Imin. = (a - a)2 = 0
4.
First diffraction minima due to a single slit of width 1.0 x 10 –5 cm is at 300. Then wavelength of light used is 0
0
(a) 400 A
(b) 500 A
0
0
(c) 600 A
(d) 700 A
Solution - (b) As a sin = n or
a sin n
or 1.0x10 5 x sin 300 1.0x10 5 x = 0.5 x 10–5 = 5 x 10–6 cm
1 2
0
= 500 A 0
5.
Light of wavelength 6000 A is normally incident on a slit. Angular position of second minimum from central maximum us 300. Width of the slit should be (a) 12 x 10–5 cm (b) 18 x 10–5 cm (c) 24 x 10–5 cm (d) 36 x 10–5 cm
Solution - (c) a sin n
n 2x6000x1010 a sin sin 300 2x6000x1010 4x6000x1010 1/ 2 = 24 x 10–7 m = 24 x 10–5 cm 0
6.
Light of wavelength 6328 A is incident normally on a slit of width 0.2 mm. Angular width of the central maximum on the screen will be (a) 0.90 (b) 0.180 (c) 0.540 (d) 0.360
Solution - (d) a sin n n 6328x1010 x1 a 0.2x103 –6 = 3164 x 10 rad = 0.003164 radian Angular width 2 sin
1800 0.360 = 2 x 0.003164 x 0
7.
Light of wavelength 5000 A is incident normally on a slit. First minimum of diffraction pattern is formed at a distance of 5 mm from the central maximum. If slit width is 0.2 mm, then distance between slit and screen will be (a) 1 m (b) 1.5 m (c) 2.0 m (d) 2.5 m
Solution - (c) a sin n x or a. n D ax 2x104 x5x103 2.0 metre D n 1x5000x1010 8.
Light of wavelength is incident on a slit. First minima of the diffraction pattern is found to lie at a distance of 6 mm from the central maximum on a screen placed at a distance of 2 m from the slit. If slit width is 0.2 mm, then wavelength of the light used will be 0
(a) 4000 A
0
(b) 6000 A
0
0
(c) 7000 A
(d) 7400 A
Solution - (b) a sin n x ax a. n or D nD 4 0 2x10 x6x103 or 6000 A 1x2
9.
A slit 5 cm wide when irradiated by waves of wavelength 10 mm results in the angular spread of the central maxima on either side of incident light by about (a) 1/2 radian (b) 1/4 radian (c) 3 radian (d) 1/5 radian.
Solution - (d) Angular spread on either side is given by 1 radians a 5 10.
In Young's double slit experiment, ten slits separated by a distance of 1 mm are illuminated by a monochromatic light source of wavelength 5 x 10 –7 m. If the distance between slit and screen is 2 metre, then separation of bright lines in the interference pattern will be (a) 0.5 mm (b) 1.0 mm (c) 1.5 mm (d) 1.75 mm
Solution - (b) D 5x107 x2 103 m 3 d 10 = 1 mm. 0
11.
A star is moving away from earth and shift in spectral line of wavelength 5700 A is 0
1.90 A . Velocity of the star is (a) 50 km s–1
(b) 70 km s–1
(c) 80 km s–1 Solution - (d) c 3x108 x1.9x1010 or 100 km s1 10 5700x10 12.
(d) 100 km s–1
c
Intensity of central bright fringe due to interference of two identical coherent monochromatic sources is I. If one of the source is switched off, then intensity of central bright fringe becomes I I (a) (b) 2 3 I (c) (d) I. 4
Solution - (c) I = (a+a)2 = 4a2, I'=(a)2 = a2 I I' a2 1 I' . 2 , 4 I 4a 4 13.
In two separate set ups of Young's double slit experiment, fringes of equal width are observed when lights of wavelengths in the ratio 1:2 are used. If the ratio of slit separation in two cases is 2:1, then ratio of distances between the plane of slits and the screen in the two set ups is (a) 1 : 2 (b) 2 : 1 (c) 1 : 4 (d) 4 : 1
Solution - (d) D D 1 1 2 2 d1 d2 D d 2 2 1 1 x 2 x 4 :1 D2 d2 1 1 1 14.
In an experiment to demonstrate interference of flight using Young's slits, separation of two slits is doubled. In order to maintain same spacing of fringes, distance D of screen from slits must be changed to D (a) D (b) 2 3D (c) 2D (d) 4 (c) D (2d) d 2d i.e., D' = 2D.
Solution-
15.
In an interference pattern by two identical slits, intensity of central maxima is I. If one slit is closed, intensity of central maxima changes to I0. Then I and I0 are related by (a) I = I0 (b) I = 2I0 (c) I = 3I0 (d) I = 4I0
Solution - (d) I1 = I 2 = a 2 Imax. = (a+a)2 = 4a2 = I If one slit is closed, Intensity, I0 = (a)2 I 4a2 2 4 or I 4I0 I0 a
16.
In a young's double slit experiment, two slits are illuminated by a mixture of two 0
0
wavelengths 12000 A and 10000 A . At 6.0 mm from the common central bright fringe on a screen 2 m away from the slits, a bright fringe of one interference pattern coincides with the bright fringe of other. Distance between the slits should be (a) 1.0 mm (b) 1.5 mm (c) 2.0 mm (d) 2.5 mm Solution - (c) D D x n (n 1) ' d d n x 12000 = (n+1) x 10000 D or n = 5 and x = n d n D 5x12000x1010 x2 d m x 6x103 = 2 x 10–3 m = 2 mm 17.
In a young's double slit experiment, slits are illuminated by a monochromatic source 0
of wavelength 6000 A and fringes are obtained. If screen is moved by a distance of 5 cm towards slits, change in fringe width is 3 x 10–5 m. Then separation between the slits will be (a) 1 mm (b) 1.2 mm (c) 1.5 mm (d) 1.63 mm Solution - (a) D' D and ' d d (D D') ' d (D D') d or ( ')
6000x1010 x5x102 m 103 m 1mm 5 3x10
18.
An unpolarized beam of light is incident on a group of three polarizing sheets which are arranged in such a way that plane of rotation of one make an angle of 40 0 with the adjacent one. The 0% of incident light transmitted by first polarizer will be (a) 33% (b) 16.6% (c) 50% (d) 25%
Solution - (c) First polariser, polarises the light and hence intensity of light reduces by 50%. 19.
80 gm of impure sugar when dissolved in a litre of water gives an optical rotation of 9.90 when placed in a tube of length 20 cm. If concentration of sugar solution is 75 gm/litre then specific rotation of sugar is (a) 440 (b) 550 0 (c) 66 (d) 730
Solution - (c) 9.9 S 660. c or S ic 2x0.075 lS 20.
In Young's double slit experiment, distance between the slits S1 and S2 is d and the distance between slits and screen is D. Then first missing wavelength on the screen in front of S1 is d2 d2 (a) (b) D 2D D (d) None of these (c) 2 d
Solution - (a) d2 2D For first missing wavelength, d2 d2 or 2D 2 D
Path difference, x
21.
In Young's double slit experiment, 12 fringes are obtained to be formed in a certain segment of the screen when light of wavelength 600 nm is used. If wavelength of light is changed to 400 nm, number of fringes observed in the same segment of the screen is given by (a) 18 (b) 24 (c) 30 (d) 36.
Solution - (a) n11 n22
n2
22.
n11 12x600 18 400 2
In Young's double slit experiment, interference pattern is found to have an intensity ratio between bright and dark fringes as 9. Then amplitude ratio will be (a) 1 (b) 2 (c) 9 (d) 3.
Solution - (b) amax. a1 a2 ; amin. a1 a2 Imax. (a1 a2 )2 9 a1 2 and a2 1 Imin. (a1 a2 )2 1
a1 2 2 a2 1
0
23.
Interference fringes from sodium light ( 5890 A) in a double slit experiment have an angular width 0.200, To increase the fringe width by 10%, new wavelength should be 0
0
(a) 5896 A
(b) 7321A
0
0
(c) 6300 A
(d) 6479 A
Solution - (d) 1 1 , 2 2 d d 1 1 2 1x 2 2 2 1 0 0.22 6479 A or 2 5890x 0.20 24.
Sol
25.
Sol
If A is the amplitude of the wave coming from a point source at distance r then (A) (B) A µ r- 2 A µ r- 1 (C) (D) A µ r2 A µ r1 1 (B) For a point source, I µ 2 r 1 1 and A µ I \ Aµ 2 or Aµ r r If A is the amplitude of the wave coming from a line source at a distance r, then Aµr (A) (B) A µ r - 1/ 2 -2 (D) (C) A µ r1/ 2 Aµr 1 (B) For a line source, I µ r \ A µ r - 1/ 2 and A µ I
26.
Phase difference between two waves having same frequency (v) and same amplitude (A) is 2/3. If these waves superimpose each other, then resultant amplitude will be (A) 2A (B) 0 (C) A (D) A2.
Sol
(C) A R = A 2 + A 2 + 2A ×A cos q
æ2p ö æ 1ö = A 2 + A 2 + 2A 2 cos ç ÷ = A 2 + A 2 + 2A 2 ç- ÷ è3 ø è 2ø A2 + A2 - A2 = A2 = A
27.
Sol
28.
Sol
Ratio of amplitudes of the waves coming from two slits having widths in the ratio 4 : 1 will be (A) 1:2 (B) 2:1 (C) 1:4 (D) 4:1 (B) Intensity slit width I1 4 = I2 1
\
a1 I = 1 = 4 = 2 :1 a2 I2
\
a1 = 2 :1 a2
Two slits S1 and S2 illuminated by a white light source give a white central maxima. A transparent sheet of refractive index 1.25 and thickness t 1 is placed in front of S1. Another transparent sheet of refractive index 1.50 and thickness t2 is placed in front of S2. If central maxima is not effected, then ratio of the thickness of the two sheets will be (A) 1:2 (B) 2:1 (C) 1:4 (D) 4:1 (B) Since there is no shift in central maxima. Therefore path difference introduced by the two sheets are equal i.e. (m1 - 1) t1 = (m2 - 1) t 2 where m1 and m2 are refraction index i.e.
29.
æp ö Two coherent waves are represented by y1 = a1 cos t and y2 = a2 cos ç - wt ÷ . è2 ø Their resultant intensity after interference will be a1 - a2 a1 + a2 (A) (B) (C)
Sol
t1 (m2 - 1) (1.5 - 1) 0.5 = = = =2 t 2 (m1 - 1) (1.25 - 1) 0.25
a12 - a22
(D)
æp ö (D) y1 = a1 cos wt ; y 2 = a2 cos ç - wt ÷ è2 ø Since the two wave differ in phase by \
30.
Sol
a12 + a22
(
p 2
)
y = a12 + a22 .
If a thin film of thickness t and refractive index is placed in the path of light coming from a source S, then increase in length of optical path is (A) t (B) /t (C) ( – 1) t (D) None of these. (C) Increase in optical path = mt - t = (m- 1) t
Temperature scales Relation between Celsius, Kelvin and Fahrenheit Scale: C K 273 F 32 5 5 9
Example 1: The electrical resistance of pure platinum increases linearly with increasing temperature. This property is used in a Platinum resistance thermometer. The relation between R (Resistance at K) and R0 (Resistance at 0K) is given as
R Ro 1 α θ θo θ
where = temperature coefficient of resistance. Now, if a Platinum resistance thermometer reads 00 Celsius when its resistance is 80 and 100o when its resistance is 90, find the temperature at which its resistance is 86. Solution:
Using the given relationship, we have 90 80Ω80Ω100
86 80Ωα[80Ω][θ] . . .
. . . (i)
(ii)
Where is the desired temperature. Taking the ratio of (i) & (ii)
CALORIMETRY Principle of Calorimetry
When two objects having different temperatures are brought in contact, heat flows from the hot object to the cold object. If the system is sufficiently thermally isolated from its surrounding, the heat lost by the hot object = the heat gained by the cold object.
Specific heat capacity
The amount of heat needed to raise the temperature of unit mass of a substance by 1 is known as its specific heat capacity. If Q amount of heat raises the temperature of m mass of a material by T , then its specific heat capacity is given as: s
Q mΔT
Q msT
Also the amount of heat supplied per unit increase in temperature for any body is known as its heat capacity, c
Q ms . T
Specific Latent Heat
In order to change the state of a substance (from solid to liquid or from liquid to gas) heat has to be supplied to it. During this process temperature remains constant. The amount of heat supplied per unit mass for such a process is known as the Specific Latent Heat of that substance for that process. Example 2: 5 gm of water at 30oC and 5 gm of ice at - 20oC are mixed together in a calorimeter. Find the final temperature of mixture and also the final masses of ice and water. Water equivalent of calorimeter is negligible, specific heat of ice = 0.5 cal/gm oC and latent heat of ice = 80 cal/gm. Solution:
In this case heat is given by water and taken by ice Heat available with water to cool from 30oC to 0oC = ms = 51 30 150 cal . Heat required by 5 gm ice to increase its temperature up to 0 oC ms 50.5 20 50 cal Out of 150 cal heat available, 50 cal is used for increasing temperature of ice from - 20oC to 0oC. The remaining heat 100 cal is used for melting the ice. if mass of ice melted is m gm then m 80 100 m 125 . gm
Thus 1.25 gm ice out of 5 gm melts and the mixture of ice and water is at 0oC. Mechanical Equivalent of Heat
1 calorie is the quantity of heat required to raise the temperature of pure water at 1 atm pressure from 14.5C to 15.5C. When work is completely converted to heat, the quantity of heat produced (Q) is found to be proportional to the quantity of work (W) that was converted into heat WQ Or,
W = JQ, where J is known as the Mechanical Equivalent of Heat. J = 4.2 107 erg cal-1 = 4.2 J/cal
Example 3: A bullet splinter of mass of 10 gm moving with a speed of 400 m/s hits an ice block of mass 990 gm kept on a frictionless floor and gets stuck in it. How much ice will melt if 50% of the lost Kinetic energy goes to ice? (Temperature of ice block = 0oC). Solution:
Velocity of bullet + ice block, V =
10 x400 m/s 1000
V =4 m/s Loss of K.E. = =
1 1 mv 2 m M V 2 2 2
1 1 [0.01(400)21(4)2] = [1600 16]= (1584/2) J 2 2
Heat generated = 1584 / 2 4.2 = 95 Cal Mass of ice melted =
95 cal = 1.2 gm. 80 Cal / gm
Thermal Expansion An increase in the temperature of a body is generally accompanied by an increase in its size. This is known as Thermal Expansion.
1 L L T
The value of linear coefficient of thermal expansion at temperature T can be found by taking limit T 0 . 1 L lim T 0 T L
or
1 dL L dT
For most of the solids the value of is small and nearly independent of T. In such cases the linear dimension of an object at a different temperature is given by: L L o (1 T)
(Here L o is the initial length and T is the change in temperature).
Example 4: A clock with an iron pendulum is made so as to keep correct time at 10oC. Given αiron12 10 6 per oC . How fast or slow does the clock move per day if the temperature rises to 250 C? Given iron = 12 106 per 0C. Solution:
When the pendulum keeps correct time, its period of vibration is 2 sec and so it makes 24 60 60 43200 Vibration /day 2
If length of pendulum at 10oC is 10 and at 25oC is 25 25 10[1 (2510)]
10 [115] g
as T 2 i.e. T i.e. n
1
n is no. of vibrations per sec.
n25 10 n10 25
115
1 / 2
1
15 2
15 n 25 n10 1 1210 6 = 43200[1 0.00009] = 43196 .12 2
That is the clock makes (43200 43196.12) = 3.88 vibration loss per day. That is clock losses 3.88 2 = 7.76 sec per day. Area and Volume expansion
If the temperature of a two-dimensional object (lamina) is changed, its area changes. If the coefficient of linear expansion of the material of lamina is small and constant, then its final area is given by A A o 1 T , where A o is the initial area, T is the change in temperature and
is the area coefficient of thermal expansion.
It can be shown that 2 . Similar relation also holds for the volume of a three-dimensional object V Vo 1 T
(Here 3 is known as the coefficient of volume expansion). For most solids, ~ 106/C
Thermal Expansion in liquids: The experimental measurement of for a liquid becomes slightly difficult due to expansion of the container, when a liquid is heated in a container. The initial level of the liquid falls due to expansion of the container. But afterwards it rises due to faster expansion of the liquid.
The actual increase in the the apparent + The increase in the volume of = volume of the liquid increase in the the container. volume of liquid liquid = apparent + container , for liquids, is of the order of 104/C
Anomalous expansion of water: If the temperature of water is increased from 0oC, it contracts until the temperature reaches 4oC and expands thereafter. In other words the density of water is highest at 4oC.
Example 5: A glass vessel of volume Vo is completely filled with a liquid and its temperature is raised by T . What volume of the liquid will overflow? Coefficient of linear expansion of glass = g and coefficient of volume expansion of the liquid = . Solution:
Volume of the liquid over flown = Increase in the volume of the liquid increase in the volume of the container = Vo (1 T) Vo Vo (1 g T) Vo
= Vo T( g ) = Vo T( 3 g )
( 3 )
Thermal Stress When a rod of length L is held between two rigid supports and the temperature of rod is increased by T , the rigid supports prevent the rod from expanding. This causes compressive stress in the rod. As the length of the rod remains unchanged, Thermal expansion = Mechanical compression LT
FL AY
F/A Y L / L
Thermal stress F / A YT , which is compressive.
KINETIC THEORY OF GASES Fluids consist of molecules which are in random motion, which collide with each other and with the walls of the container. The intermolecular separation in a gas is larger by an order of magnitude than the intermolecular separation in a liquid. The intermolecular separation in a gas increases as the temperature increases, and decreases with increasing density. it is experimentally observed that most gases follow a universal equation of state pV = n RT Gas molecules are relatively free: the interactions between them are small. This means that the total energy of the gas molecules is mostly kinetic. A simple model of a gas, with point particles representing molecules and their collisions with the walls of the confining vessel causing the pressure exerted by the gas leads to some important conclusions:
(a)
The pressure exerted by an ideal gas is given by p=
1 2 mc rms , where = number of molecules per unit volume, 3
m = mass of each molecule, crms = the rms speed of each molecule.
(b)
The total internal energy of an ideal gs is
1 2
U = Nf k B T where N = number of molecules in the gas, f = number of
degree of freedom, T = absolute temperature of the gas, kB = R NA = Boltzmann’s constant.
(C)
The rms speed is given by crms=
3RT , where M represents the molar mass of the gas (i.e. mass M
of 1 mole of the gas)
(d)
The pressure of an ideal gas is given by p =
2 U tran , where Utran is the translational energy of the molecules of 3 V
the gas.
Example6: Find r.m.s speed of Hydrogen molecules at room temperature (=300 k). Solution:
Mass of 1 mole of Hydrogen gas= 2 gm = 2 10-3 kg Vrms = =
3 RT M
3 8.3 300 2 10 3
= 1.93 103 m/s.
2.
When the container contains more than one gas, total pressure exerted by all the gases on the wall is sum of pressures exerted by each gas as it would while filling the container alone. In a way, each gas behaves independent of each other. Thus we have P = P1+P2+P3+……, where P1, P2, P3 are the partial pressures of gases 1, 2 & 3 respectively This is known as Dalton’s Law of partial pressures.
3.
One mole of any gas occupies a volume of 22.4 litre at standard temperature and Pressure which are 273.15 (=0°C) and 1.013105 Pa (=1atm) respectively,
Example 7: 4 gm Hydrogen is mixed with 11.2 litre of He at S.T.P. in a container of volume 20 litre. If the final temperature is 300 K find the pressure. Solution:
4 gm Hydrogen = 2 moles Hydrogen 11.2 He at S.T.P. = 1/2 mole of He P = PH + PHe = (nH+nHe)
8.3 (300 k ) RT = (2+½) V (20 10 3 )m3
= 3.12 105 N/m2.
Internal Energy Internal energy, of any body is sum total of kinetic energies and potential energies of its constituents (at molecular level). In case of an ideal gas, as there are no intermolecular forces, except during collision the possibility of potential energy is ruled out, so it is only kinetic energy. The kinetic energy of the molecules can be of three types. (i)
Translational
(ii)
Rotational
(iii)
Vibrational
In a way, it means that the energy of molecules is shared in various modes. These independent modes of motions are called degrees of freedom. The table given below gives the number of degrees of freedom for various types of molecules at normal temperature.
Nature of motion
Degree of Freedom (f) Translational
Rotational
Vibrational*
Total
(1) Monoatomic
3
0
0
3
(2) Diatomic
3
2
0
5
Poly
Linear
3
2
0
5
Nonlinear
3
3
0
6
Atomicity
* At room temperature the energy associated with vibrational motion is negligibly small in comparison to translational and rotational K.E. Equipartition of Energy According to the Law of equipartition of energy the average K.E. of a molecule is equally shared among different degrees of freedom. The average energy per degree of freedom of a molecule is ½ kT, where k is the Boltzmann’s constant and T is the absolute temperature. Thus, for a monoatomic ideal gas: U (the internal energy) = Also for one mole U =
3 kT 2
3 RT 2
Example 8: Find the average kinetic energy per molecule at temperature T for an equimolar mixture of two ideal gases A and B where A is monoatomic and B is diatomic. Solution:
No. of degrees of freedom per molecule for A = 3 No. of degrees of freedom per molecule for B = 5 Since the mixture is equimolar, the average kinetic energy per molecule will be given by the average of the two values i.e. 35 kT = 4kT 2
where k is Boltzmann’s constant.
Work Done In Different Processes Work done by an enclosed gas on its surroundings is given by W = p dV For different types of processes, we have got different relations between p (Pressure) and V (Volume) and accordingly we have different expressions for work.
(a)
Isochoric Process Here, the volume is constant throughout and therefore the work done by the gas, irrespective of the manner in which pressure varies, is zero Wisochoric = 0
(b)
Isobaric Process In this case, pressure of the gas remains constant throughout the process. Hence, pdV = p V = n RT(n = number of moles) ( T= change in absolute temperature)
(C)
Isothermal process The temperature remains constant throughout the process. Using ideal gas equation, we get, p=
n RT V
Hence, pdV = n RT
V2
V1
(d)
V dV = n RT ln 2 V1 V
Adiabatic Process: pV = Constant = C (say)
p = CV-
W = pdV =
V2
CV
dV
V1
=
p1V1 p 2 V2 = nCv(T2 T1) 1
The work done by a gas can also be evaluated from the p-V diagram of the process. p Area enclosed by the curve in a p-V diagram = work done by the gas
V
First Law of Thermodynamics First law of thermodynamics is simply a re-statement of the principle of conservation of energy for a thermally isolated system. If Q, U & W represent the heat given to the system, change in its internal energy and the work done by the system respectively, the first law of thermodynamics states that,
Q = U + W The heat transferred to the system (Q) is either utilised to do work (W) or increase the internal energy of the system (U). Example 9: 3000 J of heat is given to a gas at constant pressure of 2 105 N/m2. If its volume increases by 10 litres during the process find the change in the internal energy of the gas Q = 3000 J
Solution:
W = P V = (2105 N/m2) (10 10-3m3) = 2 103 J U = Q – W = 3000 2000= 1000 J.
Specific Heat Capacities of Gases S=
1 m
Q T
where Q = amount of heat required for ‘T’ temperature change. m = mass of the gas. In case of gases, the concept of a molar heat capacity is useful. Molar heat capacity is the amount of heat required to raise the temperature of one mole of the gas by one degree. So, if Q amount of heat goes to change the temperature of ‘n’ moles of a gas in a particular process, molar heat capacity ‘C’ can be mathematically given by: C=
1 Q n T
In terms of differentials, C=
1 dQ n dT
Two special cases are:(i)
If volume is kept constant during the process then CV =
1 Q n T V constant
This is the molar heat capacity of the gas at constant volume Note: Since U is independent of the process. U = n Cv T is true for all processes.
(ii)
If pressure remains constant, then Cp =
1 Q n T pconstant
This is the molar heat capacity of the gas at constant pressure
Relation Between Cp and Cv Cp Cv = R This is known as Mayer’s relation.
The Values of Cp and Cv If f is the number of degrees of freedom of a gas molecule then the internal energy of n moles of that gas is given as U = f/2 n RT
U = f/2 n RT = n CvT
Cv = f/2 R
From Mayer’s Relation Cp = Cv + R Cp = (f/2+1)R And the ratio of specific heats = =
Cp Cv
=
f /21 f /2
f 2 2 = 1 f f
Example 10: Find the molar heat capacity of an ideal gas with adiabatic exponent ‘’ for the polytropic process P V = constant. Solution:
We have, from first law of thermodynamics C = Cv +
PdV ndT
(n = number of moles)
We have, P V = constant From Ideal gas equation P V = n RT
Taking ratio, T V1 = Constant Differentiating we get
dV V = T( 1) dT
Putting it in the equation for ‘C’. C = Cv = Cv C=
PV n RT = Cv nT( 1) n T( 1) R 1
R R 1 1
Second law of thermodynamics
(i) Kelvin Statement:- It is impossible to derive a continuous supply of work by cooling a body to a temperature lower than that of the coldest of its surroundings.
(ii) Clausius Statement:-It is impossible for a self acting machine, unaided by an external agency to transfer heat from a body to another at higher temperature.
Reversible Process: A process which can be made to proceed in the reverse direction by variations in its conditions so that all changes occurring in any part of the direct process are exactly reversed in the corresponding part of the reverse process is called a reversible processes. Irreversible Process: A process which can not be made to proceed in the reverse direction is called an irreversible process. Heat Engine: It is a device which continuously converts heat energy into the mechanical energy in a cyclic process.
Efficiency of heat engine:
=
Q work output W Q1 Q 2 = 1 2 Q1 Q1 heat input Q1
Where Q1 is the heat supplied by the source and Q2 is the heat rejected to the sink.
Carnot Engine: It is an ideal heat engine which is based on Carnot's reversible cycle. It works in four steps viz. Isothermal expansion, adiabatic expansion, isothermal compression and adiabatic compression. The efficiency of a Carnot engine is given by =1
Q2 T =1 2 T1 Q1
where T1 and T2 are the temperatures of source and sink respectively. Example 11: The efficiency of a Carnot cycle is 1/6. If on reducing the temperature of the sink by 650C, the efficiency becomes 1/3, find the initial and final temperatures between which the cycle is working. Solution:
Given 1 =
1 1 , 2 = 6 3
If the temperatures of the source and the sink between which the cycle is working are T1 and T2, then the efficiency in the first case will be 1 = 1 -
T2 1 = T1 6
In the second case 2 = 1 Solving T1 = 390 K and
T2 65 T1
=
1 3
T2 = 325 K.
OBJECTIVE 1:
Calculate the root mean square speed of smoke particles of mass 5 1017 kg in Brownian motion in air at NTP. Boltzmann constant k 1.38 1023 JK 1
(A)
1.5 cm/s
(B)
2.2 cm/s
(C)
2.3 cm/s
(D)
4.4 cm/s
Ans. (a) Solution:
PV =
=
Vrms
m 1 2 RT mVrms M 3 3RT M
3NK T N
3kT
where = mass of one molecule.
Vrms
3 1.38 10 23 273 5 10 17
= 15 103 m / s 1.5 cm/s
2:
During an experiment an ideal gas is found to obey an additional law VP 2 = constant. The gas is initially at temp T and volume V. What will be the temperature of the gas when it expands to a volume 2V? (A)
T 4 T
(B)
T 2 T
(C)
T 5 T
(D)
T 6 T
Ans. (b) Solution:
According to the given problems VP2 = constant From the gas law PV = nRT
k V nRT V
nR V T K
V1 T1 , i.e, V2 T2
T 2 T
V T 2V T
Q.3-5 We have two vessels of equal volume, one filled with hydrogen and the other with equal mass of Helium. The common temperature is 27oC. 3:
What is the relative number of molecules in the two vessels ? (A)
nH 1 nHe 1
(B)
nH 5 n He 1
(C)
nH 2 n He 1
(D)
nH 3 n He 1
Ans. (C)
4:
If pressure of Hydrogen is 2 atm, what is the pressure of Helium ? (A)
pHe = 2 atm.
(B)
pHe = 3 atm.
(C)
pHe = 4 atm.
(D)
pHe = 1 atm.
Ans. (d)
5:
If the temperature of Helium is kept at 27o C and that of hydrogen is changed, at what temperature will its pressure become equal to that of helium ? The molecular weights of hydrogen and helium are 2 and 4 respectively. (A)
123oC
(B)
140oC
(C)
160oC
(D)
183oC
Ans. (a)
Solution 3-5: 3.
The masses of hydrogen and helium gases in the vessels are equal. This means that the product of the number of molecules and the mass of a molecule must be same for H2 and He gases. Since molecular masses of H2 and He are in the ratio 1: 2, their number of molecules nH and nHe in the vessels must be in the reverse ratio, that is,
4.
The equation of state for one mole of a gas is pV = RT = NkT
nH 2 nHe 1
Where N is Avogadro’s number (no. of molecules in one mole) and k is Boltzmann’s constant. If a gas has n molecules, the equation of state will be pV = nkT For a given volume and a given temperature, we have p n. Since H2 and He have same volume and same temperature (27 oC), we have pH n 2 H pHe nHe 1
Here pH = 2 atm.
5.
pHe = 1 atm.
Again, we have pV = nkT H2 and He have equal volumes. For having equal pressure, we must have nHTH = nHeTHe or
THe n H 2 TH nHe
Here THe = 27 + 273 = 300 K
TH =
1 THe = 150 K 2
= 150 273 = 123oC
6:
A vessel contains a mixture of 7 gm of nitrogen and 11 gm of carbon dioxide at temperature T = 290 K. If pressure of the mixture P = 1 atm, calculate its density (R = 8.31 J/mol k) (A)
2.5 kg/m3
(B)
1.5 kg/m3
(C)
4.5 kg/m3
(D)
7.5 kg/m3
Ans. (b) Solution:
As molecular weight of N2 and CO2 are 28 and 44, and n
m , M
nN
11 7 1 1/ 4 and n C 44 28 4
So, n nN n C
1 1 1 4 4 2
Now, according to gas law PV = nRT V
nRT 1 8.31 290 = 1.19 10 2 m3 5 P 2 1.01 10
and m = 7+11 = 18 gm = 18 10-3 kg so, =
18 10 3 kg m 1.5kg / m 3 2 3 V 1.19 10 m
Q.7-10.The pressure of a monoatomic gas increases linearly from 2 2 3 4 105 N/m to 8 105 N/m when its volume increases from 0.2 m to 3 0.5 m . Calculate 7:
work done by the gas (A)
2.8×105 J
(B)
1.8×106 J
(C)
1.8×105 J
(D)
1.8×102 J
Ans. (C)
8:
increase in internal energy (A)
U 4.8 105 J
(B)
U 4.8 104 J
(C)
U 6.8 105 J
(D)
U 4.8 106 J
Ans. (a)
9:
amount of heat supplied (A)
8.6 105 J
(B)
12.6 105 J
(C)
6.6.105 J
(D)
10.6 105 J
Ans. (C)
10:
molar heat capacity of the gas [R = 8.31 J/mol k] (A)
20.1 J/molK
(B)
17.14 J/molK
(C)
18.14 J/molK
(D)
20.14 J/molK
Ans. (b)
Solution 7- 10: 7.
Work done by the gas,
W PdV = area under P-v curve
= P1 VF VI
PF
B P
PI
A
1 PF PI VF VI 2 VI
1 = VF VI PF PI 2
=
8.
1 0.5 0.8 4 105 1.8 105 J 2
Change in internal energy of a gas is given by U nC V T
nRT PF VF PI VI 1 1
As the gas is monoatomic, = 5/3 So U
10 5 8 0.5 4 0.2 3 10 5 4 0.8 2 5 3 1
U 4.8 105 J
9.
From 1st law of thermodynamics Q U W
= 4.8 1.8 10 5 6.6. 10 5 J
10.
Molar heat capacity is defined as C
6.6 10 5 8.31 Q Q R 5 nT PF VF PI VI 10 8 0.05 4 0.2
i.e, C =
54.846 17.14 J/molK 3.2
V
VF
Q.11-13.Two moles of Helium gas ( = 5/3) are initially at temperature 27oC and occupy a volume of 20 litres. The gas is first expanded at constant pressure until the volume is doubled. Then it undergoes an adiabatic change until the temperature returns to its initial value. 11:
What are the final volume. (A)
113.13103 m 3
(B)
213.13103 m 3
(C)
313.13103 m 3
(D)
13.13103 m 3
Ans. (A)
12:
What are the final pressure of gas? (A)
0.44105 N / m 2
(B)
0.84105 N / m 2
(C)
0.94105 N / m 2
(D)
0.34105 N / m 2
Ans. (a)
13:
What is the work done by the gas? (Gas constant R = 8.3 T/mole K) (A)
13450J
(B)
14450J
(C)
16450J
(D)
12450J
Ans. (d)
Solution:11. From ideal gas equation PV = nRT initial pressure P 2.49 10 5 N / m 2
P
A
B
PA nRT 2 8.3 300 20 10 3 V
C vA
vB
When volume of gas is doubled at constant pressure, its temperature is also doubled. This process is shown on P-V curve by line AB. The gas then cools to temperature T adiabatically. This is shown by curve BC. The whole process is represented by curve ABC. At point B, pressure PB PA 2.49 10 5 N / m 2 . Volume VB 2VA 4010 3 m3 , Temperature TB 600K.
V
Now from adiabatic equation T V 1 = constant We have TA VA ( 1) TC VC( 1) 1
VC VB
VC 1 / 1 = 23/2 2 VB
TB 600 2 TC 300
Final volume
VC 2 2 VB
2 1.414 40103 113.13103 m3
12.
final pressure PC
13.
2 8.3 300 nRTC 0.44 10 5 N / m2 3 VC 113 .13 10
The work done by gas in isobaric process AB 2.49105 (40 20)103 4980 J
The work done by gas during adiabatic process BC W2
nR T2 T1 2 85.3 300 600 7470 J . 1 1 3
Net work done by gas W W1 W2
4980 7470 12450 J
14-15. When 1 gm of water changes from liquid to vapour phase at constant pressure of 1 atmosphere, the volume increases from 1 cm 3 to 1671 c.c. The heat of vaporization at this pressure is 540 cal/gm. Find 14:
The work done (in J) in change of phase (A)
170.78 Joule
(B)
200.67 Joule
(C)
190. 78 Joule
(D)
168 .67 Joule
(B)
3099.33 J
Ans. (d)
15:
Increase in internal energy of water. (A)
2099.33 J
(C)
4099.33 J
(D)
5099.33 J
Ans. (a)
Solution:14. As the process is isobaric W PdV PVF VI
= 1.01 10 6 1671 1 1688 .7 10 6 erg = 168 .67 Joule
15.
[1 erg = 10-7J]
From 1st law of thermodynamics Q U W Q mL 1 540 cal
= 2268 J, [ 1 cal = 4.2J] so, U Q W = 2268 168.67J = 2099.33 J
16:
A glass flask of volume one litre at 0 o C is filled level full of mercury at this temperature. The flask and mercury are now heated to 100 oC. How much mercury will spill out if coefficient of volume expansion of mercury is 1.82 104 / o C and linear expansion of glass is 0.1 104 / o C respectively? (A)
14.2 c.c.
(B)
15.2 c.c.
(C)
18.2 c.c.
(D)
20.2 c.c.J
Ans. (b) Solution: In case of thermal expansion of liquid, change in volume of liquid relative to container is given by V V L S
Here V = 1 litre = 1000 c.c. S = 3 glass = 0.3 104 / o C So, V 1000 1.82 0.3 10 4 100 0 = 15.2 c.c.
17:
Two cylinders A and B fitted with pistons contain equal amounts of an ideal diatomic gas at 300K. The piston A is free to move, while that of B is held fixed. The same amount of heat is given to the gas in each cylinder. If the rise in temperature of the gas in A is 30K, then the rise in temperature of the gas in B is. (A)
30K
(B)
18K
(C)
50K
(D)
42K
Solution:
For cylinder A. For cylinder B
dQ = nCPdT1 dQ = nCvdT2
nCPdT1 = nCvdT2 From (I) and (II) c v dT2 c v R30
dT2
(c v R)30 cv
5 2
For diatomic gas c v R
dT2 42K .
18:
80 gm of water at 30o C is poured on a large block of ice at 0 o C . The mass of ice that melts is (A)
30 gm
(B)
80 gm
(C)
150 gm
(D)
1600 gm
Solution: Since the block of ice at 0 o C is large, the whole of ice will not melt, hence final temperature is 0 o C .
Q1 = heat given up by water in cooling up to 0O C
= ms 80 1 30 0 = 2400 cal If m gm be the mass of ice melted, then
Q 2 = ML = m 80 Q1 Q2 m 80 2400 or m 30 gm
Here A is correct.
19:
A gas at pressure Po is contained in a vessel. If the masses of all the molecules are halved and their speeds doubled, the resulting pressure would be (A)
4Po
(B)
2Po
(C)
Po
(D)
Po 2
Solution:
P
1 mn 2 V rms where m = mass of one gas molecules 3 V
n = total no. of gas molecules i.e, P m and P Vrms Here m is halved and Vrms is doubled
pressure will be doubled Hence, (B) is correct
20:
The volume V versus temperature T graphs for a certain amount of a perfect gas at two pressures P1 and P2 are shown in the figure. Here (A) P1 < P2
(C) P1 = P2
(B) P1 > P2
(D) can’t be
Solution:
v
For a perfect gas, PV
m RT M
mR RT PM
So, the slope of the graph is Slope
1 P
mR PM
P2 P1 V
O
T
Hence P1 > P2 Hence, (C) is correct
21:
At room temperature the rms speed of the molecules of a certain diatomic gas is found to be 1930 m/s. The gas is (A)
H2
(B)
F2
(C)
O2
(D)
Cl 2
Solution:
M
Vrms 3RT Vrms
2
3RT M
3 x8.31x300 2
(1930)
2.0078x103 kg 2.00 gm
It is molecular weight of hydrogen (H2 ) .
22:
The latent heat of vaporization of water is 2240 J. If the work done in the process of vaporization of 1 gm is 168 J, then increase in internal energy is (A)
2408 J
(B)
2240 J
(C)
2072 J
(D)
1904 J
Solution:
L = 2240 J, m = 1 gm
dW = 168 J dQ = mL = dU + dW or 1 2240 dU 168 dU = 2072 J Hence, (C) is correct
23:
For a gas, y = 1.286. What is the number of degrees of freedom of the moleculas of this gas ? (A)
3
(B)
5
(C)
6
(D)
7
Ans. (d) Solution:
(D) 1
2 2 2 1.286 or 0.286 or 7 n n 0.286
24:
Which of the following temperatures is the highest? (A)
100 K
(B)
–13oF
(C)
–20oC
(D)
–30oC
Solution:
25:
(B ) –13oF is (13+32)o below ice point on F scale.
An ideal gas ( = 1.5) is expanded adiabatically. How many times has the gas to be expanded to reduce the root mean square velocity of molecules 2.0 times (A)
4 times
(B)
16 times
(C)
8 times
(D)
2 times
Ans. (B)
3RT M
Vrms
Solution: Vrms
T
Vrms is to reduce two times i.e, temperature of the gas will have to reduce four times or T 1 T 4
During adiabatic process TV 1 T V 1 1
1 V T 1 4 1.5 1 4 2 16 or, V T
V 16V
Hence, (B) is correct
26:
A thin copper wire of length L increases in length by 1% when heated from 0 o C to 100o C . If a thin copper plate of area 2L L is heated from 0 o C to 100o C , the percentage increase in its area will be (A)
1%
(B)
2%
(C)
3%
(D)
4%
Ans. (b)
L = Lo 1
Solution:
L L o L Lo
= percentage increase in length =
L L o 1
1 100 2
1 Hence 2L 2L o 1 100 2
2L2 2L2o 2L2o
or
A 2L2o
1 100
2
2 100
2 2% 100
Hence, (B) is correct
27:
Gas at pressure Po is contained in a vessel. If the masses of all the molecules are doubled and their speed is halved, the resulting pressure P will be equal to (A)
2Po
(B)
Po/4
(C)
Po
(D)
Po/2
Ans. (d) Solution: P=
Po = 1 3
1 mn 2 v rms 3 v
m n 2 v rms v
where m = 2m, v rms
v rms 2
putting the value
2 P mv rms 2m v 2 rms 1 = 2 2 Po m 4 v rms 2 mv rms
P = Po/2
28:
The molar heat capacity in a process of a diatomic gas if it does a work of Q/4, when Q amount of heat is supplied to it is (A)
2 R 5
(B)
(C)
10 R 3
(D)
5 R 2
6 R 7
Ans. (C) Solution:
5 2
dU = CV dT = R dT dT
2dU 5R
From 1st law of thermodynamics dU = dQ – dW or dU = Q Now molar heat capacity C
Q 3Q 4 4
dQ Q 5QR 10 R dT 3 3Q dU 2 2 4 5R
Hence (C) is correct
29:
For an ideal gas: (A)
the change in internal energy in a constant pressure process from temperature T1 to T2 is equal to nCv (T2 - T1), where Cv is the molar specific heat at constant volume and n the number of moles of the gas.
(B)
the change in internal energy of the gas and the work done by the gas are equal in magnitude in an adiabatic process.
(C)
the internal energy does not change in an isothermal process.
(D)
no heat is added or removed in an adiabatic process.
(A)
A, B
(B)
A, B, C
(C)
A, B, C, D
(D)
A, C
Solution: (C) Change in internal energy depends only on change in temperature since internal energy is a function of state only i.e. dU = nCv,dT. In adiabatic process, dQ = 0, Hence, dU + dW = 0 dU = dW
i.e. magnitude of change in internal energy is equal to magnitude of work done.
30:
Heat is supplied to a diatomic gas at constant pressure. The ratio of Q : U : W is (A)
5:3:2
(B)
5:2:3
(C)
7:5:2
(D)
7:2:5
Ans. (C) Q nC P dT
Solution:
U nC V T
7 nR T , 2
5 nR T , 2
7 CP 2 R 5 C V 2 R
and W Q U nRT Q : U : W 7 : 5 : 2 Hence, C is correct
31:
Two mole of argon are mixed with one mole of hydrogen, then Cp/Cv for the mixture is nearly (A)
1.2
(B)
1.3
(C)
1.4
(D)
1.5
Ans.(C) Solution: fav =
Average degree of freedom 2 3 1 5 11 23 5
mix = 1 +
1 f av
=1+
5 16 = = 1.4 11 11
Hence, C is correct Answer.
32:
An ideal gas is taken through the cycle A BC A as shown in figure. If the net heat supplied to the gas in the cycle is 5J, the work done by the gas in the process CA is, (A) -5J
(B) -10J
(C) -15J
(D) -20J
Solution:
(A) Qnet Wnet WCA WAB
5 WCA 10 (2 1) 5 10 WCA WCA 5
33:
When an ideal gas at pressure P, temperature T and volume V is isothermally compressed to a V/n, its pressure becomes Pi . If the gas is compressed adiabatically to V/n, its pressure becomes P a. The ratio Pi / Pa is (A)
1
(B)
n
(C)
n
(D)
n 1-
Solution:
(D) For isothermal process, PV = constant. Therefore
PiVi = PV or Pi
V PV or Pi = nP n
……(i)
For adiabatic process, PV = constant. Therefore Pa (Va) = PV
or
(P, V, T)
V Pa PV n
or
Pa = nP
…..(ii)
P
V Pn , ,T ' n
From (i) and (ii) we get Pi n n (1 ) Pa n
34:
V Pi , ,T n
v
When an ideal monatomic gas is heated at constant pressure, the fraction of heat energy supplied which increases the internal energy of the gas is (A)
2/5
(B)
3/5
(C)
3/7
(D)
3/4
Solution:
U
(B) U
w 1
nRT 1
w nRT
Q U 1
Q W U
U ( 1) Q U U 1 1 Q
U 1 3 [for monoatomic gas = 5/3 ] Q 5
35:
A monatomic ideal gas, initially at temperature T1 is enclosed in a cylinder fitted with a frictionless piston. The gas is allowed to expand adiabatically to a temperature T2 by releasing the piston suddenly. If L1 and L2 are the lengths of the gas column before and after expansion respectively, then T1/T2 is given by (A)
(L1/L2)2/3
(B)
L1/L2
(C)
L2/L1
(D)
(L2/L1) 2/3
Solution:
(D) TV 1 constant
Initial position T1(L1A)–1 = constant Final position T2 (L2A)–1 = constant
T L 1 1 T2 L2
1
1
T1 L2 T2 L1 L 2 L1
5 / 3 1
[for monoatomic gas
L 2 L1
36:
1
= 5/3]
2/3
An ideal mono atomic gas at 300K expands adiabatically to twice its volume. What is the fine temperature (A)
189K
(B)
289K
(C)
30Kj
(D)
Non of these
Solution:
(A) TV 1 constant
1 TV T2V2 1 1 1
300 V11 T2 2V1
1
300 T2 21 T2
37:
300 300 300 300 5 / 31 2 / 3 189K 1 2 2 2 1.587
What will be P-V graph corresponding to the P-T graph (process AB) for an ideal gas shown in figure
B
C
P
A
D
T
(A)
Hyperbolic
(B)
Circle
(C)
Straight line
(D)
Elliptical
Solution: (A)
38:
Starting with the same initial conditions, an ideal gas expands from volume V1 to V2 in three different ways. The work done by the gas is W1 if the process is purely isothermal, W2 if purely isobaric and W3 if purely adiabatic. Then (A)
W2 > W1 >W3
(B)
W2 > W3 >W1
(C)
W1 > W2 >W3
(D)
W1 > W3 >W2
Solution:
(A) Isobaric
P Isothermal Adiabatic V
39: of =
One mole of argon is heated using PV5/2 = = const. By which amount heat is obtained by the process when the temperature change by T = -26K. (A)
100J
(B)
200J
(C)
108J
(D)
208J
(C)
Here n = 1
Solution: C= C=
R R 1 1 x
For
polytropic process
R R 5 / 3 1 3 1 2
3 Q nCT 1 R 2R 26 108J 2
40: =
3 moles of an ideal monoatmic gas performs a cycle as shown in the fig. The gas temperatures T1 = = 400K, T2= = 800K, T3 = = 2400K, T4 = 1200K. What will be the net work done.
P
B
A
C D T
(A)
20J
(B)
20000J
(C)
200J
(D)
2000J
Solution:
(B) W AB = CCD = 0, because process is isochoric.
WAD = nRT = 3R(TA – TB) = 3 8.31 (2400 – 800) = 39884
Total work done W AD + W BC = 39888 – 19944 = 20 103J = 19944 20 103J
41:
How much heat is absorbed by the system in going through the process shown in the fig. (consider that value is taken in SI system)
900 P
100 800
V
400
(A)
20.4104 J
(B)
30.4104 J
(C)
21.4104 J
(D)
25.12104 J
Solution:
(D) Work done = Area of PV diagram
=
P2 P1 V2 V1 4
=
800 400 4
= 800 100 3.14 = 8 104 3.14 = 25.12 104
42:
3000J of heat is given to a gas at constant pressure of 2105 N/m2. If its volume increases by 10 litres during the process, what will be the change in the internal energy of the gas (A)
1000J
(B)
100J
(C)
200J
(D)
2000J
Solution: (A) Q 3000J P = 2 105 = Vi Vf = (Vi + 10 10–3) W = Pdv W = 2 105 10 10-3 = 2 103
Q W U 3000 = 2 103 + U
Q 1000
43:
A gas at atmospheric pressure is contained in a cylinder of volume 80 litre. When it is compressed adiabatically to 20 litre its pressure rises to 7 atm. What will be the ratio of specific heats of the gas (A)
1.33
(B)
1.4
(C)
1.67
(D)
1.5
Pi = 1atm = 1 105 N/m2
Solution: (B) Vi = 80 10 – 3m3 Vf = 20 10–3 m3 Pf = 7atm = 7 105 N/m2 PiVi = PfVf
1 105 (80 10–3) =(7 105)(20 10–3)
80 10 3 7 3 20 10
4
7
log 4 log7 log7 1.40 log 4
44:
A gas consisting of rigid diatomic molecules was initially under standard conditions. Then gas was compressed adiabatically to one fifth of its initial volume. What will be the mean kinetic energy of a rotating molecule in the final state? (A)
1.44 J
(B)
4.55J
(C)
787.98 10–23
(D)
757.310-23J
Solution: (C)
1.4
TV 1 constant T1V1 -1 = T2 V2 -1 7
(300)V1
7/5 – 1
V 5 = T2 1 5
1
T2
300 V12 / 5 1 V12 / 5 5
2/5
300 300 52 / 5 300 (1.903) = 571 2 / 5 5
Mean kinetic energy of rotating molecules = KT = 1.38 10 – 23 571 KT = 787.98 10–23
45:
Immediately after the explosion of an atom bomb, the ball of fire produced has a radius of 100m and a temperature 105K . What will be the approximate temperature when the ball expands adiabatically to a radius of 1000m (suppose mono atomic gas is there) (A)
1000K
(B)
100K
(C)
105 (10–3)2/3
(D)
200K
Solution:
Vi =
(C) r = 100 m
4 (100)3 3
Ti = 105 K after explosion r = 1000 m, Vf = 1 TV Tf Vf 1 i i
V Tf Ti i Vf
1
5
3 4 3 (100) Tf = 105 3 4 (1000)3 3
1
= 105 (10–3)2/3 = 1.05 K
46:
Which of the following is false? (A)
Enthalpy is a path function.
(B)
Work is a path function.
(C)
Heat is a path function.
4 (1000)3 3
(D) Solution:
47:
Energy is a state function (C)
A gas mixture consists of 32 gram of oxygen and 36 gram of Ar a temperature T. Neglecting all vibration modes, the total internal energy of the system is (A) 4RT
(B) 8RT
(C) 9RT
(D) 11RT
Solution: (D) F 5 3 n RT 2 RT 4 RT 11RT 2 2 2
48:
Energy
=
A mono atomic gas is supplied heat Q very slowly keeping the pressure constant. The work done by the gas is (A)
2/5 Q
(B)
3/5 Q
(C)
Q /5
(D)
2/3 Q
Solution:
(D) For monatomic gas
U 1 Q 3
or U
Q 3
From the first law of thermodynamics Q = U + W
2 W Q 3 49:
Which of the following parameters does not characterise the thermodynamic state of matter (A)
Work
(B)
Pressure
(C)
Temperature
(D)
Volume
Solution:
50:
(A) Work
Which of the following is correct (A)
For an isothermal change PV = = constant
(B)
For a isothermal process, the change in internal energy must be equal to the work done
P V For an adiabatic change 2 2 , where is the ratio of the P1 V1 two specific heats
(C)
(D)
In an adiabatic process external work done must be equal to the heat entering the system
Solution:
51:
(A)
An ideal gas goes through cyclic process ABC and following (P vs T) curve is obtained. This process can be represented by
B P C
A
T
B
B
P
P C
A
A
C V
V
(A)
(B)
B B
P
P
C
A
V
(C)
C
A
V
(D)
1 . Process B C is isobaric V and C A adiabatic. Slope of adiabatic > slope of isothermal.
Solution:(B) Process A to B isothermal. Then P
52:
A container contain 0.1 mol of H2 and 0.1 mol of O2 , If the gases are in thermal equilibrium then (A)
Only the average kinetic energy of the molecule of H2 and O2 is same.
(B)
Average speed of the molecule of H2 and O2 is same.
(C)
Only the specific heat at constant pressure of two gases is same.
(D) energy gases.
(d) The specific heat at constant pressure and the kinetic are same for both the
Solution: (D) The specific heat at constant pressure (Cp) is the amount of heat required to raise the temperature of one gram through 1°C when the pressure of the gas is kept constant. Again, the mean kinetic energy per molecule (3/2)kT depends only upon temperature. Clearly both the specific heats at constant pressure and mean kinetic energy are depending on the temperature which is again same for the two gases.
53:
Two systems are in thermal equilibrium. The quantity which is common for them is (A)
Heat
(B)
Momentum
(C)
Temperature
(D)
Specific heat
Solution:
54:
Mean molecular weight is defined as (A)
the number of free particles per positron mass
(B)
the number of free particles per electron mass
(C)
the number of free particles per neutral mass
(D)
the number of free particles per photon mass
Solution:
55:
(C)
(D)
Which one of the following statements is true about a gas undergoing an adiabatic change
(A)
The temperature of the gas remains constant
(B)
The pressure of the gas remains constant
(C)
The volume of the gas remains constant
(D)
The gas is completely insulated from the surroundings
Solution:
56:
If an ideal gas is allowed to expand adiabatically, the work done is equal to (A)
The loss in heat
(B)
The loss in internal energy
(C)
The gain in internal energy
(D)
The gain in enthalpy
Solution:
57:
(A)
Isothermal
(B)
Adiabatic
(C)
Isobaric
(D)
Isochoric
(A)
zero
(B)
infinite
(C)
positive
(D)
negative
(B)
The internal energy of the system remains constant when it undergoes (A)
a cyclic process
(B)
an adiabatic process
(C)
an isothermal process
(D)
an isobaric process
Solution:
60:
(A)
Specific heat of a gas undergoing adiabatic changes is
Solution:
59:
(B)
For the Boyle’s law to hold, the necessary condition is
Solution:
58:
(D)
(C)
The first law of thermodynamics incorporates the concepts of
(A)
conservation of energy
(B)
conservation of heat
(C)
conservation of work
(D)
equivalence of heat and work
Solution:
(D)
12-Transference of heat
SYLLABUS Thermal expansion of solids, liquids and gases; Calorimetry, latent heat; Heat conduction in one dimension; Elementary concepts of convection and radiation; Newton’s law of cooling; bulk modulus of gases; Blackbody radiation: absorptive and emissive powers; Kirchhoff’s law, Wien’s displacement law, Stefan’s law, Specific heat of a liquid using calorimeter (M and E)
1.
Thermal Expansion
When a body is heated, it expands in terms of length, area & volume and temperature rises. In a solid, molecules can only have thermal agitation (random vibrations). As temperature of a body increases, the vibrations of molecules will become fast and due to this the rate of collision among neighbouring molecules increases, it develops a thermal stress in the body and due to this the intermolecular separation increases which results in thermal expansion of body. At ordinary temperatures, the atoms in a solid oscillate about their equilibrium position with an amplitude of approximately 10–11 m. The average spacing between the atoms is about 10–10 m. As the temperature of solid increases, the atoms oscillate with greater amplitudes, as a result the average separation between them increases. Consequently the object expands. In the similar way the block diagram shown in Fig. explains the way how thermal expansion takes place. Temperature of Body Increases
Themal vibration energy of molecules increases
Rate of collision between neighbouring molecules increases
Interatomic separation increases
Thermal exp ansion takes place
Thermal expansion of a substance can be classified in three broad categories, these are (i) Linear Expansion (ii) Superficial Expansion and (iii) Cubical Expansion or Volume Expansion (i) Linear Expansion: Consider a rod of length l1 at a temperature T1. Let it be a heated to a temperature T2 and the increased length of the rod be l2 then l2 l1 (1 t) Coefficient of linear expansion and t T2 T1 Example 1. The density of substance at 0C is 10 g cm-3 and at 100C, its density is 9.7g cm-3. The coefficient of linear expansion of substance is d0 dt (1 3t) Solution: 10 9.7(1 3 100) 0.3 10 1 1 or 0.0001/ C or 9.7 300 9.7 300
(ii) Superficial Expansion (expansion in surface area): If A1 is the area of solid at T1 C and A2 is the area at T2C . Then A2 A1(1 t ) Coefficient of superficial (areal) expansion and t T2 T1
(iii) Volume expansion (Part A: expansion in solids): If V1 is the volume of solid at T1C and V1 is the volume at T2 C then V2 V1 1 t coefficient of cubical (volume) expansion and t T2 T1
Note: For isotopic solids: 2, 3
Relation between expansion coefficient are 6 3, 2
As temperature increases, density of solid decreases. If d1 is the density at T1C, d2 is the density at T2C then d1 d2 = or d2 =d1 1- t where t T2 T1 1+t (iv) Expansion of gases (Part B: expansion in gases): Pressure coefficient of a gas is the ratio of increase in pressure for 1C rise in temperature to the pressure at 0C, provided the volume of the gas is kept constant. p =
pt po po t
Where p = pressure coefficient Pt = pressure at tC Po= pressure at 0C Volume coefficient of a gas is similarly defined as (pressure being kept constant) V Vo v = t Vo t Where v = volume coefficient Vt = volume at tC Vo= volume at 0C Experiments have shown the value of p (or v) is the same for all gases and equal to 1/273 per degree celsius, i.e. 1 1 perC or per K, p = v = 273 273 Where K stands for absolute of kelvin temperature. Example 2. A grid iron pendulum consists of 5 iron rods and 4 brass rods. What will be the length of each brass rod if the length of each iron rod is 1 m ? (Fe= 12 10-6/C, brass = 20 10-6/C) Solution: There are 5 iron rods and 4 brass rods in the pendulum. The number of iron in one half of the pendulum including the central rod is 3. The number of brass rods is 2. Since the pendulum is compensated, lbrass liron or
2l2 2 t 3l11t or l2
3l11 2 2
3 1 12 10 6 m 0.9m 2 20 106
2.
ERROR in measurament DUE TO THERMAL EXPANSION
2.1
HEATING A METALLIC SCALE
A metallic scale (linear) expands in length when heated. As a result all the markings are displaced from their usual (correct) positions. A reading of l unit on a heated scale is equivalent to an actual length of l 1 t , where is coefficient of linear expansion of material of scale, and t is the temperature of the heated scale. If the reading is x, actual length x(1 t) actual length reading (1 t)
2.2
1
2
3 Scale
1
2
3 Heated scale
DIFFERENCE OF LENGTHS OF TWO RODS
Consider two rods 1 and 2 of lengths l1 and l2. Let they be heated through a temperature t . If l’1 and l’2 are their expanded lengths, then: l’2 l2 1 2 t where 2 is coefficient of linear expansion of rod 2 l’1 l1 1 1t where 1 is coefficient of linear expansion of rod 1 Since the difference of length of two rods is constant l '2 l '1 l 2 l1 l11 l 22
Example3. Two rods of length l1 and l2 are made of materials whose coefficients of linear expansion are 1 and 2 respectively. If the difference between the two lengths is independent of temperature, then relation between length and linear expansion coefficients. Solution: l1 l11t l2 l22 t
l1 l2 0
2.3
l11t l22 t or
l1 1 l2 2
TIME PERIOD OF PENDULUM
Time period (T) of a sample pendulum of length l is given by l T 2 g If there is a rise in temperature by t, length of the pendulum increases and hence the time period increase. As a result the clock slows down. If l0 be the length of the pendulum and corresponding time period be T0 then l T0 2 0 g If the pendulum be heated by t (rise in temperature), the new time period T1 is:
T1 2
l1 g
l 1 t T1 l 1 0 ln T0 l0
T1 1 as is very small 1 t 1 t T0 2 T T 1 T 1 1 0 t t T0 2 T0 2 The above relation gives the time lost per second by the pendulum clock. If t is the fall in temperature, same equation will give the time gained by the clock per second as its oscillation will become faster due to reduction in its length.
Example 4:
A clock with a metallic pendulum is 5 second fast each day at a temperature of 15 o C and 10 seconds slow each day at a temperature of 30 o C . Find coefficient of linear expansion for the metal.
Solution:
Time lost or gained per second by a pendulum clock is given by 1 ( Here T is difference of temperature) t T 2 Here temperature is higher then graduation temperature thus clock will loose time and if it is lower then graduation temperature will gain time. 1 t T 86400 [as 1 day 86400sec.] 2 If graduation temperature of clock is T0 then we have. At 15C, clock is gaining time, thus 1 5 = T0 15 86400 ……(1) 5 At 30C clock is loosing time, thus 1 ……(2) 10 = 30 T0 86400 5 Dividing equation (2) by (1), we get 2(T0 – 15) = (30 – T0) or T0 = 20C Thus from equation (1) 1 5 = 20 15 86400 2 2.31 105 C
2.4
STRESS IN OBJECTS DUE TO THERMAL EXPANSION
If cross–sectional area of wire is A and F be the tension developed in wire due to stretching. Thus the stress developed in the wire due to this tension F is given as Force Strees = = F/A …..(1) Area Strain produced in wire due to its elastic properties is L Strees = T …..(2) L If young’s modulus of the material of wire is Y, we have
Y
strees strain
F/A T or F YA T …..(3) Equation - (3) gives the expression for tension in the wire due to decrease in its temperature by T. This result gives the tension in wire if initially wire is just taught between clamps. If it already has some tension in it then this expression will give the increment in tension in the wire.
or
Y
Example 5. A uniform metal rod of 2 m m2 cross-section is heated from 0C to 20C. The coefficient of liner expansion of the rod is 12 10-6 per C, Y=1011 N/m2. The energy stored per unit volume of the rod is solution: Energy per unit volume 1 1 1 stress strain (Yt)(t) Y 2t 2 2 2 2 12 11 10 144 10 400 J m3 2 288 10 J m3 2880 J m3
3.
SPECIFIC HEAT
When heat energy flows into a substance, the temperature of the substance usually rises. The heat required to raise the temperature of unit mass of a body through 1 oC or ( 1 oK ) is called specific heat capacity or simply specific heat of the material of the body. If Q heat changes the temperature of mass m by T. 1 Q … (1) c m T The SI unit of specific heat is J/kg K. Heat is so frequently measured in calories, therefore the practical unit cal/g C is also used quite often. The specific heat capacity of water is approximately 1 cal/g C. From Eq. (1), we can define the specific heat of a substance as “the amount of energy needed to raise the temperature of unit mass of that substance by 1C (or 1 K)”. A closely related quantity is the Molar heat capacity C. It is defined as, 1 dQ … (2) C n dT where n is the number of moles of the substance. If M is the molecular mass of the m substance, then n = were m is the mass of the substance and, M M dQ C= … (3) m dT The SI units of C is J / mole K. Key points: (a) It depends on nature of material of body. Dulong and petit has found formula for elemental solids that (with few exceptions such as carbon) Atomic weight Specific heat 6 cal / oC So, heavier the element lesser will be the specific heat, i.e., CHg < CCu< CAl
Specific heat of a substance also depends on temperature (particularly at low temperatures) the variation of specific heat with temperature for wateris shown in Fig (A) for metals in Fig (B). This temperature dependence of specific heat is usually neglected. 1.008
6
1.004
4
Molar Sp. heat
Sp. heat cal / gm C
(b)
1.000
0.996
40
60
80
100
100
Temp. in C
Fig. (A) Water
(e)
(f)
(g) (h)
(i)
300 200 Temp. in K
400
500
Fig. (B) Metals
Specific heat also depends on the state of substance, i.e., solid, liquid or gas. e.g., specific heat of solid copper will be different from that of liquid copper. In case of water Specific heat of Ice (solid) Water (liquid) Steam (gas)
(d)
2
0 20
(c)
c T3
In cal/g C 0.5 1 0.47
In SI units, i.e., J/kg K 2100 4200 1970
If a substance is undergoing change of state which takes place at constant temperature (called isothermal change) , specific heat 1 dQ 1 dQ [as T = 0] c m 0 m dT i.e., specific heat of a substance at its melting pint or boiling point or isothermal change is infinite. Specific heat is found to be maximum for hydrogen (3.5 cal/gm C) then for water (1 cal/gm C = 4200 J/kg K). For all other substances specific heat is lesser than 1 cal/gm C and is minimum for radon and actinium (= 0.22 cal/gam C) If the temperature of a body changes without transfer of heat with the surroundings (adiabatic change) as in shaking a liquid or compressing a gas, Q 0 c= =0 [as Q = 0] mT mT i.e., specific heat of a substance, when it undergoes adiabatic change, is zero. Specific heat of a substance can also be negative. Negative specific heat means that in order to raise the temperature, a certain quantity of heat is to be withdrawn from the body. Specific heat of saturated water vapours is negative. When specific heats are measured, the values obtained are also found to depend on the conditions of the experiment. In general measurements made at constant pressure are different from those at constant volume. For solids and liquids this difference is very small and usually neglected. The specific heat of gases are quite different under constant pressure condition (cp) to constant volume condition (cv). As by definition c = (Q/m T), heat required to change the temperature of m gm of a substance through T: Q = mc T and as T = (Q/mc), greater the specific heat of a substance lesser will be the change in temperature for a given mass when same amount of heat is supplied. Now as specific heat of water is very large (1 cal/g C), by absorbing or releasing large
amount of heat its temperature changes by small amounts. This is why, it is used in hot water bottles or as coolant in radiators. This is also how the sea moderates the climate of nearby coastal land.
4.
WATER EQUIVALENT
Water–equivalent of a body is the mass of water which when given same amount of heat as to the body, changes the temperature of water through same range as that of the body, i.e., W = (m c) gm The unit of water equivalent W is gm while its dimensions [M]. Units and dimensions of some physical–quantities used in heat are given below in a tabular form. S. No. 1. 2. 3.
Physical quantity Heat Specific–heat Molar sp. heat
Symbol Q c C
Dimensions
1 –1
]
4. 5. 6.
5.
Latent heat Thermal capacity Water–equivalent
L Tc W
[ML2T–2] [L2T–2–1] [ML2T–2– [L2T–2] [ML2T–2–1] [M]
Units SI Joule J/kg K J/mol K
CGS calorie cal/gm C cal/mol C
J/kg J/K kg
cal/gm cal/C gm
PHASE CHANGES AND LATENT HEAT
Suppose that we slowly heat a cube of ice whose temperature is below 0C at atmospheric pressure, what changes do we observe in the ice? Initially we find that its temperature increases according to equation Q = mc(T2 – T1). Once 0C is reached, the additional heat does not increase the temperature of the ice. Instead, the ice melts and temperature remains at 0C. The temperature of the water then starts to rise and eventually reaches 100C, whereupon the water vaporizes into steam at this same temperature. During phase transitions (solid to liquid or liquid to gas) the added heat causes a change in the positions of the molecules relative to one another, without affecting the temperature. The heat necessary to change a unit mass of a substance from one phase to another is called the latent heat (L). Thus, the amount of heat required for melting and vaporizing a substance of mass m are given by, Q = mL … (1) For a solid-liquid transition, the latent heat is known as the latent heat of fusion (Lf) and for the liquid-gas transition, it is known as the latent heat of vaporization (Lv).
6.
PRINCIPLE OF CALORIMETRY
When two bodies (one being solid and other liquid or both being liquid) at different temperatures are mixed, heat will be transferred from body at higher temperature to body at lower temperature till both acquire same temperature. The body at higher temperature releases heat while body at lower temperature absorbs it, so that: Heat lost = Heat gained, i.e. principle of calorimetry represents the law of conservation of heat energy. While, using this principle always keep in mind that:
(a)
Temperature of mixture (T) is always lower temperature (TL) and higher temperature (TH), i.e., TL T TH i.e. the temperature of mixture can never be lesser than lower temperatures (as a body cannot be cooled below the temperature of cooling body) and greater than higher temperature (as a body cannot be heated above the temperature of heating body when there is no chemical reaction).
(b)
When temperature of a body changes, the body releases heat if its temperature falls and absorbs heat when its temperature rises. The heat released or absorbed by a body of mass m is given by: Q = mc T Where c is specific heat of the body and T change in its temperature in C or K.
(c)
When state of a body changes, change of state takes place at constant temperature [m.pt. or b.pt.] and heat released or absorbed is given by Q = mL Where L is latent heat. Heat is absorbed if solid converts into liquid (at m.pt.) or liquid converts into vapours (at b.pt.) and is released if liquid converts into solid or vapours converts into liquid.
HEAT TRANSFER Heat can be transferred from one place to the other by any of three possible ways: conduction, convection and radiation. In the conduction, convection processes, a medium is necessary for the heat transfer. Radiation, however, does no have this restriction. This is also the fastest mode of heat transfer, in which heat is transferred from one place to the other in the form of electromagnetic radiation. (i) Conduction Figure shows a rod whose ends are in thermal contact with a hot reservoir at temperature T1 and a cold reservoir at temperature T2. The sides of the rod are covered with insulating medium, so the transport of heat T1 T2 is along the rod, not through the sides. The molecules at the hot reservoir have greater vibrational energy. This T1 Q energy is transferred by collisions to the atoms at the end (Hot) face of the rod. These atoms in turn transfer of heat through a substance in which heat is transported without direct mass transport is called conduction. Most metals use another, more effective mechanism to conduct heat. The free electrons, which move throughout the metal can rapidly carry energy from the hotter to cooler regions, so metals are generally good conductors of heat. The presence of ‘free’ electrons also causes most metals to be good electrical conductors. A metal rod at 5C feels colder than a piece of wood at 5C because heat can flow more easily from your hand into the metal. Heat transfer occurs only between regions that are at different temperature, and the rate dQ . This rate is also called the heat current, denoted by H. Experiments of heat flow is dt show that the heat current is proportional to the cross-section area A of the rod and to the
T2 (Cold)
temperature gradient
dT , which is the rate of change of temperature with distance along dx
the bar. In general dQ dT H= kA dt dx
… (1)
dQ dT a positive quantity since is negative. The dt dx constant k, called the thermal conductivity is measure of the ability of a material to conduct heat. A substance with a large thermal conductivity k is a good heat conductor. The value of k depends on the temperature, increasing slightly with increasing temperature, but k can be taken to be practically constant throughout a substance if the temperature difference between its ends is not too great. Let us apply Eq. (1) to a rod of length L and constant cross sectional area A in which a steady state has been reached. In a steady state the temperature at each point is constant in time. Hence, dT T1 T2 dx Therefore, the heat Q transferred in time t is T T Q kA 1 2 t …(2) L
The negative sign is used to make
Thermal Resistance (R) Eq. (2) in differential form can be written as dQ T T …(3) H dt l / kA R Here, T = temperature difference (T.D) and l = thermal resistance of the rod. R kA (ii) Convection Although conduction does occur in liquids and gases also, heat is transported in these media mostly by convection. In this process, the actual motion of the material is responsible for the heat transfer. Familiar examples include hot-air and hot-water home heating systems, the cooling system of an automobile engine and the flow of blood in the body. You probably have warmed your hands by holding them over an open flame. In this situation, the air directly above the flame is heated and expands. As a result, the density of this air decreases and then air rises. When the movement results from differences in density, as with air around free, it is referred to as natural convection. Air flow at a beach is an example of natural convection. When the heated substance is forced to move by a fan or pump, the process in called forced convection. If it were not for convection currents, it would be very difficult to boil water. As water is heated in a kettle, the heated water expands and rises to the top because its density is lowered. As the same time, the denser, cool water at the surface sinks to the bottom of the kettle and is heated. Heating a room by a radiator is an example of forced convection. Ingen Hausz Experiment:
Ingen Hausz provided a method to compare the thermal conductivities of different materials. He took wax coated rods of different materials but of the same area. One end of the rods is kept in a hot water bath and the other end is kept at the temperature of surroundings. If 1, 2, 3 . . . represent the lengths upto which the wax has melted and K1, K2, K3 . . . are their thermal conductivities, then K1 21
or
K2 22
= . . . = constant
K 1 21 K 2 22
Example-6. A wall is made of two equally thick layers A and B of different materials. The thermal conductivity of A is twice that of B. In the steady state, the temperature difference across the wall is 36C. The temperature difference across the layer A will be Solution: K constant A K 1 B 2K 2 1 A 36 12C 2 1
(iii) Radiation The third means of energy transfer is radiation which does not require a medium. The best known example of this process is the radiation from sun. All objects radiate energy continuously in the form of electromagnetic waves. 1 2 Energy 2 0E The rate at which an object radiates energy is proportional to the fourth power of its absolute temperature. This is known as the Stefan’s law and is expressed in equation form as P = AeT4 Here P is the power in watts (J/s) radiated by the object, A is the surface area in m2, e is a fraction between 0 and 1 called the emissivity of the object and is a universal constant called Stefan’s constant, which has the value = 5.67 10–8 W/m2-K4 BLACK BODY RADIATION (i)
Perfectly black body
A body that absorbs all the radiation incident upon it and has an emissivity equal to 1 is called a perfectly black body. A black body is also an ideal radiator. It implies that if a black body and an identical another body are kept at the same temperature, then the black body will radiate maximum power as is obvious from equation p = eat4 also. Because e = 1 for a perfectly black body while for any other body e < 1. Materials like black velvet or lamp black come close to being ideal black bodies, but the best practical realization of an ideal black body is a small hole leading into a cavity, as this Absorbs 98% of the radiation incident on them. (ii) Absorptive power ‘a’ “It is defined as the ratio of the radiant energy absorbed by it in a given time to the total radiant energy incident on it in the same interval of time.” energy absorbed a energy incident As a perfectly black body absorbs all radiations incident on it, the absorptive power of a perfectly black body is maximum and unity. (iii) Spectral absorptive power ‘a’ The absorptive power ‘a’ refers to radiations of all wavelengths (or the total energy) while the spectral absorptive power is the ratio of radiant energy absorbed by a surface to the radiant energy incident on it for a particular wavelength . It may have different values for different wavelengths for a given surface. Let us take an example, suppose a = 0.6, a = 0.4 for 1000 Å and a = 0.7 for 2000 Å for a given surface. Then it means that this surface will absorbs only 60% of the total radiant energy incident on it. Similarly it absorbs 40% of the energy incident on it corresponding to 1000 Å and 70% corresponding to 2000 Å. The spectral absorptive power a is related to absorptive power a through the relation
a a d 0
(iv) Emissive power ‘e’ (Don’t confuse it with the emissivity e which is different from it, although both have the same symbols e). “For a given surface it is defined as the radiant energy emitted per second per unit area of the surface.” It has the units of W/m2 or J/s–m2. For a black body e=T4. (v) Spectral emissive power ‘e’ “It is emissive power for a particular wavelength .” Thus,
e e d 0
Kirchhoff’s law: “According to this law the ratio of emissive power to absorptive power is same for all surfaces at the same temperature.” T
T
1
2
Perfectly black body
T
Fig.
e1 e e 2 Perfectly black body a1 a2 a but (a)black body = 1 and (e)black body = E (say) e Then, cons tan t E a for any surface Similarly for a particular wavelength , e E a for any body Hence,
Here
E = emissive power of black body at temperature T = T4 From the above expression, we can see that e a i.e., good absorbers for a particular wavelength are also good emitters of the same wavelength. COOLING BY RADIATION Consider a hot body at temperature T placed in an environment at a lower temperature T0. The body emits more radiation than it absorbs and cools down while the surrounding absorb radiation from the body and warm up. The body is losing energy by emitting radiations at a rate. P1 = eAT4 and is receiving energy by absorbing radiations at a rate P2 = aAT04 Here ‘a’ is a pure number between 0 and 1 indicating the relative ability of the surface to absorb radiation from its surroundings. Note that this ‘a’ is different from the absorptive power ‘a’. In thermal equilibrium, both the body and the surrounding have the same temperature (say Tc) and, P1 = P2 or eATc4 = aATc4 or e=a Thus, when T > T0, the net rate of heat from the body to the surroundings is, dQ eA (T4 – T04) dt dT 4 4 or mc eA (T – T0 ) dt Rate of cooling
dT dt
or
eA 4 (T T0 4 ) mc
dT (T4 – T04) dt
NEWTON’S LAW OF COOLING According to this law, if the temperature T of the body is not very different from that dT of the surroundings T0, then rate of cooling – is proportional to the temperature dt difference between them. T0 prove it let us assume that T = T0 + T 4
T So that T = (T0 + T) = T 1 T0 4T (from binomial expansion) T04 1 T0 4
4
4 0
(T4 – T04 ) = 4 T03 (T)
or (T4 – T04 ) T (as T0 = constant) Now, we have already shown that rate of cooling dT 4 4 dt (T T0 ) and here we have shown that (T 4 T04 ) T , if the temperature difference is small. Thus, rate of cooling d dT dt T or dt as dT = d or T = Variation of temperature of a body according to Newton’s law Suppose a body has a temperature i at time t = 0. It is placed in an atmosphere whose temperature is 0. We are interested in finding the temperature of the body at time t, assuming Newton’s law of cooling to hold good or by assuming that the temperature difference is small. As per this law, rate of cooling temperature difference d eA 3 or dt mc (40 ) 0 d or dt 0 4eA30 Here = is a constant mc t d i 0 0 dt
0
0 i 0 et
cons tan t
i
t0
0 cons tan t
tt
FROM THIS EXPRESSION WE SEE THAT = I AT T = 0 AND = 0 AT T = , I.E. TEMPERATURE OF THE BODY VARIES EXPONENTIALLY WITH TIME FROM I TO 0 (<I). THE TEMPERATURE VERSUS TIME GRAPH IS AS SHOWN IN FIG.
i
0
t
NOTE: IF THE BODY COOLS BY RADIATION FROM 1 TO 2 IN TIME T, THEN TAKING THE APPROXIMATION d 1 2 and = av = 1 2 t dt 2 d The equation 0 becomes dt 1 2 1 2 0 t 2 This form of the low helps in solving numerical problems related to Newton’s law of cooling.
WEIN’S DISPLACEMENT LAW At ordinary temperatures (below about 600C) the thermal radiation emitted by a body is not visible, most of it is concentrated in wavelengths much longer than those of visible light. Figure shows how the energy of a black body radiation varies with temperature and wavelength. As the temperature of the black body increases, two distinct behaviours are observed. The first effect is that the peak of the distribution shifts to shorter wavelengths. This shift if found to obey the following relationship called Wein’s displacement law maxT = b Here b is a constant called Wein’s e constant. The value of this constant in 4000 K SI unit is 2.898 10–3 m–K. Thus, 3000 K 1 max T 2000 K Here max is the wavelength corresponding to the maximum spectral emissive power e. 0 3 1 2 4 The second effect is that the total Wavelength ( m) amount of energy the black body emits per unit area per unit time (= T4) increases with fourth power of absolute temperature T. This is also known as the emissive power. We know e =
0
ed = Area under e–
graph = T4
Area T4
or
e
A1
T m
A2
2T m 2
A2 = (2)4A1 = 16A1 Thus, if the temperature of the black body is made two fold, max remains half while the area becomes 16 times. Example-7. Estimate the surface temperature of sun. Given for solar radiations, m = 4753 A° Solution: From Wien’s displacement law m T = b T=
b (2.89 10 3 m k ) = 6097°K. m ( 4753 10 10 m)
Example-8. Experimental investigations show that the intensity of the solar radiation is maximum for wavelength m 4753 A o in the visible region. Estimate the surface temperature of the sun. Assume the sun to be a black body. Wien’s constant (b) = 2.892 10 3 mk . Solution : Wien’s law states that m T = constant = b T
b 2.892 10 3 mk m 4753 10 3 m
= 6060 K Example-9: A body cools from 62°C to 50°C in 10 minutes and to 42°C in the next 10 minutes. Find the temperature of surroundings. Solution: For the first ten minutes, 62o 50o dT = –1.2°C/min and dt 10 62 50 T = T0 (56 T0 )o C 10
1.2°C/min = KA (56T0)°C Similarly for the next ten minutes
…(1)
dT 42o 50o = 0.8 °C/min and dt 10 42 50 T = T0 ( 46 T0 )o C 2
0.8°C/min = KA (46To)°C Dividing (1) by (2) 3 56 T0 2 46 T0
T0 = 26°C.
…(2)
measurment OF SPECIFIC HEAT CAPACITY & error analysis As shown in the figure Regnault’s apparatus to determine the specific heat capacity of a solid heavier than water, and insoluble in it. A wooden partitions P separates a steam chamber O and A calorimeter C. The steam chamber O is a double walled cylindrical vessel. Steam can be passed in the space between the two walls through an inlet A and it can escape out through an outlet B. The upper part of the vessel is closed by a cork. The given solid may be suspended the vessel is closed by a cork. The given solid may be suspended in the vessel by a thread passing through the cork. A thermometer T1 is also inserted into the vessel to record the temperature of the solid. The stem chamber is kept on a wooden platform with a removable wooden disc D closing the bottom hole of the chamber. To start with, the experimental solid (in the form of a ball or a block) is weighed and then suspended in the steam chamber. Steam is prepared by boiling water in a separate boiler and is passed through the steam chamber. A calorimeter with a stirrer is weighed and sufficient amount of water is kept in it so that the solid may be completely immersed in it. The calorimeter is again weighed with water to get the mass of the water. The initial temperature of the water is noted. When the temperature of the solid becomes constant (say for 15 minutes), the partition P is removed. The calorimeter is taken below the steam chamber, the wooden disc D is removed and the thread is cut to drop the solid in the calorimeter. The calorimeter is taken to its original place and is stirred. The maximum temperature of the mixture is noted. Calculation: Let the mass of the solid =m1 mass of the calorimeter and the stirrer =m2 mass of the water =m3 specific heat capacity of the solid =s1 specific heat capacity of the material of the calorimeter(and stirrer) =s 2 specific heat capacity of water =s3 initial temperature of the solid =1 initial temperature of the calorimeter, stirrer and water =2 final temperature of the mixture = We have heat lost by the solid =m1s1(1-) heat gained by the calorimeter (and the stirrer) =m2s2(-2) Assuming no loss of heat to the surrounding the heat lost by the solid goes into the calorimeter stirrer and water. Thus m1s1(1 – ) = m2s2 (–2)+ m2s3( – 2) ……(1) (m2s2 m3 s3 )( 2 ) or s1 = m1(1 ) Knowing the specific heat capacity of water (s3 = 4186 J/kg-K) and that of the material of the calorimeter and the stirrer (s2=389 J/kg-K if the material be copper), one can calculate s1.Specific heat capacity of a liquid can also be measured with the Regnault apparatus. Here a solid of known specific heat capacity s1 is used and the
experimental liquid is taken in the calorimeter in place of water. The solid should be denser than the liquid. Using the same procedure and with the same symbols we get an equation identical to equation (1) above, that is, m1s1(1– ) = m2s2 (–2) + m3s3( – 2) in which s3 is the specific heat capacity of the liquid. We get m s (1 )m2s2 S3 = 1 1 m3 ( 2 )m3 Error analysis After correcting for systematic errors, equation (1) is used to estimate the remaining errors.
OBJECTIVE 1.
A piece of metal floats on mercury. The coefficients of volume expansion of the metal and mercury are 1 and 2 respectively. If their temperature is increased by T, the fraction of the volume of metal submerged in mercury changes by a 1 2 T 1 2 T (A) (B) 1 1T 1 1T
1 2 T 2 (D) 1 1 1T Solution: The correct choice is (A). It follows from the fact that the volumes of metal and mercury increase to V0 (1 + 1 T) and (1+ 2 T) respectively; Vo being the initial volume. (C)
2.
The coefficient of expansion of a crystal in one direction (x–axis) is 2.0 10–6 K–1 and that in the other two perpendicular (y–and z–axes) direction is 1.6 10–6 K–1. What is the coefficient of cubical expansion of the crystal? (A) 1.6 10–6 K–1 (B) 1.8 10–6 K–1 (C) 2.0 10–6 K–1 (D) 5.2 10–6 K–1 Solution: (D) Coefficient of cubical expansion is = x + y + z = x + 2 y ( y = z) –6 = 2.0 10 + 2 1.6 10–6 = 5.2 10–6 K–1 Hence the correct choice is (D). 3.
uniform metal rod of length L and mass M is rotating with angular speed about an axis passing through one of the ends and perpendicular to the rod. If the temperature increases by t C, then the change in its angular speed is proportional to (A) (B) 1 2 (C) (D) Solution: (B) At tC, the length of the rod becomes L’= L (1 + t), where is the coefficient of linear expansion. From the law of conservation of angular momentum, we have
or
I = I’’ 1 1 ML2 = ML’2 ’ 3 3 2
1 ' L or = L' (1 t)2 Now, for a given value of t, (1+ at)–2 is constant, say k. ' =k ' or k 1 or ’ – = (k – 1) i.e. (’ – ) . Hence the correct choice is (B).
4.
Solution:
o cylindrical rods or lengths l1 and l2, radii r1 and r2 have thermal conductivities k1 and k2 respectively. The ends of the rods are maintained at the same temperature difference. If l1 = 2l2 and r1 = r2/2, the rates of heat flow in them will be the same if k1/k2 is: (A) 1 (B) 2 (C) 4 (D) 8 (D) The rate of heat flow in rods A and B are Q1 k1r12 Q2 k 2 r22 and t l1 t l2 Q1 = Q2, if k1 l1 r22 k1r12 k 2r22 or 2 = 2 (2)2 = 8 k 2 l2 r1 l1 l2 Hence the correct choice is (D).
5: The temperature of the two outer surfaces of a x 4x composite slab, consisting of two materials having coefficients of thermal conductivity K and 2K and thickness x and 4x, respectively, are T2 and T1 (T2 > T1). T 2K The rate of heat transfer through the slab, in a steady 2 K A(T2 T1 )K state is f, with f equal to (see Fig.) x (A) 1 (B) 1/2 (C) 2/3 (D) 1/3 Solution: (C) Let A be the area of each slab. In the steady state, the rate of heat flow through the composite slab is given by A T2 T1 T2 T1 Q … (1) l1 l2 l1 l2 t K1A K 2 A K1 K 2 Given l1 = x, l2 = 4x, K1 = K and K2 = 4K. Using these values in (1) we get Q A T2 T1 A T2 T1 K 2 x 4x t x 3 K 2K
T1
Comparing this with the given rate of heat transfer, we get f =
2 . Hence the 3
correct choice is (C). 6.
Three identical rods A,B and C of equal lengths and equal diameters are joined is series as shown in following fig. Their thermal conductivities are 2K,K and K/2 respectively. Calculate the temperature at two junction points.
T1 100C
Solution:
7:
Solution:
or
T2
C 0C A B 0.5K 2K 85.7, 57.1C 77.33, 48.3C
(A) (B) 80.85, 50.3C (C) (D) 75.8,49.3C (A) ith 1 = ith 2 = ith 3 (100 T1 )2KA (T1 T2 )KA (T2 0)KA l l 2l 3T2 T (100 – T1) = (T1 – T2) = 2 T1 = 2 2 T (100 – T1) = 1 3 2T1 2 85.7 600 = 85.7C and T2 = = 57.1C r T1 = 7 3 3 Three rods made of the same material ane having the same cross-section have been joined as shown in. Each rod is of the same length. The left and right ends are kept at 0C and 90C respectively. The temperature of the junction of the three rods (A) 45C (B) 60C (C) 30C (D) 20C I=I1+I2 kA(90 T) (90 t)kA (t 0)kA I l l l 0 TC 3T = 180C or T = 60C
90C
0C
90C
90C
I1
I2 90C
8:
A solid copper sphere (density p and specific heat C) of radius r at an initial temperature 200 K is suspended inside a chamber whose walls are at almost 0 K. What is the required for temperature of the sphere to drop to 100 K? 7rpc ´ 10- 6 7rpc ´ 10- 8 (A) (B) 72es 72es - 10 7rpc ´ 10 7rpc ´ 10+6 (C) (D) 72es 72es
Solution:
(A) According to Stefan’s law P = eAT4
dt eAt 4 dQ mcdT eAT 4 or dr mc dt dt 2 4 t rpc dT dT e4r T or dt 3 p4rc 3e T 4 o dt 3
or
rpc 1 9e T3
100
200
7rpc 106 seconds 72e
A double pane window used for insulating a room thermally from outside consists of two glass sheets each pf area 1 m2 and thickness 0.01 m separated by a 0.05 m thick stagnant air space. In the steady state the room glass interface and glass outdoor interface are at 27C and 0C respectively. Calculate the rate of flow of heat through the windowpane. Also find the flow of heat through the windowpane. Also find the temperature of other interface if, conductivities of glass and air are 0.8 and 0.08 Wm-1 K-1 respectively. (A) 0.72C (B) 0.52C (C) 0.192C (D) 0.32C dQ KA dt L R L 0.05 1 0.01 = Req = 2 KA A 0.8 0.08 1 5 26 A = 1 m2 Req = 40 8 40 dQ (27 0) 40 = 41.5 W dt R 26 27 1 or 1 = 26.48C 41.5 = 0.8 12 0.01 0.8 12 (2 0) 41.5 = or 2 = 0.52C 0.01
(B)
Room
27C
0.01
Ce Outside
Air K 0.08
Kg 0.8
Solution:
Kg 0.8
9:
t=
0.05m
1
2
0.01
10:
A space object has the space of a sphere of radius R. Heat sources ensure that the heat evolution at a constant rate is distributed uniformly over its volume. The amount of heat liberated by a thermodynamic temperature. In what proportion would the temperature of the object change if its radius is decreased to half? (A) 2.19 (B) 5.19 (C) 1.19 (D) 4.19
Solution:
(C)
Heat liberated
T4 R T14 R Thus = 4 T2 R/2 Or
dQ dQ 2 4 R3 and RT dt dt
0C
4
T1 1 Or T2 = T1 or T2 = 1.19 2 That is , temperature decreases by a factor of 1.19.
11:
Solution:
12:
Solution:
13.
14.
The room temperature is +20C when outside temperature is 20C and room temperature room temperature is + 10C when outside temperature is -40C. Find the temperature of the radiator heating the room. (A) 30C (B) 60C (C) 90C (D) 45C (B) Applying Newton’s law In case (1) K1(T-Tr1)= K2(Tr2-Tout 1) And in case (2) K1(T-Tr2)= K2(Tr2-Tout 2) Dividing these equations T -Tr1 Tr2 -Tout1 T 20 20 ( 20) = T -Tr2 Tr2 -Tout2 T 10 10 ( 40) or T=60C Some water at 0C is placed in a large insulated enclosure (vessel). The water vapour formed is pumped out continuously. What fraction of the water will ultimately freeze, if the latent heat vaporization is seven times the latent heat of fusion? (A) 7/8 (B) 8/7 (C) 3/8 (D) 5/8 (A) m = mass of water, f = fraction which freezes L1 = latent heat of vaporization L2 = latent heat of fusion L1 = 7L2 Mass of water frozen = mf Heat lost by freezing water = m fL2 Mass of vapour formed = m(1 - f) Heat gained by vapour = m (1 – f) L1 mfL2 = m (1 - f) x 7L2 f = 7 – 7f or f = 7/8 A substance is in the solid form at 0C. The amount of heat added to this substance and its temperature are plotted in the following graph. If the relative specific heat capacity of the solid substance is 0.5, find from the graph, the mass of the substance is (Specific heat capacity of water = 1000 cal kg-1 K-1 ) (A) 0.02 kg (B) 2 kg (C) 0.04 kg (D) 0.05 kg
t
120 100 80 60 40 20
C A
B 2160
0
1000 Q
A substance is in the solid form at 0C. The amount of heat added to this substance and its temperature are plotted in the following graph. If the relative specific heat capacity of the solid substance is 0.5, find from the graph, the specific latent heat of the melting process is (Specific heat capacity of water = 1000 cal kg-1 K-1 )
2000
(calories)
(A) 6000 cal kg-1 (C) 1000 cal kg-1
4000 cal kg-1 2000 cal kg-1
(B) (D)
A substance is in the solid form at 0C. The amount of heat added to this substance and its temperature are plotted in the following graph. If the relative specific heat capacity of the solid substance is 0.5, find from the graph, the specific heat of the substance in the liquid state is (Specific heat capacity of water = 1000 cal kg-1 K-1 ) (A) 300 cal kg –1 K-1 (B) 500 cal kg –1 K-1 –1 -1 (C) 700 cal kg K (D) 100 cal kg –1 K-1
15.
Solution 13: (A) 800 calories of heat raise the temperature of the substance from 0C to 80C. 800 = m (1000 0.5 ) 80 ( specific heat = relative sp. heat x sp. heat of water) or m = 0.02 kg Solution 14: (B) Latent heat = 200 4 = 800 cal ( = 0.02 L L= 4000 cal kg-1
1 div reads 200 cal )
Solution 15: (C) In the liquid state temperature rises from 80C to 120C, that is, by 40C after absorbing (2160 – 1600) cal.
0.02 s 40 = 2160 – 1600 or s = 700 cal kg –1 K-1
16:
An earthenware vessel loses 1g of water per second due to evaporation. The water equivalent of the vessel is 0.5 kg and the vessel contain 9.5 kg of water. Find the time required for the water in the vessel to cool to 28C from 30C. Neglect radiation losses. Latent heat of vaporization of water in this range of temperature is 540 cal g-1 . (A) 2 min 5 s (B) 3 min 5 s (C) 5 min 3 s (D) 6 min 2 s
Solution:
(B) Here water at the surface is evaporated at the cost of the water in the vessel losing heat. Heat lost by the water in the vessel = (9.5 + 0.5) 1000 (30 – 20) = 105 cal Let t the required time in seconds. Heat gained by the water at the surface = (t 10-3) 540 103 ( L = 540 cal g-1 = 540 103 cal kg-1) 105 = 540 t or t = 185 s = 3 min 5 s
17:
In an industrial process 10 kg of water per hour is to be heated from 20C to 80C. To do this, steam at 150C is passed from a
boiler into a copper coil immersed in water. The steam condenses in the coil and is returned to the boiler as water at 90C. How many kg of steam are required per hour? Specific heat of steam = 1 kilo cal kg–1C–1. Latent heat of steam = steam = 540 kilo cal kg–1. (A) 1 kg (B) 2 kg (C) 3 kg (D) 4 kg Solution:
(A) Let the mass of steam required per hour be m kg. Heat gained by water in boiler per hour is = 10 kg 1 kilo cal kg–1 C–1 (80 – 20)C = 600 kilo cal … (1) Heat lost by steam per hour is = heat needed to cool m kg of steam from 150C to 100C + heat needed to convert m kg of steam at 100C into water at 100C + heat needed to cool m kg of steam at 100C to 90C = m 1 (150 – 100) + m 540 + m 1 (100 – 90) = 50 m + 540 m + 10 m = 600 m kilo cal … (2) Heat lost = heat gained. Equating (1) and (2) we have 600 m = 600 or m = 1 kg
18:
A closed cubical box made of a perfectly insulating material has walls of thickness 8 cm and the only way for the heat to enter or leave the box is through two solid cylindrical metal plugs, each of cross–sectional area 12 cm2 and length 8 cm fixed in the opposite walls of the box. The outer surface A of one plug is kept at a temperature of 100C while the outer surface of the other plug is maintained at a temperature of 4C. The thermal conductivity of the material of the plug is 0.5 cal cm–1 s–1 (C)–1. A source of energy generating 36 cal s–1 is enclosed inside the box. Find the equilibrium temperature of the inner surface of the box assuming that it is the same at all points on the inner surface. (A) 70C (B) 45C (C) 65C (D) 76C Insulator
(D) At equilibrium, the total energy generated by the source per second must Metal plug equal the heat leaving per second through the two metal plugs (Fig.). Let TC be the A equilibrium temperature. Then heat leaving atSurface 100 C the box per second through surface A k(T 100) 12 = cal s–1 8 and heat leaving the box per second through the surface B k(T 4) 12 cals1 8 12k Hence (T – 100 + T –4) = 36 8
Solution:
Metal plug Energy source Surface B at 4 C
or 2T – 104 =
36 8 36 8 = = 48 or T = 76C 12k 12 0.5
19:
Find the time during which a layer of ice of thickness 2.0 cm on the surface of a pond will have its thickness increased by 2 mm when the temperature conductivity of ice = 5 10-3 cal cm-1 s-1 (C)-1, density of ice at 0C = 0.91 g cm-3 and latent heat of fusion = 80 cal g-1 (A) 6 min 5 s (B) 2 min 6 s (C) 5 min 6 s (D) 3 min 5 s
Solution:
(C) substituting the given quantities in the expression pL t= ( x 22 - x12 ) 2kT we have 0.91 80 t= [(2.2)2-2.0)2] = 306 s = 5 min 6 s. 3 2 5 10 20
20:
Water is being boiled in a flat-bottomed kettle placed on a stove. The area of the bottom is 300 cm2 and the thickness is 2 mm. If the amount of steam produced is 1 g/min, calculate the difference of temperature between the inner and other outer surfaces of the bottom. The thermal conductivity of the material of kettle = 0.5 cal cm-1 s-1 (C)-1 and the latent heat of stem = 540 cal g-1. (A) 0.012 C (B) 0.0.04 C (C) 0.02 C (D) 0.0.08 C
Solution:
(A) If T is the temperature difference between the inner ad outer surface of the bottom of the kettle, then the amount of heat flowing through the bottom per second is Q kAT 0.5 300T = 750 cal per second … (1) t d 0.2 Q mL 1g 540calg1 But … (2) t t 60s Equating (1) and (2), we have 9 750 T = 9 or T = = 0.012 C 750
21:
If an anisotropic solid has coefficients of linear expansion x, y and z for three mutually perpendicular directions in the solid, what is the coefficient of volume expansion for the solid? (A) x + y + z (B) 2x + y + z 2 2 2 (C) x+ y+ z (D) 3x + 3y + 3z
Solution:
(A) Consider a cube, with edges parallel to X, Y, Z of dimension L 0 at T = 0. After a change in temperature T = (T –0), the dimensions change to
given by
Lx = L0 (1 + xT) Ly = L0(1 + yT) Lz = L0(1 + zT) and the volume of the parallelepiped is V = V0 (1 + xT)(1 + yT)(1 + zT) V0 [1 + (x + y + z)T] Where V0 = L30 . Therefore, the coefficient of volume expansion is x + y + z.
22.
In aluminum sheet there is a hole of diameter 2m and is horizontally mounted on a stand. Onto this hole an iron sphere of radius 2.004 m is resting. Initial temperature of this system is 25° C. Find at what temperature, the iron sphere will fall down through the hole in sheet. The coefficients of linear expansion for aluminum and iron are 2.4 10–4and 8.6 10 –5 respectively. (A) 82C (B) 43C (C) 45C (D) 15C
Solution:
(B) As value of coefficient of linear expansion for aluminum is more than that for ion, it expends faster then iron. So at some higher temperature when diameter of hole will exactly become equal to that of iron sphere, the sphere will pass through the hole. Let it happen at some higher temperature T. Thus we have at this temperature T, (diameter of hole)Al = (diameter of sphere)iron 2[1 + Al (T – 25)] = 2.004[1 + iron(T – 25)] 2Al (T – 25) = 0.004 + 2.004 iron (T – 25) 0.004 25 C or T 2 AL 2.004iron 0.004 or T 25 2 2.4 104 2.004 8.6 105 or T = 43C
23.
An iron ball has a diameter of 6 cm and is 0.010 mm too large to pass through a hole in a brass plate when the ball and plate are at a temperature of 30C. At what temperature, the same for ball and plate, will the ball just pass through the hole? (A) 23.8C (B) 13.8C (C) 53.8C (D) 83.8C (C) We let I stand for the iron ball and B stand for the brass plate. LI = 6 cm and LI – LB = 0.001 cm at t = 30C. Since the brass plate expands uniformly, the hole must expand in the same proportion. Then heating both the ball and the plate leads to increases in the diameters of the ball and the hole, with the hole increasing more, since B > I. We require LB – LI = 0.001 cm. LB = BLB t; LI = ILI t. We can approximate LB in this formula by 6 cm = LI. Then LB –LI = (B – I)LI t = 0.001 cm solving, t = 23.8C, and finally t = 30C + 23.8C = 53.8C
Solution:
24.
It is desired to put an iron rim on a wooden wheel. The diameter of the wheel is 1.1000m and the inside diameter of the rim is
1.0980 m. If the rim is at 20C initially, to what temperature must it be heated to just fit onto the wheel? (A) 52C (B) 72C (C) 102C (D) 152C (D) for iron is found to be 1.2 10–5 C–1. L = 1.1000 – 1.0980 = 0.0020 m = L t 0.0020 = (1.2 10–5)(1.098) t t + 20 = 172C t = 152C
Solution:
25.
Find the coefficient of volume expansion for an ideal gas at constant pressure. 1 (A) (B) g =T T 1 1 (C) (D) g= 3 g= 2 T T (A) For an idea gas PV= nRT As P is constant, we have P.dV = nRdT dV nR dT P 1 dV nR nR 1 V dT PV nRT T 1 T
Solution:
26.
What should be the lengths of steel and copper rod so that the length of steel rod is 5cm longer then the copper rod at all the temperatures. Coefficients of linear expansion for copper and steel are 1.7 and 1.1 (A) 2.17, 14.17 cm (B) 9.17, 14.17 cm (C) 9.17, 18.17 cm (D) 3.17, 5.17 cm (B) It is given that the difference in length of the two rods is always 5 cm. Thus the expansion in both the rods must be same for all temperatures. Thus we can say that at all temperature differences, we have LCu Lsteel or Cu l1t st l 2t [If l1 and l2 are the initial lengths of Cu and steel rods] or Cu l1 st l 2 or 1.7 l1 1.1 l 2 …..(1) It is given that l2 – l1 = 5cm ….(2) 1.7 1.1 1 l1 5cm 5 1.1 9.17cm or l1 0.6
Solution:
Now from equation (2) l2 = 14.17 cm A steel wire of cross-sectional area 0.5 mm2 is held between two rigid clamps so that it is just taut at 20 o C . Find the tension in the wire at 0 o C . Given that Young’s modulus of steel is Yst = 2.1 1012 dynes / cm2 and coefficient of linear expansion of steel is st 1.1 105 C .
27.
(A) (C)
2.31 10–4 9.31 10–6
(B) (D)
2.31 10–2 2.31 10–6
Solution:
(D) We know that due to drop in temperature, then tension increment in a clamped wire is T = YA T = 2.1 1012 0.5 10–2 1.1 10–5 20 = 2.31 10–6
28.
Two bodies have the same heat capacity. If they are combined to form a single composite body, show that the equivalent specific heat of this composite body is independent of the masses of the individual bodies. 2s1s2 s1s2 (A) (B) s2 - s1 s2 + s1 s2 2s1s2 (C) (D) s2 + s1 s2 + s1
Solution:
(C) Let the two bodies have masses m 1, m2 and specific heats s1 and s2 then m1s1 = m2s2 or m1/m2 = s2/s1 Let s = specific heat of the composite body. Then (m1 + m2) s = m1s1 + m2s2 = 2m1s1 2m1s1 2m1s1 2s1s2 s= m1 m2 m1 m1(s1 / s2 ) s2 s1
29.
20 gm steam at 100C is let into a closed calorimeter of water equivalent 10 gm containing 100 gm ice at – 10C. Find the final temperature of the calorimeter and its contents. Latent heat of steam is 540 cal/gm, latent heat of fusion of ice=80 cal/gm, specific heat of ice = 0.5 cal/C gm. (A) 13C (B) 63C (C) 93C (D) 33C
Solution: (D) Heat lost by steam = mL + ms (100 - ) where, is the equilibrium temperature Heat lost by steam = 20 540 + 20 1 (100 - ) = 10800 + 2000 - 20 Heat gained by (ice + calorimeter) = 100 80 + 100 0.5 10 + 100 Now Heat lost = Heat gained or
10800 + 2000 – 20 = 8000 + 500 + 110 130 = 4300
or
=
4300 = 33C. 130
30.
Victoria Falls in Africa is 122 m in height. Calculate the rise in temperature of the water if all the potential energy lost in the fall is converted to heat. (A) 0.29 K (B) 29 K (C) 0.69 K (D) 0.99 K
Solution:
(A) Consider mass m of water falling. mgy = mc t gy = c t We express both sides in joules by noting c = 1 kcal/kg K = 4184 J/kg K Then 9.8(122) = 4184 t and t = 0.29 K
31.
An electric heater supplies 1.8 kW of power in the form of heat to a tank of water. How long will it take to heat the 200 kg of water in the tank from 10 to 70C? Assume heat losses to the surroundings to be negligible. (A) 1.75 h (B) 7.75 h (C) 4.75 h (D) 5.75 h
Solution: (B) The heat added is (1.8 kJ/s)t and the heat absorbed is cm T = (4.184 kJ/kg K) (200 kg)(60 K) = 5.0 104 kJ Equating heats, t = 2.78 104 s = 7.75 h. 32.
What will be the final temperature if 50 g of water at 0C is added to 250 g of water at 90C? (A) 15C (B) 30C (C) 45C (D) 75C
Solution: (D) Heat gained = heat lost. (We assume no hat transfer to or from container.) (50 g)(1.00 cal/g C)(t – 0C) = (250 g)(1.00 cal/g C)(90C – t) where t is the final equilibrium temperature. 50t = 22500 – 250t or 300t = 22500 t = 75C 33.
A 500 g piece of iron at 400C is dropped into 800 g of oil at 20C. If c = 0.40 cal/gC for the oil, what will be the final temperature of the system. Assume no loss to the surroundings. (A) 15.7C (B) 30.7C (C) 45.7C (D) 75.7C
Solution:
(D) Heat lost = heat gained is written as 0.11(500)(400 – t) = 0.40(800)(t – 20) from which t = 75.7C
34.
A copper rod 2 m long has a circular cross section of radius 1 cm. One end is kept at 100C and the other at 0C, and the surface is insulated so that negligible heat is lost through the surface. Thermal conductivity of copper is 401 W/m–K. Find the thermal resistance of the bar. (A) 15.9 K/W (B) 20.9 K/W (C) 40.9 K/W (D) 50.9 K/W
35.
A copper rod 2 m long has a circular cross section of radius 1 cm. One end is kept at 100C and the other at 0C, and the surface is insulated so that negligible heat is lost through the surface. Thermal conductivity of copper is 401 W/m–K. Find the thermal current H. (A) 1.3 watt (B) 6.3 watt (C) 12.3 watt (D) 18.3 watt
36 .
A copper rod 2 m long has a circular cross section of radius 1 cm. One end is kept at 100C and the other at 0C, and the surface is insulated so that negligible heat is lost through the surface. Thermal conductivity of copper is 401 W/m–K. Find the dT temperature gradient . dx (A) –10C/m (B) –40C/m (C) –50C/m (D) –90C/m
37 .
A copper rod 2 m long has a circular cross section of radius 1 cm. One end is kept at 100C and the other at 0C, and the surface is insulated so that negligible heat is lost through the surface. Thermal conductivity of copper is 401 W/m–K. Find the temperature 25 cm from the hot end. (A) 20.5C (B) 40.5C (C) 77.5C (D) 87.5C
Solution 34 (A) Thermal resistance R = or R = Solution 35 (B) or
2
401 10 2
2
l l kA k ( r 2 )
= 15.9 K/W
Thermal current, H =
T 100 R R 15.9
H = 6.3 watt
0 100 = – 50 K/m = –50C/m 2 Solution 37 (D) Let 0c be the temperature at 25 cm from the hot 100 C end, then ( – 100) = (temperature gradient) (distance) or – 100 = (– 50) (0.25) or = 87.5C
Solution 36 (C) Temperature gradient =
38.
C
0.25 m 2.0 m
Two metal cubes with 3 cm–edges of copper and aluminium are arranged as shown in figure. Thermal conductivity of copper is
0 C
401 W/m–K and that of aluminium is the total thermal current from one other. (A) 0.14 K/W (B) (C) 0.74 K/W (D) 39.
237 W/m–K. Find reservoir to the 100 C
0.34 K/W 0.94 K/W
Al Cu
wo metal cubes with 3 cm–edges of copper and aluminium are arranged as shown in figure. Thermal conductivity of copper is 401 W/m–K and that of aluminium is 237 W/m–K. Find the ratio of the thermal current carried by the copper cube to that carried by the aluminium cube. (A) 0.75 K/W (B) 0.05 K/W (C) 0.15 K/W (D) 0.35 K/W
Solution: 38 (A) Thermal resistance of aluminium cube R1 = or R1 =
(3.0 10 2 )
(237) 3.0 10
2
2
= 0.14 K/W
Solution: 39 (B) Thermal resistance of copper cube R2 = or R2 =
(3.0 10 2 )
(401) 3.0 10 2
2
l kA
l kA
= 0.08 K/W
As these two resistances are in parallel, their equivalent resistance will be
R1R2 R1 R2 (0.14)(0.08) = = 0.05 K/W (0.14) (0.08)
R=
40.
What temperature gradient must exist in an aluminum rod in order to transmit 8 cal/s per square centimeter of cross section down the rod? k for aluminum is 0.50 cal/s cm C. (A) 36C/cm (B) 46C/cm (C) 86C/cm (D) 16C/cm
Solution:
(D) H = kA
T x
or 8 cal/s = (0.50 cal/s cm C)(1 cm2)
T x
T = magnitude of temperature gradient = 16C/cm. x The actual sign of the gradient depends on whether the heat flow is in the positive or negative x direction, being negative or positive for the two cases, respectively.
41.
What is heat conduction. (A) (A T)/d (C) (kA T)/d
(B) (D)
(k T)/d (T)/d
20 C
Solution: (C) The conduction equation, H Q/t = (kA T)/d can be reexpressed as H = T/R, where R d/(kA) is called the thermal resistance of the slab. It has units of K/W in SI. 42.
Consider the two insulating sheets with resistances R1 and R2 shown in Fig. What is the value of T' ? R T + R2T2 R T R1T2 (A) (B) T ¢= 1 1 T 2 1 R1 + R2 R1 R2 R T + R1T2 R T + R1T2 (C) (D) T ¢= 2 1 T ¢= 2 21 R1 - R2 R1 + R22
Solution:
(B) For the two sheets, H1 = (T1 – T)/R1, H2 = (T – T2)/R2. Noting that H1 = H2 = H, we have (T1 – T)/R1 = (T – T2)/R2. Cross multiplying we get R2(T1 – T) = R1(T – T2), or rearranging terms, R2T1 + R1T2 = (R1 + R2)T, and T’ = (R2T1 + R1T2)/(R1 + R2).
T2 R2
R1
H
43.
A spherical blackbody of 5 cm radius is maintained at a temperature of 327C. What is the power radiated? (A) 231 W (B) 431 W (C) 531 W (D) 631 W
Solution:
(A) The surface area of a sphere is 4r2. In this case, then, the area is 4(25 10–4) = 0.01m2. The power radiated is given by Stefan’s law: P= T4A = (5.67 10–8 W/m2.K4)(600 K)4(0.01 m2) = 231 W.
44..
A spherical blackbody of 5 cm radius is maintained at a temperature of 327C. What wavelength is the maximum wavelength radiated. (A) 2.82 m (B) 4.82 m (C) 3.82 m (D) 5.82 m
Solution:
(B) The surface area of a sphere is 4r2. In this case, then, the area is 4(25 10–4) = 0.01m2. The power radiated is given by Stefan’s law: P= T4A = (5.67 10–8 W/m2.K4)(600 K)4(0.01 m2) = 231 W By Wien’s law, m(600 K) = 2898 m.K and m = 4.82 m.
45.
A sphere of 3 radius acts like a blackbody. It is equilibrium with its surroundings and absorbs 30 kW of power radiated to it from the surroundings. What is the temperature of the sphere? (A) 2300 K (B) 2800 K (C) 3000 K (D) 2600 K
Solution:
(D) The power absorbed by a blackbody is Pa = A Ta4 , or (30 103 W)= (5.67 10–8 W/m2.K4) 4(0.03 m)2 Ta4 .
T T1
Ta4 = 4.68 1013; Ta = temperature of surroundings = 2600 K. Since the body is in equilibrium with its surroundings, it is at the same temperature, 2600 K.
46 .
A blackbody is at a temperature of 527C. To radiate twice as much energy per second, what temperature must be increased ? (A) 951 K (B) 451 K (C) 551 K (D) 651 K
Solution:
(A) Since P T4, the temperature must be increased to 21/4(800 K) = 951 K.
47.
Use Stefan’s law to calculate the total power radiated per square meter by a filament at 1727C having an absorption factor of 0.4. (A) 0.16 MW/m2 (B) 0.26 MW/m2 (C) 0.36 MW/m2 (D) 0.46 MW/m2
Solution:
(C) Stefan’s law gives R = T4 = 0.4(5.67 10–8)(2000)4 = 0.36 MW/m2
48.
A blackbody is at a temperature of 527C. To radiate twice as much energy per second, how many times of increase in radiated power when the temperature of a blackbody is increased from 7 to 287C. (A) 5 times (B) 16 times (C) 6 times (D) 20 times
Solution:
(B) Since P T4, the temperature must be increased to 21/4(800 K) = 951 K. 4
P(560 K) 560 And 16 times P(280 K) 280
49.
The initial and final temperature of water as recorded by an observer are (40.6 0.2) °C and (78.3 0.3) °C. Calculate the rise in temperature with proper error limit. (A) (27.7 0.5)°C (B) (17.7 0.5)°C (C) (37.7 0.9)°C (D) (37.7 0.5)°C
Solution:
(D) Let 1 = 40.6°C, 1 = 0.2 °C 2 = 78.3°C, 2 = 0.3 °C = 2 – 1 = 78.3 – 40.6 = 37.7°C & = (1 + 2) = (0.2 + 0.3) = 0.5°C Hence rise in temperature = (37.7 0.5)°C
50.
A cylinder of radius R made of a material of thermal conductivity k1 is surrounded by a cylindrical sheet of inner radius R and outer radius 2R made of material of thermal conductivity k 2 . The two ends of the combined system are maintained at two different
temperatures. There is no loss of surface and the system is in steady thermal conductivity of the system. (A) 4K = 2K1 + 3K2 (C) 4K = 6K1 + 4K2 Solution :
heat across the cylindrical state. Calculate the effective (B) (D)
4K = 5K1 + 3K2 4K = K1 + 3K2
(D) Two cylinders are in parallel, therefore equivalent thermal resistance R is given by 1 1 1 R R1 R 2
kA kA k 1A 1 k 2 A 2 1 2
R
But
Here A 2 2R R2 3R2
1 2 , A1 R2
2
and
A 2R 4R 2
K 4R 2 K 1R 2 K 2 3R 2
i.e.
4K = K1 + 3K2
2
51.
A lake is covered with ice 2 cm thick. The temperature of ambient air is 15oC. Find the rate of thickening of ice. For ice k = 4 104k-cal-m1s1(C)1. Density =9103 kg/m3and latent heat L = 80 Kilo Cal/Kg. (A) 2.5 cm/ hour (B) 1.5 cm/ hour (C) 3.5 cm/ hour (D) 4.5 cm/ hour
Solution :
(B) Heat energy flowing per sec is given by H
Q KA x t
. . . (I)
if dm mass of ice is increased in time dt, then dm A dx dx A. . dt dt dt
Since,
dm H L dt dx L . . . (ii) H A dt
From eq. (I) and (II) A L
dx KA dt x
Rate of thickening of ice = dx/dt 0( 15) 410 4 dx KA = K = 3 dt AL x L x 0.910 80 210 2
= 4.166 106 m/s = 1.5 cm/ hour.
52.
A body cools from 60C to 50C in 10 minutes. If the room temperature is 25C and assuming Newton’s law of cooling to hold good, find the temperature of the body at the end of the next 10 minutes. (A) 42.85C (B) 12.85C (C) 20.85C (D) 52.85C
Solution :
(A) According to Newton’s law of cooling rate of loss of heat (T To) where T = mean temperature of the body
60 50 10
Also,
60 50 25 2
. . . (1)
50 T 50 T 25 10
2
. . . (2)
where T is the required temperature Solving both equation, we get, T = 42.85C 53.
Two bodies A and B have thermal emissivities of 0.01 and 0.81 respectively. The outer surface areas of the two bodies are same. The two bodies emit total radiant power at the same rate. The wavelength B corresponding to maximum spectral radiancy in the radiation from B is shifted from the wavelength corresponding to maximum spectral radiancy in the radiation from A by 1.00 m. If the temperature of A is 5802 K, calculate (a) the temperature of B and (b) wavelength B. (A) 2000K, 1.6 m (B) 3000K, 1.8 m (C) 1934K, 1.5 m (D) 4000K, 1.9 m
Solution :
(C) (a) According to stefan’s law the power radiated by a body is give by P eAT4
According to the given problem, PA PB , with A A A B So that e A TA 4 eB TB 4 i.e, 0.01 5802 4 0.81TB 4 1 3
or TB 5802 1934K (b)
According to Wien’s displacement law. A TA B TB ; i.e, B 5802 / 1934 A i.e, B 3 A and also B - A = 1m(given) 1 3
so B B 1m i.e. B 1.5m
A body which has a surface area 2.00 cm2 and a temperature of 727oC radiates energy at the rate of 5 W. What is its emissivity? (Stefan boltzman constant 5.67 10 8 W / m2k 4 ). (A) 0.14 (B) 0.44 (C) 0.34 (D) 0.24
54
(B) P e AT4
solution:
5 e 5.67 10 8 2 10 4 1000 5 e 8 5.67 10 2 10 4 1012 4
5 0.44 11.34
55. Two rods of equal cross-sectional area A and length L L L are joined and what is the temperature of junction as shown A A k 2k in the figure. The temperature of one end of the composite o o 100 C 0 C o o rod is 0 C and the other end of the rod is at 100 C . Draw a graph showing the variation of the temperature with the distance x from the end maintained at 0 o C . Assume that the flow is longitudinal and there is no loss of heat from the lateral surface. 100 (x + L ) 100x L (A) (B) T T= 3L 4L 100 (x + L ) 100 (x + L ) (D) (C) T= T= L 5L L L 3L kA 2kA 2kA 1002kA 200kA H 3L 3L 200 x T 200kA T 3L x / kA 3L
solution:
(A) R
L o
0C
L 2k
k
o
100 C
when x L
x L x L L kA 2kA 2kA T2kA 200kA x L 3L 100x L T 3L R
56.
The graph is as shown
T o
100 C
O
L
2L
x
Two spherical soap bubbles coalesce to form a bigger bubble without any leakage of air. If V is the total change in volume of the contained air and S the total change in surface area, what is the value of 3PV + 4ST, where T is the surface tension of the soap solution and P is the atmospheric pressure. (A) 0 (B) 1 (C) 2 (D) 3
solution:
(A) Let r1 and r2 be the initial radii of two bubbles and r is the final radius of the bubble after they coalesce. Then 4T 4 3 rT 4 4T 4 3 P r1 P r23 P r r1 3 r2 3 r 3 P r13 r23 r 3 4T r12 r22 r 2 0 …(1) 4 also V r13 r23 r 3 = change in total volume 3 and S 4 r12 r22 r 2 = change in surface area
…(2) …(3)
from (1), (2) & (3) we get 3PV 4ST 0
57.
The graphs gives the variation of temperature of two bodies having the same surface area with time where Ax and Ay represent absorptivity and x and y represent emissivity then (A) x > y and Ax < Ay (B) x < y and Ax > Ay (C) x > y and Ax > Ay (D) x < y and Ax < Ay
T
x y t
solution: (D)According to Kirchoff’s law good emitter is a good absorber Ey > Ex and Ay > Ax The temperature gradient in a rod of 0.5 m length is 80 oC/m. If the temperature of hotter end of the rod is 30oC, then the temperature of the cooler end is (A) 40oC (B) –10oC o (C) 10 C (D) 0oC. 30 Solution: (B) 80 0.5
58.
59. A body at 300oC radiates 10 J cm–2 s–1. If Sun radiates 105 J cm–2 s–1, then its temperature is (A) 3000oC (B) 5457oC 4o (C) 300 × 10 C (D) 5730oC Solution : (B) (573)4 105 T4 10 The ratio of energy of radiation emitted by a black body at 27oC and 927oC is (A) 1:4 (B) 1 : 16 (C) 1 : 64 (D) 1 : 256. 4 Solution:(B) E=T
60.
13-Electrostatics
ELECTROSTATICS ELECTRIC CHARGE Charge is the physical property of certain fundamental particles (like electron, proton). Charges are divided into two parts (i) positive (ii) negative. Like charges repel and unlike charges attract. SI unit of charge is coulomb and CGS unit is esu. 1C = 3 109esu. Magnitude of the smallest known charge is e = 1.6 10-19C (charge of one electron or proton).
PROPERTIES OF ELECTRIC CHARGE QUANTIZATION OF CHARGE charge on any body is the integral multiple of the charge on an electron Q = ne, where n = 0, 1, 2, ................. Distribution of Charges
The concentration of the charges is maximum on a surface with greater curvature.
COULOMB’S LAW The force between two point charges is directly proportional to the product of their charges and inversely proportional to the square of the distance between them, In free space F=
1 q1 q 2 4 o r2
- permittivity of the medium
In material medium F=
1 q1 q 2 4 r 2
Where = 0 in vector form F
r
q1 q 2 1 r 4 0 r3
0 – permittivity of free space (Vacuum) r = relative permittivity of the medium 1 4 0
9 10 9
Nm 2 coul 2
0 = 8.85410-12 1. 2. 3. 4.
coul 2 Nm 2
This is a fundamental law and is based on physical observation The force is an action - reaction pair. The direction of force is always along the line joining the two charges. Electrical force between two point charges is independent of presence or absence of other charges.
Example 1.
A particle ‘A’ having a charge of 2 10 C and a mass of 100g is fixed at the bottom of a smooth inclined plane of inclination 30 . Where should another particle B, having same charge and mass be placed on the incline so that it may remain in equilibrium ?
Solution :
First of all draw the F.B.D. of the masses. For equilibrium F = 0 N = mg cos30 . . . (1) Fe = mg sin30 . . . (2)
-6
From (2)
x=
Electric Field Intensity F q0 q q 1 r E 4 0 r 3 q 1 E 4 0 r 2
kq 2 mg x2 2
N
Fe B
x mg sin30 mg cos30
2kq 2 = 27cm. mg
A 30
It is the force experienced by a unit positive charge placed at a point in an electric field.
E lim
[In vacuum or free space] N/coul or Volt/m
Example 2.:
Two particles A and B having charges 8 x10 C and –2 x10 C respectively are held fixed with a separation of 20 cm. Where should a third charged particle be placed so that it does not experience a net electric force ?
Solution :
As the net electric force on C should be equal to zero, the force due to A and B must be opposite in direction. Hence, the particle should be placed on the line AB. As A and B have charges of opposite signs, C cannot be between A and B
-6
-6
Also A has larger magnitude of charge than B. Hence, C should be placed closer to B than A. The situation is shown in figure. Suppose BC=x and the charge on C is Q y 1 (8.0 106 ) Q ˆ i 2 4π 0 (0.2 x ) A B 6 20 cm 1 (2.0 10 )Q ˆ and FCB i 2 4 0 x 1 (8.0 10 6 ) Q (2.0 10 6 ) Q FC FCA FCB i 4 0 (0.2 x )2 x2 But | FC | 0
FCA
Hence
FCB
C
x
1 (8.0 10 6 ) Q (2.0 10 6 ) Q 0 4 0 (0.2 x )2 x2
Which gives x = 0.2 m Superposition principle Force experienced by a given charge in the field of a number of point charges is the vector sum of all the forces. Fnet F1 F2 F3 F4
Similarly for electric field
Enet E1 E 2 E 3 ...........
Lines of Force Lines of force originate from positive charges and terminate on negative charges Lines of force originate or terminate perpendicular to the surface Tangent to the lines of force at any point gives the direction of the electric field. Lines of force do not intersect.
1.
Electric Field Intensity
Due to a point charge E=
q 1 4 0 r 2
Due to linear distribution of charge = linear charge density of rod
FCA x
(i)
, L
At a point on its axis.
a
dx
1 1 E= 4 0 a a L
x
= linear charge density (ii)
At a point on the line perpendicular to one end 1 1 Ex = 4 0 y y 2 L2 1 L Ey = 2 4 0 y [L y 2 ]1/ 2
Due to ring of uniform charge distribution dq
At a point on its axis
(R2 + x2)1/2 a
dE cos
P
qx 1 Ex 4 0 [a 2 x 2 ] 3 / 2
x
dE dE sin
Due to uniformly charged disc. At a point on its axis. Ex 2 0
x 1 [ x 2 a 2 ]1/ 2
= surface charge density
dr
r x
P
Thin spherical shell E
q 1 4 0 r 2
2.
Non conducting solid sphere having uniform volume distribution of charge
(i)
Outside point E=
1 4 0
q r2
P
(ii)
Inside point E
q = total charge
qr 1 4 0 a 3
= volume density of charge
Cylindrical Conductor of Infinite length Outside point E
2 0 r
= linear density of charge (as charges reside only on the surface)
Inside point E=0
Non-conducting cylinder having uniform volume density of charge 2 0 r r E 2 0 E
Outside point Inside point
Infinite plane sheet of charge E=
2 0
= surface charge density
Two oppositely charged sheets (Infinite) (i) Electric field at points outside the charged sheets EP = E R = 0 (ii) Electric field at point in between the charged sheets EQ =
0
Example 3.
Find electric field intensity due to long uniformly charged wire. (charge per unit length is )
Solution:
Elemental charge dq = d. Field at point P is dE = K .
R 4 0
Also
R
dq R 2
dq R 2 cos d Ex 3
dE x
2
dE y
2
d
X
2
2 2
dq sin R 2 2
Y
4 0R
R
dEX P dEY
dE
3.
On integration EY
4.
4 0 R
E E 2X E 2Y
tan
Example 4.
2 2 0 R
EX EY
45
What is the electric field at any point on the axis of a charged rod of length `L' and linear charge density`'? The point is separated from the nearer end by a. a
dx
P
x
Solution :
Consider an element, dx at a distance x from the point, P, where we seek to find the electric field. The elemental charge, dq =dx Then, dE k.
dx x2 aL
or
E k
a
Thus , E
Example 5.
1 dx k 2 x
aL
1 x a
1 1 k a L a
1 1 4 o a L a
[ k=
A ring shaped conductor with radius R carries a
y
total charge q uniformly distributed around it as
dq
shown in the figure. Find the electric field at a
R
point P that lies on the axis of the ring at a
1 ] 4 0
O
(R2 + x2)1/2
dE cos
P
x
x
distance x from its centre. dE dE sin
Solution:
Consider a differential element of the ring of length q ds . 2πR This element sets up a differential electric field d E at point P. The resultant field E at P is found by integrating the effects of all
ds. Charge on this element is
dq =
the elements that make up the ring. From symmetry this resultant field must lie along the right axis. Thus, only the component of d E parallel to this axis contributes to the final result. E = dE E = dE cos
1 dq 1 4 0 r 2 4 0 x cos = 2 (R x 2 )1 / 2
dE =
1 q ds . 2 2 2 R R x
To find the total x-component Ex of the field at P, we integrate this expression over all segment of the ring. Ex =
dE cos =
1 qx 4 0 2R. R2 x 2
ds 3/2
The integral is simply the circumference of the ring = 2R E=
1 qx 2 4 0 R x 2
3/2
As q is positive charge, field is directed away from the centre of the ring, along its axis.
5.
DIPOLE
Two equal and opposite charges separated by a small distance constitute an electric dipole P q 2a
P is directed from –ve to the +ve charge
(i)
Electric field at an Axial Point 1 2Px 4 0 [ x 2 a 2 ] 2
E=
x +q
A
-q
For x >> a E
(ii)
1 2P . 4 0 x 3
At a point on its perpendicular bisector E=
1 P 2 4 0 [a y 2 ] 3 / 2
For y >> a E
(iii)
A
1 4 0
P
y
+q
-q
y3
At any point r >> a
2P cos P sin 1 1 and E Er 3 4 0 4 0 r r3 1 P E 3 cos 2 1 3 4 0 r
Angle = tan-1 (1/2 tan )
A r
+q
-q
6.
GAUSS THEOREM
The flux of an electric field E through an arbitrary closed surface S is equal to the algebraic sum of the charge enclosed by this surface divided by 0.
q E. ds in 0
Example 6.
Figure shows a section of an infinite rod of charge having linear charge density which is constant for all points on the line. Find electric field E at a distance r from the line.
r E + + + ++ + + + + + + + + + + + + + + + +
Solution : From symmetry, E due to a uniform linear charge can only be radially directed. As a Gaussian surface, we can choose a circular cylinder of radius r and length l, closed at each end by plane caps normal to the axis. o E.ds qin o E.ds E.ds = qin
Cylindrical
Plane Surface
o E (2rl) + E.ds. cos 90 o l λl λ ε o 2πrl 2πε 0 r The direction of E is radially outward for a line of positive charge.
E=
Example 7. Figure shows a sphericaly symmetric distribution of charge of radius R. Find electric field E for points A and B which are lying outside and inside the charge distribution respectively. Solution:
The spherically symmetric distribution of charge means that the charge density at any point depends only on the distance of the point from the centre and not on the direction. Secondly, the object can not be a conductor, or else the excess charge will reside on its surface.
R
r
B
A
Now, apply Gauss’s law to a spherical Gaussian surface of radius r ( r > R for point A) o E .ds qin E=
ε oE( 4πr 2 ) = q
1 q where q is the total charge 4 0 r 2
For point B (r
3
r q 1 R 1 qr E= 40 40 R3 r2
7.
Electric Potential
Amount of work done in moving unit positive charge from infinity upto the point under consideration, against the field of a given charge q. In other words, it is negative of the work done by the internal forces. r
V = E. d r
=
1 q unit-volt 4 0 r
It is a scalar quantity (i)
Potential at a point due to several charges V = V1 + V2 + V3 + …
(ii)
Potential of a point due to an electric dipole V= For r >> a V=
P cos 1 4 0 (r 2 a 2 cos 2 )
P cos 1 4 0 r2
8.
Potential due to a uniformly charged disc
(i)
At a point on its axis
V= (ii)
(R
2
x 2 )1 / 2 x
= surface charge density
on disc At the centre of the disc V=
9.
2 0
R 2 0
Potential at the edge of a uniformly charged disc V=
R 0
Potential at the apex of a cone having charge Q distributed uniformly on its curved surface and having slanting length L V=
Q 2 0 L
Potential of a conducting sphere of radius R at a distance r from the centre (i)
r>R
(ii)
r=R
(iii)
r
Kq r Kq V= R
V=
V
q = charge on the sphere K=
1 4 0
kq R
10. Potential of a dielectric sphere (i)
q – total charge in the dielectric sphere
r>R V=
Kq r
(ii)
r=R
V=
Kq R
(iii)
r
V=
Kq r2 3 2R R 2
Torque on a dipole in presence of external electric field Torque = qE. 2a sin = PE sin PE
Potential Energy U = P . E
Example 8.
Two conducting spheres having radii a and b charged to q1 & q2 respectively. Find the potential difference between 1 & 2 ?
q2 q1 2
1
a b
Solution :
The potential on the surface of the sphere 1 is given by v1 = The
1 q1 1 q2 40 a 40 b
. . . . (A)
potential on the surface of the sphere 2 is given by, v2 =
1 q1 1 q2 , 4 o b 4 o b
V1 V2
1 q1 1 q1 4 o a 4 o b
V1 V2
q1 4 o
1 1 a b
11.
Electric Potential Energy The electric potential energy of a system of point charges is the amount of work done in bringing the charges from infinity in order to form the system. For point charges q1 and q2 separated by a distance r12, Electric potential energy of the system q1 and q2 is given by U=
1 q1 q 2 4 0 r12
For three particle system q1, q2 and q3 U=
1 4 0
q1 q 2 q1 q 3 q 2 q 3 r r13 r23 12
We can define electric potential (VP) at any point P in a electric field as , VP = UP ; where Up, is the change in electric potential energy in bringing the test q
charge q0 from infinity to point P. Example 9.
Determine the interaction energy of the point charges of the following setup
Solution :
U = U12 + U13 + U14 + U23 + U24 + U34
-q
a
+q 1
kq 2 kq 2 2 a 2a
= 3
kq 2 a2
2
kq 2 kq 2 kq 2 2 2 a a 2a
2
kq 2 2 a
2
a
a 4
(Here k = 1/ 40)
3
-q
a
+q
12. CAPACITORS 13.
Capacitance If Q is the charge given to a conductor and V is the potential to which it is raised by this amount of charge, then it is found that Q V. Or Q = CV, where C is a constant called capacitance of the conductor. C=
Q V
For parallel plate capacitor C =
0 r A d
i.e. the capacitance depends only on geometrical factors namely, the plate area and plate separation.
14. Spherical Capacitor C=
4 0 R1 R 2 (R 2 R1 )
15. Isolated Capacitor An isolated sphere can be thought of as a capacitor where other plate is at infinity. R1 = R and R2 =
C=
4 0 R1 4 0 R R1 1 R2
16. Cylindrical Capacitor C=
17.
2 0 R n 2 R1
R2 R1
18.
Combination of Capacitors Series Combination: In series combination, each capacitors has equal charge for any value of capacitance. Equivalent capacitance C is given by 1 C
C1 q
-q
1 1 1 C1 C2 C3
C2
C3
q -q
q -q
V
Parallel Combination: In parallel combination the potential differences of the capacitor connected in parallel are equal for any of capacitor. Equivalent capacitance is given by C = C1 + C2 + C3
q1
C1
q2
C2
q3
C3
V
Example 10.
In the above circuit, find the potential difference across AB.
8f
8f P
A
1
3 8f
10 V
2 Q
Solution:
8f
4 B
Let us mark the capacitors as 1, 2, 3 and 4 for identifications. As is clear, 3 and 4 are in series, and they are in parallel with 2. Then 2, 3, 4 combine is in series with 1. C34
C3 .C4 4f , C2,34 8 4 12f C3 C 4
Ceq.
8x12 4.8f , q Ceq. V 4.8 x10 48C 8 12
The `q’ on 1 is 48 C, thus V1=q/c=6v
[v1 =
48c 6v ] 8F
VPQ = 10 6 = 4V
By symmetry of 3 and 4, we say, VAB = 2V. Dielectric Substances having polar atoms/molecules intrinsically or being polarized are called dielectrics.
19.
Polar Dielectric
Substances which have polar atoms/molecules intrinsically but are randomly arranged. On application of external electric field they get polarized parallel to the external electric field. Net electric field inside dielectric. E E 0 Ei 1 E i E 0 1 K
E i = electric field due to induced charges
– surface charge density of capacitor plates
1 qi q 1 K
i – induced charge density
1 i 1 K
20. Capacitance of a parallel plate capacitor having dielectric slabs in series. C=
A 0 t1 t 2 ... K1 K 2
21. Energy stored in a Capacitor UC
Q2 1 1 CV 2 Q . V 2C 2 2
Energy supplied by Source US = Q. V. Energy lost in form of heat =
1 Q. V 2
22. Energy density of the electric field u
1 in free space 0 E 2 2 1 u K 0 E 2 in presence of dielectric medium ; K - dielectric 2
const.
23. Force on a Dielectric in a Capacitor F
Q 2 dC 2C 2 dx
when source is connected
Q C 1 2 dc V F dx 2
V=
Example 11. A 4f capacitor is charged to 150 V and another 6f capacitor is charged to 200 V. Then they are connected across each other. Find the potential difference across them. Calculate the heat produced. Solution :
4f charged to 150 V would have q1 = C1V1 = 600C
C1
q1
6f charged to 200 V would have q2 = C2V2 = 1200C After connecting them across each other, they will have a common potential difference V. Charges will readjust as q1’ and q2’ q 1 ' q 2 ' q 1 ' q 2 ' q q2 1800C 1 C 1 C 2 C 1 C 2 C 1 C 2 ( 4 6)f
V
V q2
C2
[conservation of ch arg e]
V 180 volt. Initial energy Ui
1 1 1 1 2 2 C 1 V1 C 2 V2 ( 4f )(150 V ) 2 (6f ).(200 V ) 2 0.165J 2 2 2 2
Final energy Uf
1 C1 C 2 .V 2 2
1 ( 4f 6f ).(180) 2 2
= 0.161J
Heat produced = |Uf Ui| = 0.003 J OBJECTIVE 1:
A long string with a charge of per unit length passes through an imaginary cube of edge a. The maximum flux of the electric field through the cube will be (A)
a o
(B)
2 a o
(C)
6 a 2 o
(D)
3a o
Solution :
The maximum length of the string which can fit into the cube is 3 a, equal to its body diagonal. The total charge inside the cube is
3 a , and hence the total flux through the cube is
3 a . o
(D) 2
2:
Potential in the x-y plane is given as V = 5(x + xy) volts. The electric field at the point (1, 2) will be (A) 3J V/m (A) 5 J V/m (C) 5J V/m (D) 3J V/m
Solution :
Ex =
V = (10 x + 5y) = 10 + 10 = 0 x V = 5x = 5 Ey = x E 5ˆj V / m .
(B) 3:
In the circuit shown, the equivalent capacitance between the points A and B is (A) C/5 (B) C/3 (C) C/2 (D) C
C C
A
C
B
C C
Solution:
Rearranging the circuit, the points E and D are at the same potentital (by symmetry). Then the capacity between E and D can be removed. 1 1 1 C C C
C = C/2 C C and are in parallel. 2 2
Hence Ceq =
(D)
C C =C 2 2
C
E C
A A C
C D
B 6C
4:
Electric field in a region is given by E (2iˆ 3jˆ 4kˆ) V/m. The potential difference between points (0, 0, 0) and (1,2,3) will be (A) 2 V (B) 5 V (C) 4V (D) 6 V
Solution:
P.d. across the points = E. r V2 – V1 = (2 ˆi 3ˆj 4kˆ ).(ˆi 2ˆj 3kˆ ) = 2 6 + 12 = 4 volts. (C)
5:
What is the mechanical work done in pulling the slab out of the capacitor after disconnecting it from the battery (A)
1 2 E C(r 1) 2
(C) E C (r 1) 2
Solution:
2
(B) E C (D) none of these
Work done = change in potential energy = U2 – U1 v1 = (1/2) E2C v2 =
1 (EC )2 1 E2C2 2 r = (1/2) E Cr 2 C' 2 C
Work done = (1/2) E2C (r – 1). (A). 6:
Solution:
Two point charges each of charge +q are fixed at (+a, 0) and (-a, 0). Another positive point charge q placed at the origin is free to move along x- axis. The charge q at origin in equilibrium will have (A) maximum force and minimum potential energy. (B) minimum force & maximum potential energy. (C) maximum force & maximum potential energy. (D) minimum force & minimum potential energy.
Y q (-a, 0)
The net force on q at origin is F F1 F2
=
1 q2 ˆ 1 q2 ˆ . 2 i . ( i ) 0ˆ 40 r 40 r 2
The P.E. of the charge q in between the extreme charges at a distance x from the origin along +ve x axis is U = =
1 q2 1 q2 . . 2 40 x 40 (a x ) 1 1 1 .q2 . 40 a x a x
dU q2 dx 40
1 1 2 (a x )2 (a x )
q q
X (+a, 0)
For U to be minimum dU 0, dx
d2U dx 2
0,
(a-x)2 = (a + x)2 a + x = (a – x) x = 0, because other solution is irrelevent. Thus the charged particle at the origin will have minimum force and minimum P.E. (D). 7:
In a parallel plate capacitor of capacitance C, a metal sheet is inserted between the plates, parallel to them. The thickness of the sheet is half of the separation between the plates. The capacitance now becomes (A) 4C (B) 2C (C) C/2 (D) C/4
Solution :
Before the metal sheet is inserted, C = o A / d After the sheet is inserted, the system is equivalent to two capacitors in series, each of capacitance C
o A 4C d/ 4
The equivalent capacity is now 2C. (B) 8:
The value of q if it floats in air is mgε r (A)
(C)
Solution :
2mgε r
(B)
q
m
mgε r
r
(D) none of these
qE = q(/r)
The magnitude of electric field due to the charged conducting plate is E =
σ
r
As the charged particle is floating in air (neglecting the buoyant force due to air we obtain)
q
mg
mg = qE mg E mg r q=
q=
(A) 9:
The capacitance of an filled parallel plate capacitor is 20 F. The separation between the plates is doubled and the space between the plates
is then filled with wax giving the capacitance a new value of 40 10 farads. The dielectric constant of wax is (A) 12.0 (B) 10.0 (C) 8.0 (D) 4.2
Solution :
101012 = K=8 (C)
10 :
o A K o A . 401012 = d 2d
The ratio of the time periods of small oscillation of the insulated spring and mass system before and after charging the masses is (A) 1 (C) 1
Solution :
q
k
+ m
(B) > 1 (D) = 1
In equilibrium of the charged small bodies
1 q2 = kx0 . 4 ( 0 x 0 )2
where x0 is the elongration in the spring in equilibrium. Let a further small elongation of x is made in the spring. Then net restoring force on any of the charged particle is given by,
F = k( x0 x )
1 q2 2 4 ( 0 x0 x )
= kx. a=
Since x < < x0 from (1) 2k x m
As F = a where = Hence =
2k m
mm a = 2 x, mm
T=2
m 2k
In absence of charge is T0 = 2 Therefore (D)
12
T 1 T0
m . 2k
q + m
11:
The charge flowing across the cell on closing the key k is equal to (A) CV (B) CV/2 (C) 2CV (D) zero
C
C 2
1
V
Solution:
When the key is kept open, the charge drawn from the source is Q = CeqV =
C V 2
When the key is closed the capacitor 2 gets short circuited And Ceq = C Q = CV charge flown through cell Q Q =
C V 2
(B) is correct choice. 12:
The conducting spherical shells shown in the figure are connected by a conductor. The capacitance of the system is (A) 40
ab
(B) 40 a
ba
(C) 40 b
Solution:
b
(D) 40
a
a2 ba
Hence, the capacitance of the system is the capacitance due to outer sphere of radius b. C = 4ob (C) is correct choice.
13:
A point charge q moves from point P to S along the path PQRS in a uniform electric field E pointing parallel to the positive direction of the x-axis. The coordinate of the point P, Q, R
y
P(a, b, 0)
and S are (a, b, 0), (2a, 0, 0), (a, b, 0) and (0, 0, 0) respectively. The work done by the field in the above process is given by the expression (A) qaE (C) q
a
2
b
2
E
(B) qaE (D) 3qE
a
Q(2a, 0, 0) x 100V
S(0, 0, 0) R(a, b, 0)
2
b2
E
Solution:
The work done is independent of the path followed and is equal to
qE . r ,
where r = displacement from P to S
Here, r = a ˆi bˆj , while E = E ˆi
work = (qE ˆi ). a ˆi bˆj = qaE Hence, B is correct choice. 14:
Charge Q is given to the upper plate of an isolated parallel plate capacitor of capacitance C. The potential difference between the plates (A)
(C)
Solution:
Q
(B)
C Q 2
Q
C 2
(D) zero
C
In general, for charge Q1 and Q2on upper and lower plate respectively the charge distributions on outer and inner part of the plates are shown in figure. Here Q1 =Q, Q2 = 0 Q1 Charge on inner side of Q1+Q2 plate
Q 2
are
and
respectively. Hence V =
Q 2
Q 2
2 Q1Q2 2 Q2
C
(Q1Q2) 2 Q1+Q2 2
Hence (C) is correct choice. 15:
A particle A of mass m and charge Q moves directly towards a fixed particle B. Which has charge Q. The speed of A is v when it is far away from B. The minimum separation between the particle is proportional to 2 (A) v (B) v 1 2 (D) v (C) v
Solution:
From Conservation of energy (KE+EPE)minimum separation = (KE+EPE)far away 1 Q2 1 mv 2 0 4 o rmin 2 1 rmin 2 v
0+
Hence, (D) is correct choice.
Q
A dipole of dipole moment p is kept along an electric field E such that E and p are in the same direction. Find the work done in rotating the
16:
dipole by an angle .
(B) W = –2Ep (D) W = –Ep
(A) W = 2Ep (C) W = Ep Ans. (A) Solution :
W = U = U2 – U1 Now U2 = (Ep cos ) = Ep U1 = (Ep cos 0) = Ep W = 2Ep.
17:
A simple pendulum of mass m and length carries a charge q. Find its time period when it is suspended in a uniform electric field region as shown in figure.
(A) T = 4
g 2 ( Eq / m) 2
(C) T = 2 2
g2 (Eq / m)2
(B) T = 2 (D) T = 2
E
g2 (Eq / m)2
2 g 2 ( Eq / m) 2
Ans. (B) Solution:
Time period of the pendulum = 2
geff
Here, geff = =
(mg )2 (Eq )2 m
=
g2 Eq / m
T=2
18:
Tension in the string in equibliriu m position mass of bob
2
g2 (Eq / m)2
.
A particle of mass 100 gm and charge 2 C is released from a distance of 50 cm from a fixed charge of 5C. Find the speed of the particle when its distance from the fixed charge becomes 3 m. Neglect any other force.
(A) –1.73 m/s
(B) 2.73 m/s
(C) –2.73 m/s Ans. (D) Solution:
Loss of potential energy = gain in kinetic energy U1 – U2 = K. 1
KQq r1
1 2 = ½ mv – 0 r2
2kQq m
v
r2 r1 r1 r2
2 9 109 5 10 6 2 10 6 2.5 = 1.73 m/s. 0.1 3 0.5
=
19:
(D) 1.73 m/s.
If a point charge q is placed at the centre of a cube what is the flux linked with the cube? (A)
(C)
1 q o
(B)
6 q o
(D)
3 q o
1 q 6 o
Ans. (A) Solution :
1 o
From gauss’s law, flux linked with a closed body is
times the
charge enclosed. The cube encloses a charge q so flux linked with cube, 20:
1 q o
If a point charge q is placed at the centre of a cube what is the flux linked with each face of the cube?
1 q o
(A)
(C)
6 q o
(B)
3 q o
(D)
1 q 6 o
Ans. (B) Now as cube is a symmetrical body with a faces and the point charge is at its centre, so electric flux linked with each face will be F
1 q 6 6 o
21:
2
Supposing that the earth has a surface charge density of 1 electron/m ; calculate earth's potential (A) 0.115Volt (C) 0.105Volt Ans. (A)
Solution:
0.110 Volt 0.112 Volt
(B) (D)
If R be the radius and the surface charge density of earth, then the total charge q on its surface is given by Q = 4R2 (i) The potential V at a point on earth's surface is same as if the entire charge q were concentrated as its centre. Thus V=
1 4R 2 R 1 q = 4 0 R o 4 0 R
Substituting the given value: V=
6.4 10 6 1.6 10 19 8.9 10 12
= 0.115 N-m/C = 0.115 J/C = 0.115Volt. 2
22:
Supposing that the earth has a surface charge density of 1 electron/m ; calculate electric field just outside earth's surface. 19 6 The electronic charge is 1.610 C and earth's radius is 6.410 m. 2 12 2 (0 = 8.910 C /N-m ) 8 (A) 1.810 N/C (B) + 1.8108 N/C 9 (C) 1.810 N/C (D) + 1.8109 N/C Ans. (A)
Solution:
(ii) Again, the electric field E just outside the earth's surface is same is if the entire charge q were concentrated at this cetnre. Thus E=
1 4R 2 1 q = 2 2 4 0 R 4 0 R o
Substituting the given value:
1.6 10 C / m 19
E=
2
8.9 10 12 C 2 / N m 2
= 1.8108 N/C The minus sign indicates that E is radially inward.
23 :
A circular wire of radius R carries a total charge Q distributed uniformly over its circumference. A small length of the wire subtending angle at the centre is cut off. Find the electric field at the centre due to the remaining portion .
(A)
(C)
Q
sin
4 0 R Q sin 2 2 4 0 R 4 2
2
(B)
(D)
sin 4 0 R 2 Q sin 2 2 4 0 R 8
Q
2
2
Ans. (B)
Solution :
Electric field due to an arc at its centre is k 1 , 2 sin , Where k = R 40 2
= angle subtended by the wire at the centre, = Linear density of charge. Let E be the electric field due to remaining portion. Since intensity at the centre due to the circular wire is zero. Applying principle of superposition.
k 2 sin nˆ E 0 R 2
E
Q .2 sin 4 0 R 2R 2 1
.
sin 4 0 R 2 Q
2
2
24: An electric dipole, made up of a positive and a negative charge, each of 1 C and 5 placed at a distance 2 cm apart, is placed in an electric field 10 N/C. Compute the maximum torque which the field can exert on the dipole, and the work that must be done to turn the dipole from a position = 0 to = 180
(A) 4106N-m or Joule (C) 410+6N-m or Joule Ans. (D)
Solution :
(B) (D)
4109N-m or Joule 4103N-m or Joule
The torque exerted by an electric field E on a dipole of moment p is given by = pE sin, Where is the angle which the dipole is making with the field.
is a maximum, when = 90. That is max = pE Here p = q(2) = 1106 0.02C/m and E = 105N/C max = 1106 0.02 105 = 2103N-m The work done in rotating the dipole from an angle o to is given by W=
o
pE sin d = pE (coso cos)
Here o = 0 and = 180 W = pE (cos0 cos180) = 2pE = 4103N-m or Joule 3
25:
A parallel-plate air capacitor has a plate area of 100cm and separation 5mm. A potential difference of 300V is established between its plates by a battery. After disconnecting a battery, the space between the plates is filled by ebonite (K = 2.6). Find out new potential difference between plates of the capacitor. (A) 115V (B) 120V (C) 110V (D) 125V Ans. (A)
26:
A parallel-plate air capacitor has a plate area of 100cm and separation 5mm. A potential difference of 300V is established between its plates by a battery. After disconnecting a battery, the space between the plates is filled by ebonite (K = 2.6). Find out initial and final capacitances of the capacitor.
3
(A) (C)
2.771011F, 4.61011F 1.771011F, 6.61011F
(B) (D)
1.771011F, 4.61011F 1.771013F, 4.61011F
Ans. (B) 27:
3
A parallel-plate air capacitor has a plate area of 100cm and separation 5mm. A potential difference of 300V is established between its plates by a battery. After disconnecting a battery, the space between the plates is filled by ebonite (K = 2.6). Find out initial and final surface-density of charge on the plates. +7
2
(A) 5.3110 C/m 7 2 (C) 5.3110 C/m Ans. (C)
Solution:
(B) 6.31107C/m2 (D) 9.31107C/m2
Capacity of the parallel plate air capacitor =
o A 8.86 10 12 100 10 4 = 1.771011F 3 d 5 10
Final capacity of the capacitor with dielectric between the plates is C = KC = 2.61.771011, C = 4.61011F
Initial charge on the capacitor 1.771011 300 = 5.31109C Since, the battery has been disconnected, the charge remains the same, therefore the new potential difference is V =
q 5.31 10 9 = 115V C 4.6 10 11
The surface density of charge remains the same in both the cases, i.e., = 28:
q 5.31 10 9 = = 5.31107C/m2 4 A 100 10
A spherical condenser has 10cm and 12 cm as the radii of inner and outer spheres. The space between the two is filled with a dielectric of dielectric constant 3. Find the capacity when the outer sphere is earthed.
(A) 41010F (C) 21010F Ans. (C)
q2 q1
R
r
(B) 11010F (D) 61010F
3 0.1 0.12 ab = 21010F = 9 0 . 12 0 . 1 b a 9 10
Solution:
C = 4k0
29:
A spherical condenser has 10cm and 12 cm as the radii of inner and outer spheres. The space between the two is filled with a dielectric of dielectric constant 3. Find the capacity when the inner sphere is earthed.
15 10 10 F 32 32 (C) 10 9 F 15 Ans. (D)
(A)
(B)
32 10 10 F 15
(D)
32 10 10 F 15
q2 q1
R
r
By choice of reference potential at infinity and the potential of earthed conductor zero, the given system can be visualised as combination of two spherical capacitors, both being at same potential difference. Connection wise these may be considered to be in parallel connection. 32 10 10 F C = 4ob + 4k0 ab = b a
15
30:
A charge Q is distributed over two concentric hollow spheres of radii r and R (R >r) such that the surface densities are equal. Find the potential at the common centre.
(A)
1 Q R r 4 0 R 2 r 2
Q R r (C) 4 0 R 2 r 2 Ans. (A) 1
Solution:
q1 + q2 = Q =
q1 4r
2
(B)
Q R r 4 0 R 2 r 2
(D)
Q R r 4 0 R 2 r 2
1
1
2
. . . (i) q2
. . . (ii)
4R 2
from (i) and(ii) q1 =
Qr 2 r 2 R 2
q2 =
Vcentre = V1 + V2 = = 31:
1 Q R r 4 0 R 2 r 2
1 4 0
QR 2 r 2 R 2
q1 q 2 R r
An electric dipole of dipole moment P is placed in a uniform electric field E is stable equilibrium position. Its moment of inertia about the centroidal axis is I. If it is displaced slightly from its mean position find the period of small oscillation.
T 2
PE I
(B)
(C) T 4
I PE
(D) T 4 2
(A)
I PE
T 2
I PE
Ans. (B)
Solution:
When displaced at an angle , from its mean position the mean position the magnitude of restoring torque is
q
PE
For small angular displacement sin PE
The angular acceleration is,
pE 2 cos I I
E P
+q
+q q
PE I I T 2 PE
Where 2
32:
Figure shows two conducting thin concentric shells of
q
radii r and 3r. The outer shell carries charge q. Inner
R2
shell is neutral. Find the charge that will flow from inner shell to earth after the switch S is closed.
r s
3r
q 3 q (C) 3 Ans. (D)
(A)
Solution:
(B)
q 2
(D)
q 3
Let q be the charge on inner shell when it is earthed.
Vinner = 0
1 q q 0 4 o r 3r
q q / 3 i.e. 33:
q charge will flow froms inner shell to earth. 3
A capacitor stores 10C charge when connected across a battery. When the gap between the plates is filled with a dielectric, a charge of 20 C flows through the battery. Find the dielectric constant of the dielectric.
(A) k = 2 (C) k = 3 Ans. (C) Solution:
In absence of dielectric Q = CV = 10C With dielectric Q = kCV = 30C From (i) and (2)
(B) k = 4 (D) k = 1
. . . (1)
. . . (2)
k=3 34:
Two infinitely large sheets S1 and S2 having surface charge densities 1 and 2 (1 > 2) respectively are placed at a distance d. Find the work done by the electric field when a charge particle ‘ Q’ is displaced by a 0 distance ‘a’ (a < d) at an angle 45 with the normal of the sheets. It is assume that the charge does not affect the surface charge densities of the two plates.
(A)
WPQ =
(C)
WPQ =
q (s 2 - s 1 ) 2 2 q (s 1 - s 2 ) 2 2
d
(B)
WPQ =
d
(D)
WPQ =
q (s 1 - s 2 ) -2 2 q (s 2 - s 1 ) -2 2
d d
Ans. (C) Ans.
Electric field at any point between the plate
æs s ö =ç 1 + 2 ÷ è2 Î o 2 Î o ø P WPQ q E.dr
Q
æs s ö = - qòç 1 - 2 ÷dr cos 45o 2Îo 2Îoø Qè P
0 æs s ö = - qòç 1 - 2 ÷cos 45o dr 2Îo 2Îoø dè
1 1 o q (s 1 - s 2 ) r 2Îo 2 d q (s 1 - s 2 ) WPQ = d 2 2
= -
35.:
A positive charge (+q) is located at the centre of a circle as shown in figure. W1 is the work done in taking a unit positive charge from A to B and W2 is the work done in taking the same charge from A to C. Then. (A) W1 > W2 (B) W1 < W2 (C) W1 = W2 (D) W1 = W2 = 0
Solution:
Point A, B and C are at the same distance from charge +q; hence electrical potential is the same at these points, i.e. There is no potential difference between A, B and C. Hence W 1 = W 2 = 0.
36.
A point Charge q is placed inside the cavity of a metallic shell.
Which one of the diagram correctly represents the electric lines of force.
(A)
(B)
(C)
(D)
Solution:
(C) Because electric field inside the conductor is zero and electric field lines are perpendicular to Gaussian surface.
37.:
A soap bubble has radius R, charge Q, surface tension T. Find the excess pressure in it. 32π 2 R 2 ε0T - q 2 64π 2 R 3ε0T - q 2 (A) (A) 32π 2 R 4 ε0 32π 2 R 4 ε0 (C)
Solution:
128π 2 R 3ε0T - q 2 32π 2 R 4 ε0
none of these
(C) Pexces
38..
(D)
q2 4T R 32 2R 4 0
A charge of 2C is brought from B to C along the path as shown by arrow in the figure. The work done is (A) 0.75 J (B) 0.6J (C) 0.06J (D) 0.075J
Solution:
(D)
w
B
q1q2 4 0
1 1 rf ri
1 1 10 2 1012 9 109 0.6 0.8 = 0.075 J 39.
Solution:
40.
2 F
80 cm
The electric field intensity at a point is 20 i + 15 j N/C . Considering potential at origin to be zero, the potential at P (2, 2) is (B) (A) 40 i + 60 j V (C) –100V (D)
60 cm
10 i + 15 j V 20V
x y (C) V = Ex dx Ey dy Ex x Ey y 100V 0 0
The equivalent capacitance between A and B is (each of the capacitors obtained is of capacitance equal to C) B’
B A’
A
1 C 2 5 (C) C 3 Ans. (C)
3 C 5 2 (D) C 5
(B)
(A)
1 2
B
A
4
D
5 A
D A=E
1 2 3 2 4 3
5 4
3
B
E
B
C
41.
CAB
2C C 5C C 3C 3
28. Page 426 Four large metal plates are located a small distance apart from one another as shown in the Fig. 24. The extreme plates are connected by means of a conductor, while a potential difference V is applied to the internal plates. Find the electric fields between the neighbouring plates
(A) E23
V V , E34 d 2d
V V , E23 E34 2d 2 (D) None of these
(B) E12
(C) zero Ans.
(A) We may take the plates to be connected as shown in the Fig. 56. Let E23 be the electric field between plates 2 and 3. Then from V the fact that electric field = rate of change of potential, E22 . d 1
2
4
4
– + + –– + + – – + + –– + + – – + + –– +
A
B
Applying `loop rule' to the loop `A – 1 – 2 – 3 – 4 – B – A' V V' – V – + V' V' = 2 where V' is the numerical value of the potential difference between plates 1 and 2 or 3 and 4 V E12 or E34 2d E 0 1 (density of charge on the plate 1) V 0 E12 0 4 (numerically) 2d The surface density of charge of the right surface of plate 2 is ' 0 ,E23 0 Vd
2 (surface density of charge on plate 2)
0 V 0 V 3 0 V 2d d 2 d
Thus V V E12 or E34 d 2d V 3 V (ii) 1 4 0 2 0 0 . 2d 2 d
(i) E23
42.
A simple pendulum consists of a small sphere of mass m suspended by a thread of length l. The sphere carries a positive charge q. The pendulum is placed in a uniform electric field of strength E directed vertically upwards. With what period will the pendulum oscillate if the electrostatic force acting on the sphere is less that the gravitational force? Assume the oscillations to be small 1/2
1/2
l (A) T = 2π g
l (C) T = 2π g - qE m Solution:
ml (B) T = 2π qE 1/2
1/2
Let x be the small displacement given to the pendulum such that the angle is small. The forces acting at A are (i) tension T along the thread (ii) weight mg acting vertically downwards. qE θ (iii) electrical force qE vertically T l upwards. The resultant force vertically down A wards is (mg – qE). Therefore θ B qE Net acceleration g ' g m mg mg qE cosθ mg qE sinθ Time period l l 2 T 2 qE g' g m
43.
l (D) T = 2π g + qE m
1/ 2
A ring has charge Q and radius R. If a charge q is placed at its centre then the increase in tension in the ring is Qq (A) (B) zero 4πε0 R 2
(C)
Solution:
44.
Solution:
45.
Qq 4π 2 ε0 R 2
(D)
T cos
T cos
Consider a small element AB, is very small. Then AB = R(2) Q Q Change on AB is dQ = 2R 2 R dQ q Qqθ 2T sinθ = = 2 4π 0 R 4π 2 0 R 2 Qq Qq 2Tθ = or T= 2 2 2 4π 0 R 8π 0 R 2
T
A
B
q T sin T cos
A charge Q is uniformly distributed over a large plastic plate. The electric field at point P, close to the centre of the plate is 10Vm-1 . If the plastic plate is replaced by a copper plate of the same dimensions and carrying same charge Q then the electric field at the point P will be (A) 5Vm-1 (B) zero -1 (C) 10Vm (D) 20Vm-1 (C) because E = 0
A hemisphere of radius r is placed in a uniform electric field of strength E. The electric flux through the hemisphere is (A (C)
2Er2 -2Er2
(B) (D) x
A
B
E
Solution:
Qq 8π 2 ε0 R 2
E.dS
B
E.dS
A
X B
E.dS
E r 2 E r 2 E r 2 E r 2
A X
E.dS
–Er2 zero
T
x
A
dS
B
E
short cut AXB is symmetrical surface Electric flux due to this part is zero. However, electric flux due to AB part is –ER2. 46.
1. What is the potential difference at the centre C +q (A) Zero (B) Kq a 2 a Kq (D) none of these (C) 2 a -q Solution: (A) We know that potential is a scalar quantity 1 q V= 40 r Where r is the distance between charge and point where potential has to be found. total potential at center C 1 1 (q1 + q2 + q3 + q4). = r 40 1 = (+ q – q + q – q) 40r =0
-q
C
47.
MeV is (A) 1.6 1019 J (B) 1.6 1022 J (D) 1.6 1013 J (C) 1.6 1013 J Solution: (C) 1 MeV = 1.6 10–16 106 Joules = 1.6 10–13 Joules 48.
Solution: field 49.
What is the angle between electric dipole moment and the electric field strength due to it on the axial line (A) 0 (B) 90 (C) 180 (D) none of these (D) Direction of dipole movement will be opposite to the electric
The total electric flux, leaving spherical surface of radius 1 cm, and surrounding an electric dipole is
+q
(A) q (C) 2q
0 0
(B) zero (D) 8 r 2
q
0
+q 4pe 0 -q Electric flux passes due to – q charge of dipole 4pe 0
Solution: Electric flux passes as due to q charge of dipole
\ Net flux passes due to both the charges =
50.
Solution:
q q 0 4pe 0 4pe 0
An isolated sphere of radius R contains a uniform Electric field magnitude volume distribution of positive charge. Which of the A curve on the graph below correctly illustrates the B dependence of the magnitude of the electric field of A,B,C,D C the sphere as a function of the distance r from its D r centre O R (A) A (B) B (C) C (D) D KQTOTAL (C) E r for inside point and E for outside point r2 3 0
and it will be
represented by C.
51.
The variation of potential with distance R from a fixed point is as shown in figure 4 (A) the electric field at R =5m is V m 1 5 æ4 ö (B) the electric field at R =5m is ç ÷V - m +1 , è5 ø (C) the potential at R = 5m is + 2.5 V- m– 1 is continuous and a decreasing function of distance. (D) the segment AB corresponds to an equipotent surface.
Solution:
(D) (The segment AB corresponding to an equipotential surface)
52.
An electric dipole is placed inside a conducting shell. Mark the correct statement(s) (A) the flux of the electric field through the shell is zero (B) the electric field is zero at every point on the shell (C) the electric field is not zero anywhere any where on the shell
Solution:
53.
(D) the electric field is zero on a circle on the shell. (A) Inside the surface the total charge is zero. So flux must be zero.
A network of six identical capacitors, each of value C is made as shown in the figure. Equivalent capacitance between points A and B is (A) C/4 (B) 3C/4 (C) 4C/3 (D) 3C A
B Solution:
(C) The network is equivalent to – Therefore equivalent capacitance = [2C series C] // [C series 2C] 2C×C 4C = 2 = 2C +C 3
54.
The charge flowing across the cell on closing the key k is equal to (A) CV (B) CV/2 (C) 2CV (D) zero
C
C 2
1
V
Solution:
When the key is kept open, the charge drawn from the source is Q = CeqV =
C V 2
When the key is closed the capacitor 2 gets short circuited And Ceq = C Q = CV charge flown through cell Q Q =
C V 2
(B) is correct choice.
55.
The capacitance of the system of parallel plate capacitor shown in the figure is (A) (C)
2 0 A1 A2 ( A1 A2 )d
0 A1 d
(B) (D)
2 0 A1 A2 ( A2 A1 )d
0 A2 d
A1
d A2
Solution :
Since the electric field between the parallel charge plates is uniform and independent of the distance, neglecting the edge effect, the effective area of the plate of area A2 is A1. Thus the capacitance between the plates is A C= 0 1 d
+
A1
0
d
E -
A2
(C) 56.
Solution:
57.
The three capacitors in figure, store a total energy in J of (A) 12 (B) 36 (C) 48 (D) 80
6μF
Total capacitance 6 F Applied voltage 4v 1 1 Stored energy U CV 2 6 106 (4)2 48F 2 2 In the circuit shown in figure C = 6 F. The charge stored in capacitor of capacity C is
6μF
3μF
(C)
C
(A) (C)
2C
zero 40 C
4V
10V
(B) (D)
90 C 60 C
Solution: (C) Both the capacitors are in series. Therefore charge stored on them will be same. (C)(2C) 2 2 Net capacity C 6F 4F C 2C 3 3 Potential difference 10V q CV 40C 58.
Figure shows two parallel plates, R and S joined to a battery of voltage V and with charges +Q and –Q. 1. the energy stored is QV. 2. the electric field strength between the plates increases uniformly from S to R. 3. the electric potential between the plates decreases uniformly from R to S. Which of the statement is/are correct (A) 3 only (C) 2 and 3 only
(B) 1 and 2 only (D) 1 only
Solution: uniformly
(A)
The electric potential between the plates decreases from R to S.
59.
In the circuit shown in figure potential difference between A and B E = 190V C
3C
A
Solution:
B
C
is (A) 30 V (C) 10 V
3C
(B) 60 V (D) 90 V
(C) Potential difference is divided among two capacitance C1 and C2 in the inverse ratio of their capacities when it is final in series. Therefore E = 190V C
3C
A
C
B 3C
Voltage across P and Q is C =190 C CapacitanceacrossPandQ =190
C
P
3C
Q
C + C
3C
equivalent capacitance between P and Q C =190 40 volt 15C C 4 Potential difference between P and Q (i.e. 40 volt) Will divide between two capacitor C and 3C which is in series. Therefore voltage across 3C capacitance. C 10Volt volt 40 C 3C Hence (C) is correct choice
60.
A parallel plate capacitor has two layers of dielectrics as shown in figure. This capacitor is connected across a battery, then the ratio of potential difference across the dielectric layers is k 1= 2
k2 = 6
d
2d
(A) 4/3 (C) 1/3 Solution:
(D)
(B) 1/2 (D) 3/2
Capacitance (for k 2) kA C1 0 d C1 2C
Where C
0 A d
Capacitance (for k 6) 6 A C2 0 2d C2 3C Therefore ratio of potential difference across the dielectric layer is 3/2
CURRENT ELECTRICITY CURRENT ELECTRICITY Flow of electric charge constitutes electric current. For a given conductor, if 'Q' charge flows through a cross-section of area A in time 't', then the average electric current through the conductor is given as Q dQ I= and its instantaneous value is . dt t
Crosssection
MECHANISM OF CURRENT FLOW IN METALLIC CONDUCTOR When an external potential difference is applied across a metallic conductor then an electric field is set up within the conductor. Applied electric field Force on electrons drift of electrons Due to the externally applied electric field electrons drift with an average velocity called drift velocity. This causes an electric current Total charge crossing a cross-section in one second is equal to I = neAvd. Here Avd is the volume of a cylinder of cross-section A length vd and ne is charge density of charge carriers (e.g. electrons).
vd I S
The current density is defined by J = /A Example1 : A steady current passes through a cylindrical conductor. Is there an electric field inside the conductor ? Solution : Yes; No doubt under steady state conditions in electrostatics when a conductor is charged, electric field inside it is zero as metal is an equipotential surface. However when a potential difference is applied across a conductor and a steady current flows though it, the condition no longer remains static and there exists an electric field inside the conductor. OHM’S LAW It states that "the potential difference across a conductor is directly proportional to the current flowing through it at a given temperature". At constant temperature
V constant(R) I
the constant 'R' is called resistance of the conductor. Resistivity () and conductivity ():
The resistance R of a given conductor is directly proptinal to length () and inversitional proptional cross-sectional area (A) such that R =
, where = A
resistivity of the material of the given conductor. Its S.I. unit is m. Reciprocal of resistivity is called the electrical conductivity () of the material, thus =
1 = whereas reciprocal of resistance is called conductance of the RA
given conductor. S.I. unit of conductivity is ( - m)-1 and is usually written as mho/m. Temperature Dependence of Resistivity: The conductivity of a metal decreases as its temperature is increased. Thus resistivity increases with the rise in temperature. If T and 0 represent the resistivities at temperatures T and T0 respectively, then for small temperature variations, T= 0 [1 + (T T0)] Where is called the temperature coefficient of resistivity. The resistivity varies over a very wide range. For metals (good conductor) 10-8 -m and for insulators 1017 -m Semiconductors (silicon, germanium ) have intermediate value much smaller than insulator but much larger than metals. Temperature coefficient of resistivity is negative for semiconductors and positive for the metals. For superconductors resistivity is zero. Thermistor: A thermistor is a semiconductor electronic device in which the resistance decreases as its temperature increases. This is used as a thermometer. The temperature coefficient of resistivity is negative for semiconductors, hence thermistors are usually prepared from oxides of various metals such as nickel, iron, cobalt and copper etc. A thermistor is used to detect small changes in temperature of the order of even 10-3 0C. Colour code for carbon Resistors: tolerance
The four bands indicate digit -1, digit-2, multiplier and tolerance respectively and the values of different colours are given in the following table.
digit 1 digit 2
Resistance code (in )
multiplier
Colour
Digit
Multiplier
Black
0
1
Brown
1
10
Red
2
102
Orange
3
103
Yellow
4
104
Green
5
105
Blue
6
106
Violet
7
107
Gray
8
108
White
9
109
Tolerance
Gold
0.1
5%
Silver
0.01
10%
Sometimes the carbon resistor indicates only three bands and the tolerance is missing from the colour code. This means tolerance has to be taken as 20%. Example2: Find the resistance of a carbon resistor if the colour code from left to right indicates brown, yellow, green and gold. Solution:
Use diagram 5 1 4 10 5% 5 R = (14 10 5%) = (1.4 106 + 0.07 106 ) = (1.4 0.07)M
KIRCHHOFF’S LAWS Junction Rule: It is based on the law of conservation of charge. At a junction in a circuit the sum of incoming currents is always equal to the sum of outgoing currents. In otherwords the algebraic sum of the currents at a junction is zero. Loop rule The algebraic sum of the changes in potential around any closed path is zero. It is based on the law of conservation of energy. In case of a resistor of resistance 'R' potential will decrease in the direction of current. Hence, for the shown conductor Va – Vb = IR
For an emf source, the potential changes will be obtained as illustrated below, Emf = , internal resistance = r i
a
b
I
a
b
R
Emf = , internal resistance = r i
a
b
Va – Vb = + ir Va – Vb = + ir Students can use any sign convention which they find easy. Example3: In the series circuit shown, E,F,G,H are cells of emf 2V,1V,3V and 1V respectively, and their internal resistance are 2, 1, 3 and 1 respectively. Calculate (i) the potential difference between B and D and (ii) the potential difference across the terminals of each of the cells G and H. Solution: Let us redraw the circuit. I At junction D, we have applied the A I1 junction rule, whereby we get current in DB as shown. + 1V F Loop BADB 1 2I1 - 2 + 1 + I1 + 2 (I1 I2) = 0 D 5I1 2I2 = 1 Loop DCBD 3+3I2+I2+12(I1I2)=0 6I2 – 2I1 = 2
E A
+ G
D
2V +
B I2
2 2
I1- I2
1V
+
1
3
3V
G
2 5 V 1 13 13
VBD = 2(I1) 2 + 1 + I1= 3 I1 1 = 3
(ii)
Terminal voltage of G = |3 + 3.I2| = 3 3
21 6 V = 13 13
GROUPING OF RESISTANCES
C
E
(i)
6 19 (I) 1 V . 13 13
H
+
2
5 6 A, I2 A 13 13
Terminal voltage of H =
B
+ F
1
I1
+
C
H
Resistance in series Let the equivalent resistance between A & B equals Req , by definition. Req =
V I
A
R1
R2
R3
B
I V
. . . (1)
Using Kirchoff's 2nd rule for the loop shown in figure, V = IR1 + IR2 + IR3 . . . (2) From (1) and (2) Req = R1 + R2 + R3 Resistance in parallel Here again, Req =
R1 i1
V I
. . . (1)
R2
A
I = i1 + i2 + i3 =
V V V R1 R2 R3
. . . (2)
I
i2
B
R3
i3
I
From (1) and (2) V
1 1 1 1 Req R1 R2 R3
Example 4: Find the equivalent resistance between A and B in the circuit shown here. Every resistance shown here has a magnitude of 2 .
C 2
2 O
A 2
B 2
D
Solution:
Points C, O & D are at the same potential. Therefore, resistances AO, AC and AD are in parallel . Similarly BC, BO and BD are in parallel. Similarly BC, BO and BD are in parallel.
=
RAB =
1 1 (2) + (2) 3 3
4 = 1.33 . 3
ENERGY, POWER AND HEATING EFFECT When a current I flows for time t from a source of emf E, then the amount of charge that flows in time t is Q = It. Electrical energy delivered W = Q.V = VIt Thus, Power given to the circuit, = W/t = VI or V2/R or I2R E r In the circuit 2 2 E. I = I R + I r, where I EI is the rate at which chemical energy is converted to 2 R electrical energy, I R is power supplied to the external 2 resistance R and I r is the power dissipated in the internal resistance of the battery.
An electrical current flowing through conductor produces heat in it. This is known as Joule's effect. The heat developed is given by H = I2.R.t joule, where I = current in ampere , R = resistance in , t = time in second. Maximum Power Theorem In a circuit, for what value of the external resistance the maximum power be drawn from a battery? For the shown network power developed in resistance R equals P
2
r
I R
E .R E ( I = and P = I2R ) 2 (R r ) Rr
Now, for dP/dR = 0 (for P to be maximum
E
dP 0) dR
(R r )2 2(R)(R r ) 0 E2. (R r ) 4
R + r = 2R R = r The power output is maximum, when the external resistance equals the internal resistance. R=r Example5 : A copper wire having a cross-sectional area of 0.5 mm2 and a length of 0.1 m is initially at 25oC and is thermally insulated from the surroundings. If a current of 10 A is set up in this wire (a) Find the time in which the wire starts melting. The change of resistance of the wire with temperature may be neglected. (b) What will this time be, if the length of the wire is doubled? Density of Cu = 9 103 Kg m-3 specific heat of Cu = 9 10-2 Cal Kg-1 oC-1, M.P. (Cu) 1075 oC and specific resistance = 1.6 10-8m. Solution : (a) Mass of Cu = Volume density = 0.5 10-6 0.1 9 103 = 45 10-5 Kg. Rise in temperature = = 1075-25 = 1050 oC. Specific heat = 9 10-2 Kg-1 oC 4.2 J
I2Rt = mS
but
R
t
mS I2 .R
L 1.610 8 0.1 3.2 10 3 A 0.5 x10 6
t
454.210 5 10500.09 558 s 10103.210 3
(b) When the length of wire is doubled, R is doubled, but correspondingly mass is also doubled. Therefore, wire will start melting in the same time. WHEATSTONE BRIDGE
For a certain adjustment of Q, VBD = 0, then no current flows through the galvanometer. VB = VD or VAB= VAD I1.P = I2.R Likewise, VBC = VDC I1.Q = I2.S
BI
1
Q
P
A
G
C
I2
R
P R Dividing, we get, Q S
S D
I2
GROUPING OF IDENTICAL CELLS 1. Series Grouping emf of the cell is and internal resistance is r. R is the external resistance, I is the current passing through the circuit and n is the total number of cells. Applying Kirchhoff’s law ir + ir + ........ (to n times ) iR = 0 i
n R nr
r
I R
2. Parallel grouping Applying Kirchhoff’s law I r IR 0 n n I r nR
r
r
r
R
3. Mixed grouping Number of rows is m and number of cells in each row is n Apllying kirchhoff’s law I r IR 0 m mn I= mR nr n n
For current through R to be maximum, mR = nr R=
I/m I/m I/m
R
nr = one line total internal m
resistance/No. of Line RC-CIRCUIT Charging: Let us assume that the capacitor in the shown network is uncharged for t < 0. The switch is connected to position 1 at t = 0. Now, 'C' is getting charged. If the charge on capacitor at time 't' is q.
r
writing the loop rule, q + IR E = 0 C
1
S
R
2
I
I
q dq R E c dt dq 1 dt EC q RC
C
dq RC EC q dt I
Integrating
q
o
dq 1 t dt EC q RC o
1 .t RC
ln | EC q |oq EC q t EC RC
ln
q EC 1 e t / RC At t = 0, q = 0
and at t = , q = E C (the maximum charge.) = qmax Thus,
qmax = c
q qmax 1 e
t RC
q
dq qmax t / RC E t / RC e e dt RC R E i imax e t / RC where imax R
t
i
i imax = /R
t
Discharging Flip the switch to 2 Consider the same arrangement as we had in the previous case with one difference that the capacitor has charge qo for t<0 and the switch is connected to position 2 at t = 0. If the charge on capacitor is q at any later moment t then the loop equation is given as
q IR 0 c dq q R dt c
Integrating, at t = t, q = q
1 S
2
C
R
dq 1 dt q RC
t = 0, q = q0
I
q
q0
i
dq 1 t dt q RC o
ln
q t or qo RC
q q0 .e t / RC
q0 t / RC e RC
EC t / RC e RC i i0e t / RC i
'-ve' sign indicates that the discharging current flows in a direction opposite to the charging current. Discharging
O i
q
t
-imax = /R
O t
Example6: Calculate the steady-state current in the 2 resistor shown. The internal resistance of the battery is negligible and the capacitance of the capacitor is 0.2 F.
2 B
A 3 0.2F
6V
4
2.8
Solution : The resistance of the parallel combination of 2 and 3 resistors is given by 1 1 1 5 R 2 3 6
R 1.2
This resistance is in series with 2.8 giving a total effective resistance = 1.2 + 2.8 = 4 . In the steady state, charge on the capacitor C has stablised and hence no current passes through 4 resistor which is in series with the capacitor. Thus the current through the circuit = 6/4 = 1.5 A, VAB = 1.5 1.2 = 1.8 V, I through 2 resistor = 1.8/2 = 0.9 A. AMMETER An Ammeter is an instrument used for measuring current in electrical circuits. A galvanometer is a low resistance instrument. A large current passing through it may damage the instrument Changing the range of an ammeter Suppose the ammeter gives full scale deflection when a current Ig flows through it. Now if we want to convert the reading of the ammeter in such a
manner that it gives full scale deflection for a higher current I in the branch of the circuit, we connect a small resistance S in parallel to the coil of the galvanometer, which has a resistance G. The resistance value is so chosen that out of the total current I only Ig flows through the coil and the remaining current flows through S. As potential difference across S = potential difference across G.
(I – Ig)S = IgG
G
Ig
I
(I-Ig)
IgG I Ig
S
S=
So, effectively this ammeter will measure current up to I ampere and its effective resistance GS . S G
=
In practice G is large as compared to S. Therefore the effective resistance of the ammeter equals RA S, which is small. An ideal ammeter has zero resistance. VOLTMETER A voltmeter is an instrument used for measuring potential difference across the two ends of a current carrying conductor. It is connected in parallel with the conductor across which the potential difference is to be measured. The current through the conductor should not change on connecting the voltmeter, and so the voltmeter should draw a very small current, i.e. its resistance has to be high. When a galvanometer is used to measure potential difference across the ends of a current carrying conductor, a high resistance R is connected in series with the galvanometer. Consider the diagram VOLTMETER shown. Suppose the galvanometer gives full scale G Ig deflection when a current Ig passes through its coil. If G is resistance of the galvanometer coil then:
V
R
Potential difference to be measured = V = Ig (G R) So, effectively the voltmeter has resistance = Rv = (G + R)
R=
V -G Ig
I
In practice Rv is very large compared to G. An ideal voltmeter should possess infinite resistance.
POTENTIOMETER A potentiometer is used to compare electromotive forces of two cells or to measure the internal resistance of a cell. Principle: The potentiometer is based upon the principle that when a constant current is passed through a wire of uniform cross sectional area, the potential drop across any portion of the wire is directly proportional to its length. R Consider the network shown in the E diagram : I
E R1 R R R 2 1
Here, VA VB
I A
I
I
R1
B
R2 (Fig. i)
Now, let us consider that an EMF source having EMF same as (V A – VB) calculated above and internal resistance ‘r’ connected in parallel to R1. Suppose that a current ‘I2’ passes through the EMF source R
E
A
I1
B
R1
I2
R2
E', r
(I1+I2)
C
(Fig. ii)
We have, applying Kirchoff’s rules, the following equations R2 (I1+I2) + R1 I1 + R (I1+ I2) = E R1 I1 = I2 r + E
. . . (i) . . . (ii)
Solving (i) & (ii) we get I2 =
ER1 E' (R1 R2 R) R1R R2 r R1 R2 R
E R1 we get I2 = 0 R1 R2 R
For the chosen value of E' that is
Thus, taking a length of uniform resistance wire between A and C instead of the two resistors R1 & R2, it is possible to get zero (null) deflection in the galvanometer in the circuit shown below, provided that. E1 < IRAC
E R
A
B
1 P
I
C (I1+I2)
G
R2
E1
R2 (Fig. iii)
Let 1 be the length AB of the wire, corresponding to zero deflection in galvanometer, and let B be the point where the movable pointer ‘P’ (also called Jockey) makes contact with wire AC. If the same experiment is repeated but with another cell of EMF E2, we will be getting a different length ‘2’ at the instant galvanometer shows null deflection. If in both the cases the rheostat ‘R’ is kept unchanged, E1 1 . E2 2
The arrangement shown in figure (iii) is called Potentiometer and we use it to measure EMF using the above relationship. Application : A potentiometer can be used to compare emfs of two cells or to measure internal resistance of a cell. The method to compare the emfs of two cells has already been explained. When we want to measure the internal resistance of a cell, consider the following formula : E 1S 1 2 S , V 2
r=
where r = internal resistance of the cell, E = EMF, V = potential difference across the cell during current flow, S = resistance of resistance box, 1 , 2 are two balancing lengths. Example7:
A potentiometer wire of length 100 cm has a total resistance of 10. It is connected in series with a resistance R and a cell of emf 2 volts and of negligible internal resistance. A cell of emf 10 mV is balanced against a length of 40cm of potentiometer wire. What is the value of the external resistance R ?
Solution :
As shown in the figure, if R is the unknown resistance, the current in the circuit V 2 I= (r R ) (10 R )
10 mV G
A
B 40 cm 100 cm
2V
C
R
Now as the 100 cm wire has a resistance of 10 , the resistance of 40 cm of wire will be 40 (10/100) = 4 ohm.
Potential drop across 40 cm wire will be V = I 4 but here V = 10 mv (given) 10 10 3
Hence,
2 4 (10 R )
=
i.e. R = 790 . OBJECTIVE 1:
Solution:
A copper wire of diameter 1.02 mm carries a current of 1.7 amp. Find the drift velocity (vd) of electrons in the wire. Given n, number density of electrons in copper = 8.5 1028 /m3. (A) 1.75 mm/sec (B) 1.25 mm/sec (C) 2.5 mm/sec (D) 1.5 mm/sec Ans. (d) I = 1.7 A J = current density =
I r
2
1.7 (0.51 10 3 )2
= nevd = 8.5 1028 (1.6 10-19 ) vd
vd =
1.7 (0.51 10
3 2
) 8.5 10 28 1.6 10 19
= 1.5 10-3 m/sec. = 1.5 mm/sec. 2:
A cylindrical conductor of length and inner radius R1 and outer radius R2 has specific resistance . A cell of emf is connected across the two lateral faces of the conductor. Find the current drown from the cell.
(A) I =
(C) I =
2l R ln 2 R1
2l
R ln 2 R1 Ans. (a)
2l R ln 1 R2 2 (D) I = R ln 2 R1 (B) I =
2
R2 x
R1
dx
Solution: figure.
Consider the differential element of the cylinder as shown in the
R2
R
dx
l ( R ) a
2xl
R1
R2 ln 2l R1 I= R 2l I= R ln 2 R1
R=
3:
A copper wire of resistance 4 is melted and redrawn to thrice its original length. Find the resistance of stretched wire. (A) 40 (B) 60 (C) 45 (D) 36 Ans. (d)
Solution :
Since volume of the wire does not change. 1A1 = 2 A2 where 1 and A1 are the initial length and crosssection of the wire, and 2 and A2 are the final length and crosssection.
4:
l1 A 2 l2 A1
R1 =
R1 l1 A 2 l1 l1 R2 l2 A1 l2 l2
R1 l12 R2 l22
R2 = 9R1 = (9 4) = 36
. . . (1)
l1 A1
and
as
R2 =
l2 A2
. . . (2) 2 = 31
An electric bulb rated 220 v and 60 W is connected in series with another electric bulb rated 220 v and 40 W. The combination is connected across 220 volt source of e.m.f. Which bulb will glow more? (A) P1 P2 (B) P1 P2 (C) P1 P2 (D) P1 / P2 Ans. (b)
Solution :
R
V2 P
resistance of first bulb is R1
V2 P1
and resistance of the second bulb is R2
V2 P2
In series same current will pass through each bulb Power developed across first is P1 I2 and that across second is P2 I2
V2 P2
P1' P2 P2' P1
as P2 P1
V2 P1
P1 1 P2
P2 1 P1
P1 P2
The bulb rated 220 V & 40 W will glow more. 5:
A battery of emf 1.4 V and internal resistance 2 is connected to a 100 resistor through an ammeter. The resistance of the ammeter is 4/3 . A voltmeter is also connected to find the potential difference across the resistor. The ammeter reads 0.02A. What is the resistance of the voltmeter? (B) 300 (A) 200 (D) 150 (C) 400 Ans. (a)
6:
A battery of emf 1.4 V and internal resistance 2 is connected to a 100 resistor through an ammeter. The resistance of the ammeter is 4/3 . A voltmeter is also connected to find the potential difference across the resistor. (iii) The voltmeter reads 1.10 V. What is the error in reading? (A) 0. 3 V (B) 0.43 V (C) 0.53 V (D) 0.23 V Ans. (d)
Solution:
5&6 E = 1.4V r =2
A
100
4/3
V
(ii)
100R v 4 Total resistance in the circuit = 2 + 3 100 R v Emf I Total resis tan ce 1 .4 V 0.02A 100R v 4 2 + 3 100 R v
(iii)
Rv = 200 Potential difference
across
the
voltmeter
100 x 200 0.02. 100 200
=1.33 V. Voltmeter reading = 1.10 V Error = 1.33 - 1.10 = 0.23 V.
7:
B
Five equal resistances each of R value R are connected to form a network as shown in figure. Calculate the equivalent A resistance of the network between the points B and D.
R
R
C
R
R D
(A) Re q BD
3 R 2
(C) Re qBD R 1 2
(B) Re q BD (D) Re q BD
5 R 2 7 R 2
Ans. (c) Solution :
The circuit can be redrawn as in the figure. P=R (a) for points B and D, resistance (P+R), G an (Q+S) are in A parallel. With P = Q = R = S = G = R 1 1 1 1 i.e. Re qBD 2R 2R R Re qBD 1 R 2
B Q=R G=R
R S=R
So,
D
C
=
8:
Five equal resistances each of R value R are connected to form a network as shown in figure. Calculate the equivalent A resistance of the network between the points A and C, and A R
B R
R
C
R D
(A) R (C) 2 R Ans. (a) Solution :
(B) R/2 (D) 5 R
for points A and C, the given network is balanced wheatstone bridge as
P R . So excluding G, (P+Q) is in parallel with Q S
(R+S) i.e.
1 1 1 Re qAC 2R 2R
i.e. (Req)AC = R 9:
Five equal resistances each of R value R are connected to form a network as shown in figure. Calculate the equivalent A resistance of the network between the points A and C
B R
R
C
R
R D
(A)
5 R 8
(C)
7 R 8
6 R 8 9 (D) R 8
(B)
Ans. (a) Solution :
for points A and B, starting from opposite side of AB, (Q+S) in parallel with G gives RBCD =
2R R 2 R 2R R 3
RBCD in series with R gives RBDO =
2 5 RR R 3 3
And RBDO in parallel with P gives the equivalent resistance between AB, i.e. (Req)AB =
5 / 3R R i.e. (Req) = AB 5 / 3R R
5 R 8
10:
E Figure shows a potentiometer circuit for determining the internal resistance of a cell. When switch S is open, Athe E G balance point is found to be at 76.3 cm of the wire. When switch S is closed and the value of R is R S 4.0 , the balance point shifts to 60.0 cm. Find the internal resistance of cell E. (A) 1 (B) 1.1 (C) 2.1 (D) 5.1 Ans. (b)
Solution :
r
J J B
EV I E r 1R , V
But
E / , hence V
76.3 60.0 r R 4 .0 1.1 60.0
11:
In the circuit shown in figure each cell has emf 5V and has an internal resistance of 0.2 ohm. What is the reading of ideal voltmeter V. (A) 0V (C) 2V Ans. (a)
Solution:
+ V
(B) 1V (D) 5V
As internal resistance of an ideal voltmeter is infinite, the resistance of the battery across which it is connected will not change by its presence as 1 1 1 r1 r
r'
=
r
Now as the given 8 batteries are discharging in series i.e. Eeq = 8 5 = 40 V and req = 8 0.2 = 1.6 so current in the circuit I = Hence
= 25 A potential difference across the v = E – Ir = 5 – 25 0.2 = 0V.
Eeq req
required
40 1.6
battery
12:
When two resistances X and Y are put in the left hand and right hand gaps in a wheatstone meter bridge, the null point is at 60cm. If X is shunted by a resistance equal to half of itself then find the shift in the null point. (A) 26.7 cm (B) 36.7 cm (C) 46.7 cm (D) 96.7 cm Ans. (a)
Solution:
Arrangement is shown in the figure. X 60 3 Y 40 2
X
Y
A
. . . . (1)
When X is shunted then resistance in the left gap becomes X 2 X X' X 3 X 2
X/2
X.
Now
13 :
. . . (2)
40 cm
X 1 3 l l 3 l = 33.3cm Y (100 l ) 3 2 (100 l )
Shift = 60 – 33.3 = 26.7 cm.
Each resistor in the network of the given figure has a resistance of 10. The resistance between points A and B is (A) 10 (C) 30
Solution:
60 cm
A
C
D
E
F
B
(B) 20 (D) 40
Resistors in arms CD and EF are in series which add up to 10+10=20 . Resistors in arms CE and EF are also in series which add upto 20 Hence resistance between points C and F, is given by 1 1 1 or R 20 20
R = 10
Hence, resistance between A and B = 10 + 10 + 10 = 30 (C)
14:
In a gas discharge tube if 3 1018 electrons are flowing per sec from left to right and 2 1018 protons are flowing per second from right to left through a given cross section. The magnitude and direction of current through the cross section (A) 0.48, left to right (B) 0.48 A, right to left (C) 0.80A, left to right (D) 0.80 A, right to left
Solution :
As current is rate of flow of charge in the direction in which positive charge will move, the current due to electron will be ie =
neqe = 3 1018 1.6 1019 t
= 0.48 A (Opposite to the motion of electrons, i.e. right to left) Current due to protons ip =
np qp t
= 2 1018 1.6 1019
= 0.32 A (Right to left) so total I = ie + ip = 0.48 + 0.32 = 0.80 A (Right to left) Hence correct answer is (D) 15 :
The current in a wire varies with time according to the equation I = 4 + 2t, where I is in ampere and t is in sec. The quantity of charge which has passed through a cross-section of the wire during the time t = 2 sec to t = 6 sec will be (A) 60 coulomb (B) 24coulomb (C) 48 coulomb (D) 30 coulomb
Solution :
Let dq be the charge which has passed in a small interval of time dt, then dq = idt = (4 + 2t)dt Hence total charge passed between interval t = 2 sec and t = 6 sec 6
q = ( 4 2t )dt = 48 coulomb 2
(C) 16 :
A uniform copper wire of length 1 m and cross-sectional area 5 10 7 m 2 carries a current of 1 A. Assuming that there are 8 10 28 free electrons per m3 in copper, how long will an electron take to drift from one end of the wire to the other? (A) 0.8 10 3 sec (B) 1.6 10 3 sec (C) 3.2 10 3 sec (D) 6.4 10 3 sec Solution : The drift velocity of electrons is given by Vd
I enA
If is the length of the wire, the time taken is t
e n A 1 1.6 10 19 8 10 28 5 10 7 Vd I 1
= 6.4 10 3 sec (D) 17 :
A electric current of 16 A exists in a metal wire of cross section 106 m2 and length 1m. Assuming one free electrons per atom, the drift speed of the free electrons in the wire will be (Density of metal =5 103 kg/m3, atomic weight = 60) (A) 5 103 m/s (B) 2 103 m/s (C) 4 103 m/s (D) 7.5 103 m/s
Solution :
According to Avogadro's hypothesis N m NA M N m N so n = NA A v VM M
Hence total number of atoms n =
6 10 23 5 10 3 60 10 3
= 5 10 /m As I = ne eA vd 28
3
Hence drift velocity vd = vd =
I neeA
16 5 10
= 2 10 (B)
28
3
1.6 10 19 10 6
m/s
18:
A copper wire is stretched to make it 0.1 % longer. The percentage change in its resistance is (A) 0.2 % increase (B) 0.2% decrease (C) 0.1 % increase (D) 0.1 % decrease
Solution :
For a given wire, R =
L s
with L s = volume = V = constant so that R =
L2 V
R L =2 = 2 (0.1 %) L R
= 0.2 % (increase) (A) 19 :
The driver cell of a potentiometer has an emf of 2V and negligible internal resistance. The potentiometer wire has a resistance of 5 and is 1m long. The resistance which
must be connected in series with the wire so as to have a potential difference of 5mV across the whole wire is (A) 1985 (B) 1990 (C) 1995 (D) 2000 Solution :
In order to have a potential drop of 5 mV = 5 10 3 V across a wire of resistance 5 , the current flowing in the wire should be I
5 10 3 1 10 3 A 5
If R is the resistance to be connected in series with the wire, then 2 1 10 3 R5
which gives R = 1995 (C) 20 :
A battery of 10 volt is connected to a resistance of 20 ohm through a variable resistance R. The amount of charge which has passed in the circuit in 4 minutes, if the variable resistance R is increased at the rate of 5 ohm/min. (A) 120 coulomb (B) 120 loge2 coulomb (C)
Solution :
120 loge 2
coulomb
(D)
60 loge 2
coulomb
dq V dt R dq dR V . dR dt R dR dq = 12 V R 40 dR q = 12 V = 12 V (loge 40 – loge20) R 20
I=
= 12 10 loge2 (B)
21:
In the given circuit, R1 = 10 , R2 =6 and E = 10 V. Then effective resistance (A) Effective resistance of the circuit is 20 (B) Effective resistance of the circuit is 30 (C) Effective resistance of the circuit is 40 (D) Effective resistance of the circuit is
R1
A2
R1
R1
R2 R1
A3 R1
A1
R2 R1
R1 R2 R1
50 Solution:
Potential difference across R2 resistances is zero, therefore current in three branches is zero, therefore current in two branch containing R1 will be same, simplified circuit will be Effective resistance of the circuit Reff = current through the circuit I =
10
A2
10
10
10
10
A1
10
10
10
10 V
40 40 = 20 40 40
10 = (1/2) amp 20
Hence reading of A1 = ½ amp. Hence reading of A2 =1/4 amp. (A)
22:
Solution:
In the given circuit, R1 = 10 , R2 =6 and E = 10 V. Then reading of A1 (A) Reading of A1 is 1 amp. (B) Reading of A1 is 1/2 amp. (C) Reading of A1 is 2 amp. (D) Reading of A1 is 3 amp. Potential difference across R2 resistances is zero, therefore current in three branches is zero, therefore current in two branch containing R1 will be same, simplified circuit will be
R1
R1
R2 R1
R1 R2
R2
A3
R1
R1
R1
A1
10
A2
10
10
10
10
A1
Effective resistance of the circuit Reff = current through the circuit I =
R1
A2
10
10
10
10 V
40 40 = 20 40 40
10 = (1/2) amp 20
Hence reading of A1 = ½ amp. Hence reading of A2 =1/4 amp. (B)
23:
In the given circuit, R1 = 10 , R2 =6 and E = 10 V. Then reading of A2 (A) Reading of A2 is 1 amp (B) Reading of A2 is 2 amp (C) Reading of A2 is 1/4 amp. (D) Reading of A2 is 3 amp
R1
A2
R1
R1
R2 R1
A3 R1
A1
R2 R1
R1 R2 R1
Solution:
Potential difference across R2 resistances is zero, therefore current in three branches is zero, therefore current in two branch containing R1 will be same, simplified circuit will be
10
A2
10
A1
Effective resistance of the circuit Reff = current through the circuit I =
10
10
10
10
10 V
40 40 = 20 40 40
10 = (1/2) amp 20
Hence reading of A1 = ½ amp. Hence reading of A2 =1/4 amp. (c) 24:
Solution :
One billion electrons pass from A to B in 1 ms. What is the direction and magnitude of current? (A) 1.6 A (B) 0.8 A (C) 0.16 A (D) 1.6 A i
Ne (109 )(1.6 x10 19 C) 1.6 x10 7 A 3 t 10
i = 0.16 A. The current flows from B to A. 25:
Solution:
26:
Solution:
27:
(C)
An electron gun in a TV set shoots out a beam of electrons. The beam current is 10A. How many electrons strike the TV screen each second? (A) 2.78 1014 (B) 6.3 1013 (C) 6.78 104 (D) electron will not reach (B) The number of electrons per second N = I/e = (1.0 10–5 C/s)/(1.6 10–19 C) = 6.3 1013 electrons per second. In above question, how much charge strikes the screen in a minute? (A) –600C (B) 600C (C) 1300C (D) –1300C (A) The charge Q striking the screen obeys |Q| = IT = (10 C/s)(60 s) = 600 C. Since the charges are electrons, the actual charge is Q = – 600 C. A copper bus bar carrying 1200 A has a potential drop of 1.2 mV along 24 cm of its length. What is the resistance per m of the car?
10
10
Solution:
28:
Solution:
29:
(A) 5.2 (B) 3.2 (C) 8 (D) 4.2 (D) From Ohm’s law, applied to 24 cm of the bar, V24 = IR24, or (1.2 10–3V) = (1200 A) R, and R24 = 1.0 . By proportion, R100 = (100/24) R24 = 4.2 A 20–cm–long copper tube has an inner diameter of 0.85 cm and an outer diameter of 1.10 cm. Find its electric resistance when used lengthwise.( = 1.7 10–8) (B) 89 (A) 51.2 (C) 80 (D) 42 (B) R= (L / A). The cross–sectional area is [(1.102 – 0.852)]/(4 104)] = 3.83 10–5 m2; then with L = 0.20 m and = 1.7 10–8, then we get R = 89. At what value of resistance Rx in the circuit in the figure will the total resistance between points A and B is R? 2R
A
R
R B
(A) (C) Solution: If
Rx =
R R
3 1
3 1
2
R
(B) (D)
Rx
R
R R
3 1 3 1
3 1
R x 2R R or Rx2 + 2RRx – 2R2 = 0 On solving and rejecting the negative root of the quadratic equation, we have Rx = R 3 1
A
2R
A Rx
R B
–0.444 0.222
3
3
(B) (D)
0A –0.222
The loop equations are: –6 + 6 – 10I4 = 0, so I4 = 0. (This result tells us that I1 and I2 flow through their respective 3 ) +6 –3I1 + 12I3 = 0 –6 –12I3 –3I2 = 0
Rx
B
What will be the I1 for the following circuit. (A) (C)
Solution:
2R
(A) R x 2R R
30:
2R
12 6V
I1
E
I4
10
I3
6V
I2
The point equation is I2 = I1 + I3. Solve simultaneously to find I1, I2, I3 and I4, 0.222, –0.222, –0.444 and 0 A. Hence the correct choice is (C) 31:
Solution:
What will be the I2 for the following circuit.
3 3 (A) –0.222 (B) –0.212 12 (C) –0.222 (D) 0.222 6V 6V I The loop equations are: –6 + 6 – 10I4 = 0, so I I4 = 0. (This result tells us that I1 and I2 flow 10 E through their respective 3 ) +6 –3I1 + 12I3 = 0 I –6 –12I3 –3I2 = 0 The point equation is I2 = I1 + I3. Solve simultaneously to find I1, I2, I3 and I4, 0.222, –0.222, –0.444 and 0 A. Hence the correct choice is (C)
3
1
I2
4
32:
3 What will be the I3 for the following circuit. 3 12 (A) 0.222 (B) –0.444 6V 6V I (C) –0.444 (D) 0.444 I The loop equations are: –6 + 6 – 10I4 = 0, so I4 = 0. (This result tells us that I1 and I2 flow 10 E through their respective 3 ) I +6 –3I1 + 12I3 = 0 –6 –12I3 –3I2 = 0 The point equation is I2 = I1 + I3. Solve simultaneously to find I1, I2, I3 and I4, 0.222, –0.222, –0.444 and 0 A. Hence the correct choice is (C)
3
Solution:
1
I2
4
33:
Solution:
What will be the I4 for the following circuit. (A) 1A 3 (B) 0.222 (C) 0A 6V (D) 2A The loop equations are: –6 + 6 – 10I4 = 0, so I4 = 0. (This result tells us that I1 and I2 flow through their E respective 3 ) I +6 –3I1 + 12I3 = 0 –6 –12I3 –3I2 = 0 The point equation is I2 = I1 + I3. Solve simultaneously to find I1, I2, I3 and I4, 0.222, –0.222, –0.444 and 0 A. Hence the correct choice is (C)
3
12
I3
6V
I1
I2
10
4
34:
Three resistances of 4 each are connected as shown in figure. If the point D divides the resistance into two equal halves, the resistance between points B and C will be (A) 12 (B) 6
4
4 4
B
4
C
8 3 The circuit can be rearranged as shown in Fig. because it is Wheatstone’s Bridge circuit The resistance between points B and C is given by 1 1 1 R 4 4 2 2 8 or R = . Hence the correct choice is (d) 3
3
(C)
Solution:
(D)
4
2
C
4
A
2
D
B
R
35:
Solution:
What will be the equivalent resistance of the network shown in the figure between the R terminals 1 and 2. R 3 2 8 (A) R (B) R 5 5 R 7 R R (D) none (C) 1 5 (C) This is the familiar problem, which can be solved using Kirchhoff’s rule. Here, the problem is going to be simplified using Delta–star Transformation. Before going through the solution, let us understand this Delta–Star Transformation. This figure (a) and (b) show the Delta and Star respectively. a
a
Ra
R1
R2 Rc
b
R3
c
b
Rb
c
(a) (b) For obtaining the expression for equivalent we first derive expression for resistance between points of terminals and then equating the expression for resistance of the corresponding terminals of two circuits. Now for Delta Configuration the equivalent resistance between a and c, R1R2 Ra = R1 R2 R3 R1R3 Rb = R1 R2 R3 R2R3 Rc = R1 R2 R3 The given network can be reduced in the form as shown using series and parallel grouping.
2
R 4 R
2
2 R
2R R
3 R
R
2R 4
3
2R
4
R/4
R/2
1
R/2 1
(c) (d) Using Delta–Star Transformation, the figure (c) can be reduced to figure (d). Now the reduced network is very simple and using series and parallel grouping, the equivalent resistance between 7 terminal 1 and 2 becomes R. 5 CMP I:
A cell having a steady emf of 2volt is connected across the potentiometer wire of length 10m. The potentiometer wire is of manganine and having of 11.5 /m. A new resistance of 5 with negligible length is put in series with the potentiometer wire.
36:
What is total resistance of wire? (B) 115 (A) 11.5 (C) 1150m (D) 120 Solution: Resistance per unit length of wire = 11.5/m Length of wire = 10 m Resistance of wire = 11.5 10 = 115 Total resistance = 115 + 5 = 120 Hence correct choice is (D) 37:
Solution:
What is potential gradient when 5 resistance is not connected (A) 0.2 V/m (B) 0.4 V/m (C) 0 V/m (D) 1.38 V/m appliedvoltage Fall in potential per meter = = 2/10 = lengthof wireof potential
0.2 v/m potential gradient = 0.2 V/m Hence correct choice is (A) 38:
Solution:
What is potential gradient after connect 5 resistance. (A) 0.2 V/m (B) 0.15 V/m (C) 0.104 V/m (D) 2.0 V/m
5
115
wire length 10 m
2V
2 115 230 23 V 115 5 220 22 voltage across potentiometer Potential gradient = lengthof wireof potential 23 / 22 0.104 V / m = 10 Hence correct choice is (C)
Voltage across 115 resistance =
CMP II:
An ammeter and a voltmeter are connected in series to a battery with an emf E = 6 volt when a certain resistance is connected in parallel with voltmeter, the reading of latter decreases two times, where as the reading of the ammeter increasing the same number of times.
39:
What is ratio of resistance of voltmeter to resistance of ammeter (A) 2 (B) 1/2 (C) 1/3 (D) 3
40:
What will be voltmeter reading before the connecting of the resistance. (A) 1V (B) 2V (C) 3V (D) 4V
41:
What will be voltmeter reading after the connecting of the resistance. (A) 1 V (B) 2V (C) 4V (D) 3V
Solution:
Suppose RA = Resistance of ammeter, Rv = resistance of Voltmeter In the first case current is the circuit 6 i ….(1) ( RA RV ) And voltage across voltmeter V = 6 – Voltage across ammeter V = 6 – iRA 6 V 6 RA ….(2) ( RA RV ) In the second case reading of ammeter becomes two times i.e. the total resistance become half while the resistance of ammeter remains unchanged. Hence 6 12 i =i ( RA RV ) / 2 ( RA RV )
and voltage across voltmeter V ’ = 6 – i RA 12 V' 6 …..(3) ( RA RV ) It is given that V V' … (4) 2 6 6 R R 12RA A V [From (2) and (3)] 6 ( RA RV ) 2 R RA 1 V 2 (Ans. of Question 39) RA RA RV 3 Substituting this value into equation (3) V’ 6 12 1/3 V’ 2 (Ans. of Question 40) From Eq, (4) V 4 (Ans. of Question 41)
42:
Solution:
43:
Solution:
In the Bohr model, the electron of a hydrogen atom moves in a circular orbit of radius 5.3 10–11 m with a speed of 2.2 106 m/s. Determine its frequency f and the current I in the orbit. (A) 1.06 mA (B) 2.06 mA (C) 0.06 mA (D) 3.06 mA 6 2.2 10 m / s = 6.6 1015 rev/s f= 11 2 r 2(5.3 10 m) Each time the electron goes around the orbit, it carries a charge e around the loop. The charge passing a point on the loop each second is current = I = ef = (1.6 10–19C) (6.6 1015s–1) = 1.06 mA. Note that the current flows in the opposite direction to the electron, which is negatively charged. (A)
A wire carries a current of 2.0 A. What is the charge that has flowed through its cross–section in 1.0 s. How many electrons does this correspond to? (A) 3.0 C, 1.25 1019 (B) 2.0 C, 1.25 1019 (C) 4.0 C, 1.25 1019 (D) 2.0 C, 5.25 1019 q i= t q = it = (2.0 a) (1.0s) = 2.0 C q = ne q 2.0 n= = 1.25 1019 19 e 1.6 10
(B) 44:
Solution:
45:
A current of 7.5 A is maintained in a wire for 45s. In this time (a) how much charge and (b) how many electrons flow through the wire? (A) 400.5C, 2.1 1021 (B) 300.5C, 2.1 1021 (C) 337.5C, 2.1 1021 (D) 700.5C, 2.1 1021 (a) q = It = (7.5A) (45 s) = 337.5C (b) The number of electrons N is given by 337.5 C q N= = 2.1 1021 19 e 1.6 10 C where e = 1.6 10–19 C is the charge of an electron. (C)
What will be the current if charge q revolves with the frequency f?
(A) f (C) q Solution:i = qf (D)
(B) (D)
q/f qf
46:
Calculate the mean free time between collision in copper at room temperature? and electron average collision per second. (A) 2.4 10–14 s (B) 4 10–14 s (C) 5.4 10–14 s (D) .4 10–14 s Solution: n = 8.5 1028 m–3 = 1.72 10–8 e = 1.60 10–19 c and m = 9.11 10–31 kg m = 2.4 10–14 s 2 ne Taking the reciprocal of this time, we find that each electron average about 4 1013 collision every second. (A)
T=
47:
Solution:
48:
A coil of wire has a resistance of 25.00 at 20C and a resistance of resistance of 25.17 at 35C. What is its temperature coefficient of resistance? (A) 4.5 10–4C–1 (B) 5 10–4C–1 (C) 0.5 10–4C–1 (D) 4.0 10–4C–1. R = Ro[1+ (T–T0)], or = R/(R0T), with R = R–R0 = 0.17 and T = T – T0 = 15C. Then = (0.17)/(25.00 15) = 4.5 10–4C–1. (A) The three resistors in Fig. are R1 = 25 , R2 = 50 , and R3 = 100 . What is the total resistance of the circuit? (A) 50.3 (B) 60.3
Solution:
(C) 58.3 (D) 80.3 The sum of R2 and R3 in parallel is E 12V
I R1
R2
I1 R3 I2
1 1 1 1 1 3 R R 2 R3 50 100 100
or R = 33.3
Since R is in series with R1, the total resistance of the circuit is R = R + R1 = 33.3 + 25 = 58.3 (C) 49:
Solution :
The three resistors in Fig. are R1 = 25 , R2 = 50 , and R3 = 100 . What are the currents l1, l2 and l3 for a 12–V battery? E 12V I= = 0.206 A E 12V R 58.3 I (A) 0.206 A, 0.137, 0.0685 A R (B) 0.506 A, 0.137, 0.0685 A R (C) 0.606 A, 0.137, 0.0685 A I (D) 0.706 A, 0.137, 0.0685 A 1
2
1
R3 I and R3 is The potential V across R2 V = E – R1I = 12 V –(25)(0.206 A) = 6.85 V. Therefore, V 6.85V V 6.85V = 0.137 A I3 = = 0.0685 A I2 = R2 50 R3 100 (A) 2
50:
What is total resistance across AB in the following network. C
8
4
A
B
4
8
2
D
(A) 6.4 (C) 7.4 Solution:
(B) 2.4 (D) 5.4
We can’t apply Wheatstone’s bridge condition because 4 2 8 8
Break delta ACD into star Y connection, For that assign 1,2 and 3 like following fig. 2C 8
4 1 A
B
4
8
2 3
D
R12 R13 R12 R13 R23 42 8 = = 0.8 4 2 4 10 R12 R23 44 16 R2 = = = 1.6 4 2 4 10 R12 R23 R13 R23 R13 42 8 = = 0.8 R3 = 4 2 4 10 R12 R23 R13 Connect R1, R2 and R3 like following fig. R1 =
2
C
4 R
2
1 .6
8
1
A
R1 0.8
4
B
R 3 0.8
8
2
3
D
remove delta circuit from above fig., we get 1.6
A
8
0.8
B
0.8
8
resistance across A and B is R = 0.8 + [(8 + 1.6) || (0.8 + 8)] 9.6 8.8 = 0.8 = 5.4 9.6 8.8 51:
(D)
Find the effective resistance between points A and B of the network shown if Fig.
3
E
D
3
C
6
6
3
6
F
3
3
A
3
(A) 3 (C) 5
B
(B) 4 (D) 7
Resistors AF and FE of 3 each are in series with each other. Therefore, the network AEF is a parallel combination of two 6 resistors. Thus the resistance between points A and E is given by
Solution:
3
E
D
3
C
6
6
3
6
F
3
3
A
3
B
1 1 1 R AE 6 6
giving RAE = 3 . The network reduces to that shown in Fig. (a). Similarly the resistances between points A (A) 52:
Kirchhoff’s current law obeys conservation of (A) charge (B) momentum (C) energy (D) none of these. Solution: (A)
53:
Solution:
The slide wire Wheatstone bridge shown in Fig. is balanced when the uniform slide wire AB is divided as shown. Find the value of the resistance X. (A) 3 (B) 4 (C) 2 (D) 7
N 3
X
G A
L 40 cm D
M
B
60 cm
X L 40cm or 3 M 60cm
54:
X = 2
(C)
A dry cell delivering 2 A has terminal voltage 1.14V. What is the internal resistance of the cell if its open–circuit voltage is 1.59 V? (A) 5.09 (B) 6.09 (C) 7.09 (D) 0.09
Solution:
55.
Solution :
56.
Solution :
57.
Solution : 58.
Solution :
The open–circuit voltage is simply the emf of the cell, so V = E – ir with V = 1.41 V, i = 2 A, E = 1.59 V. 1.41 = 1.159 – 2r, and r = 0.09 The sensitivity of a galvanometer of resistance 406 ohm is increased by 30 times. The shunt used is (A) 88 (B) 14 (C) 6 (D) 16 . Gi g 226 i 14 . (B) = 30. The shunt S = i ig 29 ig The sensitivity of a galvanometer of resistance 8722 is decreased by 90 times. The shunt used is (A) 88 (B) 90 (C) 94 (D) 98 . i (D) Here g (90)–1 . i Gi g Gi g G (90)–1 S 98 89 i ig 1 ig / i 90 A wire l = 8.00 m long, of uniform cross-sectional area A = 8.00 mm2, has a conductance of G = 2.45 –1. What is the resistivity of the material of the wire ? (A) 2.1 × 10–7 S (B) 3.1 × 10–7 S –7 (C) 4.1 × 10 S (D) 5.1 × 10–7 S. R A 8.00 106 4.1 107 S (C) A Gl 2.45 8.00 A wire 250 cm long and 1 mm2 in cross-section carries a current of 4 A when connected to a 2 V battery. The resistivity of the wire is (A) 0.2 × 10–6 m (B) 2 × 10–7 m (C) 5 × 10–6 m (D) 4 × 10–6 m l (A) R = A
1 106 A 2 P 100S l 250
59.
Solution :
An electrical cable of copper has just one wire of radius 9 mm. Its resistance is 5 . This single wire of cable is replaced by 6 different well-insulated copper wires each of radius 3 mm. The total resistance of the cable will now be equal to? (A) 7.5 (B) 45 (C) 90 (D) 270 Sl 5 11 81 or S (A) R l Sl 5 l 81 45 R1 l 9 1 6 45 or R 7.5 . R 45 5
60.
When cells are arranged in parallel (A) the current capacity decreases (B) the current capacity increases (C) the e.m.f. increases (D) the e.m.f. decreases.
Solution :
(B) When cells are connected in parallel, the current capacity increases.
THERMAL & CHEMICAL &THERMOELECTRICITY
EFFECTS
OF
CURRENT
HEATING EFFECT OF CURRENT, JOULE'S LAW (i) (ii)
(iii)
Whenever work is transformed into heat or heat into work, the quantity of work done is mechanically equivalent to the quantity of heat. Thus Work Heat. or Work = a constant × heat or W = JH Where J is known as Joule's constant or mechanical equivalent of heat. It is defined as that mechanical work which produces unit calorie of heat. Expression for heat produced in a conductor due to current flow through itA
B
i R
(a) (b)
(c)
Let the potential difference between the points A and B of a conductor is V, on account of which a current i flows through it. As potential difference is the work done per unit test charge W = QV But Q = it W = Vit W Vit i2RT But heat developed H = J J J Where R is the resistance of wire. If i i is in ampere, V is in volt and R is in ohm, then W = Vit Joule = Vit × 107 ergs. Vit Heat developed H = calories J Vit 0.24 Vit cals. 4.2 = 0.24 i2Rt cals.
JOULE'S LAWS OF HEATING EFFECTS OF CURRENT (i) (ii) (iii)
(iv)
The heat developed in a conductor is given by H = i2Rt Joule = 0.24 i2Rt cals The amount of heat developed in a conductor, in a given time, is directly proportional to the square of the current. i.e. H i2 , when R and t are constants. The amount of heat developed in a conductor by a given current in a given time is directly proportional to the resistance of the conductor. i.e. H R , when i and t are constants. The amount of heat developed in a given conductor due to a given current is proportional to the time of flow of the current. i.e. H t , when i and R are constants.
ELECTRICAL ENERGY OR WORK If Q units of charge be carried between two points differing in potential by V, then electrical work done is W = Q × V Joule = Q × V × 107 ergs
POWER (i)
(ii)
(iii) (iv)
The rate at which work is done is defined as power Q V Power Vi t = i2R V2 R Thus electric power = potential difference × current i.e. 1 watt = 1 Volt × 1 Ampere = 1 Joule/sec = 107 ergs/sec The practical unit of power is kilowatt 1 kilowatt = 1000 watt Horse-power = 550 ft-lbs per sec. = 550 × 12 × 2.54 ×453.6 gm-cm/sec = 746 × 107 ergs/sec = 746 watt.
Illustration 1:
Solution:
A coil of resistance 2 is immersed in 1 kgm of water and is connected to the terminals of the battery of internal resistance 4 and emf 6 volt for 3 minutes. The increase in temperature of water is : (A) 0.93C (B) 0.085C (C) 1.92C (D) 4.31C Heat developed in coil = Heat absorbed by water Rt i2 = ms Dt J 2 i2Rt æ E ö Rt \ Dt = =ç ÷ Jms è R + r ø Jms 2 ´ 180 æ 6 ö =ç ´ » 0.085°C ÷ è 4 + 2 ø 4.2 ´ 1 ´ 103 2
Illustration 2:
Solution:
A Daniel cell has an emf of 1.08 volt and internal resistance 0.5 . It is successively connected to two wires whose resistances are 2 and 3. The ratio of the amounts of heat developed in two wires will approximately be : (A) 4:3 (B) 3:4 (C) 2:3 (D) 3:2 Heat developed in a wire H = i2Rt E2Rt or (R + r)2 R \ Hµ (R + r)2 2
\
H1 æR1 ö æR2 + r ö = ´ H2 çè R2 ÷ ø çè R1 + r ÷ ø 2
2 æ3.5 ö 2 49 = ´ ç = ´ » 4:3 ÷ 3 è 2.5 ø 3 25
Illustration 3:
Solution:
Illustration 4:
Solution:
Illustration 5:
Solution:
(v)
For tungsten wire, the average temperature coefficient above 20 C is 5.1 10–3C–1. The filament of an electric lamp has a cold resistance of 9.7 at 20C. When glowing it has a resistance of 121. Its temperature while glowing is : (A) 2250C (B) 2190C (C) 2270C (D) 2430C R1 = R20 (1 + t) or 121 = 9.7 (1 + 5 10–3 t) \ t = 2250°C above 20C the temperature of glowing = 2250 + 20 = 2270C If a cell has an e.m.f. of 1.08 volt and internal resistance 0.5 . Its terminals are connected to two wires of 1 and 2 in parallel. The current in each wire will be : (in ampere) (A) 0.617, 0.308 (B) 0.412, 0.206 (C) 0.824, 0.412 (D) 2.618, 1.309 Total current E 1.08 1.08 ´ 6 i= = = R1R2 1 2 7 + +r 2 3 R1 + R2 1 iµ R 2 2 1.08 ´ 6 current in 1 resistance = i = ´ = 0.617 amp. 3 3 7 i 1.08 ´ 6 current in 2 resistance = = = 0.308 amp. 3 7´ 3 In the above problem the ratio heats generated in the two wires will be : (A) 1:2 (B) 2:1 (C) 1:3 (D) 3:1 2 V Heat generated H = t R In parallel combination V = constant H1 R2 2 \ = = = 2 :1 H2 R1 1
Kilo-watt-hour or Board of Trade (B.O.T.) unit(a) Energy consumed in a given time is the product of power and time (b) When power of one watt is consumed for an interval of one hour, then energy consumed = 1 watt × 1 hour = 1 watt – hour = 0.001 kilo-watt-hour (c) Board of Trade (B.O.T.) unit of electric power is kilo-watt-hour by which the consumption of electric energy is measured and charged by power supply authorities. 1 kilo watt-hour = 103 × 60 × 60 watt-sec = 36 × 105 Joule
(d)
Rule for calculation of cost(i) Number of B.O.T. units = Sum of wattages of all systems × time (in hour) 1000 (ii) Total cost = Number of B.O.T. units × rate of charge per unit (iii) Total cost = V2 ( P)or ( Vi) or ( i2R)or time in hour rate per unit R 1000
Illustration 6:
Solution:
Illustration 7:
Solution:
Illustration 8:
Solution:
A dwelling house is installed with 15 lamps, each of resistance 10 3 and 4ceiling fans each driven by 1/8th horse-power motor. If the lamps and fans are run on an average for 6 hours daily, then the number of B.O.T. units consumed by lamps in a month of 31 days will be: (A) 135 (B) 150 (C) 165 (D) 180 Number of B.O.T. units consumed by 15 lamps V 2 ´ 15 ´ 6 ´ 31 = R ´ 1000 220 ´ 220 ´ 15 ´ 6 ´ 31 = » 135 103 ´ 1000 In the above problem the number of B.O.T. units consumed by the fans in the month will be : (A) 193 (B) 173 (C) 143 (D) 113 No. of B.O.T. unit consumed by 4 fans 1 ´ 746 ´ 6 ´ 31 ´ 4 = » 143 3 ´ 1000 In Illustration 6, the total cost of electric power consumed in the month, at the rate of Rs. 1.40 per B.O.T. unit, will be : (A) Rs. 279/(B) Rs. 389/(C) Rs. 421/(D) Rs. 538/Total number of B.O.T. units consumed = 135 + 143 = 278 cost of electric power consumed = 278 1.40 = Rs. 389.20 or 389/-
CHEMICAL EFFECTS OF CURRENT Definitions of various terms (i) Electrolysis: The process of splitting up or decomposing a liquid by passing an electric current through it, is defined as electrolysis. (ii) Electrolyte: The compound, whether fused or in solution, which undergoes decomposition by an electric current, is defined as electrolyte. (iii) Anions and Kations: The decomposed substance appears in the form of ions. The ions appearing at the anode are known as anions and those at the cathode are known as cations.
(iv) (v) (vi)
(vii) (viii)
(ix)
(x) (xi) (xii)
Ionisation: The phenomenon of separation of a molecule into oppositely charged ions is known as ionisation. Equivalent weight- The ratio of the atomic weight of an element to its valency is defined as its equivalent weight. Atomic weight: The ratio of the average mass of an atom of an element of the mass of an atom of hydrogen, taken as 1.008, is defined as the atomic weight of that element. Valency: The valency of an element is the number of atoms of hydrogen or chlorine which combines with or is displaced by one atom of the element. Molecular weight: The molecular weight of a substance is the ratio of the mass of one molecule of the substance to the mass of an atom of oxygen which is taken as 16. Gram-equivalent: It is the weight in grams of an element which will combine with or replace 1 gm of hydrogen. Gram - atom Gram-equivalent = valency Gram-molecule: The molecular weight of any substance. expressed in grams is defined as gram-molecule of that substance. Avogadro number: The number of molecules of a substance in its one-grammolecule is known as Avogadro number. Normal solution: A solution containing one gram-equivalent of the solute per litre is called a normal solution.
FARADAY'S LAWS OF ELECTROLYSIS (i)
(ii)
(iii)
First law: The towal mass of ions liberated at an electrode, during electrolysis, is proportional to the quantity of electricity which passes through the electrolyte. But Q = it i.e. m Q m it Hence the first law may also be stated as follows The mass of ions liberated at an electrode during electrolysis is proportional to (a) the current following through the electrolyte, and (b) the time for which the current flows Second law: If same quantity of electricity is passed through different electrolytes, the masses of the substances (ions) deposited at the respective cathodes are directly proportional to their chemical equivalents (equivalent weights). i.e. m E (chemical equivalent) Electro-chemical equivalent (E.C.E.): (a) The electro-chemical equivalent of an element is its mass in grams deposited on the electrode by the passage of 1 coulomb of charge through it i.e. by the passage of 1 ampere current for 1 second. (b) According to Faraday's first lawm i.t or m = Z i.t Where Z is the constant of proportionality known as the electro- chemical equivalent of the substance. It is numerically equal to the mass in grams of the element deposited when a unit current flows in unit time. (c) According to Faraday's first law m = ZQ If same charge is passed two electrolytes, the m1 Z1 m2 Z2
(iv)
According to Faraday's second law m1 E1 m2 E2 m Z 1 1 m2 Z2 E or Z2 = Z1 2 E1 (d) E.C.E. of any substance = E.C.E. of hydrogen × chemical equivalent of the substance. (e) If silver is taken as the standard substance for which E.C.E. is 0.001118 gm per coulomb, then E.C.E. of any substance = Chemical equivalent of subs tance 0.001118 chemical equivalent of silver Faraday: (a) The quantity of electricity (i.e. charge) required to liberate a gram equivalent of a substance during electrolysis. (b) As derived above E1 Z1 E or cons tant F E2 Z2 Z The constant F is known as one Faraday. For example for copper E = 31.5 gm and Z = 0.000329 gmC–1 31.5 F 96500 coulomb 0.000329 (c) According to Faraday's first law m = Z.i.t. = ZQ If p is the valency of the element, then electrons has to flow through the solution to deposit one atom. charge required to deposit 1 mol of the substance = Npe, where N = Avogadro number and m = M M = ZNpe M 1 E E Z p Ne Ne F Where F = Ne = Faraday constant or F = 6.0229 × 1023 × 1.602 × 10–19 = 96487C 96500C
Illustration 9:
Solution:
In producing chlorine through electrolysis 100 KW power at 125 volt is being consumed. If the E.C.E. of chlorine is 0.367 10–6 kg/coul, then the mass of chlorine liberated per minute will be: (A) 16.3 10–4 kg (B) 17.61 10–3 kg (C) 18.2 10–3 kg (D) 10–4 kg P Mass of chlorine liberated = Zit = Z t V -6 5 0.367 ´ 10 ´ 10 ´ 60 = = 17.61 ´ 10- 3 125
Illustration 10: The chemical equivalents of copper and silver are 32 and 108 respectively. When copper and silver voltammeters are connected in series and electric current is passed through them for sometime, 1.6 gm of copper is deposited. The mass of silver deposited will be :
Solution:
(A) 3.5 gm (B) 2.8 gm (C) 5.4 gm (D) none of these E1 m1 = E2 m2 E 108 \ m2 = 2 ´ m1 = ´ 1.6 = 5.4 gm E1 32
Illustration 11: A silver and a copper voltmeters are connected across a 6 volt battery of negligible resistance. In half an hour, 1 gm of copper and 2 gm of silver are deposited. The rate at which energy is supplied by the battery will approximately be : (Given E.C.E. of copper = 3.294 10–4 g/C & E.C.E. of silver = 1.118 10–3 g/C) (A) 64 watt (B) 32 watt (C) 96 watt (D) 16 watt m Solution: m = Zit or i = Zt For silver voltmeter m 2 i1 = 1 = = 0.994 amp Z1t 1.118 ´ 10- 3 ´ 1800 For copper voltmeter m 1 i2 = 2 = = 1.687 amp. Z2 t 3.294 ´ 10- 4 ´ 1800 power of circuit = V(i1 + i2) = 6 (0.994 + 1.687) = 6 2.681 16 W.
SOME IMPORTANT POINTS (i) (ii) (iii)
Capacity of a cell- It is expressed in kilowatt hours which involves both current and the voltage. Energy capacity- It is expressed in kilowatt hours which involves both current and the voltage. Efficiency- (a) It is the ratio of discharging capacity to the charging capacity. (b) If discharge takes place slowly, the maximum available energy is 80% of that spent in charging. (c) If the cell is short circuited, the discharge takes place suddenly and the cell is spoiled.
APPLICATIONS OF ELECTROLYSIS The phenomenon of electrolysis has been used as a valuable tool in (i) identifying the components of certain liquids (ii) making accurate measurement of current (iii) calibrating an ammeter (iv) determining the electro-chemical equivalents of elements (v) electroplating (vi) producing pure metals (vii) electrolying
THERMOELECTRICITY-SEEBECK EFFECT
(i) (ii)
Thermocoule: If two wires of different metals are joined at their ends so as to from two junctions, then the resulting arrangement is called a thermocouple. Seebeck effect: (a) When the two junctions of a thermocouple are maintained at different temperatures, then a current starts flowing through the wires. This is called Seebeck effect and the e.m.f. developed in the circuit is called thermo-emf. (b) The Seebeck effect is perfectly reversible, i.e. if the hot and the cold junctions are interchanged, the direction of current is reversed. (c) The thermo-emf is of the order of a few micro-volts per degree temperature difference.
THERMOELECTRIC SERIES (i)
(ii) (iii)
Seebeck arranged a number of metals in the form of a series called thermoelectic series. The arrangement of some of the metals forming the series areSb, Fe, Zn, Ag, Au, Mo, Cr, Sn, Pb, Hg, Mn, Cu, Co, Ni, Bi. In the above series, current flows through the cold junction from the metal which appear earlier to the metal which appears later. Greater the separation of the two metals in the series, greater is the thermo emf generated.
VARIATION OF DIFFERENCE
WITH
TEMPERATURE-
If the cold junction is kept at 0o and the temperature of hot junction (t) is gradually increased, it is found that the thermo emf e first increases, attains a maximum value and then decreases to become zero again (Fig.) Thermo emf V
(i)
THERMO-EMF
tn 0oC
(ii) (iii) (iv)
(v)
ti
Temperature of hot junction t
If the temperature is further increased, the emf changes direction. Netural temperature (tn)- The temperature at which thermo-emf is maximum is defined as neutral temperature. tn is fixed for a given thermo-couple. Temperature of inversion (ti)- The temperature at which thermo-emf changes its direction is called the temperature of inversion. ti is an much above tn as the tenoeratyre if cikd hybctuib us bekiw tn. Mathematically 1 e t t 2 2 where and are constants for a given thermo- couple.
Illustration 12: The e.m.f. of a thermocouple, one junction of which is kept at 0 C, is given by e = at + bt2 The neutral temperature will be: a a (A) (B) b b
a a (D) 2b 2b At neutral temperature e = maximum de \ =0 dt or a + 2btn = 0 (C)
Solution:
\
tn = -
a . 2b
Illustration 13: In the above problem, the temperature of inversion will be : a a (A) (B) b b a a (D) (C) 2b 2b t +t Solution: tn = o i 2 \ ti = 2tn - to or ti = 2 ´ -
a a - 0=2b b
Illustration 14: In Illustration 12, the Peltier co-efficient will be : (A) (t – 273) (a + 2bt) (B) (t + 273) (a – 2bt) (C) (t – 273) (a – 2bt) (D) (t – 273) (a – 2bt) de Solution: Peltier coeff. p = T dT tC = T – 273 \ e = a(T - 273) + b(T - 273)2 d.w.r.t. T de = a + 2b(T - 273) dT de \ p=T = T[a + 2b(T - 273)] dT or p = (t + 273) (a + 2bt) . Illustration 15: In Illustration 12, the Thomson co-efficient will be : (A) 2a(t + 273) (B) 2a(t – 273) (C) 2b(t + 273) (D) 2b(t – 273) 2 de Solution: Thomson co-efficient s = T 2 dT de = a + 2b(T - 273) dT d.w.r.t. T d2e = 2b dT 2 d2e \ s = T 2 = 2b(t + 273) dT
SEEBECK COEFFICIENT The rate of change of thermo emf with temperature is called thermo-electric power or Seebeck coefficient (S). de X t dt When t = tn, e is maximum de 0 dt tn 0 or tn
PELTIER EFFECT (i) (ii) (iii)
(iv)
This effect is the converse of Seebeck effect. If a current is passed through a junction of two dissimilar metals, heat is eigher absorbed or evolved at the junction. On reversing the direction of current, the heating effect is also reversed. If the seebeck current is in a certain direction through the hot junction, then an external current sent in the same direction through this junction produces a cooling at this junction and a heating at the other junction. Peltier coefficient ()- It is the amount of heat absorbed or evolved at a junction per second when a current of one ampere flows through it. de T dT
THOMSON EFFECT (i) (ii)
(iii)
An emf is developed between two parts of a single metal if they are at different temperatures. This is called Thomson effect. Thomson coefficient- If de is the potential difference between two points in a metal which have a temperature difference dt, then the ratio de d2e T 2 dT dT is defined as the thomson coefficient. If one part of a metal is at a higher temperature than the other, the free electrons at hot part will have more kinetic energy and as a result these electrons will diffuse faster towards colder part than the electrons from cold to the hot part. This would result in the net transfer of electrons setting an electric current in the metal.
APPLICATIONS OF THERMOELECTRIC EFFECTS (i) (ii) (iii) (iv)
Measurement of temperature Detection of heat radiations Refrigeration Power generation.
OBJECTIVE ASSIGNMENT 1.
Sol: 2.
Sol:
3.
Sol:
4.
Sol:
5.
Sol: 6.
Consider the two statements A and B (i) the neutral temperature does not depend on temperature of cold junction (ii) the inversion temperature does not depend on temperature of cold junction (A) both A and B are correct (B) A is correct but B is wrong (C) A is wrong but B is correct (D) both A and B are wrong. (B) The neutral temperature of a thermocouple with cold junction at 20 oC is 220oC. Its temperature of inversion is (A) 420oC (B) 120oC (C) 110oC (D) 440oC (A) We know tn - t c = ti - tn tc = neutral temperature ti = inversion temperature \ ti = 2tn - tc = 440° - 20° = 420° C For a thermocouple if the cold junction is maintained at 0oC the inversion temperature is 680oC. Its Neutral temperature is (A) 1360oC (B) 680oC (C) 340oC (D) 170oC (C) We know inversion temperature = 2(neutral temperature) 680 Neutral temperature = = 340° C 2 Consider the following two statements A and B and identify the correct choice in the given answers. (i) Thermo e.m.f. is minimum at neutral temperature of a themocouple. (ii) When two junctions made of two different metallic wires are maintained at different temperatures, an electric current is generated in the circuit (A) A is false and B is true (B) A is true and B is false (C) Both A and B are false (D) Both A and B are true (A) At neutral temperature of a thermocouple, thermo emf should be maximum, but not minimum A is false and B is true. According to the Faraday's Law of electrolysis, the mass deposited or liberated at an electrode is proportional to : m Q2 (B) (A) mQ (C) (D) m does not depend on Q m l2 (A) Two resistors having equal resistances are joined in series and a current is passed through the combination. Neglect any variation in resistance as the temperature changes. In a given time interval. (A) equal amounts of thermal energy must be produced in the resistors (B) unequal amounts of tgernak ebergt nat be oridyced (C) the temperature must rise equally in the resistors
Sol: 7.
Sol:
(D) (A)
the temperature must rise unequally in the resistors.
A constant voltage is applied between the two ends of a uniform metallic wire. Some heat is developed in it. The heat developed is doubled if (A) both the length and radius of the wire are halved (B) both the length and radius of the wire are doubled (C) the radius of the wire is doubled (D) the length of the wire is doubled Where the time remains the same in both the cases. V 2 V 2 pr 2 r 2 (B) H = = µ rL R L 2
H ær1 ö æL 2 ö = ´ H' çè r2 ÷ ø çè L1 ÷ ø H 1 2 1 = ´ = H' 4 1 2 Þ H' = 2H Þ
8.
Which of the following plots may represent the thermal energy produced in a resistor in a given time as a function of the electric current? (A) a (B) b (C) c (D) d
d
c b a
i
Sol:
(A) H = I 2 Rt = U
Þ U µ I2 It represents a parabola. 9.
Sol:
Two heater wires, made of the same material and having the same length and the same radius, are first connected in series and then in parallel to a constant potential difference. If the rates of heat produced in the two cases are H s and Hp respectively, H then s will be : Hp 1 1 (A) (B) 2 (C) (D) 4 2 4 (C) Since l1 = l2 , r1 = r2 & r 1 = r 2 So resistance R1 = R2 = R Hs =
V2 V2 = Req 2R
V2 V2 = Hp = R Req 2
æV 2 ö çè 2R ÷ ø
Hs 1 = = Hp æ ö 4 2 çV ÷ çR ÷ çè ÷ 2 ø
10.
Sol:
11.
Sol:
12.
Sol:
13.
Sol:
A hot electric iron has a resistance of 80 and is used on a 200 V source. The electrical energy spent, if it is used for 2hr. will be (A) 800 Wh (B) 1000 Wh (C) 2000 Wh (D) 8000 Wh 2 2 V (200) (B) (P) = = = 500W R 80 Therefore energy spent = power time = 500 2 = 1000 Wh. Antimony and Bismuth are usually used in a thermocouple, because (A) higher thermo e.m.f. is produced (B) lower thermo e.m.f. is produced (C) constant thermo e.m.f. is produced (D) negative thermo e.m.f. is produced Since antimony and bismuth form the pair of metals producing higher thermo emf in comparison with other pairs of thermocouple. Therefore pair is usually used in thermocouple. At 40C resistance of platinum is 3.14 and at 100C its resistance is 3.76. Its temperature coefficient is (A) 3.75 10–1/C (B) 3.75 × 10–2/C (C) 0.00329/C (D) None of these R2 - R1 (C) a = R1 ´ (t 2 - t1 ) 3.76 - 3.14 Þ a = = 0.00329 3.14 ´ (100 - 40) Two resistance filaments of same length are connected first in series and then in parallel. Fine the ratio of power dissipated in both cases assuming that equal current flows in the main circuit. (A) 1:4 (B) 4:1 (C) 1:2 (D) 2:1 (D) Required ratio i/2
R
i i/2 R Parallel Combination
R R Series Combination
i2R + i2R 2
2
æi ö æi ö R+ç ÷ R çè ÷ è2 ø 2ø
14.
Sol: 15.
Sol:
16.
Sol:
17.
Sol:
18.
Sol:
19.
= 2 :1
Thermocouple thermometer is based on (A) Seeback effect (B) (C) Peltier effect (D) (A)
Crompton effect Photoelectric effect.
A wire has a resistance of 3.1 at 30C and a resistance 4.5 at 100oC. The temperature coefficient of resistance of the wire (A) 0.0064oC–1 (B) 0.0034oC–1 o –1 (C) 0.0025 C (D) 0.0012oC–1 R - R30 4.5 - 3.1 (A) a = 100 = R30 ´ Dt 3.1 ´ (100 - 30) 1.4 1.4 = = = 0.0064°C- 1 3.1 ´ 70 217 Faraday constant (F), chemical equivalent (E) and electrochemical equivalent (Z) are related as : E E Z F F 2 (C) (D) (A) F = EZ (B) F= Z Z E (C) We know frmo the Faraday's second law of electrolysis that electrochemical equivalent 1 M 1 E (Z) = ´ = ´ E or F = F P F Z Two heater coils separately take 10 minutes to boil a certain amount of water. If both coils are connected in series, time taken to boil the same amount of water will be : (A) 15 min. (B) 20 min. (C) 7.5 min. (D) 2.5 min. 2 V t Þ tµR (B) Q = R t R \ 1 = 1 t 2 R2 10 R Þ = Þ t 2 = 20 min t 2 2R Resistance of a copper coil is 4.64 at 40oC and 5.6 at 100oC. Then its resistance at 0oC is (in ): (A) 4 (B) 0.96 (C) 5.12 (D) 4.2 R - R0 5.6 - R0 4.64 - R0 (A) a = t = = R0 ´ 4t R0 ´ 100 R0 ´ 40 Þ R0 = 4W Resistance of a coil at 100oC is 4.2 and the temperature coefficient of resistance of its material is 0.004oC–1. Then its resistance at 0oC is () : (A) 5 (B) 3 (C) 4 (D) 3.5
Sol:
(B) Rt = R0 (1 + a t) Þ 4.2 = R0 (1 + 0.004 ´ 100)
Þ R0 = 20.
Sol:
4.2 = 3W 1.4
An electric kettle has two heating elements. One brings it to boil in ten minutes and the other in fifteen minutes. If the two heating filaments are connected in parallel, the water in the kettle will boil in (A) 5 minutes (B) 25 minutes (C) 8 minutes (D) 6 minutes. V2 V2 (D) Q = ´ 10 = ´ 15 R1 R2
Þ
R1 2 = R2 3
V2 V2 ´ 10 = Q= t R1 æ R1 R2 ö çè R + R ÷ 1 2 ø
æ R ö 5 Þ 10 = ç1 + 1 ÷t = t è R2 ø 3 Þ 21.
Sol:
22.
Sol: 23.
Sol: 24.
t = 6 min
The resistance of a conductor is 5 ohm at 50oC and 6 ohm at 100oC. Its resistance at 0oC is (A) 2 ohm (B) 1 ohm (C) 4 ohm (D) 3 ohm. R - R0 (C) a = 1 R0 t 5 - R0 6 - R0 Þ a = = Þ 10 - 2R0 = 6 - R0 50R0 100R0 R0 = 4W The mass liberated at any electrode when electricity is passed through it is directly proportional to 1 (A) q2 (B) q (C) (D) q3 q (B) m = ZIt = Zq Þ m µ q Which one of the following causes production of heat when current is set up in a wire? (A) fall of electrons from higher orbits to lower orbits (B) interatomic collisions (C) interelectron collisions (D) collisions of conduction electrons with atoms. (D) In the Seeback series bismuth occurs first followed by Cu and Fe among other. The Sb is the last in the series. If V1 be the thermo-emf at the given temperature difference for Bi-Sb themocouple and V2 be hat for Cu-Fe thermocouple, then
Sol: 25.
Sol: 26.
Sol:
27.
Sol:
28.
Sol: 29. Sol: 30.
Sol:
(A) (C) (D)
V1 = V 2 V1 > V 2
(B) (D)
V1 < V 2 data insufficient.
Which of the following is a characteristic temperature for the themocouple ? (A) temperature of cold junction (B) temperature of hot junction (C) temperature of inversion (D) neutral temperature. (D) When a copper voltameter is connected with a battery of electromagnetic field 12V, 2gm of copper is deposited in 30 minutes. If the same voltmeter is connected across a 6V battery, the mass of copper deposited in 45 minutes would be (A) 1 gm (B) 1.5 gm (C) 2 gm (D) 2.5 gm (B) m µ zIt, m µ zvt m2 v 2 t 2 = m1 v1t1 v t 6 45 Þ m2 = 2 2 ´ m1 = ´ ´ 2 = 1.5 gm v1t1 12 30 The electro-magnetic field of a Cu-Fe thermo couple varies with temperature of the hot junction (cold junction at 0oC) as E = 14 – 0.022. The neutral temperature is given by (A) 350oC (B) 400oC (C) 450oC (D) 500oC. dE (A) E = 14q - 0.02q2, =0 dq Þ q = 350° C Same current is being passed through a copper voltmeter and a silver voltmeter. The rate of increase in weights of the cathodes in the two voltameters will proportional to: (A) atomic masses (B) atomic numbers (C) relative densities (D) none of these. (A) Which of the following acts a depolariser in Leclanche cell? (A) ZnCl2 (B) Mn2O3 (C) MnO2 (D) NH4Cl (C) In the circuit shown the 5 resistor develops 20W power due to current flowing through it, then power dissipated in 4 resistance is (A) 4W (B) 6W (C) 10 W (D) 20 W (A) If the current in 5 branch is i, then current in 4 branch will be i/2. P = i2R = i2 ´ 5 = 20 \ i = 2A
P' = i2R' = (1)2 ´ 4 = 4W 4Rt = R0 (1+ µ t) 1. 3R0 = R0 (1 + 4 ´ 10- 3 t)
1 = 500° C (B) 2 ´ 10- 3 H = Pt = 210 ´ 300 J 210 ´ 300 = cal = 15000 cal 4.2 Þ t=
2.
3.
Q = It = 1.6 ´ 60 = 96C 96 No. of Cu++ions = = 3 ´ 1020 3.2 ´ 10- 19
For Test 1.
Sol: 2.
Sol: 3.
Sol: 4.
Sol:
5.
Sol: 6.
Sol: 7.
At what temperature will the resistance of a copper wire become three times its value at 0oC? [Temperature coefficient of resistance for copper = 4 × 10–3 per oC] (A) 400oC (B) 450oC (C) 500oC (D) 550oC. (C) How much heat is developed in a 210 watt electric bulb in 5 minutes ? [Mechanical equivalent of heat = 4.2 joule/calorie] (A) 7500 calories (B) 15000 calories (B) 22500 calories (D) 30000 calories. (B) A current of 1.6 A is passed through a solution of CuSO 4. How many Cu++ ions are liberated in one minute ? [Electronic charge = 1.6 × 10–19 coulomb]. (A) 3 × 1020 (B) 3 × 1019 (C) 6 × 1020 (D) 6 × 1019 (A) The smallest temperature difference that can be measurred with a combination of a thermocouple of thermo e.f.f. 30V per degree and a galvanometer of 50 ohm resistance, capable of measuring a minimum current of 3 × 10–7 amp, is : (A) 0.5 degree (B) 1.0 degree (C) 1.5 degree (D) 2.0 degree. (A) Minimum voltage that can be measured by the galvanometer V = ig G = 3 ´ 10- 7 ´ 50 = 1.5 ´ 10- 5 volt Smallest temperature difference that can be mesured 1.5 ´ 10- 5 Dt = = 0.5°C 30 ´ 10- 6 If 2.2 kilowatt power is transmitted through a 10 ohm line at 22000 volt, the power loss in the form of heat will be(A) 0.1 watt (B) 1 watt (C) 10 watt (D) 100 watt. 2 P R 2200 ´ 2200 ´ 10 = 0.1 W (A) H = I 2R = 2 = V 22000 ´ 22000 For electroplating a spoon, it is placed in the voltameter at (A) the position of anode (B) the position of cathode (C) exactly in the middle of anode and the cathode (D) anywhere in the clectrolyte (B) Positive ios get deposited on cathode. The quantity of electricity needed to liberate one gram equivalent of an element is(A) 1 ampere (B) 96500 amperes
Sol:
8.
Sol: 9.
Sol:
10.
Sol:
11.
Sol:
(C) 96500 coulomb (D) 96500 faradays. (C) The amount of electricity required to liberate 1 gm equivalent is 1 Faraday or 96500 coulomb. When current is passed through a junction of two dissimilar metals, heat is evolved or absorbed at the junction. This process is called(A) Seebeck effect (B) Joule effect (C) Peltier effect (D) Thomson effect. (C) Peltier effect A 5oC rese in temperature is observed in a conductor by passing a current. When the current is doubled the rise in temperature will be approximately(A) 10oC (B) 16oC (C) 20oC (D) 12oC. (C) As H µ I2 , on doubling the current, heat produced and hence rise in temperature becomes fuor times. Rise in temperature 4 5 = 20C Three equal resistors connected across a source of e.m.f. together dissipate 10 watt of power. What will be the power dissipated in watts if the same resistors are connected in parallel across the same source of e.m.f.? 10 (A) 10 (B) (C) 30 (D) 90. 3 V2 (D) In series Power = = 10W \ V 2 = 30R 3R 2 V 30R ´ 3 = = 90W In parallel P' = R æR ö çè ÷ 3ø If nearly 105 coulombs liberte 1 gm-equivalent of aluminium, hen the amount of aluminum, (equivalent weight 9) deposited through electrolysis in 20 minutes by a current of 50 amp. will be(A) 0.6 gm (B) 0.09 gm (C) 5.4 gm (D) 10.8 gm. (C) m = zit or m µ it
m1 i1 t1 = m2 i2 t 2 or m2 = 5.4 gm \
12.
Sol:
13.
or
9 105 = m2 50 ´ 20 ´ 60
Two filaments of same length are connected first in series and then in parallel. For the same amount of main current flowing, the ratio of the heat produced is : (A) 2:1 (B) 1:2 (C) 4:1 (D) 1:4 2 (C) Q = I R or Q µ R Qs Rs 2R \ = = = 4 :1 Qp Rp æR ö çè ÷ 2ø Constantan wire is used in making standard resistances, because its (A) specific resistance is low (B) density is high (C) temperature coeff. of resistance is negligible
Sol:
14.
Sol:
(D) melting point is high. (C) The temperature coefficient of resistance of constantan is quite negligible, hence its resistance almost remains constant. As the temperature of hot junction increases, the thermo cmf : (A) always increases (B) always decreases (C) may increase of decrease (D) always remains constant (C) Thermo e.m.f.
0o C
Temp. of hot junction
t
The variation of thermo-e.m.f. with temperature of hot junction is as shown in the figure. 15.
Sol:
16.
Sol:
17.
A certain charge liberates 0.8 gm of Oxygen. The same charge will liberate how many gms of Silver? 108 gm . (A) 108 gm (B) 10.8 gm (C) 0.8 gm (D) 0.8 MAg EAg (B) = M0 E0 108 ´ 0.8 or mAg = = 10.8 g 8 An electric kettle has two heating coils. When one coil is used, water in the kettle boils in 5 minutes, while when second coil is used, same water boils in 10 minutes. If the two coils, connected in parallel are used simultaenously, the same water will boil in time. (A) 3 min. 20 sec. (B) 5 min. (C) 7 min. 30 sec. (D) 2 min. 30 sec. 2 V t Þ tµR (A) Q = R t 5 R1 \ 1 = = Þ R2 = 2R1 t 2 10 R2 In second case R1 R 5 = = 1 R1R2 2 t R R1 + R2 3 1 10 Þ t= min = 3 min 20 sec. 3 An external resistance R is connected to a battery of e.m.f. V and internal resistance r. The joule heat produced in resistor R is maximum when R is equal to
r (C) 2r (D) infinitely large. 2 (A) According to maximum power transfer theroem, transfer of power from the surce to the load will be maximum when load resistance is equal to the internal resistance of the source. R=r
(A)
Sol:
18.
Sol:
19.
Sol:
20.
Sol:
21.
Sol: 22.
r
(B)
If Tc, Tn and Ti denote the temperatures of cold junction, neutral temperature and inversion temperature of a thermo-couple respectively, then : (A) Tc + Ti = Tn (B) Ti + Tc = 2Tn (C) Tc + Ti = 2Tn (D) Tc – Ti = 2Tn T + Ti (A) Tn = c 2 Þ Tc + Ti = 2Tn The electrochemical equivalent of a material in an electrolyte depends on (A) The nature of the material (B) The current through the electrolyte (C) The amount of charge passes through electrolyte (D) The amount of material present in electrolyte. (A) m = Zit = ZQ m \ Z= Q It depend on the nature of material. A wire of length I and 3 identical cells of negligible internal resistances are connected in series. Due to the current, the temperature of the wire is raised by T in a time t. A number of N similar cells is now connected in series with a wire of the same material and cross section but the length 2L. The temperature of the wire is raised by the same amount T in the same time t. The value of N is(A) 4 (B) 6 (C) 8 (D) 9. 2 V t (B) The heat generated in the wire is givein by H = R 2 (3E) t In the first case, we have mCp DT = . . . (1) R in the second case, we have (NE)2 t (2m) Cp Dt = (\ R µ L) . . . (2) 2R Dividing Equation (2) by Equation, (1) we get N2 / 2 2= Þ N2 = 36 Þ N = 6 9 Two wires having resistance of 2 and 4 are connected to same voltage. Ratio of heat dissipated in the resistances is(A) 1:2 (B) 4:3 (C) 2:1 (D) 5 : 2. H1 R2 2 (C) As heat dissipated = V 2 /R , so it is inversely proportional to 1/ R \ = = H2 R1 1 If thermo e.m.f. is given by E = at + bt2, then neutral temperature is-
a 2b (B) 2b a (A) At neutral temperature dE/dt = 0 or a + 2bTN = 0 a TN = 2b
(A) Sol:
23.
Sol:
24.
Sol:
(C)
a b
(D)
b a
Two parallel and straight conductors are at a distance d apart, each have current i following through them. The force per unit length is m0i2 m0i (A) (B) 2p d 4pd 2 m0i m0i (C) (D) 4pd d2 (B) Force due to magnetic field induction = B12, where m 2I B1 is the magnetic field due to first = 0 × 1 4p d m0 i2 \ F= × 2p d Thomson coefficient of a conductor is 10V/K. The two ends of it are kept at 50C and 60C respectively. Amout of heat abosrbed by the conductor when a charge of 10 C flows through it is : (A) 1000 J (B) 100 J (C) 100 mJ (D) 1 mJ (D) Amount of heat absorbed by the conductor H = Q dt H = (10 10–6) (10) (10) = 10–3 J = 1 mJ
25_57. In an electroplating experiment 4A current for 2 minutes deposits m gram of silver. When 6A current flown for 40 seconds then the deposited silver will be (A) 4m (B) m/2 (C) m/4 (D) 2m Sol: (B) m i t m1 i1 t1 m 4 ´ 120 \ = = or m2 i2 t 2 m2 6 ´ 40 m or m2 = . 2 26_58. The temperature coefficient of a resistance wire is 0.00125 per degree. The resistance at 300K is 1 ohm. At what temperature its resistance will be 2 ohm. (A) at 1154 K (B) at 1100 K (C) at 600 K (D) at 1400 K Sol: (B) Rt2 = Rt1 [1+ a (t 2 - t1)] 2 = 1[1 + 0.00125(t 2 - 300)]
or or
1 = 800 0.00125 t 2 = 1100 K t 2 - 300 =
27_59. Two electric heater wires are made from the same material and have the same dimensions. First these are connectted in series and then these are connected in parallel (to a 220 V AC source) then the ratio of heat generated will be (A) 2:1 (B) 1:2 (C) 4:1 (D) 1:4 2 V 1 Sol: (D) H = µ R R R H1 R2 \ = = 2 = 1: 4 H2 R1 2R 28_60. The mass of a product liberated on anode in an electrochemical cell depends on (A) ( t)1/2 (B) t (C) /t (D) 2 t (Where t is the time period for which the current is passed). Sol: (B) m Q But Q = t m t or m = Z t 29_61. If i is the inversion temperature, n is the neutral temperature, c is the temperature of the cold junction, then (A) i + c = n (B) i – c = 2n qi + qc (D) c – i = 2n (C) = qn 2 Sol: (C) Neutral temperature is the mean of temperature of inversion and temperature of q + qc cold junction, i.e., qn = i . 2 30_75. The thermo e.m.f. of a thermo-couple is 25V/C at room temperature. A galvanometer of 40 ohm resistance, capable of detecting current as low as 10 –5 A, is connected with the thermocouple. The smallest temperature difference that can be detected by this system is (A) 16C (B) 12C (C) 8C (D) 20C Sol: (A) Minimum p.d. to be measured V = iG = 10 - 5 ´ 40 = 4 ´ 10 - 4 V 4 ´ 10- 4 = 16°C . Minimum temp. diff. detected = 25 ´ 10- 6
1.
VARIOUS DEFINITIONS & FEATURES RELATED TO THEM
1.1
BAR MAGNET
A system composed of two poles, equal in magnitude but opposite in polarity, placed at a small displacement apart is known as a bar magnet. It is also known as a magnetic dipole. –mp
+mp N
S
N
S (a)
(b)
1.2
POLE STRENGTH (MP)
(Ia)
The strength of a magnetic pole to attract magnetic materials towards it, is known as pole strength. Greater the number of unit poles in a magnetic pole, greater will be its strength. F Magnetic force mp B Magneticinduction Its unit is ampere-meter.
(b) (c) (d)
Illustration 1:
Solution :
The magnetic flux density of magnetic field is 1.5 Wb/m2. A proton enters this field with a velocity of 2 107 m/s in a direction making an angle of 30 with the field. The force acting on the proton will be : (A) 2.4 10–12 Newton (B) 0.24 10–12 Newton (C) 24 10–12 Newton (D) 0.024 10–12 Newton (A) F = qvBsin q = 1.6 ´ 10- 19 ´ 2 ´ 107 ´ 1.5 ´ 0.5 = 2.4 ´ 10- 12
1.3
MAGNETIC LINES OF FORCE
(a)
The imaginary lines which represent the direction of magnetic field, are known as magnetic lines of force. The imaginary path traced by an isolated (imaginary) unit north pole is defined as a line of force. Magnetic lines of force are closed curves. Outside the magnet their direction is from north pole to south pole and inside the magnet these are from south to north pole.
(b) (c)
N
(d) (e) (f) (g)
S
They neither have origin nor end. These lines do not intersect, because if they do so then it would mean two value of magnetic field at a single point, which is not possible. The tangent drawn at any point to the line of force indicates the direction of magnetic field at that point. At the poles of the magnet the magnetic field is stronger because the lines of force there are crowded together and away from the poles the magnetic field is week. i.e. magnetic field intensity number of lines of force.
(h)
(i)
The number of magnetic lines of force passing through unit normal area is defined as magnetic induction (B) whereas the number of lines of force passing through any area is known as magnetic flux. The lines of force can emerge out of the north pole of magnet at any angle and these can merge into the south pole at any angle.
1.4
MAGNETIC FIELD
(a)
The space around a magnet in which a torque acts on a magnetic needle is known as magnetic field. The space around a magnet in which a net force acts on a magnetic test pole is known as magnetic field. The space around a magnet in which its effect is experienced is known as magnetic field. There are four types of magnetic field : (i) Uniform magnetic field: (a) The magnetic field, in which the intensity of magnetic field is same at all points, is known as uniform magnetic field. (b) In such a magnetic field the magnetic lines of force are parallel and equidistant. e.g. the magnetic dines of force of earth's magnetic field. (ii) Non-uniform magnetic field: (a) The magnetic field, in which the intensity of magnetic field at different points is different, is known as non-uniform magnetic field. (b) It is represented by non-parallel lines of force. (iii) Varying magnetic field: (a) The magnetic field, which keeps on changing with respect to time is known as a variable magnetic field. (b) Example :– B = B0 sin t or B = B0 cos t (iv) Non-varying magnetic field: (a) The magnetic field which does not change with time is known as a constant magnetic field. (b) The direction of magnetic field is that in which a force acts on a unit test pole. (c) It can be produced by moving charges, current carrying loops, and variations in electric currents. F F N S
(b) (c) (d)
B
B
1.5
MAGNETIC DIPOLE
(a)
The structures, which tend to align along the direction of magnetic field, are known as magnetic dipoles. Bar magnet, current carrying solenoid, current carrying coil, current carrying coil, current loop, magnetic needle etc. are the examples of magnetic dipole. It is not possible to separate out the two poles of a magnet. Magnetic dipole is not a system composed of two poles because the existence of monopoles is not possible. A loop of single turn is also a magnetic dipole. One face of the loop behaves as north pole and the other face behaves as south pole. The face of the coil, in which current is anticlockwise, behaves as north pole and the face in which current is clockwise, behaves as south pole.
(b) (c) (d)
1.6
MAGNETIC FLUX
(a) (b) (c)
The number of lines of force passing through a given area is defined as magnetic flux. The magnetic flux passing through unit normal area is defined as magnetic induction (B). When the magnetic field is normal to the plane then = BA, when A = 1m2 then = B. B nˆ Normal to surface
A
Vibrative lines Of force Surface
(d)
When magnetic field makes an angle with the normal to the plane : (i) Magnetic flux linked with the plane = Area of the plane (A) × Component of magnetic field normal to the plane (B cos ) i.e. = AB cos If the number of turns in the coil is N. then = NAB cos os Bc
B
Bcos
A
B
nˆ
or A
(ii)
= magnetic field normal to the plane (B) × component of A in the direction of magnetic field (A cos ) i.e. = BA cos
A
os Ac
B
In both cases q is the angle between B and nˆ .
(e)
(iii)
When B and A are mutually parallel then = 0o and = BA.
(iv)
When B and A are mutually perpendicular, = 90o and = 0.
(v) When B and A are antiparallel, then = 180o and = BA. When the angle between B and the plane of coil is then = BA sin . If the number of turns in the coil is N then = NBA sin . B O
B sin nˆ
180o A
B
B cos
(i) (ii)
When the plane of coil is parallel to B then = 0o and = 0. When B and the plane of coil are mutually perpendicular i.e. = 90, then = BA.
(f)
(iii) When B and the plane of coil are mutually antiparallel i.e. = 180o, then = 0. Magnetic flux linked with a small surface element dA d B.d A B dA cos where d A = area of small element. nˆ = unit normal vector. B
DA
nˆ
(g)
The flux linked with total area of the surface A d B.dA B dA cos
(h)
Positive magnetic flux: When the magnetic induction B and the unit normal vector nˆ are in the same direction then is called the positive magnetic flux. B A = BA Negative magnetic flux: When the magnetic induction B and unit normal vector are mutually in opposite directions then is called negative magnetic flux = – BA The flux emanating out of a surface is positive and the flux entering the surface is negative. is a scalar quantity. The net magnetic flux coming out of a closed surface is always zero, i.e. B.dA 0
(i)
(j) (k) (l)
A
or
.B 0
1.7
MAGNETIC FLUX DENSITY OR MAGNETIC INDUCTION
(a)
The magnetic lines of force passing through unit normal area in a magnetic field is defined as magnetic induction. B when A 1m.2 then B A
(b)
F B
i
i
(c)
(d) (e)
L
Direction of magnetic induction: The direction in the magnetic field in which if a current carrying conductor is placed then no force acts on it, is known as the direction of magnetic induction. Magnetic induction is a vector quantity. The magnetic induction due to a bar magnet 2KM (i) In axial position B 3 r KM (ii) In equatorial position B 3 r m Here K = 0 = 10- 7 Weber / A.m 4p
1.8
MAGNETISING FIELD OR INTENSITY OF MAGNETIC FIELD H
(a)
The ratio of magnetic induction produced in vacum (B o) and magnetic permeability of Bo vacuum is defined as magnetising field (H), i.e. H o The intensity of magnetic field due to a pole of strength m p at a distance r from it is m H p2 r M Due to a small magnet H = 3 1 3 cos2 r
(b)
(c)
Illustration 2:
Solution: Illustration 3:
1000 turns per meter are wound over a Rowland ring of ferromagnetic material. On passing a current of 2 ampere in the coil, a magnetic field of 10 Wb/m2 is produced in it. The magnetising force generated in the material will be : (A) 1.2 10–3 A/m (B) 2.6 10–3 A/m –4 (C) 2.6 10 A/m (D) 2 103 A/m Ni = ni = 103 ´ 2 = 2 ´ 103 A / m (D) H = 2p r In the above problem, the value of intensity of magnetisation in A/m will be : (A) 7.96 106 (B) 7.96 10–6 3 (C) 3.98 10 (D) zero
Solution:
(A) B = m0 (I + H)
\ I= Illustration 4:
Solution:
B 10 - H= - 2 ´ 103 = 7.96 ´ 106 A / m m0 4p ´ 10- 7
In the above question 115 the relative permeability of the material will be : (A) 4.98 103 (B) 4.98 10–3 (C) 2.98 10–3 (D) 3.98 103 7.96 ´ 106 I (D) mr = 1 + X = 1 + = 1 + = 3.98 ´ 103 3 H 2 ´ 10
1.9
MAGNETIC MOMENT (M)
(a)
If a magnet of length l and magnetic moment M is bent in the form of a semicircular 2M are then its new magnetic moment will be M' = mp
mp
N
S
l = r
l l N
S
N
2r
S
M = mpI
(b)
The magnetic moment of an electron due to its orbital motion is 1B whereas that due to its spin motion it is B . 2 eh eh i.e. Morbital = l 4ml 2ml
eh eh B Mspin = s s sB 2 4ms 2ms Here B = Bohr magneton eh (i) The value of Bohr magneton B = 4 m (ii) B = 0.93 × 10–23 Amp-m2 and
(c)
Other formulae of M: (i) M = nir2 eVr er 2 er 2 2f er 2 M (ii) 2 2 2 T e (iii) 2mJ (iv) M = nB
(d)
Resultant magnetic moment : (i) When two bar magnets are lying mutually perpendicular to each other, then
M M12 M22 2mpl
–mp S l mp N S –mp
l
N mp
M1
M 2
M 1
M 2
2
M2
(ii)
When two coils, each of radius r and carrying current i, are lying concentrically with their planes at right angles to each other, then
M M12 M22 2ir 2 if M1 = M2
Illustration 5: Solution:
A square loop OABCO of side carries a current i. It is placed as shown in figure. Find the magnetic moment of the loop. Magnetic moment of the loop can be written as,
M i BC CO
Here, BC k
CO cos 60 i sin 60 j
2
i
3 j 2
3 j M i ( k) i 2 2 i 2 or M j 3 i Ans. 2
Illustration 6:
Find the magnitude of magnetic moment of the current carrying loop ABCDEFA. Each side of the loop is 10 cm long and current in the loop is i 2.0 A.
C D
i B
A
E
F
Solution :
By assuming two equal and opposite currents in BE, two current carrying loop (ABEFA and BCDEB) are formed. Their magnetic moments are equal in magnitude but perpendicular to each other. Hence
Mnet M2 M2 2 M
Where M ia (2.0)(0.1)(0.1) 0.02 A-m2 Mnet ( 2 ) (0.02) A – m2 0.028 A-m2 C D B
A
E
F
Illustration 7:
Solution:
Two identical bar magnets each of length L and pole–strength m are placed at right angles to each other with the north pole of one touching the south pole of the other. Evaluate the magnetic moment of the system. As magnetic moment is a vector M2 sin and as here M1 = MR (M12 M22 2M1M2 cos )1/ 2 with tan M1 M2 cos M2 = mL and = 90°, so MR = (M2 + M2 + 2MM cos 90)1/2 = ( 2)mL Msin90 tan 1, i.e., = tan–1(1) = 45° And, M Mcos90 S
M2
M1
M1 N
N
MR
M2 (A)
(B)
1.10 INTENSITY OF MAGNETISATION (I) (a)
(b) (c)
(d) (e) (f) (g) (h)
The magnetic moment per unit volume of a material is defined as intensity of M magnetisation (I). i.e. I = , when V V = 1m3 then I = M. The unit of intensity of magnetisation is ampere/meter and its dimensions are M 0L– 1 0 1 TA The pole strength per unit area of cross-section is defined as intensity of m magnetisation. i.e. I p when A = 1m2 then A=1m2 then I = mp A It is a vector quantity whose direction is along the magnetic field. In para- and ferro-magnetic materials its direction is in the direction of H and in diamagnetic materials it is opposite to that of H. I is produced in materials due to spin motion of electrons. The value of I and its direction in a material depend on the nature of that material. I - M curve
(i)
For para magnetic materials
(ii)
I
For diamagnetic materials I
Para
dia M
M
(iii)
For ferromagnetic materials I
Ferro
(i)
I - H curve Ferromagnetic I
H Diamagnetic
(j) (k) (l) (m) (n)
Its value depends on temperature. It is produced on account of induction in a material. For low magnetising field I H. i.e. I H or I = Ht is a dimensionless constant i.e. it carries no unit. Other types of intensity of magnetisation: (i) Mass intensity of magnetisation (Imass)Intensity of magnetisation 1 Imass orIm density of magnetic material (ii) Molar intensity of magnetisationImolar = IMW = Mo. Wt.× mass intensity of magnetisation = MW × Im (iii) Molecular intensity of magnetisationMolar int ensity of magnetisation IMW Imolecular IM Avogadro No. N
1.11 MAGNETIC SUSCEPTIBILITY OR ( OR K) (a)
(b) (c) (d) (e)
(f) (g) (h)
(i) (j)
The ratio intensity of magnetisation (I) in a material and the magnetising field (H) is defined as magnetic susceptibility (). I i.e. H When H = 1 Oersted then = I. The intensity of magnetisation induced in a material by unit magnetising field is defined as magnetic susceptibility. It has no unit and no dimensions. It is a measure of ease with which a material can be magnetised by a magnetised by a magnetising field (H). Magnetic susceptibility of various materials(i) For diamagnetic materials – = low and negative (ii) For paramagnetic materials – = low but positive (iii) For ferromagnetic materials – = high and positive I For paramagnetic substances it is inversely proportional to temperature i.e. T For low magnetising field the value of is constant. Different types of magnetic susceptibility(i) Volume susceptibility I V X H (ii) Mass or specific susceptibility m Volume susceptibility m v density of meterial (iii) Molar susceptibility MW MW = m × MW = Specific susceptibility × Mol. Wt. (iv) Molecular susceptibility m m = Atomic wt. × specific susceptibility AI = A m H –T curve 1 – : T
(a)
T
For paramagnetic substances Illustration 8:
(b)
T–1
For paramagnetic substances
The magnetic induction along the axis of an air solenoid is 0.03 Tesla. On placing an iron core inside the solenoid the magnetic induction becomes 15 Tesla. The relative permeability of iron core will be : (A) 300 (B) 500
(C) Solution:
700
(D) 900 15 m Bm (B) Bm = mH, B = m0H \ mr = = = = 500 B 0.03 m0
1.12 ABSOLUTE MAGNETIC PERMEABILITY () (a) (b) (c) (d) (e) (f) (h) (i) (j)
(k)
The ratio of magnetic induction (B) to magnetising field (H) is defined as magnetic permeability (). The extent to which magnetic permeability of that medium. It is the characteristic property of a magnetic material because it represents the amplification of magnetising field in that material. Its value is always positive and is different for different materials. For materials its value can be greater or less than 0. Its value depends on H and T. = 0 [1 + ] = 0r (i) For feeromagnetic materials = high (ii) For paramagnetic materials = low (iii) For diamagnetic materials = very low The unit of magnetic permeability is Weber per ammere-meter or Henry per meter and its dimensions are M1L0T–2A–2.
1.13 RELATIVE PERMEABILITY (R) (a)
(b) (c) (d) (e) (f) (g)
The ratio of magnetic permeability of medium () to the magnetic permeabiliy of free space (0) is defined as relative permeability (r). i.e. r 0 Number of lines of force passing throughunit area in medium r Number of lines of force passing throughunit area in vacuum magnetic flux density in medium r Magnetic flux density in vacuum The limit unto which a magnetic field penetrates matter, is known as relative permeability of that material. It has no unit and no dimensions. r = 1 + Relative permeability of various substances(i) For diamagnetic substances the value of r is slightly less than one i.e. r < 1. (ii) For paramagnetic substances the value of r is slightly greater than one i.e. r > 1. (iii) For ferromagnetic substance the value of r is much greater than one i.e. r >> 1.
2.
MAGNETIC DIPOLE IN A UNIFORM MAGNETIC FIELD
(i)
When a dipole is placed (or suspended) in a uniform magnetic field, then no net force acts on it i.e. the resultant force acting on it is zero. When a dipole is placed in a uniform magnetic field then a torque acts on it which is given by M B Newton-meter The magnitude of torque is given by MB sin 2 ml B sin Where m is the pole strength and 2l is the length of dipole.
(ii)
(iii)
(iv) (v)
B mB q
N
2l S mB
The units of M are ampere-meter2 or Joule/Tesla and its dimensions are M0L2A1 When a magnetic dipole is placed in a non-uniform magnetic field, then a torque as well as a force both act on it. The force acting on the dipole is given by
(vi)
(vii)
dB F M Newton dr (a) The work done in rotating a dipole in a magnetic field from an angle 1 to angle 2 with the field direction is given byW = MB (cos 1 – cos 2) Joule (b) If initially the dipole is lying along the field direction, then 1 = 0 W = MB (1 – cos ) Potential energy (U): (a) The potential energy of a dipole in a magnetic field is given by U = – M.B Joule (b) Special cases: When = 0, then U = – MB When = 90o, U = 0 When = 180o, U = – MB (–1) = MB
Illustration 9:
Solution:
A Magnet is suspended in the magnetic meridian with a untwisted wire. The upper end of the wire is rotated through 180 to deflect the magnet by 30 from magnetic meridian. Now this magnet is replaced by another magnet. Now the upper end of the wire is rotated through 270 to deflect the magnet by 30 from magnetic meridian. Compare the magnetic moments of magnets. If be the twist of the wire, then C , where C being restoring couple per unit twist of wire. Here 1 180 – 30 150 (150 /150) radian 2 (270 – 30 ) 240 (240 /180) radian If M be the magnetic moment and H, the horizontal component of earth’s field, then MH sin C MH sin If M1 and M2 be the magnetic moments of the two magnets respectively, then C 1 M1H sin for first magnet
C 2 M2H sin
for second magnet
M 150 /180 5 1 M1 or 1 M2 240 /180 8 2 M2
M1 : M2 5 : 8
3.
CLASSIFICATION OF MATERIALS
(i) (ii)
The root cause of magnetism in matter is the motion of electric charges. The motion of electrons and protons in atoms is responsible for their magnetic properties. The variation in the number of fundamental charged particles and variation in their arrangement in different materials are responsible for differences in their magnetic properties. On the basis of mutual interactions or behaviour of various materials in an external magnetic field, the materials are divided in three main categories. (1) Diamagnetic substances (2) Paramagnetic substances (3) Ferromagnetic substances
(iii)
(iv)
Comparative study of these materials: Diamagnetic Paramagnetic Property substances substances Cause of Orbital motion Spin motion of electrons magnetism substances Explanation On the basis of On the basis of spin and of magnetism orbital motion of orbital motion of electrons electrons Principle Electron principle Electron principle Behaviour in These are repelled These are feebly a non-uniform in an external attracted in an external magnetic field magnetic field i.e. magnetic field i.e. have a have a tendency to tendency to move from move from high to low to high field region. low field region. State of These are weakly These get weekly magnetisation magnetised in a magnetised in the direction opposite direction of applied to that of applied magnetic field magnetic field When a rod The materials align The materials align of the themselves at right themselves along the material is angles to the direction of magnetic suspended direction of field. between the magnetic field. pole pieces of a magnet. N
S (a)
N
S (b)
Ferromagnetic substances Ferromagnetic substances On the basis of domains formed. Domain principle These are strongly attracted in an external magnetic field. i.e. have an easy tendency to move from low to high field region. These get strongly magnetised in the direction of applied magnetic field. The materials easily align themselves in the direction of magnetic field.
N
S (a)
N
S (b)
Liquid or power in a watch glass when placed between the pole pieces (a) when poles are far apart. (b) When poles are close to each other. When the material in the form of liquid is filled in the U-tube and placed between pole pieces.
(a) The liquid gets bulged in the middle N
S
(b) The liquid gets depressed in the middle. N
(a) The liquid gets depressed in the middle N
S
(b) The liquid gets bulged in the middle. N
S
(a) The liquid is very much depressed in the middle. N
(b) The liquid gets very much bulged in the middle.
S
Liquid level in that limb gets depressed
S
N
Liquid level in that limb rises up.
S
Liquid level in that limb rises up very much.
N S
N S
N S
liquid
liquid
liquid
On placing the gaseous materials between pole pieces. The value of magnetic induction B Magnetic susceptibility Dependence of on temperature
The gas expands at right angles to the magnetic field.
The gas expands in the direction of magnetic field.
The gas rapidly expands in the direction of magnetic field.
B < B0
B > B0 Here B0 = magnetic induction in vacuum Low but positive 1
B > > B0
1 C or c = T - TC T - TC This is called Cure-Weiss law. TC = Curie temperature.
Dependence of on H
does not depend
Inversely proportional to temperature 1 C cµ or c = . This T T is called Curie law where C = Curie constant. does not depend
Low and negative 1 Does not depend on temperature (except Bi at low temperature)
Positive and high 102.
cµ
does not depend
Relative permeability (r) Intensity of magnetisation ()
r < 1
r > 1
r > > 1 r 102
is a direction opposite to that of H and its value is very low.
is in the direction of H but value is low.
is in the direction of H and value is very high.
-H Curves I
I
I
H
H
Magnetic moment (M)
Transition of materials (at Curie temperature) –T Curve
The value of M is very low ( 0 and is in opposite direction to H.) These do not change
c
The value of M is very low and is in the direction of H.
The value of M is very high and is in the direction of H.
On cooling, these get converted to ferromagnetic materials at Curie temperature.
These get converted into paramagnetic materials above Curie temperature.
c
T
The property of magnetism
Examples
Nature of effect
Diamagnetism is found in those materials in the atoms of which the number electrons is even. Cu, Ag, Au, Zn, Bi, Sb, NaCl, H2O air and diamond etc. Distortion effect
H
c
T
TC
T
Paramagnetism is found in those materials in the atoms of which the majority of electron spins are in the same direction. Al, Mn, Pt, Na, CuCl2, O2 and crown glass.
Ferro-magnetism is found in those materials which when placed in an external magnetic field are strongly magnetised.
Orientation effect
Hysteresis effect
Fe, Co, Ni, Gd, Fe3O4 etc.
4.
CURIE LAW AND CURIE TEMPERATURE
(a)
Curie law: (i) The magnetic susceptibility of paramagnetic substances in inversely 1 proportional to its absolute temperature i.e. c µ T C (ii) c µ Where C = Curie constant, T = absolute temperature T (iii) On increasing temperature, the magnetic susceptibility of paramagnetic materials decreases and vice versa.
(iv)
(b)
The magnetic susceptibility of ferromagnetic substances does not change according to curie law. Curie temperature (TC): (i) The temperature above which a ferromagnetic material behaves like a paramagnetic material is defined as curie temperature (TC). (ii) The minimum temperature at which a ferromagnetic substance is converted into paramagnetic substance is defined as curie temperature. (iii) For various ferromagnetic materials its values are different. e.g. for Ni TCNi = 358°C for Fe TCFe = 770°C for (iv)
5.
CO
TCCO = 1120°C
At this temperature the ferromagnetism of the substances suddenly vanishes.
CURIE-WEISS LAW
–T Cure c
T
TC
6.
OTHER IMPORTANT FEATURES AND FORMULAE
(A)
Angle of declination and Geographical meridian: Geographical meridian
BH
Angle of declinetion
f q
Angle of dip
B
BV Magnetic meridian
(i) (ii)
The horizontal component of earth's magnetic field is from S to N. If at any place the angle of dip is and magnetic latitude is then, tan = 2 tan The total intensity of earth's magnetic field I = I 0 1 + 3 sin2 l
M R3 Here M and R are the magnetic moment of bar magnet of earth and radius of earth respectively. At magnetic equator of earth = 0 and at poles = 90. Ipole = 2I equator here
(iii)
I0 =
(B)
(iv)
At the poles and equator of earth, the values of total intensity are 0.66 and 0.33 oersted respectively.
(i) (ii) (iii)
In vacuum :– B0 = 0H In medium :– B = H Resultant magnetic field due to and H :– (a) B = B1 + BH = 0 + 0H B = 0 ( + H) B = 0H (1 + /H) or B = 0 (1 + )H + = H (a) = 0 [1 + ] (b) = 0r B (c) m= H m (a) mr = m0 (b) mr = 1 + c (a) c = mr - 1
(iv)
(v)
(vi)
I H (vii) The magnetic potential due to a small magnet at a point distant r is given by :– Mcos q V= r2 (viii) The mutual interaction force between two small magnets of moments M 1 and M2 is given by 6M1M2 F=K r4 Magnetic torque (a) = MB sin (b) = BiNA sin t =M´ B (c) (b)
(C)
(D)
Magnetic potential energy (UB) (a) UB = M ×B (b)
7.
c=
UB = M ×B (1- cos q)
ASSIGNMENT
1.
The magnetism of atomic magnet is due to (A) only spin motion of electrons (B) only orbital motion of electrons (C) both spin and orbital motion of electrons (D) the motion of protons. Solution: (C) The magnetism of atomic magnet is due to both spin an orbital motions of electrons. 2.
The earth's magnetic field inside an iron box as compared to that outside the box, is-
(A) Solution:
less
(B)
more
(C)
zero
(D)
same.
(A) The earths magnetic field inside an iron box is less than that outside the box.
The magnetic susceptibility of a paramagnetic materials varies with absolute temperature T as cons tan t . T (A) (B) (C) (D) eT T 1 Solution: (C) The magnetic susceptibility of a paramagnetic substance is inversely 1 proportional to the absolute temperature i.e. T 3.
4.
There are two points A and B on the extended axis of a 2 cm long bar magnet. Their distances from the centre of the magnet are x and 2x respectively. The ratio of magnetic fields at points A and B will be(A) 8 : 1 approximately (B) 4 : 1 (approximately) (C) 4:1 (D) 8 : 1.
Solution:
(A) The intensity of magnetic field on the axis 2m B 0 3 4 r 3
B r 2x 1 2 8 :1 B2 r1 x 5.
3
The resultant magnetic moment of neon atom will be(A)
Solution:
B . 2 (B) Neon atom is diamagnetic, hence its net magnetic moment is zero.
infinity
(B)
zero
(C)
mB
(D)
A magnetic material of volume 30 cm3 is placed in a magnetic field of intensity 5 oersted. The magnetic moment produced due to it is 6 amp-m2. The value of magnetic induction will be(A) 0.2517 Tesla (B) 0.025 Tesla (C) 0.0025 Tesla (D) 25 Tesla. Solution: (A) B = m0 (I + H) M 6 2 155 amp / m I= 6 V 30 10 5 amp / m H = 5 oersted = 4 10 3 –7 m0 Wb/amp-m 5 B 4 10 7 2 105 3 4 10 = 0.2517 Tesla 6.
The mass of an iron rod is 80 gm and its magnetic moment is 10 A–m2. If the density of iron is 8gm/C.C. hen the value of intensity of magnetisation will be(A) 106 A/m (B) 104 A/m (C) 102 A/m (D) 10 A/m M Md Magnetic moment density I Solution: (A) I V m mass
7.
10 8 103 80 103 = 106 A/m
8.
The main difference between electric lines of force and magnetic lines of force is(A) Electric lines of force are closed curves whereas magnetic lines of force are open curves. (B) Electric lines of force are open curves whereas magnetic lines of force are closed curves. (C) Magnetic lines of force cut each other whereas electric lines of force do not cut. (D) Electric lines of force cut each other whereas magnetic lines of force do not cut. Solution: (B) The magnetic lines of force are in the form of closed curves whereas electric lines of force are open curves.
9.
The mass of a specimen of a ferromagnetic material is 0.6 kg. and its density is 7.8×103 kg/m3. If the area of hysteresis loop of alternating magnetising field of frequency 50 Hz is 0.722 MKS units then the hysteresis loss per second will be(A) 277.7×10–5 Joule (B) 277.7×10–6 Joule (C) 277.7×10–4 Joule (D) 277.7×10–4 Joule. Solution: (A) W H = VAft m Aft d 0.6 0.722 50 or WH 7.8 103 = 277.7 × 10–5 Joule
The horizontal component of flux density of earth's magnetic field is 1.7×10 –5 tesla. The value of horizontal component of intensity of earth's magnetic field will be? (A) 24.5 A/m (B) 13/5A/m (C) 0.135 A/m (D) 1.35 A/m. 5 2 B 1.7 10 Wb / m = 13.5 A/m Solution: (B) H 0 4 10 7 Wb / A m 10.
A magnetising field of 2×103 an iron rod. The relative permeability of the rod will be(A) 102 (B) 10o (C) 104 B Solution: (C) r 0 H0 8 r 104 or 3 2 10 4 10 7
11.
(D)
101
In a hydrogen atom the electron is revolving in a circular path of radius 5.1×10 –11m with a frequency 6.8×1015 Hz. The equivalent magnetic moment will be(A) 8 × 10–24 A–m2 (B) 8 × 10–22 A–m2 –20 2 (C) 8.9 × 10 A–m (D) 8 × 10–18 A–m2 2 Solution: (A) –19 or M = 1.6×10 ×6.8×1015×3.14×(5.1×10–11)2 = 8.9 ×10–24 A–m2
12.
13.
A cube of side l is placed in a magnetic field of intensity B. The magnetic flux emanating out of it will be(A) zero (B) Bl2 (C) 2Bl2 (D) 6Bl2
Solution: (A) The net flux coming out of a closed surface is zero. Hence the flux coming out of the cube will be zero. 14.
A circular disc of area (4iˆ 5ˆj) 10 3 m2 is placed in a uniform magnetic field of intensity (0.2iˆ 0.3ˆj) Tesla. The flux crossing the disc will be-
(A) (C) Solution:
23 Weber 23×10–3 Weber (C) B.A (0.2iˆ 0.3ˆj).[0.004iˆ 0.005ˆj]
(B) (D)
23×10–2 Weber 23×10–4 Weber
= 0.0008 + 0.0015 = 0.0023 Weber = 23×10–4 Weber 15.
The total magnetic flux in a material, which produces a pole of strength mp when a magnetic material of cross-sectional area A is placed in a magnetic field of strength H, will be(A) 0 (AH + mp) (B) 0 AH (C) 0 mp (D) 0 [mp AH + A] Solution: (A) = BA ....(A) ....(B) B 0 (I H) From eqs. (A) and (B) 0 (I H)A m I p .....(C) A From eqs. (B) and (C) m 0 A p H 0 [mp AH] A
The relation between and H for a specimen of iron is as follows 0.4 12 10 –4 Henry / meter H The value of H which produces flux density of 1 Tesla will be(A) 250 A/m (B) 500 A/m (C) 750 A/m (D) 103 A/m Solution: (B) B = H 0.4 12 1004 H or B H –4 or 1 = 0.4 + 12×10 H H = 500 A/m
16.
17.
The correct curve between X and
1 for paramagnetic materials isT
(A)
(B)
1/T
(C)
1/T
(D)
1/T
Solution:
(A)
X
1/T
1 T
18.
The SI unit of magnetic flux is(A) Weber (B) Maxwel (C) 2 m Solution: (A) BA Weber 2 Weber m
Tesla
(D)
Gauss
19.
The intensity of magnetic field due to an isolated pole of strength m p at a point distant r from it will bemp mp r2 2 (A) (B) m r (C) (D) p mp r2 r Solution: (A) The magnetic intensity due to an isolated pole of strength mp at a distance r = mp r2
20.
A uniform magnetic field is directed from left towards right in the plane of paper. When a piece of soft iron is placed parallel to the field. The magnetic lines of force passing through it will be(A)
(B)
(C)
(D)
Solution: (C) The magnetic lines of force in a ferrmagnetic material are crowded together. 21.
A bar magnet of magnetic moment M is cut into two equal parts. The magnetic moment of either of the parts will beM (A) (B) 2M (C) 2M (D) I H2 2 l M Solution: (A) M ml M' m 2 2 22.
A loop of area 0.5 m2 is placed in a magnetic field of strength 2 Tesla in direction making an angle of 60o with the field. The magnetic flux linked with the loop will be1 3 3 Weber Weber (A) (B) 2 Weber (D) Weber (C) 2 2 1 1 Solution: (A) BA cos , 2 cos 60o 2 2 23.
The force experienced by a pole of strength 100 A-m at a distance of 0.2m from a short magnet of length 5 cm and pole strength of 200A-m on its axial line will be (A) 2.5×10–2 N (B) 2.5×10–3N (C) 5.0×10–2N (D) 5.0×10–3N. 2m'l Solution: (A) F mB 0 3 m 4 x 7 10 2 200 0.05 100 8 103 –2 = 2.5×10 N
The magnetic susceptibility of a paramagnetic substance is 3×10 –4. It is placed in a magnetising field of 4×104 amp/m. The intensity of magnetisation will be(A) 3 108 A/M (B) 12 108 A/M (C) 12 A/M (D) 24 A/M –4 3 Solution: (C) I = XH = 3×10 ×4×10 = 12 A/m
24.
25.
Volt-second is the unit of(A) B (B) d or et volt sec Solution: (B) e dt
(C)
I
(D)
x
The volume susceptibility of a magnetic material is 30×10–4. Its relative permeability will be(A) 31 10–4 (B) 1.003 (C) 1.0003 (D) 29 10–4 1 X 1 30 10 4 1.003 Solution: (B) r 0
26.
27.
A current of 2 ampere is passed in a coil of radius 0.5 m and number of turns 20. The magnetic moment of the coil is(A) 0.314 A–m2 (B) 3.14 A–m2 (C) 314 A–m2 (D) 31.4 A–m2 2 Solution: (D) N = 2×3.14×0.25×20 = 31.4 A–m2
28.
The value of magnetic susceptibility for super-conductors is(A) zero (B) infinity (C) +1 (D) Solution: (D) For superconductor r = 1 + X = 0 X = –1
–1
29.
A current of 1 ampere is flowing in a coil of 10 turns and with radius 10 cm. Its magnetic moment will be(A) 0.314 A-m2 (B) 3140 A-m2 (C) 100 A-m2 (D) 0 A-m2 Solution: (A) M = iA = R2 Ni = 3.14×0.01×10×1=0.314 Am2
The magnetic moment of a magnet of mass 75 gm is 9×10 –7 A-m2. If the density of the material of magnet is 7.5×103 kg/m3 then intensity of magnetisation will be(A) 0.9 A/m (B) 0.09 A/m (C) 9 A/m (D) 90 A/m –7 3 M Md 9 10 7.5 10 Solution: (B) I 0.09 A / m V m 75 10 –3 30.
For test 1.
The area of hysteresis loop of a material is equivalent to 250 Joule. When 10 kg material is magnetised by an alternating field of 50Hz then energy lost in one hour will be if the density of material is 7.5 gm/cm3. (A) 6×104 Joule (B) 6×104 Erg (C) 3×102 Joule (D) 3×102 Erg. m Solution: (A) E nA Vt nA t d 50 250 10 3600 6 10 4 Joule 3 7.5 10
2.
The corecivity of a bar magnet is 100 A/m. It is to be demagnetised by placing it inside a solenoid of length 100 cm and number of turns 50. The current flowing the solenoid will be(A) 4A (B) 2A (C) 1A (D) zero. H 100 i 2A Solution: (B) H = ni n 50
3.
A current i is flowing in a conductor of length l. When it is bent in the form of a loop then its magnetic moment will be4 l2i l2 (A) (B) (C) (D) 4l2i. l2i 4 4 l2 il2 Solution: (A) M = iA = i r2 But 2r = l M i 2 4 4
4.
A rod of ferromagnetic material with dimensions 10 cm×0.5 cm×0.2 cm is placed in a magnetic field of strength 0.5×104 amp/m as a result of which a magnetic moment of 5 amp-m2 is produced in the rod. The value of magnetic induction will be(A) 0.54 Tesla (B) 0.358 Tesla (C) 2.519 Tesla (D) 6.28 Tesla. M Solution: (D) B 0 (I H) 0 H V 5 4 3.14 10 7 6 5000 6.28 Tesla . 10
5.
The ratio of intensities of magnetic field in the axial and equatorial positions of a magnet will be(A) 1:4 (B) 4:1 (C) 1:2 (D) 2:1 Solution: (D) 0 2M M2 Ba 4 3 2 :1 0 M Bc 4 r 3 r O
r
M
6.
The resultant magnetic moment due to two current (i) carrying concentric coils of radius r, mutually perpendicular to each other will be(B) (C) (D) (A) 2ir 2 2r 2 2ir 2 2ir Solution: (A) M' = 2M = 2r 2i The magnetic susceptibility of a paramagnetic material at –73oC is 0.0075 then its value at –173oC will be(A) 0.0045 (B) 0.0030 (C) 0.015 (D) 0.0075. 1 Solution: (C) For paramagnetic materials X T X T 0.0075 200 2 1 or X2 0.015 X1 T2 100 7.
8.
The area of cross-section of three magnets of same length are A, 2A and 6A respectively. The ratio of their magnetic moments will be(A) 6:2:1 (B) 1:2:6 (C) 1 : 4 : 36 (D) 36 : 4 : 1 Solution: (B) M µ m µ A (Area of cross-section) M1 : M2 : M3 = 1: 2 : 6 9.
If the radius of a circular coil is doubled and the current flowing in it is halved then the new magnetic moment will be if its initial magnetic moment is 4 units(A) 8 units (B) 4 units (C) 2 units (D) zero. 2 M = p R Ni Solution: (A) 2
M1 æR1 ö = M2 çè R2 ÷ ø
æi1 ö 4 1 çè i ø ÷ or M = 4 ´ 2 \ M2 = 8 units 2 2
10.
Newton/Weber is the unit of
(A) Solution:
(B) M F i f = =H (A) F = Bil = il or A f l H
(C)
B
(D)
11.
The inner and the outer radii of a toroid are 9cm and 11cm respectively and the number of turns in it is 3140. A magnetic field of 2.5 Tesla is produced in it when a current of 0.5 ampere is passed in it. The permeability of core material is (in Henry/meter)(A) 10–1 (B) 10–2 (C) 10–3 (D) 10–4 B B 2.5 ´ 2 ´ 3.14 ´ 0.1 Solution: (C) m= = = = 10- 3 Henry / meter Ni H 3140 ´ 0.5 2p r 12.
In the above problem the relative permeability of the core will be(A) 684.5 (B) 864.7 (C) 369.4 (D) -3 m 10 = = 796.2 Solution: (D) mr = m0 12.57 ´ 10- 7
796.2
A magnetising field of 5000 A/m produces a magnetic flux of 5×10 –5 Weber in an iron rod. If the area of cross-section of the rod is 0.5 cm2, then the permeability of the rod will be (in henry/m)(A) 1 10–3 (B) 2 10–4 (C) 3 10–5 (D) 4 10–6 f B = = mH A Solution: (B) 5 ´ 10- 5 f \ m= = = 2 ´ 10- 4 Henry / m 3 -4 AH 0.5 ´ 10 ´ 5 ´ 10 13.
14.
In the above problem the magnetic susceptibility of the rod will be(A) 158.2 (B) 198.0 (C) 295.3 (D) 343.6 Solution: (A) m mr = = 1+ X m0 \
X=
m 2 ´ 10- 4 - 1= - 1 = 159.2 - 1 = 158.2 m0 12.56 ´ 10- 7
15.
In the above problem the intensity of magnetisation will be (in A/m)(A) 7.9×102 (B) 7.9×102 (C) 7.9×102 (D) 7.9×105 Solution: (D) I = XH = 158.2 ´ 5000 = 7.9 ´ 106 Amp / m
16.
At any place on earth, the horizontal component of earth's magnetic field is the vertical component. The angle of dip at that place will be(A) 60o (B) 45o (C) 90o (D) 30o
3 times
Solution: =
(D) tan q = 1 3
= tan30° \
BV BV = BH 3 BV
BH q
q = 30°
BV
The horizontal component of earth's magnetic field at any place is 0.36×10 –4 Weber/m2. If the angle of dip at that place is 60o then the value of vertical component of earth's magnetic field will be-(in Wb/m2)(A) 0.12 10–4 (B) 0.24 10–4 (C) 0.40 10–4 (D) 0.62 10–4 Solution: (D) BV = BH tan q = 0.36 ´ 10 - 4 tan60° 17.
= 0.36 ´ 10- 4 ´
18.
3 = 0.62 ´ 10 - 4 Wb / m2
The value of angle of dip at a place on earth is 45 o. If the horizontal component of earth's magnetic field is 5×10–5 Telsa then the total magnetic field of earth will be(A) (B) 5 2 105 Tesla 10 2 105 Tesla (D) zero. (C) 15 2 105 Tesla
Solution:
(A) B = B2V + BH2 = BH2 tan2 45° + BH2 = BH 2
= 5 2 ´ 10- 5 Wb / m2
19.
The ratio of intensities of magnetic field, at distances x and 2x from the centre of a magnet of length 2cm on its axis, will be(A) 4:1 (B) 4 : 1 approx (C) 8:1 (D) 8 : 1 approx. m m 2M 2M » 0 3 Solution: (D) B = 0 2 2 3/2 4 p (l + x ) 4p x 3
\
B1 æx 2 ö = = 8 : 1 approximately B2 çè x1 ÷ ø
The length of a bar magnet is 10 cm and its pole strength is 10 –3 Weber. It is placed in a magnetic field of induction 4 10–3 Tesla in a direction making an angle of 30o with the field direction. The value of orque acting on the magnet will be(A) 2 10–7 N-m (B) 2 10–5 N-m (C) 0.5 102 N-m (D) 0.5 N-m. Solution: (A) t = MBsin q = m/Bsin q -3 = 10 ´ 0.1 ´ 4p ´ 10- 3 ´ 0.5 = 2p ´ 10- 7 N - m
20.
A magnetic needle of magnetic moment 60 amp-m2 experiences a torque of 1.210–3 N-m directed in geographical north. If the horizontal intensity of earth's magnetic field 2 , then the angle of declination will be(A) 30 (B) 45 (C) 60 (D) 90 Solution: (A) t = MBsin q 1.2 ´ 10- 3 1 t \ sin q = = = \ q = 30° -6 MB 60 ´ 40 ´ 10 2
21.
22.
The ratio of total intensities of magnetic field at the equator and the poles will be-
(A) 1:1 (B) 1:2 (C) 2:1 (D) Solution: (A) Total intensity of magnetic field remains constant Be \ = 1: 1 Bp
1:4
The magnetic flux density of magnetic field is 1.5 Wb/m 2. A proton enters this field with a velocity of 2×107 m/s in a direction making an angle of 30o with the field. The force acting on the proton will be(A) 2.4 10–12 Newton (B) 0.24 10–12 Newton (C) 24 10–12 Newton (D) 0.24 10–12 Newton Solution: (A) F = qvBsin q = 1.6 ´ 10- 19 ´ 2 ´ 107 ´ 1.5 ´ 0.5 = 2.4 ´ 10 - 12
23.
24.
The intensity of magnetic field at a point X on the axis of a small magnet is equal to the field intensity at another point Y on its equatorial axis. The ratio of distances of X and Y from the centre of the magnet will be(A) (B)–3 (B) (B)–3 (C) 23 (D) 21/3 m 2M m M dx = 21/ 3 Solution: (D) 0 3 = 0 3 \ 4p d x 4p d y dy
25.
Two magnets A and B are equal in length, breadth and mass, but their magnetic moments are different. If the time period of B in a vibration magnetometer is twice that of A, then the ratio of magnetic moments will be1 (B) 2 (C) 4 (D) 12 (A) 2 1 I Solution: (C) T = 2p \ Mµ 2 MB T 2
\ 26.
MA æTB ö æ2 ö = = ç ÷= 4 MB çè TA ÷ ø è1ø
The period of oscillation of a freely suspended bar magnet is 4 second. If it is cut into two equal parts lengthwise then the time period of each part will be(A) 4 sec. (B) 2 sec. (C) 0.5 sec. (D) 0.25 sec.
Solution:
I ml2 = 2p = 4 sec (A) T = 2p MB 12 ´ m / B
27.
The time period of a small magnet in a horizontal plane is T. Another magnet B oscillates at the same place in a similar manner. The size of two magnets is the same but the magnetic moment of B is four times that of A. The time period of B will beT T (A) (B) (C) 2T (D) 4T 4 2 MA TB MA T Solution: (B) = or TB = T ´ = MB TA 4MA 2
28.
The value of current enclosed by a circular path of radius 0.30 cm is 9.42 ampere. The value of magnetic field along the path will be(A) 500 amp/m (B) 1000 Amp/m
(C) Solution:
5 104 Amp/m (D) Zero i 9.42 (A) H = = = 500 amp / m 2pr 2 ´ 3.14 ´ 3 ´ 10- 3
A magnetic wire is bent at its midpoint at an angle of 60o. If the length of the wire is l and its magnetic moment is M then the magnetic moment of new shape of wire will beM M (A) 2M (B) M (C) (D) 2 2 1 M Solution: (C) M' = m × = 2 2 l/2 l/2 29.
l/2
30.
A current of 2 ampere is flowing in a coil of radius 50 cm and number of turns 20. The magnetic moment of the coil will be(A) 3.14 amp-m2 (B) 31.4 amp-m2 (C) 314 amp-m2 (D) 0.314amp-m2 Solution: (B) M = p R2 Ni = 3.14 ´ 0.25 ´ 20 ´ 2 = 3.14 amp / m2
1.
1000 turns per meter are wound over a Rowland ring of ferromagnetic material. On passing a current of 2 ampere in the coil, a magnetic field of 10 Wb/m2 is produced in it. The magnetising force generated in the material will be(A) 1.2 10–3 A/m (B) 2.6 10–3 A/m –4 (C) 0.6 10 A/m (D) 2 10–3 A/m Ni = ni = 103 ´ 2 = 2 ´ 103 amp / m Solution: (D) H = 2p r
2.
A magnet makes 10 oscillation per minute at a place where the horizontal component of earth's magnetic filed (H) is 0.33 oersted. The time period of the magnet at a place where the value of H is 0.62 oersted will be(A) 4.38 S (B) 0.38 S (C) 2.38 S (D) 8.38 S H 1 0.33 Solution: (A) T µ \ T2 = T1 1 = 6 ´ = 4.38 s H2 0.62 H
The magnetic induction inside a solenoid is 6.5×10 –4T. When it is filled with iron medium then the induction becomes 1.4T. The relative permeability of iron will be(A) 1578 (B) 2355 (C) 1836 (D) 2154. B 1.4 Solution: (D) mr = = = 2154 B0 6.5 ´ 10- 4 3.
1.
FARADAY’S LAWS OF ELECTROMAGNETIC INDUCTION
According to Faraday’s Law, whenever the magnetic flux through a circuit changes, an emf is induced in the circuit. The magnitude of the induced emf is equal to the rate at which flux changes with time. d Magnitude of the induced emf = dt If represents the flux through a single turn, and the loop has N turns, then d Induced emf = N dt This induced emf creates an induced current in the circuit whose magnitude is given as 1 d induced emf i= = R dt net resis tance of circuit
2.
LENZ’S LAW
According to Lenz’s law, induced emf in a circuit opposes the cause due to which it was induced. Consider the following examples. (a) Suppose that the north-pole of a bar magnet is moved N towards a conducting wire loop as shown in the figure. s Due to a change in the magnetic flux associated with the loop, a current is induced. Due to induced current, a magnetic field is induced and this magnetic field opposes the motion of bar magnet. The direction of the induced current can be deduced by the following argument: the north pole is moving towards the loop; therefore to oppose the motion of the bar magnet only a north pole will be induced on that face of the loop which faces the magnet. (b) A rectangular loop ABCD is being pulled out of the x x x x x x OBx x B x x x x x xA x x magnetic field which is directed into the plane of the x x x x x x x x paper. Perpendicular to the plane of the paper. As the xx xx xx xx xx xx xx xx loop is dragged out of the field, the flux associated with x x x x x xD x x C the loop decreases. The induced current flows in the loop in a sense so as to oppose the decrease in this flux. For this to happen the magnetic field due to the induced current in the loop must be directed into the plane of the paper. Thus the current in the loop must flow be clockwise.
Illustration 1:
Solution:
Magnetic field is increasing into the page with time when a conducting loop of definite radius is placed on the plane of the paper. The find the direction of current in the loop.
x
x
x
x x
x
x B x
x
x
x
x
x
x x
As the flux is increasing inside then the current in the loop will be such that it will be opposing the increase in magnetic field, i.e., the induced current in the loop will create such a magnetic field which is directed out ward.
Thus the direction of current will be anticlockwise. Illustration 2:
Solution:
A square loop ACDE of area 20cm2 and X XB X resistance 5 is rotated in a magnetic field B = A C 2T through 180 X X X (a) in 0.01 s and E (b) in 0.02 s D X X X Find the magnitude of e, i and q in both the cases. Let us take the area vector S perpendicular to plane of loop inwards. So initially, S B and when it is rotated by 180, S B Hence, initial flux passing through the loop, i = BS cos 0 = (2) (20 10-4) (1) = 4.0 10-3 Wb Flux passing through the loop when it is rotated by 180, f = BS cos 180 = (2) (20 10-4) (-1) = - 4.0 10-3 Wb Therefore, change in flux, B = f - i = - 8.0 10-3 Wb (a) Given t = 0.01 s, R=5 8.0 103 e B 0.8 volt 0.01 t i
e
R and
(b)
0.8 0.16 A 5 q = it = 0.16 0.01 = 1.6 10-3 C
t = 0.02 s
e e
B 8.0 103 0.4 volt 0.02 t
0.4 0.08 A R 5 and q = it = (0.08) (0.02) = 1.6 10-3 C Note: Time interval t in part (b) is two limes the time interval in part (a), so e and i are half while q is same. i
Illustration 3: A conducting circular loop with variable radius r is placed in a uniform magnetic field B=0.020T with plane perpendicular to the field. While the radius of the loop is contracting at a constant rate of 1.0 mm/s; find the induced emf in the loop when the radius is 2 mm. Solution:
Radius (variable) = r B = 0.02 T Induced emf E = dr 1.0 10 3 mt dt
dr d dA = B 2r (BA ) B dt dt dt
r = 2 10-3 mt
E = 0.02 2 2 10-3 1 10-3
= 6 10-8 = 18.85 10-8 V. Illustration 4:
Solution:
A rectangular loop of N turns of area A and resistance R rotates at a uniform angular velocity about Y-axis. The loop lies in a uniform magnetic field B in the direction of X-axis. Assuming that at t = 0, the plane of the loop is normal to the lines of force, find an expression for the peak value of the emf and current Induced in the loop. What is the magnitude of torque required on the loop to keep it moving with constant ? As is maximum at t = 0, (t) = BA cos t Magnitude of induced emf = N d / dt
BAN Sin t Magnitude of induced current =
BAN Sin t R
(a) peak value of emf = BAN peak value of induced current = BAN/R (b) Power input = heat dissipation per sec I2R B2 A 22N2 sin2 t R
3.
MOTIONAL EMF
Induced emf due to motion of a conductor in magnetic field Suppose a conductor is moving with velocity v in a uniform magnetic field B . We take a small element PQ = d of the conductor, then induced e.m.f. in the element. B e (v B).d VQ VP P d v Q or, e = (vB sin). d cos Where is the angle between v and B and is the angle between element and direction of force Illustration 5:
Solution:
A metal rod 1.5 m long rotates about its one end in a vertical plane at right angles to the magnetic meridian. If the frequency of rotation is 20 rev/s. Find the emf induced between the ends of rod. BH = 0.32 10-4T: Horizontal component of earth’s magnetic field. Length of rod = 1.5 mt. Frequency of rotation = 20 rev/sec. BH = 0.32 10-4 T Angular velocity of rod = 20 2 rad/sec. emf induced
=B
dA = 0.32 10-4 20 (1.5)2 dt
= 4.5 10-3 V
4.
INDUCTANCE
4.1
SELF INDUCTION
We have already discussed capacitors – devices that store energy using electric fields. Like a capacitor, an inductor is also quite a commonly used element in electric circuits. It stores magnetic energy. As we know that when current flows through a conductor a magnetic field is set-up in surrounding of it, and hence it is associated with magnetic flux. If magnetic flux associated with a coil is and current flowing through it is I, then its inductance is given by the expression L
. I
The quantity 'L' is called self-inductance of
the coil. It does not depend on the current, but it depends on the permeability of the core and the dimensions of the coil. S.I. unit of inductance is Henry. Consider the circuit, in which a solenoid is connected across a cell through a resistor. When the switch is open, the current in the circuit is zero. When the switch is closed, a current flow in it. Since current in the circuit increases from zero to a certain value, N magnetic field associated with it changes that causes induction of an emf across the solenoid. Induction of an emf due to variation in current flowing through the coil itself is known as self induction. Since B = LI, and =
dB dt
L
dI . dt
Inductance of an ideal solenoid: Let a current I flow through a solenoid. The magnetic field due to the current flowing within the solenoid is, B = 0nI, where n is the number of turns per unit length. If area of cross section of the solenoid is A then flux associated with length is equal to = nBA. (Assuming that the solenoid is ideal and long) where is the length of the solenoid. Now B = nI (n )(0n A) = 0 n2 A Self Inductance of a Coil Consider a coil of N turns and area of crossection A carrying a current i. The length of the coil is ( A). iN N B A N 0 A 2 NA 0 i
Comparing with = L i, we get: L =
4.2
R - L CIRCUIT
GROWTH OF CURRENT
i
0N2 A
i
L
R
When switch is on at t = 0. Then current starts from 0 to I. And self inductor opposes the grow of current L
LdI IR 0 dt I t LdI L I 1 LdI IR dI t dt 0 dt R 0I IR 0 R
I
I
L I / R Rt In I t In L R R 0 / R I Rt Rt R L I = 1 e L e /R R Rt dI L e dt L Rt LdI potential difference across inductor = e L dt Rt I = I e L R At t = 0, I = 0 At t I R
Rt
V= e L At t = 0 potential difference across inductor = At t potential difference across inductor = 0. So inductor will behave like plane wire. Consider graph betn I and time. (Amp.)I
R t (sec.)
Consider graph betn potential difference across Inductor and time
R
(volt.)V
t (sec.)
Illustration 6:
A coil of inductance 1.0 H and resistance 100 is connected to a battery of emf 12 V. Find the energy stored in the magnetic field of the coil 10 ms after the circuit is switched on.
Solution:
L = 1.0 H E = 12 V
R = 100
Rt E i(t) = 1 e L R
Energy stored in the magnetic field is 1/2 L i2. (12)2 1 = 1.0 2 10 4
= Illustration 7: Solution:
100 10 10 3 1 e 1
2
2
1 1 28.8 mJ 4 e 2 10 144
A coil of metal wire is kept stationary in a non-uniform magnetic field. An e.m.f. is induced in the coil. For induced emf to develop in a coil the magnetic flux through it must change. But in this case the number of magnetic lines of force through the coil is not changing. Therefore the statement is false.
4.3
INDUCTANCE OF COMMON ELEMENTS
1.
Wire L= 0 8 2a
2.
3.
Hollow cylinder 2 L = 0 In 1 2 a >> a Parallel wires d L = 0 In a >> d, d >> a
2a
2a d
4.
Coaxial conductor b L = 0 In a a b
5.
o
6.
Circular loop 4 L = 0 In 2.45 2 d = –2 o,o >> d
d
Solenoid 0N2S L= a
>> a 7.
Torus (of circular cross section) L = 0N2[o –
02 a2 ] o
8.
Sheet L = o 2
2 In b t 0.5 t b
4.4
MUTUAL INDUCTION
Consider two coils C1 and C2 placed as shown. By varying current i1 in coil C1, we change the flux not only through C1 but also through coil C2. The change in flux 2 through C2 (due to change in current i1) induces an emf in the coil C2. This emf is known as mutually induced emf and the process is known as mutual induction. 2 i1 2 = M i1
C1 I1
primary
Where M is called as mutual inductance of the pair of coils. The coil C 1 in which i varies is often called primary coil and the coil C2 in which the emf is C1 2 induced is called secondary coil. I1 induced emf in coil C2 = E2 E2
d2 di M dt dt
a secondary
The mutual inductance is maximum when the coils arc wound up on the same axis. It is minimum when the axes of coils are normal to each other. Note down the following points regarding the mutual inductance:
1.
The SI unit of mutual inductance is henry (1H).
2.
M depends upon closeness of the two circuits, their orientations and sizes and the number of turns etc.
3.
Reciprocity theorem: M21 = M12 = M e2 = - M(di1/dt) and N2B2 M12 and i1
4.
e1 = - M(di2/dt) N1B1 M21 i2
A good approach for calculating the mutual inductance of two circuits consists of the following steps: (a) Assume any one of the circuits as primary (first) and the other as secondary (second). (b) Suppose a current i1 flows through the primary circuit. (c) Determine the magnetic field B produced by the current i1. (d) Obtain the magnetic flux B2 . (e)
With the flux known, the mutual inductance can be found from, N2B2 M i1
Illustration 8:
Solution:
A long solenoid of length 1 m, cross sectional area 10 cm2, having 1000 turns has wound about its centre a small coil of 20 turns. Compute the mutual inductance of the two circuits. What is the emf in the coil when the current in the solenoid changes at the rate of 10 Amp/s? Let N1 = number of turns in solenoid; N2 = number of turns in coil A1 and A2 be their respective areas of crossection. (A1 = A2 in this problem) Flux 2 through coil crated by current i1 in solenoid is 2 = N2(B1A2) iN NN A 2 N2 0 1 1 A 2 2 0 1 2 2 i2 Comparing with 2 = M i1; we get NN A Mutual inductance = M = 0 1 2 2 4 107 1000 20 10 104 2.51 105 H 1 di Magnitude of induced emf = E2 = M 1 dt M
Illustration 9:
Solution:
The coefficient of mutual induction between the primary and secondary of a transformer is 5 H calculate the induced emf in the secondary when 3 Amp current in the primary is cut-off in 2.5 10-4 sec. M = coefficient of mutual induction = 5 H.
3 Amp current is a cut off in 2.5 10-4 sec. E = M
4.5
di 3 = -6 104 V 5 4 dt 2.5 10
L–C CIRCUIT
L.C. OSCILLATIONS A capacitor is charged to a potential difference of V o by connecting it across a battery and then is allowed to discharge through an inductor of inductance L. Initial charge on the plates of the capacitor qo = CVo At any instant, let the charge flown through the circuit be q and the current in the circuit be i. Applying Kirchhoff’s law di q0 q L =0 C dt
qo-q -(qo-q)
C
Differentiating w.r.t. time we get dq d2 i LC 2 = 0 dt dt 2 1 di i = 2i, = 2 LC dt
I
f=
1 2 LC
The charge q on the plates of the capacitor and current I in the circuit vary sinusoidally as q = q0 sin (t + ) and I = q0 cos (t + ). where is the initial phase and it depends on initial conditions of the circuit and
=
1 LC
.
The total energy of the system remains conserved 1 1 1 1 CV 2 Li 2 = constant = CV02 Li 02 2 2 2 2
Illustration 10: A–2–F capacitor is initially charged to 20 V and then shorted across a 6–H inductor. What are the frequency of oscillation and the maximum value of the current? Solution: The frequency of oscillation is independent of the initial charge and depends only on the values of the capacitance and inductance. The frequency is 1 1 1 f= 2 2 LC 2 LC 1 = = 4.59 104 Hz 6 6 2 (6 10 )(2 10 ) According to equation the maximum value of the current is related to the maximum value of the charge by Q0 Im = Q0 = LC The initial charge on the capacitor is Q0 = CV0 = (2 F)(20 V) = 40 C Thus 40 C Im = = 11.5 A (6H)(2 F)
4.6
MASS–SPRING SYSTEM VS INDUCTOR–CAPACITOR CIRCUIT
A comparison of oscillations of a mass spring system and an L - C circuit. S. No.
Mass Spring System
1. 2.
Displacement (x) Velocity (v)
3.
Acceleration (a)
11.
d2 x k 2 x, where 2 dt m x = A sin (t ) or x = A cos (t ) dx v A 2 x2 dt dv a 2 x dt 1 Kinetic energy = mv2 2 1 Potential energy = kx 2 1 1 1 mv2 + kx2 = constant = kA2 = 2 2 2 1 2 mv max 2 vmax A
12.
amax 2 A
4. 5. 6. 7. 8. 9.
10.
13. 14.
5.
Inductor-Capacitor Circuit Charge (q) Current (i) di Rate of change of current dt 2 dq 1 2q, where 2 dt LC q = q0 sin (t ) or q = q0 cos (t ) dx i q02 q2 dt di 2 q Rate of change of current dt 1 Magnetic energy = Li 2 1 q2 Potential energy = 2 C 2 1 2 1 q 1 q02 Li + = constant = = 2 2 C 2 C 1 2 Limax 2 imax = q0
di 2q0 dt max
1 k m
C L
ALTERNATING CURRENTS AND CIRCUITS
The basic principle of the ac generator is a direct consequence of Faraday’s law of induction. When a conducting loop is rotated in a magnetic field at constant angular frequency a sinusoidal voltage (emf) is induced in the loop. This instantaneous voltage is, V = V0 sin t
(i)
The usual circuit diagram symbol for an ac source is shown in Figure.
In Equation (i) V0 is the maximum output voltage of the ac generator, or the voltage amplitude and is the angular frequency, equal to 2 times the frequency f. = 2f
The frequency of ac in India is 50 Hz, i.e., f = 50 Hz So, = 2f 314 rad/s The time of one cycle is known as time period T, the number of cycles per second the frequency f. 1 2 T or T f A sinusoidal current might be described as, i = i0 sin t If an alternating current is passed through an ordinary ammeter or voltmeter, it will record the mean value for the complete cycle, as the quantity to be measured varies with time. The average value of current for one cycle is, T
i
(iav )T
one cycle
0 T 0
Thus, i
idt
2 /
(i0 sin t)dt
0
dt
2 /
0
0
dt
0
one cycle
Similarly, the average value of the voltage (or emf) for one cycle is zero. V
one cycle
0
Since, these averages for the whole cycle are zero, the dc instrument will indicate zero deflection. In ac, the average value of current is defined as its average taken over half the cycle. Hence, T/2
i
Half cycle
(iav )T / 2
idt
0 T/2 0
/
0
(i0 sin t)dt
dt
/
0
dt
2 i0
This is sometimes simply written as, iav. Hence, 2 iav i Half cycle i0 0.637 i0 2 Vav V0 0.637V0 Similarly, A dc meter can be used in an ac circuit if it is connected in the full wave rectifier circuit. The average value of the rectified current is the same as the average current in any half cycle, 2 i.e., time the maximum current i0 . A more useful way to describe a quantity is the root mean square (rms) value. We square the instantaneous current, take the average (mean) value of i2 and finally take the square root of that average. This procedure defines the rootmean-square current denoted as irms. Even when i is negative, i2 is always positive so irms is never zero (unless i is zero at every instant). Hence, T
2
i
0
one cycle
T
0
irms
i2
i2dt
2 /
0
dt
one cycle
(i02 sin2 t)dt
2 /
0
i0 2
dt
0.707i0
i02 2
Thus, irms
i0
0.707i0 2 Similarly, we get V Vrms 0 0.707V0 2 The square root of the mean square value is called the virtual value and is the value give by ac instruments. Thus, when we speak of our house hold power supply as 220 volts ac, this means that the rms voltage is 220 volts and its voltage amplitude is, V0 = 2 Vrms = 311 volt
Form Factor
i0
V / 2 rms value 0 1.11 The ratio, average value 2V0 / 2 2 is known as or factor,
irms 0.707i0 iav 0.637i0 t
Note: 1. The average value of sin t, cos t, sin2t, cos2t, etc. is zero because it is positive half of the time and negative rest half of the time. Thus,
sin t cos t sin2 t cos2t 0 then 2.
i = i0 sin t i i0 sin t i0 sin t 0 1 2 1 sin2 t cos2 t 2 2 2 2 i i0 sin t
The average value of sin2 t and cos2 t is or If then
3.
If
i2 i02 sin2 t i02 sin2 t
i02 2
Like SHM, general expressions of current/voltage in an sinusoidal ac are,
or
i = i0 sin (t ), i = i0 cos (t ),
and
V = V0 sin (t ) V = V0 cos (t )
Illustration 11: If the current in an ac circuit is represented by the equation, i = 5 sin (300t - /4) Here, t is in second and i in ampere. Calculate, (a) peak and rms value of current. (b) frequency of ac (c) average current Solution: (a) An in case of ac,
i = i0 sin (t ) The peak value
(b)
5
3.535 A 2 2 Angular frequency = 300 rad/s 300 47.75 HZ f 2 2 2 2 iav i02 (5) 3.18 A and
irms
i02
i0 = 5A
(c)
Capacitor in an AC Circuit If a capacitor of capacitance C is connected across the alternating source, the instantaneous charge on the capacitor, q = CVC = CV0 sin t and the instantaneous current i passing through it, is given by: V V0 sin t dq i CV0 cos t dt V0 q -q sin(t / 2) 1/ C a b or
i
i = i0 sin (t + /2)
Here, V0
i0 C
1 is the effective ac resistance or the capacitive C reactance of the capacitor and is represented as XC. It has unit as ohm. Thus, 1 XC C i, V It is clear that the current leads the voltage by 90 or the potential i drop across the capacitor lags the current passing it by 90. V i Figure shows V and i as functions of time t.
This relation shows that the quantity
0
0
t
Vc
Inductor in an AC Circuit Consider a pure inductor of self inductance L and zero resistance connected to an alternation source. Again we assume that an instantaneous current i = i 0 sin t flows through the inductor. Although there is no resistance, there is a potential difference VL between the inductor terminals a and b because the current varies with time, giving rise to self induced emf. L di VL = Vab = - (induced emg) = L a b dt i di Li0 cos t Or VL = L dt
VL= V0 sin t (i) 2 Here V0 = i0(L) (ii) V Or i0 = 0 L V0 (iii) i sin(t) L Equation (iii) shows that effective ac resistance, i.e., inductive reactance of inductor is,
Or
XL = L And the maximum current, V i0 0 XL The unit of XL is also ohm. From Equations. (i) and (iii) we see that the voltage across the inductor leads the current passing through it by 90. Figure shows VL and i as functions of time. i, VL V0 i0 i
t
VL
Illustration 12: A 100 resistance is connected in series with a 4H inductor. The voltage across the resistor is, VR = (2.0V) sin (103 rad/s)t: (a) Find the expression of circuit current (b) Find the inductive reactance (c) Derive an expression for the voltage across the inductor. V (2.0V)sin(103 rad / s)t Solution: (a) i R R 100 -2 (2.0 10 A) sin (103 rad/s)t (b) XL = L = (103 rad/s) (4H) = 4.0 103 ohm (c) The amplitude of voltage across inductor, V0 = i0XL = (2.0 10-2 A) (4.0 103 ohm) = 80 volt In an ac voltage across the inductor leads the current by 90 or /2 rad. Hence, VL = V0 sin (t + /2)= (80 volt) sin (103 rad / s)t rad 2 Note: That the amplitude of voltage across the resistor (=2.0 volt) is not same as the amplitude of the voltage across the inductor (=80 volt), even though the amplitude of the current through both devices is the same.
5.1
SERIES L–R CIRCUIT
As we know potential difference across a resistance in ac is in phase with current and it leads in phase by 90 with current across the inductor.
VR
5.2
VL
SERIES C–R CIRCUIT
Potential difference across a capacitor in ac lags in phase by 90 with the current in the circuit. Suppose in phasor diagram current is taken along positive x-direction. Then VR is also along positive x-direction but VC is along negative y-direction. So we can write.
VC
VR
y VR
i
x
VC
V
V = VR – jVC = iR – j(iXC) i =iR – j =iZ C 1 Z = R - j C The medulus of impedance is,
Here, impedence is,
2
1 Z R C and the potential difference lags the current by an angle, V X 1/ C 1 tan1 or tan1 C tan1 C = tan1 VR R R CR 2
5.3
SERIES L–C–R CIRCUIT
Potential difference across an inductor leads the current by 90 in phase while that across a capacitor, it lags in phase by 90. VL
VC
VR
Suppose in a phasor diagram current is taken along positive x-direction. Then VR is along positive x-direction, Then VR is along positive x-direction, VL along positive y-direction and VC along negative y-direction.
V VR2 (VL VC )2
VL i
So, we can write, Here impedence is,
VL VC VR
y
x VR
VC
V = VR + jVL – jVC = iR + j(iXL) – j(iXC) = iR + j [ i (XL – XC)] = iZ 1 Z = R + j(XL – XC) = R + j L C 2
1 Z R2 L C And the potential difference leads the current by an angle, 1 L V V X X C C C or tan1 tan1 L tan1 L R VR R The modulus of impedence is,
Illustration 13: An alternating emf 200 virtual volts at 50 Hz is connected to a circuit of resistance 1 and inductance 0.01 H. What is the phase difference between the current and the emf in the circuit. Also find the virtual current in the circuit. Solution: In case of an ac, the voltage leads the current in phase by an angle, X tan1 L R Here, XL = L = (2fL) = (2) (50) (0.01) = and R = 1 = tan-1 () 72.3 V Vrms Further, irms = rms Z R2 XL2 Substituting the values we have, 200 irms = 60.67 amp. (1)2 ( )2 Illustration 14: Find the voltage across the various elements, i.e., resistance, capacitance and inductance which are in series and having values 1000, 1F nd 2. henry respectively. Given emf as, V = 100 2 sin 1000 t volt Solution: The rms value of voltage across the source, 100 2 Vrms 100 volt 2 = 1000 rad/s irms =
Vrms Vrms 2 Z R (XL XC )2
Vrms
1 R L C 2
2
100
= 0.0707 amp 2 1 (1000)2 1000 2 1000 1 106 The current will be same every where in the circuit, therefore, P.D across resistor VR = irmsR = 0.0707 1000 = 70.7 volt P.D across inductor VL = irmsXL = 0.0707 1000 2 = 141.4 volt and 1 P.D across capacitor VC = irmsXC = 0.0707 = 70.7 volt 1 1000 10 6 Note: (i) The rms voltages do not add directly as, VR + VL + VC = 282.8 volt Which is not the source voltage 100 volts. The reason is that these voltages are not in phase and can be added by vector or by phasor algebra.
5.4
CIRCUIT ELEMENTS WITH AC
Circuit elements
Amplitude relation
Circuit quantity
Phase of V
Resistor
V0 = i0R
R
In phase with i
Capacitor
V0 = i0XC
XC
1 C
Lags i by 90
Inductor
V0 = i0XL
XL = L
Leads i by 90
6.
ASSIGNMENT
1.
In a series resonant circuit the ac voltages across resistance R, inductance L and capacitance C are 5V, 10V and 10V respectively. The ac voltage applied to the circuit will be (A) 25 V (B) 20 V (C) 10 V (D) 5 V. Solution: (D) V2 V = [ R + (VL – VC)2]1/2. 2.
There is a current of 20 milliampere through an ideal choke coil of 1.5 H when connected to 220 volt, 50 hertz ac supply. The power consumed is: (A) Zero (B) 220 20 10–3 watt (C) 220 50 watt (D) 220 20 1.5 watt. Solution: (A) Ideal choke is a wattles device. Because the phase angle between the current and p p efm is . This gives power factor cos = cos = 0. 2 2
3.
The ratio of induced emf in a coil of 50 turns and area A oscillating at frequency 50 Hz to that in a coil of 100 turns and same area oscillating at frequency 100 Hz is : (A) 0.25 (B) 0.50 (C) 0.75 (D) 1.00 Solution: (A)
E nAB. Therefore
E1 n1w1 50 ´ 50 . = = E2 n2w 2 100 ´ 100
4.
In the above questions what is the potential drop across the capacitor? (A) 110 V (B) 110 2 V (C) 220 V (D) 220 2 V. Solution: (A) AT resonance, the potential drop across the capacitance is equal and opposite to that inductance. The rms current in an a.c. circuit is 2 A. If the wattles current be 3 A, what is the power factor? 1 1 1 1 (B) (C) (D) . (A) 2 3 3 2 Solution: (C) Iwattless =Irms sin 3 Hence sin = 2 = 60 1 Therefore cos = 2 . 6. A circular wire loop of radius r can withstand a radial force T before breaking. A particle of mass m and charge q(q > 0) is V sliding over the wire. A magnetic field B is applied normal to the q plane of the wire. The maximum speed Vmax the particle can have m before the loop breaks is B rT (A) Zero (B) m r é qB q2B2 4T ù ré 4T ù 2 2 + + + + qB q B (C) (D) ê ú ê ú 2 êë m m2 mr ú 2ë mr û û Solution: (D) The following forces act on the particle. Force T acting radially inwards. mv 2 Centrifugal force acting radially outwards. r Magnetic force qvB acting radially inwards. mv 2 T + qvB = r 2 mv - qvB - T = 0 r Tr æqBr ö =0 v2 = ç v÷ è m ø m
5.
v=
1 é qBr q2B2r 2 4Tr ù + + ê ú 2 ëê m m2 m ú û
v=
r é qB q2m2 4T ù + + ê ú 2 êë m m2 mr ú û
7.
A rectangular loop of sides a and b, has a resistance R and lies at a distance c from an infinite straight wire carrying current I0. The current decreases to zero in time (t - t) ,0
a
b c I0
dx
x
c
I0
Consider an infinitesimal element of length b. thickness dx at a distance x from the wire. Since d = BdA mI df = 0 (bdx0 2p x c +a m0Ib dx f = òdf = 2p ò x c
m0bI æc + a ö loge ç è c ÷ ø 2p mb æc + a ö é æ t öùì æ t öü f = 0 loge ç I0 ç1 - ÷úí I = I0 ç1 - ÷ý ÷ ê è c ø ë è t øûî è t øþ 2p
f =
Since
df dt m bI æc + a ö e = 0 0 loge ç è c ÷ ø 2pt dq m0bI0 æc + a ö = loge ç è c ÷ ø dt 2pt R
m bI æc + a ö q = 0 0 loge ç dt è c ÷ øò 2pt R 0
e=-
m0bI0 æc + a öæ 1 ö - ÷ loge ç è c ÷ç øè t ø 2p m bI e æc + a ö I = = 0 0 loge ç è c ÷ ø R 2pt R m bI æc + a ö dq = 0 0 loge ç dt è c ÷ ø 2pt R e =-
t
8.
q=
m0bI0 æc + a ö loge ç è c ÷ ø 2pt R
A network of inductances, each of value 1H, is shown in figure. The equivalent inductance of the circuit between points A and B is
A
B
(A) 6.218 H (B) 0.268 H (C) 8.162 H (D) Solution: (D) 1 and 2 in series in parallel with 3. The result of this is in series with 4, which is in parallel with 5, again in series with A 6, which again is in parallel with 7, in series with 8 and in parallel with 9. Finally result of above steps, 10 & 11 are in B series to get 2.618 H.
2.618 H 10
6
4 7
9 11
5 3
8
2
1
A uniformly wound solenoid coil of self inductance 1.8 10–4 henry and resistance 6 ohm is broken up into two identical coils. These identical coils are then connected in parallel across a 15-voll battery of negligible resistance. The time constant for the current in the circuit and the steady state current through the battery is (A) 0.3 × 10-4S, 8A (B) 1.3 × 10-4S, 8A -4 (C) 0.3 × 10 S, 5A (D) none of these Solution: (A) (i) L, R The coil is broken into two identical coils. L/2´ L/2 L Leq = = = 0.45 ´ 10- 4 H , L/2 +L/2 4 R/2´ R/2 R = = 1.5W Req = R/2 +R/2 4 Leq. 0.45 ´ 10- 4 Time constant = = = 0.3 ´ 10- 4 s Req. 1.5 E 12 Steady current I = = = 8 A. R 1.5
9.
10.
If 0 and 0 are, respectively, the electric permittivity and magnetic permeability of free space, and the corresponding quantities in a medium, the index of refraction of the medium in terms of the above parameters is me me (A) 2 (B) m0e 0 m0e 0 (C)
Solution:
1 2
me m0e 0
(D)
none of these
(B) We know that the velocity of light in vaccum c =
And the velocity of light in a medium n =
1 m0e 0
1 me
Also the refractive index n=
11.
Velocity of light in medium n 1/ m0e 0 = = = Velocity of light in vaccum c 1/ me
me m0e 0
The network shown in figure is part of a complete circuit. If at a certain instant the current (I) is 5A, and is decreasing at a rate of 103 A/s then VB – VA =
I A
(A) Solution:
15V (A)
(B)
A
1 20V
15 V
1
15 V
5mH
(C)
B
25V
5mH
VB + e - 15 + I ´ 1 = VA
Here, I = 5A,
e=
(D)
30V
B
VB – VA = 15 – e – I
df é - dI ù =Lê = - 5 ´ 10- 3 ´ 103 = - 5V ú dt ë dt û
VB – VA = 15V 12. What is the nature of RCL series circuit for frequency less than the resonant frequency? (A) Resistive (B) Inductive (C) Capacitive (D) None of the above Solution: (C) For < 0, XL < XC. So, the circuit is capacitive. 1 1 13. If the power factor changes from to then what is the increase in impedance in A.C.? 2 4 (A) 20 % (B) 50 % (C) 25 % (D) 100 % Solution: (D) 1 cos Z Z1 cosq1 1/ 4 1 = = = Z2 cosq 2 1/ 2 2 Z2 = 2Z1 2Z1 - Z1 Percentage change = 100 = 100 %. Z1 14.
What is the impedance of a purely antiresonant circuit at resonance? (A) Zero (B) R (C) 2R (D) . Solution: (D) L For antiresonant circuit Z = . For purely antiresonant circuit R = 0. Hence Z = . CR
15.
A small dc motors operating at 200 V draws a current of 3 A at its full speed of 2500 r.p.m. If the resistance of the armature is 7, the back e.m.f. of the motor is (A) 149 V (B) 159 V (C) 169 V (D) 179 V Solution : (D) E = V-I R = 200-3x7 = 179 V. 16.
A motor (A) converts electrical energy into mechanical energy (B) converts mechanical energy into electrical energy (C) converts electrical energy into magnetic energy (D) produces mechanical energy. Solution : (A)
17.
For maximum output power in D.C. motor, the induced back emf (E) should be (A) applied voltage (B) double applied voltage (C) half of applied voltage (D) one third of applied voltage
Solution (C) For maximum output, E = V / 2 18.
In ac motor, a capacitor is used to (A) reduce ripple
(B)
reduce A.C.
(C)
(D)
increase A.C.
introduce phase difference of 2
Solution : (C) The capacitor introduces a phase difference of /2 which is necessary for motors. 19.
A generator develops an e.m.f. of 130 V and has a terminal potential difference of 125 V, when the armature current is 25 A. The resistance of the armature is (A) 0.5 (B) 0.2 (C) 1.5 (D) 2.4 Solution : (B) E V 130 125 0.2 I R 25 20.
A coil is wound of a frame of rectangular cross-section. If the linear dimensions of the frame are doubled and the number of turns per unit length of the coil remains the same, then the self inductance increases by a factor of (A) 6 (B) 12 (C) 8 (D) 16. Solution : (C) The number of turns has become twice and area has gone up four times. Therefore self inductance is 4x2 = 8 times. XL = L = 2 vL with iron core XL increases. Since E 1 , I decreases. XL
21.
A Circuit contains two inductors of self inductance L 1 and L2 in series. If M is the mutual inductance, then the effective inductance of the circuit shown will be (A) L1+L2 (B) L1+L2 -2M (C) L1+L2 +M (D) L1+L2 +2M Solution : (D) when inductances are connected like this in series, then L = L1+L2+2M. 22.
A circuit contains two inductors of self inductance L 1 and L2 in series as shown in fig. If M is the mutual inductance then the effective inductance of the circuit is (A) L1+L2 (B) L1+L2 -2M (C) L1+L2 +M (D) L1+L2 +2M
Solution : (B) in this type of combination, L = L1+L2-2M
L2
L1
L2
L1
23.
Two coils of self inductance L1 and L2 are in parallels as shown in fig. If their Mutual inductance is M, then the effective inductance will be given by L1L 2M2 L1L2M (A) (B) L1 L 2 2M L1 L2 2M
L1
L2
L1L2 M L1L2 M (D) L1 L 2 2M L1 L2 2M Solution : (D) When two inductances are in parallels then L1L2 M2 L= L1 L 2 2M 2
(C)
24.
Air cored chokes are used for reducing high frequency ac because (A) air is free (B) air core has high permeability (C) with air core self inductance increases (D) with air core self inductance is not much. Solution : (D) XL = L = 2 vL, since v is large therefore L should not be large. Hence air cored chokes are used for high frequency ac. 25.
In an ac circuit V and I are given by V = 150 sin (150 t) V and I = 150 sin 150 t A 3 The power dissipated in the circuit is (A) 5625 W (B) 4825 W (C) 7450 W Solution : (A) E0 = 150 V and I0 = 150 x 10–3 mA and = 600 P
E0 2
x
I0 2
cos
(D)
3425 W
150x150 1 x 5625 watt 2 2
26.
The induced e.m.f. in a coil rotating in a uniform magnetic field depends upon (A) only on total number of turns in the coil (B) only on magnetic field (C) area of coil and speed of rotation (D) all of the above Solution : (D) E = n AB , so it depends on all these. 27.
A resistance and an inductor are connected in series to a 220 V A.C. supply. When measured with an A.C. voltmeter, the P.D. across resistor is 140 V. The P.D. across the terminals of the inductor will be nearly (A) 150 V (B) 160 V (C) 170 V (D) 180 V
solution (C) 220 = VL =
1402 VL 2 or 48400 = 19600 + VL2
28800 = 170 V
28.
If q0 is the maximum charge then while charging, the charge on a capacitor at time t is given by q0 (1 - et / RC ) (D) q0 (1 - e - t / RC ) (A) q0 t (B) q0 (1 + e - t / RC ) (C) Solution : (D) q = q0(1 e t / RC ) where RC is the time constant. 29.
In the case of charging of a capacitor, at t = RC, the charge on the capacitor (q0 is the maximum charge) is (A) 0.432 q0 (B) 0.532 q0 (C) 0.632 q0 (D) 0.732 q0 Solution : (C) When t = RC, q = q0 (1-e–1) = 0.6321 q0
30.
In the case of charging or discharging of a capacitor RC has the units of (A) Ohm (B) A-F (C) Ohm - m (D) sec. solution : (D) The quantity RC has units of volt current sec ond x sec ond current volt
For test 1.
The capacitor gets almost fully charged after time t equal to (A) RC (B) 3 RC (C) 4 RC (D) Solution : (D) After t = 5 RC, the capacitor is about 99.3% charged.
5 RC
The time constant for the given RC circuit where R = 1 K and C = 103 pF will be (A) 2 m second (B) 1 second (C) 1.5 m second (D) 0.5 second. Solution : (B) t = RC = 103 x 103 x 10–12 = 10–6 s = 1s
2.
3.
A capacitor of 4F is connected to a 15 V supply through 1 mega ohm resistance. The time taken by the capacitor to charge upto 63.2 % of its final charge will be (A) 2s (B) 3s (C) 4s (D) 5 s. Solution : (A) 63.2 % = RC = 106 x 4 x 10–6 = 4 s If is a real number, cos + j is a complex number where j is equal to 1 (C) –1 (D) (A) 1 (B) 1 Solution : (D) i = 1
4.
The Exponential function e j is equal to (A) cos (B) (C) cos - j sin (D) Solution : (D) e j = cos + j sin
5.
cos + sin cos + j sin
6.
If LCR are connected is series, then the total effective resistance called impedance in terms of complex number is giver by 1 1 jR L R jL (A) (B) C C 1 1 (D) (C) RL R jL jC jC Solution : (D) 1 Z = R jL jC
7.
The magnetic flux through a coil perpendicular to its plane and directed into paper is varying according to relation = 5t2 + 10t + 5 milli weber. The e.m.f. induced in the loop after 5 sec 15 (A) 0.03 volt (B) 0.06 volt (C) 0.08 volt (D) 0.02 volt Solution : (B) = (5t2 + 10t + 5) x 10–3 Wb d as e = (in magnitude) dt d (5t2 + 10t + 5) x 10–3 Wb sec–1 dt = (10t + 10) 10–3 volt e = (10 x 5 + 10) x 10-3 = 0.06 volt
A train is moving with a speed of 30 m s–1 NS on the rails separated by 2 m. If the vertical component of earth's field is 8 x 10–5 T, the e.m.f. is (A) 0.0048 V (B) 0.048 V (C) 0.48 V (D) 4.8 V Solution : (A) e = Blv = 30 x 2 x 8 x 10-5 = 0.0048 V 8.
An aircraft with a wingspan of 40 m flies with a speed of 1080 km h –1 in the eastward direction at a constant altitude in the northern hemisphere, where the vertical component of earth's magnetic field is 1.75 x 10 –5 T. Then the e.m.f. that develops between the tips of the wings is (A) 0.5 V (B) 0.34 V (C) 0.21 V (D) 2.1 V Solution : (C) L = 40 mv = 1080 km h-1 = 300 m sec-1 B = 1.75 x 10-5 T e = Blv = 1.75 x 10-5 x 40 x300 = 0.21 V
9.
A Uniform flux density of 0.1 Wb m2 extends over a plane circuit of area 1 m 2, and is normal to it. How quickly must the field be reduced to zero if an emf of 100 V is to be produced? 1 sec (A) 10–2 Sec (B) 10–3 sec (C) (D) 1 micro sec. 10 Solution : (B) 1 = BA = 0.1 x 1 = 0.1 Wb; 2 = 0 10.
e=-
1 d 2 dt t
0 0.1 100 = , t = 10-3 s t
11.
Current in a 10 milli henry coil increases uniformly form zero to one to one ampere in 0.01 second, then the value of induced e.m.f. is (A) +1V (B) 2V (C) –1 V (D) –2 V Solution : (C) L = 10 x 10-3 = 10-2 henry change in current = 1-0 = 1 amp. dt = 0.01 sec dl 1 e== -10-2 x = -1 volt dt 0.01
12.
A step up transformer is used on 120 V line to provide a P.D. of 2400 V. If the number of turns in primary is 75, then the number of turns in the secondary shall be (A) 25 (B) 150 (C) 1500 (D) 500 Solution : (C) Here EP = 120 V, ES = 24000 V, np = 75 and ns = ? E ns 2400 ns As S = np 120 75 EP 2400 ns x75 1500 120 13.
The current drawn from the primary of a transformer which steps down 220 V to 22 V at 0.1 amp. shall be (A) 1 amp. (B) 0.1 amp (C) 0.2 amp. (D) 0.01 amp Solution : (D) Here np = 220 V, ns = 22 V, Is = 0.1 amp. and Ip = ? E I As s P EP IS E 22 IP S xIS x0.1 0.01amp EP 220
14.
What is the r.m.s. value of current ? I = I 2 sin t A 3 (A)
2A
(B)
Solution : (B) Peak value of current, I0 = 15.
1A
(C)
1 2
A
(D)
2 A.
2 A
A hot wire instrument reads 15 A. The peak value of the current is 15 15 2 A (A) (B) (B) 15 A (D) A 2
15 A
Solution : (A) The hot wire instruments measure virtual values only. Irms 15A Ipeak 2 I r.m.s. 2x15 15 2A
16.
The exponential function ex is equal to (A)
x+x2+x3+....
x x2 (C) 1 x .... 2! 3! x 2 x3 Solution : (B) ex = 1 x .... 2! 3!
(B) (D)
x 2 x3 .... 2! 3! x 2 x3 1 x .... 2! 3! 1 x
17.
An A.C. voltmeter reads 20 V across a resister, 30 V across an inductor and 15 V across a capacitor in LCR series circuit. In this case the supply voltage will be (A) 100 V (B) 60 V (C) 50 V (D) 25 V Solution : (D) e= =
VR2 (VL VC )2 400 225 25 V
18.
An dc voltage, E (in volt) = 200 2 sin 100 t is connected through an dc ammeter. The reading of the ammeter shall be (A) 10 mA (B) 20 mA (C) 40 mA (D) 80 mA Solution : (B) 1 1 E V 200V, XC 10 4 6 C 100x10 200 I 20 mA 10 4 19.
An LCR series circuit its connected to a source of A.C. at resonance, the applied voltage and the current flowing through the circuit will have a phase difference of (A) (B) (C) (D) 0. / 2 / 4 Solution : (D) 1 At resonance L C Z R or cos 1or 0 20.
'
The Capacitor shown in the Fig. is initially charged to a voltage V and the circuit is then closed. The charge on C is found to oscillate with frequency w. The frequency will be doubled if 1 1 2 2 LC LC 22 L (A) the voltage V is doubled C (B) both L and C are halved (C) both L and C are doubled
(D) both L and C are creased by a factor Solution : (B) 1 1 4 ' 2 2 . LC L / 2.C / 2 LC 21.
2
In a parallel LCR circuit shown in the Fig. at resonance (A) the source current is maximum (B) the impedance of the circuit is minimum and is equal to R (C)
E
R
C
L
the resonance frequency will be the same as for a series resonance circuit with the same value of L, C and R the voltage across L and C are in phase.
(D) Solution : (C) In a parallel resonance circuit, at resonance
L / C RL 2 2 LC L / C RC for a good qual;ity capacitor Rc = 0 and we put RL = R. Then
1
1 R2 LC L2 If resistance is neglected then condition of resonance in both series and parallel circuit become the same, namely (LC)1/ 2
22.
In a series LCR ac circuit, off resonance, the current is (A) always in phase with the generator voltage. (B) always lags the generator voltage (C) always leads the generator voltage (D) may lead or lag behind the generator voltage. Solution : (D) In a series LCR circuit, off resonance, the power factor is R R cos 2 Z 1 2 R L C 1 R So, tan L C 1 Now depending on whether L is greater than or less, the lead or lag will C occur. 23.
Who discovered laws of EMI ? (A) Faraday (C) P.A.M. Dirac
(B) (D)
Einstein R.P. Feynman.
(B)
MLT-2A-1
Solution : (A) 24.
The Dimensions of flux is (A) MLT-2 A
(C) ML-1T-3A-1 Solution : (D) F BA cos V A cos q
25.
(D) (
ML2T-2A-1 F = Bqv)
(MLT 2 )(L2 ) ML2T 2 A 1 1 (AT)(LT )
When a current of 4 A through primary gives rise to a flux of magnitude 1.35 Wb through secondary, what is the coefficient of mutual induction (M)? (A) 0.43 H (B) 0.38 H (B) 0.34 H (D) None of these.
Solution : (C) MI M M
I
1.35 0.34 henry 4
26.
When current changes from 13 A to 7 A in 0.5 second through a coil, the e.m.f. induced is 3 x 10-4 V. What is the coefficient of self induction? (A) 25 x 10-3 H (B) 25 x 10-4 H (C) 25 x 10-5 H (D) 25 x 10-6 H Solution : (D) e = 300 x 10–6 V dl 13 7 6A dt = 0.5 s dI e=L in magnitude dt dt 300x106 x0.5 L e = 25 x 10-6 H dI 6
27.
For two coils inductively coupled, the mutual inductance is 1.9 H. The current through primary changes by 6 A in 1 ms. The induced e.m.f. is 1.4 V (A) 11.4 mV (B) 11.4 V (B) 11.4 kV (D) Solution : (A) M = 1.9 H, dI = 6 A, dt = 10-3 sec dI 6 e m 1.9x 3 dt 10 = 11.40 x 10-3 V 28.
In the middle of a long solenoid there is a coaxial ring of square cross-section, made of conductivity material of resistivity p. The thickness of the ring is equal to h, its inside and outside radii are equal to 'a' and 'b' respectively. What is the current induced in a radial width (dr), where the magnetic field varies with time as B t ? hdr hrdr hrdr hrdr (A) (B) (C) (D) 4p p 2p 2p
Solution : (D) According to the problem, the e.m.f. induced in an elementary ring of radius r 1 h dr and width dr is r 2 . The conductance of this ring is d R p 2r thus, the current induced is dI = hrdr / 2p . Upon integration we get the total current
I
b
a
hrdr h(b2 a2 ) 2p 4p
29.
In the Q.95, what is the total current ? h(b2 a2 ) h(b2 a2 ) h(b2 a2 ) h(b2 a2 ) (A) (B) (C) (D) 4p 8p 16p 2p Solution : (A) According to the problem, the e.m.f. induced in an elementary ring of radius r 1 h dr and width dr is r 2 . The conductance of this ring is d R p 2r thus, the current induced is dI = hrdr / 2p . Upon integration we get the total current
I
b
a
hrdr h(b2 a2 ) 2p 4p
30.
Suppose p is the mass density, po is the resistivity, l 0 is the length and S is the cross-section of a wire. The mass of the wire is m. What is the active resistance of the wire ? l2 2l2 (B) (D) (A) l2 / m l2 / 2m (C) 4m m Solution: (A) The time constant is the ratio L/R, where L is the inductance of the solenoid and R is the active resistance. The latter is given by R = 0l0/S where 0 is the resistivity, l0 is the length and S is the cross-section of the wire. Also, if is the mass density of the material of the wire, mass of the wire m = l0S. Upon eliminating S one can write R = 0 l0 / S The time constant () is therefore given by L mL t = = 2 ær r 0l0 ö r r 0l02 çè m ÷ ø The length of the wire l0, in terms of the length of the solenoid l, is given by 1/ 2
æ4p lL ö l0 = ç è m0 ÷ ø
Thus, the time constant is given by m æm ö mL t = = 0 ç ÷ 4 p lL 4 p è r r 0l ø rr0 ´ m0
1.
SOME IMPORTANT DEFINITIONS USED IN OPTICS
We define a varied collection of useful terms. Some of these definitions will be revisited when the appropriate topics are taken up. (i)
Ray: The path of light, as determined within the approximations of geometric optics, is a ray. In a homogeneous medium, it is a straight line.
(ii)
Beam: A collection of rays, usually referred to as a bundle of rays, forms a beam. One may have a convergent, divergent or, a parallel beam. A convergent beam may converge to a point or, a line, a divergent beam may diverge from a point or, a line. A parallel beam consists of parallel rays.
(iii)
Collimation: A process whereby a divergent (or, convergent) beam is rendered parallel usually through the use of lenses and/or mirrors.
(iv)
Object & Image: The term object is used to refer to any object (being photographed or observed) that is a source of light, and its likeness (usually two dimensional) formed or observed by an optical system is the image.
(v)
Optical System: An optical system consists of elements like lenses, mirrors, prisms, etc.
(vi)
Axis: The axis of an optical system is frequently an axis of symmetry such that a ray directed along the axis continues in the same direction or, returns backwards (if reflected within the system)
(vii)
Centre of Curvature: Most lenses and curved mirrors being manufactured spherical (i.e., their surfaces are spherical), the centre of curvature of the curved mirror or, the curved surface of a lens is important and is frequently denoted by the letter C. The radius of curvature is also equally important in the analysis.
(viii) Pole: The pole of a spherical surface (refracting or, reflecting) is the central point of the surface involved in the formation of the image. It is denoted by O or, P. The axis, for a single spherical surface, is the join of the pole with the centre of curvature; it is known as the principal axis. (ix)
Optical Centre: The optical centre of a thin lens is a point on the axis of the lens such at a ray directed towards that point emerges parallel to itself after passing through the lens.
(x)
Paraxial Rays: It is observed that the formation of clear images by spherical surfaces takes place only with rays which are close to the principal axis and make very small angles with it.
(xi)
Wave speed: It is the distance travelled by the wave disturbance in a unit time. It is denoted by the letter .
(xii)
Frequency: It is the number of vibrations made by the medium particle in 1 s. In other words, it is number of waves passing through a point of the medium in 1 second. It is generally represented by the letter n or . Its unit is hertz and it is represented as Hz or s–1.
(xiii) Wavelength: It is the distance travelled by the wave in one complete period of a medium particle. In other words, distance between two consecutive crest or trough. Relationship between the wavelength, wave speed and frequency: If the wave speed is and period is T, then by definition
Wavelength T or But T 1/ x1/ Or
Distance travelled in one period Distance travelled in T s x T
i.e., Wave speed Frequency X Wavelength.
2.
REFLECTION AND REFRACTION AT PLANE AND SPHERICAL SURFACES
2.1
REFLECTION AT PLANE SURFACE
(i) Reflection of Light When light rays strike the boundary of two media such as air and glass, a part of light is turned back into the same medium. This is called Reflection of Light. The wavelength and the velocity of the light wave remains the same. Incident
N
Reflected
N N
i r
ir i r
Laws of reflection are obeyed at every reflecting surface, i.e. (ii) Laws of Reflection (a) The incident ray (AB), the reflected ray (BC) and normal (BN) to the surface (SS') of reflection at the point of incidence (B) lie in the same plane. This plane is called the plane of incidence (or plane of reflection). (b) The angle of incidence (the angle between normal and between the reflected ray and the normal are equal, i.e. i r A
N
C
i r S
S B
(iii) Types of reflection (a) Regular Reflection (a)
(b)
Diffused Reflection or Scattering or Diffusion
Regular Reflection: When the reflection takes place from a perfect plane surface it is called Regular Reflection.
(a) Regular reflection
(b)
Diffused Reflection or Scattering or Diffusion: When the surface is rough light is reflected from the surface from bits of its plane surfaces in irregular directions. This is called diffusion. This process enables us to see an object from any position.
(b) Irregular reflection
Characteristics of Reflection by a Plane Mirror (i) Distance of object from mirror = Distance of image from the mirror. (ii) The line joining the object point with its image is normal to the reflecting surface. (iii) The image is laterally inverted (left right inversion). (iv) The size of the image is the same as that of the object. (v) For a real object the image is virtual and for a virtual object the image is real. (vi) The minimum size of a plane mirror, required to see the full self image, is half the size of that person. (vii) For a light ray incident at an angle 'i' after reflection angle of deviation d = p - 2i (viii) (ix) (x) (xi) (xii)
(xiii)
(xiv)
If i = 0 then r = 0, this implies that a ray of light incident normally on a mirror retraces its path. The eye always observes an object in the direction in which the rays enter the eye The laws of reflection holds good for all kinds of reflection. Image of an object is the point at which rays after reflection (or reflection) actually converge or appear to diverge from that point. If the direction of the incident ray is kept constant and the mirror is rotated through an angle about an axis in the plane mirror then the reflected ray rotates through an angle 2. If an object moves towards (or away from) a plane mirror at a speed v, the image will also approach (or recede) at the same speed. Further the relative speed of image to the object will be v – (–v) = 2v. If two plane mirror are inclined to each other at 90, the emergent ray is always antiparallel to incident ray, if reflected from each mirror, irrespective of angle of incident.
M1
M2 (xv)
When two plane mirror, inclined to each other at an angle , the number of images formed can be determined as follows: 360 (a) If is an even integer, say ‘p’, then number of image formed say q = p – 1, for all position.
360 (b) If is an odd integer say ‘q’ then number of image formed say p = q, if the object is not on the bisector of the angle between mirrors. Also, p = q – 1, if the object is on the bisector. 360 (c) If is a fraction, then the number of image formed will be equal to its integral part. (xvi) The images are laterally inverted. (xvii) The linear magnification is unity. ∧
∧
∧
Illustration 1:
A ray of light on a plane mirror along a vector i + j - k . The normal on
Solution:
incident point is along i + j . Find a unit vector along the reflected ray. Reflection of a ray of light is just like an elastic collision of a ball with a horizontal ground. Component of incident ray along the inside normal gets reversed while the component perpendicular to it remains unchanged.
∧
∧
Thus the component of incident ray vector A i j k parallel to normal,
i.e., i j gets reversed while perpendicular to it, i.e., k remains unchanged. Thus, the reflected ray can be written as
R i j k A unit vector along the reflected ray will be
R i j k = r R 3
2.2
REFLECTION-AT SPHERICAL MIRRORS
A spherical mirror is a reflecting surface which forms a part of a sphere (as shown in following a and b diagram). When the reflection takes place from the inner surface and outer surface is polished or silvered the mirror is known as concave mirror. Vice- versa, it is convex.
(a) concave mirror
C
Concave Mirror
F
(b) convex mirror
P
P
F
C
Convex Mirror
Ray diagram for concave and convex mirror
2.3
CHARACTERISTICS OF REFLECTION BY A REFLECTING SURFACES (FOR SMALL APERTURE)
(i)
Pole (P) is generally taken as the mid point of reflecting surface.
SPHERICAL
(ii) (iii) (iv) (v)
(vi) (vii) (viii)
Centre of curvature (C) is the centre of the sphere of which the mirror is a part. Radius of curvature is the radius of the sphere of which the mirror is a part. Distance between P and C. Principal Axis is the straight line connecting pole P and centre of curvature C. Principal focus (F) is the point of intersection of all the reflected rays which strike the mirror (with small aperture) parallel to the principal axis. In concave mirror it is real and in the convex mirror it is virtual. Focal length (f) is the distance from pole to focus. Aperture is the diameter of the mirror. If the incident ray is parallel to the principal axis, the reflected ray passes through the focus. (Fig (a)) C
FP
(a)
(ix)
If the incident ray passes through the focus, then the reflected ray is parallel to the principal axis (Fig.(b)) C
F
P
(b)
(x)
Incident ray passing through centre of curvature will be reflected back through the centre of curvature. (Fig.(c)) C F P (c)
(xi) (xii)
(xiii)
R where f = focal length R = Radius of curvature. 2 Sign convention and magnification. We follow Cartesian coordinate system convention according to which: (a) The pole of the mirror is the origin (b) The direction of the incident rays is considered as positive x-axis. (c) Vertically up is positive y-axis Note: Radius of Curvature and Focal Length of: (a) Concave mirror is positive (b) Convex mirror is positive (a) Linear Magnification or lateral magnification or transverse magnification h v =M= 2 = h1 u h2 = y coordinate of image; h1 = y coordinate of the object (both perpendicular to the principle axis of mirror) v 2 - v1 v2 (b) Longitudinal Magnification = (for any size of the object) = - 2 (for u u2 - u1 short object) f=
Tracing for spherical mirror Concave
Convex
(1) A ray going through centre of curvature is reflected back along the same direction.
(2) A ray parallel to principal axis is reflected through the focus, and vice-versa. Also, mutually parallel rays after reflection intersect on the focal plane.
(3) A ray going to the pole and the reflected ray from it make equal angles with the principal axis.
2.4
C
C
F
F
F
i i
P F
i i
REAL AND VIRTUAL SPACES
A mirror, plane or spherical divides the space into two: (i) A side, where the reflected rays exist. This is real space. (ii) The other side where the reflected rays do not exist. This is virtual space Real Space
virtual Space
Real Space
virtual Real Space Space
virtual Space
Object Object is decided by incident rays only. The point object for mirror is that point. (a) From which the rays actually diverge to be incident on the mirror (Real object). Or (b) Towards which the incident rays appear to converge (virtual object). Point Object Real
Point Point Object Object Virtual (Real)
Image Image is decided by reflected or refracted rays only. The point image for a mirror is that point. (a) Towards which the rays reflected from the mirror, actually converge (real image). Or (b) From which the reflected rays appear to diverge (virtual image). NEW CARTESIAN SIGN CONVENTION – FOR MIRRORS AND LENSES There are several sign-conventions. Students should try to follow any one of these. In this package, the new Cartesian sign convention has been used wherever required.
Incident Ray +ve
-ve
P
Mirror / Lens
(i) (ii) (iii) (iv) (v) (vi)
All distance are measured from the pole. Distances in the direction of incident rays are taken as positive. Distances in the direction of incident rays are taken as negative. Distances above the principal axis are taken as positive. Distances below the principal axis are taken as negative. Angles measured from the normal, in anti-clockwise direction are positive, while in clockwise direction are negative.
2.5
MIRROR FORMULA
Consider the shown figure where O is a point object and I is corresponding image. CB is normal to the mirror at B. B q a O
C
q g
b I
P’
P
By laws of reflection, OBC = PBC = + = , + = + = 2 For small aperture of the mirror, , , 0 tan , tan , tan , P’ P tan + tan = 2 tan BP' BP' BP' Þ + =2 OP' IP' CP' Applying sign convention, u = -OP, v = -IP, R = -CP 1 æ 1ö 2 Þ - + ç- ÷ = u è vø R 1 2 If u = , , but by definition, if u = , v = f. v R R 1 1 1 f and Hence, 2 u f For convex mirrors, an exactly similar formula emerges. Illustration 2: An object is placed 21cm infront of a concave mirror of radius of curvature 10cm. A glass slab of thickness 3cm and refractive index 1.5 is then placed closed to the mirror in the space between the object and mirror. Find the position of the final image formed. (You may take the distance of the near surface of the slab from the mirror to be 1cm) Solution: Given that radius of curvature = 10 cm focal length f = 5 cm 1 1 Here u x t 1 21 3 20cm. 3 Where 1.5 = refractive index of slab Using the formula for concave mirror
1 1 1 f u v 1 1 1 5 20 v 20 cm. 3 Applying the correction due to the slab, we have final image distance = |v| + 1 20 23 1 cm. 3 3
Solving we get v
2.6
REFRACTION AT A PLANE SURFACE
Laws of Refraction (Snell’s Law) (i) The incident ray, the refracted ray and the normal to the refracting surface at the point of incidence all lie in the same plane. (ii) The ratio of the sines of the angle of incidence (i) and of the angle of refraction (r) is a constant quantity for two given media, which is called the refractive index of the second medium with respect to the first. sin i cons tan t sin r When light propagates through a series of layers of different medium as shown in the figure, then the Snell’s law may be written as 1 sin 1 = 2 sin 2 = 3 sin 3 = 4 sin 4 = constant In general, sin = constant 1
2
3 4
1 1
2
2
1
2
water
3
4
Fig. A light ray passing from air to water Fig. A series of transparent layers of bends toward the normal different refractive indices When light passes from rarer to denser medium it bends toward the normal as shown in the fig. According to Snell’s law 1 sin 1 = 2 sin 2 When a light ray passes from denser to rarer medium it bends away from the normal as shown in the fig. 1
1
2
2
air water
Fig. A light ray passing from water to air bends away from the normal. For a given point object, the image formed by refraction at plane surface is illustrated by the following diagrams.
Case - I
Case - II O I1
I2
I
O
Denser medium
I2
Rarer medium
I1
I
R2
R1
R1
R2
Determination of the image formed for the cases given becomes much simpler when we restrict ourselves to nearly normal incident rays. CASE – I (Object in the denser medium) Rarer medium m2 A B I
q2 q1
q1
q2
Denser medium m1
O
For nearly normally incident rays 1 and 2 will be very small. AB tan 1 sin 1 Object dis tance from therefractingsurface Similarly, AB sin 2 Image dis tance from therefractingsurface Imagesdist ance from therefractingsurface Object dis tance from therefractingsurface sin 1 1 2 2 sin 2 1 The same result is obtained for the other case also. The image distance from the refracting surface is also known as Apparent depth or height. Apparent Shift Apparent shift = Object distance from refracting surface – image distance from refracting surface. 1 y (apparent shift) t 1 where t is the object distance and 1 2 If there are a number of slabs with different refractive indices placed between the observer and the object. Total apparent shift = yi Illustration 3:
A person looking through a telescope T just sees the point A on the rim at the bottom of a cylindrical vessel when the vessel is empty. When the vessel is completely filled with a liquid ( = 1.5), he observes a mark at the centre, B, of the vessel. What is the height of the vessel if the diameter of its cross-section is 10cm?
Solution:
It is mentioned in the problem that on filling the vessel with the liquid, point B is observed for the same setting; this means that the images of point B, is observed at A, because of refraction of the ray at C. N
T r
C Liquid m=1.5
i h
A
D 5 cm
For refraction at C, sinr a l 1.5 sini AD 10 Now sin r AC 102 h2 Where h is the height of vessel. BD 5 sini = = BC 5 2 + h2 \
100 + h2
.
25 + h2 = 1.5 5
\
25 + h2 9 = 2 100 + h 16 2 100 + 16h = 900 + 9h2
\
7h2 = 500
\
2.7
10
\ h = 8.45 cm
REFRACTION AT SPHERICAL SURFACE
When light strikes a spherical (curved) surface, the position of object (O), the image (I) and the radius of curvature (R) are related as: i
P r
O
1
2
I
Q
2 1 2 1 ; here 2 > 1 v u R Consider another case, a spherical surface of radius R separating two media with refractive indices m1 & m2 . The radius of curative R will be taken positive when incident rays strike the convex side of the surface. If u is the object distance and v is the image distance,
i r
O
C
R
u
μ1
μ2
I
v
For light rays going from medium 1 (1) to medium 2 (2): m1 m2 m2 - m1 + = u v R The corresponding equation for refraction at a plane surface will be: m1 m2 (taking R = ) + =0 u v Illustration 4:
Solution:
A transparent rod 40 cm long is cut flat at one end and rounded to a hemispherical surface of 12 cm radius at the other end. A small object is embedded with in the rod along its axis and half way between its ends. When viewed from the flat end of the rod, the object appears 12.5 cm deep. What is its apparent depth when viewed from the curved end? For the flat surface:
I1
O
Real depth of the object = 20 cm Apparent depth = 12.5 cm real depth Using m= apparent depth 20 m= = 1.6 12.5 For the curved surface:
I2
O
m1 m2 m2 - m1 + = u v R u = 20 cm, R = –12 cm, 1.6 1 1 - 1.6 Þ + = - 12 20 V 1 1 1.6 Þ = V 20 20 Þ V = - 33.3 cm. Hence the object appears 33.3 cm deep from the curved side. We will use
3.
CRITICAL ANGLE AND TOTAL INTERNAL REFLECTION
Consider a ray of light that travels from a denser medium to a rarer medium. The angle of incidence for which the angle of refraction becomes 90 is called critical angle. This is used for constructing totally reflecting prisms. m1
m2
c
m sin c =2 m1 = 2 sin 90° m1 1 Þ sin c = 1 m2 and if i > c When the angle of incidence of a ray travelling from denser to rarer medium is greater then the critical angle, no refraction occurs. The incident ray is totally reflected back into the same medium. Here the laws of reflection hold good. Some light is also reflected before the critical angle is achieved, but not totally. Illustration 5:
Solution:
A point source of light a placed at the bottom of a tank containing a liquid of refractive index The level of the liquid is at a height h above the bottom of the tank. A bright circular spot is seen on the surface of the liquid when viewed from above. Find the radius of the spot. The rays of light emerging from the source, which are incident at an angle q > qc . (where q c is the critical angle) will be reflected. 1 ...(i) sin qc = m R = tan qc Further, h sin qc = 1 - sin2 qc 1 = m2 - 1 h ...(ii) or, R = m2 - 1 R air
h
N
4.
THIN LENS
Consider the thin lens shown here with the two refracting surfaces having radii of curvature equal to R1 and R2 respectively. The refractive indices of this surrounding medium and of the material of the lens are 1 and 2 respectively.
O
m1
m2 (1)
m1
I
(2)
Now using the result that we obtained for refraction at single spherical surface we get, m2 m1 m2 - m1 For first surface, ...(1) = v1 u R1 m m m - m2 For second surface, 1 - 1 = 1 ...(2) v v1 R2
é1 m2 m1 1ù = (m1 - m2 ) ê ú v u ëR1 R2 û öé 1 1 1 æm2 1ù - = ç - 1÷ê ú v u è m1 ø ëR1 R2 û
Adding (1) and (2),
Þ
When u = , v = f, Also,
1 1 1 - = v u f
öé 1 1 æm2 1ù = ç - 1÷ê ú f è m1 ø ëR1 R2 û
Lensmaker's Formula
Thin Lens Formula
EQUIVALENT FOCAL LENGTH OF TWO OR MORE THIN LENSES IN CONTACT
1 1 1 = + + ... F f1 f2 Special cases (a) If the lens is immersed in a liquid whose refractive index is greater than refractive index of material of the lens then 1 1 1 (I g 1) f(liquid) R1 R2 fliquid is the focal length of lens in the liquid Ig is the R.1. of glass w.r.t. liquid. The focal length of the lens in the liquid becomes negative. (b)
If the lens is immersed in a liquid whose R.1. is equal to the R.1. of the material of the lens then a b 1 I g a I
1 1 (1 1) 0 f(liquid) R1 R2 The focal length becomes infinite
(c)
1
If the lens be immersed in a liquid whose R.1. is less than the R.1. of the material of the lens 1 1 1 (I g 1) f(liquid) R1 R2
The focal length of the lens increases. In general, if decreases, f increases Note: A lens is converging if its focal length is positive and diverging if focal length is negative (in neo-Cartesian sign convention). From this we can conclude that a convex lens need not necessarily be a converging and a concave lens diverging. Ray tracing for lens Convex (i)
Concave
A ray passing through optic centre goes undeviated. O
(ii) When a ray is incident parallel to principal axis, the refracted ray passes or appears to pass through the focus.
Illustration 6:
F
F
A glass rod has ends as shown in figure, The refractive index of glass is . The object point O is at a distance 2R from the surface of larger radius of curvature. The distance between the apexes of the ends is 3R. Show that the image point of O is formed at a distance of
Solution:
O
(9 - 4μ )R (10μ - 9 )(μ - 2 )
From the right hand vertex. R Glass AIR
R/2
O
A
B
m AIR 2R
For the first surface m2 m1 m2 - m1 = v' u R1 Here 2 =, 1 =1, = –2R, R1 = +R u 1 m- 1 + = +R v ' 2R u m- 1 1 2m- 3 = = v' R 2R 2R 2mR \ v' = 2m- 3 For the second surface é æ 2mR öù u = - ê3R - ç ú è 2m- 3 ÷ øû ë
3R
Here
2 = 1, 1 = , R2 = +
R 2
m2 m1 m2 - m1 = v u R2 m 1 1- m + = v é 2mR ù R / 2 3R - ê ú ë2m- 3û m(2m- 3) 1 2 - 2m = v R (6mR - 9R - 2mR)
1 2 - 2m m(2m- 3) - 10m2 + 29m- 18 = = v R R [4m- 9] R [4m- 9] v=
5.
(9 -
4m)R (9 - 4m)R = 10m2 - 29m+ 18 (10m- 9)(m- 2)
MAGNIFICATION
(i) Magnification in Mirror In new Cartesian sign convention, we define magnification is such a way that a negative sign (of m) implies inverted image and vice-versa. A real image is always inverted one and a virtual one is always erect. Keeping these points in mind and that the real object and its real image would lie on the same sides in case of mirror and on opposite sides in case of lenses, we define m as in case of reflection by spherical mirror as: B A’ F A
C
m=(ii)
P
B’
v u
Lateral Magnification for Refracting Spherical Surface B
m1
m2 A’
A
P B’
M=
- (A 'B') . AB
As i, r ® 0, Þ
AB
Now,
Þ
PA = m2 A 'B' m1 PA '
sini m2 = sinr m1
tani m2 = tanr m1 Þ
- PA ' m2 A 'B' = PA AB m1
v
Hence,
(iii)
m=
u
m2 m1
Magnification of A Thin Lens v - (OA ') M= = u (AO) B
A’ A
B’
Illustration 7:
Solution:
An object is placed at a distance of 10 cm to the left on the axis of a convex lens A of focal length 20 cm. A second convex lens of focal length 10 cm is placed co-axially to the right of the lens A at a distance of 5 cm from A. Find the position of the final image and its magnification. Trace the path of the rays. Here for 1st lens, u1 = –10 cm f1 = 20 cm. 1 1 1 1 1 1 = Þ = v1 u1 f1 v1 20 10 Þ v1 = - 20 cm
Q’
L1
L2
O1
O2
f1
f1
Q
i.e. the image is virtual and hence lies on the same side of the object. This will behave as an object for the second lens. For 2nd lens 1 1 1 = v 2 u2 f2 Here
u2 = - (20 + 5)
f2 = 10 cm
1 1 1 + = v 2 25 10
50 2 = 16 cm 3 3 2 i.e. the final image is at a distance of 16 cm on the right of the second lens. 3 The magnification of the image is given by, v v 20 50 4 m= 1× 2 = × = = 1.33 u1 u2 10 3 ´ 25 3 Þ
v2 =
6.
POWER OF A LENS
1 f If two lens in are separated by a distance d then power of combination of lens is d P = P1 + P2 - PP m 1 2 Where P1 and P2 are optical powers of the two lenses, and is the refractive index of the medium in between them. If f is in metres then the power P of the lens in dioptres is given by, P =
Illustration 8:
A lens has a power of +5 diopters in air. What will be its power if completely immersed in water? and a μ g = 3/2. a μ w = 4/3
Solution:
Let fa and fw be the focal lengths of the lens in air and water respectively, then 1 1 Pa = or +5 = fa fa fa = 0.2 m = 20 cm é1 1 1ù Now ...(1) = a mg - 1 ê ú fa R R ë 1 2û
(
)
é1 1 1ù = w mg - 1 ê ú fw ëR1 R2 û Dividing equation (1) by equation (2), we get, fw é a mg - 1 ù =ê ú fa êë w mg - 1ú û m 3/2 9 a g = = Again, w mg = 4/3 8 w mg
(
and
\
)
...(2)
fw (3 / 2) - 1 (1/ 2) = = =4 fa (9 / 8) - 1 (1/ 8) fw = fa ´ 4 = 20 ´ 4 = 80 cm = 0.8 m
Pw =
7.
1 1 = = 1.25 diopter. fw 0.8
SILVERING AT ONE SURFACE OF LENS
When one surface of a thin lens is silvered, then the focal length F of the effective lensmirror combination is expressed as, 1/F = å 1/ f i , where fi is the focal length of the lens or mirror to be repeated as many times as the refraction or reflection respectively is repeated. Some Cases: (i) Focal length of plano–convex lens when silvered at its plane surface When an object is placed in front of such a lens. The ray first of all are refracted from the convex surface, then reflected from the polished plane surface and again refracted out from the convex surface. If f and fm be the focal lengths of lens and mirror.
R
R
Þ
Þ (ii)
1 1 1 1 1 1 2 2 é1ù = + + , Here, = ( m - 1) ê úand = = =0 F f fm f f fm Rm ¥ ëR û f 1 2 R = Þ F= = F f 2 2( m - 1)
Focal length of plano –convex lens when silvered at convex surface 1 1 2 é1ù = ( m - 1) ê úand = f fm R ëR û 1 2 2 1 R é( m - 1) ù 2 Þ = + Þ = 2ê + Þ F= ú F f R F 2m ë R û R
Þ
(iii)
1 1 1 1 = + + , F f fm f
Here,
Focal length of convex lens where convex surface of radius R2 is silvered
é1 1 1ù 1 2 = ( m - 1) ê úand = f fm R2 ëR1 R2 û é æ1 1 2 2 1 1 öù 2 Þ = + Þ = 2 ê( m - 1) ç ú+ ÷ F f R2 F R R è ø 1 2 ë û R2 Þ
1 1 1 1 = + + , F fl fm f
Illustration 9:
Here,
An object is 1 metre in front of the curved surface of a plano–convex lens whose flat surface is silvered. A real image is formed 120 cm in front of the lens. What is the focal length of the lens?
Solution: I Q 100cm 120cm
Here u = 100 cm v = 120 cm 1 1 1 1 1 11 F v u 120 100 600 1 1 1 1 Now F FL FM FL Where FL – focal length of the lens. FM – focal length of the mirror. 2 1 = + ( FM = ¥ ) FL FM
1 2 = F FL 11 2 \ = 600 FL 1200 FL = = 109.1cm 11
8.
ANGLE OF DEVIATION OF A RAY WHEN IT PASSES A LENS
O is the object and is the image is the angle of deviation. d = a + b » tan a + tan b
h h é 1 1ù + = hê - ú -u v ëv u û h d= f
= Þ
9.
DETERMINATION OF FOCAL LENGTH OF A CONCAVE MIRROR BY U–V METHOD
The relation between object distance u and the image v from the pole of the mirror is given by, 1 1 1 v u f Where f is the focal length of the mirror. The focal length of the concave mirror is obtained 1 1 either from versus graph or from u–v graph. v u 1 1 and graph u v When the image is real (of course only upon then it can be obtained on screen), the object lies between focus (F) and infinity. In such a situation u, v and f all are negative. Hence the mirror formula 1 1 1 v u f 1 1 1 Becomes, v u f 1 1 1 or again, v u f 1 1 1 or, v u f Comparing with y = mx + c, the desired graph will be a straight line with slope –1 and 1 intercept equal to . f
1 1 versus graph is as shown in figure. The intercepts on the v u 1 horizontal and vertical axes are equal. It is equal to . A straight line OC at an angle 45o f 1 1 with the horizontal axis intersects line AB at C. The coordinates of point C are , . 2f 2f The focal length of the mirror can be calculated by measuring the coordinates of either of the points A, B or C.
The corresponding
u–v graph The u–v graph comes out to be a hyperbola as shown in figure. A line drawn at angle 45o from the origin intersects the hyperbola at point C. The coordinates of point C are (2f, 2f). The focal length of mirror can be calculated by measuring the coordinates of point C.
Determination of focal length of convex lens using u–v method The relation between u, v and f for a convex lens is, 1 1 1 v u f 1 1 and graph u v Using the proper sign convention, u is negative, u negative, v and f are positive. So we have, 1 1 1 1 1 1 or v u f v u f 1 1 Comparing with y = mx + c, graph between and is negative with v u 1 slope –1 and intercept . The corresponding graph is as shown in figure. f Proceeding in the similar manner as discussed in case of a concave mirror the focal length of the lens can be calculated by measuring the coordinates of either of the points A, B and C.
u–v Graph The u versus v graph is a hyperbola as shown in figure. By measuring the coordinates of point C whose coordinates are (2f, 2f) we can calculate the focal length of the lens.
10. PRISM A prism has two plane surfaces AB and AC inclined to each other as shown in following figure. A is called the angle of prism or refracting angle. (i) Refraction Through A A X i
D r1
r2
Y r
emergent ray
incident ray
A light ray striking at one face of a triangular glass prism gets refracted twice and emerges out from the other face as shown above. The angle between the emergent and the incident rays is called the angle of deviation (D). The angle between the two refracting faces involved is called the refracting angle (A) of the prism. From AXY, we have: A + (90° – r1) + (90° – r2) = 180° (i) Þ r1 + r2 = A Deviation D = (i – r1) + (e – r2) D = (i + e) – (r1 + r2) D=i+e–A Þ A+D=i+e (ii) Þ We also have two equations from Snell's Law at X & Y. sinr2 1 sinr =m & = sinr1 sine m (i) Angle of Deviation It can be easily seen that if we reverse the emergent ray, it goes back along the same path. The angles of incidence and emergence get interchanged but the angle of deviation remains same. A
A
1
2
angle of incidence = 1 angle of emergence = 2 q1 + q2 = A + D
1
2
angle of incidence = 2 angle of emergence = 1 q2 + q1 = A + D
Hence the same angle of deviation D is possible for two different angles of incidence : 1 and 2, where 1 + 2 = A + D. (ii) Minimum Deviation The angle of deviation is minimum when the path of light ray through the prism is symmetrical. i.e., angle of incidence = angle of emergence A
r1=r2=r
i=e Þ r1 = r2 and hence A + Dmin = i + i = 2i also A = r1 + r2 = 2r from
sini = m, we have: sinr1 æA + Dmin ö sin ç ÷ è ø 2 =m A sin 2
(iii) Grazing Incidence When i = 90°, the incident ray grazes along the surface of the prism and the angle of refraction inside the prism becomes equal to the critical angle for glass - air. This is known as grazing incidence. A i 90
r
e
(iv) Grazing Emergence When e = 90°, the emergent ray grazes along the prism surface. This happens when the light ray strikes the second face of the prism at the critical angle for glass - air. This is known as grazing emergence. A i
e = 90
(v) Maximum Deviation The angle of deviation is same for both the above cases (grazing incidence & grazing emergence) and it is also the maximum possible deviation if the light ray is to emerge out from the other face without any total internal reflection. (vi) Dispersion of Light When a ray of light passes through a prism, it splits up into rays of constituent colours or wavelengths. This phenomenon is called dispersion of light. The refractive index of a medium is different for light rays of different wavelengths. Larger the wavelengths, the lesser is the refractive index. mred < mviolet because l red > l violet. i.e.,
A ray of white light passing through a prism gets splits into different colours because the deviation is more for violet and less for red. This phenomenon is known as dispersion. Dr
white
Dv red
angle of dispertion
violet
For a prism with very small refracting angle A, the deviation D is given by: d » (m- 1) A Hence deviation of violet = Dv = (v, – 1)A and for red = Dr = (r – 1)A The angle between the red and violet rays is known as the angle of dispersion. The angle of dispersion = f = Dv - Dr = (mv - mr ) A
Dv + Dr æmv + mr ö =ç - 1÷A è 2 ø 2 mv - mr f Dispersive power of glass = w = = Dm mv + mr -1 2
Angle of mean deviation = Dm =
This phenomenon arises due to the fact that refractive index varies with wavelength. It has been observed for a prism that decreases with the increase of wavelength, i.e. mblue > mred . Angular dispersion, q = dv - dr (vii)
Dispersive power d - dr where is deviation of mean ray (yellow) w= v d
As
dv = (mv - 1) A,
\
w=
dr = (mr - 1) A,
mv - mr m- 1
(viii) Deviation without Dispersion This mean an achromatic combination of two prisms in which net or resultant dispersion is zero and deviation is produced. For the two prisms,
White
White
Þ
(mv - mr ) A + (m'v - m'r ) A ' = 0 (m - mr ) A and wd + w ' d ' = 0 A' = - v
Spectrum
m'v - m'r
Where and ' are the dispersive powers of the two prisms and and ' their mean deviations.
(ix) Dispersion without deviation A combination of two prisms in which deviation produced for the mean ray by the first prism is equal and opposite to that produced by the second prism is called a direct vision prism. This combination produces dispersion without deviation. For deviation to be zero, ( + ') = 0 Þ (m- 1) A + (m'- 1) A ' = 0 Þ
A' = -
(m- 1) A (- ve sign (m'- 1)
Þ
prism A' has to be kept inverted)
Illustration 10: A prism of crown glass refracting angle of 50 and mean refractive index = 1.51 is combined with one flint glass prism of refractive index = 1.65 to produce no deviation. Find the angle of flint glass and net dispersion. Given: v = 1.523, R = 1.513 (for crown glass) 'v = 1.665, 'R = 1.645 (for flint glass) Solution: Let A' be the angle of flint glass prism. Here A = 5 and = 1.51 for crown glass prism. d = (m- 1) A = (1.51 - 1) ´ 5 = 2.55° Deviation produced by flint glass d = (m'- 1) A ' = (1.65 - 1) A ' = 0.65 A ' For no deviation ' = or 0.65 A' = 2.55 2.55 A' = = 3.92° 0.65 Net dispersion,
(mv - mR ) A - (m'v - m'R ) A '
= (1.523 - 1.513)5 - (1.655 - 1.645)3.92 = 0.05 - 0.0784 = - 0.0284
11. ASSIGNMENT 1.
Two plane mirrors A and B are aligned parallel to each other, as shown in the figure. A light ray is incident at an angle of 30 at a point just inside one end of A. The plane of incidence coincides with the plane of the figure. The maximum number of time the ray undergoes reflections (including the first one) before it emerges out is: 2 3m 0.3mm
B
30
A
(A) Solution:
28 (B)
No. of ref =
(B) 2 3 0.2 3
= 30
30
(C)
32
(D)
34
30 30
02 2 3
2.
Consider the metal coin in figure. The right half is polished, while left half is unpolished. Polished Unpolished
In a dark room, this coin is placed face up on the floor and illuminated with a thin-beamed flashlight, as shown in figure. A and B look at the coin from the position as indicated in figure B eye Flashlight beam
A eye
Floor Coin
Solution:
According to A, which half of the coin (if either) is brighter? (A) The polished half (B) The unpolished half (C) Both halves look equally bright (D) Both halves look completely dark (A) The light reflects of the coin and reaches A’s eye. B's eye
A's eye
As the polished half is smoother than the unpolished half, hence the polished half behaves more like a mirror. So, most of the light striking the polished half of the coin reflects instead of going diffusion reflection. Hence, A sees a high percentage of the light that hits the polished part of the coin. Light hitting the unpolished half undergoes more diffuse reflection, bouncing off the coin in all different directions. Hence, a smaller percentage of that light reaches A’s eyes. So, for A the polished half looks more brighter than the unpolished half.
3.
A glass prism of refractive index 1.5 is immersed in water (R.I. = 4/3). The beam of light incident normally on the face AB is totally reflected to reach the face BC, if (A) sin 8/9 (B) sin < 2/3 (C) 2/3 < sin < 8/9 (D) None of these Solution: (A) The light ray is totally reflected internally at the point P (say). Therefore the critical angle C . Also, sin c n w sin 90 ng
sin c =
4/3 8 3/2 9
B
A
C
B
A
p
sin 8/9
C
4.
A diminished image of an object is to be obtained on a screen 1.0 m away from it. This can be achieved by approximately placing: (A) a convex mirror of suitable focal length (B) a concave mirror of suitable focal length (C) a convex lens of focal length less than 0.25 m (D) a concave lens of suitable focal length. Solution: (C) mage can be formed on the screen if it is real. Real image of reduced size can be formed by a concave mirror or a convex lens. Let u = 2f + x, then 1 1 1 + = u v f 1 1 1 Þ + = 2f + x v f 1 1 1 f +x Þ = = v f 2f + x f (2f + x ) f (2f + x ) f +x It is given that u + v = 1m. Þ
v=
2f + x +
f (2f + x ) f ù é = (2f + x ) ê1 + < 1m f +x ë f + xú û
2
f (2f + x ) < 1m or, f +x or, (2f + x)2 < (f + x ) This will be true only when f < 0.25 m.
5.
A screen beaming a real image of magnification m 1 formed by a convex lens is moved through a distance x. The object is the moved until a new image of magnification m2 is formed on the screen. The focal length of the lens is: (A) (C)
x m2 m1 x
m1m2
(B)
x m1 m2
(D)
None of these
Solution:
(A) In first case, 1+m1 =
q f
In the second case And
q 1 1 1 and =m1 p p q f
q x m2 p
…(1) 1 1 1 q x p f q x m2 = f
…(2)
(1) and (2) m2 – m1 = x/f 6.
f=
x m2 m1
If incident ray MP and reflected ray QN are parallel to each other, find the angle between the mirrors. Q
P i2
i i1 1
i2
M
N
(A) 60 (B) 45 (C) 90 (D) 180 Solution: (C) The deviation produced by first mirror = 180 – 2i1 and the deviation produced by second mirror = 180 – 2i2. The total deviation is 180 as MP and QN are anti parallel So, 180 = 360 – 2(i1 + i2) i1 + i2 = = 90 7.
A 16 cm long image of an object is formed by a convex lens on a screen placed normal to its principal axis. On moving the lens towards the screen, without changing the positions of the object and the screen, a 9 cm long image is formed again on the screen. The size of the object is: (A) 9 cm (B) 11 cm (C) 12 cm (D) 13 cm. Solution: (C) y = y1 ´ y 2
= 16 ´ 9 = 4 ´ 3 = 12cm. 8.
A small mirror of area ‘A’ and mass ‘m’ is suspended in a vertical plane by means of a weightless string. A beam of light of intensity ‘I’ falls normally on the mirror and the string is deflected form the vertical through a very small angle ‘’. Assuming the mirror to be perfectly reflecting, obtain an expression for .
2I/ C
mg
(A) (C)
2IA c mg 3I c mg
(B)
I c mg
(D)
none of the above
Solution: (A) When the light is incident on the mirror is totally reflected and there is no loss of energy. Change in momentum = 2 incident momentum Energy Intensity = Time Area Intensity Momentum = Velocity of light Force Rate of change in momentum Pressure, P = = Area A 2IA = (where c = speed of light) c P A 2IA = = tan = mg c mg 9. If x and y be the distances of the object and image formed by a concave mirror from its focus and f be the focal length then (A) xf = y2 (B) xy = f2 (C) x/y = f (D) x/y = f2 2 Solution: (B) According to Newton's formula xy = f . Note that m = or 10.
f y x x
f f v f u f
xy = f2
An observer can see, through a pin-hole, the top end of a thin rod of height h, placed as shown in the figure. The beaker height is 3h and its radius is h. When the beaker is filled with a liquid up to a height 2h, he can see the lower end of the rod. Then the refractive index of the liquid is:
3h
h 2h
(A) 5/2 Solution: (B) sin i 1 = sin r m
(B)
5/2
(C)
3/2
(D)
3 / 2.
Eye r
h
x
2h-x 2h
x =
1 m
...(i)
=
x h
...(ii)
x 2 + 4h2 2h - x
(2h - x )2 + h2 2h - x
sin r =
2
(2h - x)
+ h2
5 2 11. The sun (diameter d) subtends an angle radians at the pole of a concave mirror of focal length f. What is the diameter of the image of the sun formed by the mirror? (A) f (B) f/2 (C) f2 (D) f2 B Solution: (A) Since the sun is at a very large distance, u is very large and so (I/u) is practically zero. Using (i) and (ii), m=
So,
1 1 0 v f
i.e., v = f
A
i.e., the image of sun will be formed at the focus and will be real, inverted and diminished. Now as the rays from the sun subtend an angle radians at the pole, then in accordance with figure. = i.e. 12.
A B d FP f
A
C F B
M
P M
[where d is the diameter of the image of the sun]
d = f
A ray of light is incident on the left vertical face of a glass cube of refractive index 2, as shown in figure. The plane of incident is the plane of the page, and the cube is surrounded by liquid (1). What is the largest angle of incidence 1 for which total internal reflection occurs at the top surface? 1
B A 1
2
c
c
2
2
(A)
sin1 =
μ2 -1 μ1
2
(B)
sin1 =
μ2 +1 μ1
2
(C) Solution:
μ sin1 = 1 +1 (D) sin1 = μ2 (A) Consider A Point Applying Snell’s law, 1sin1 = 2sin2 …(1) But 2 = 90 – c cos2 = sinc = 1 …(2) 2 Elimination of 2 between (1) and (2), we get
2
μ1 -1 μ2
2
sin1 = 2 1 1 13. A ray incident at a point as an angle of incidence of 60° enters a glass sphere of R.I. n = 3 and is reflected and refracted at the farther surface of the sphere. The angle between the reflected and refracted rays at this surface is: (A) 50° (B) 60° (C) 90° (D) 40° P Solution: (C) Refraction at P. 600 Sin 600 3 Sin r1
Sinr1
1 r1= 30° 2
r1
Since r2 = r1 r2 = 30° Refraction at Q
r2 r’2
Q
i2
Sin r2 1 Sin i2 3
Putting R2= 30° we obtain i2 = 60° Reflection at Q r2 r2 30o
180 0 (r2 i2 )
= 180°- (30°+60°) = 90° 14.
What is the number of images of an object formed by two mirrors if the angle between the mirrors is in between 2/4 and 2/5? (A) 10 (B) 8 (C) 6 (D) 4 2 2 Solution: (D) As we know, if angle between two mirrors is between and then n n 1 the no. of images formed is ‘n’ 2 2 So here in the question it is and 4 4 1 no. of images will be 4. If one face of a prism of prism angle 30 and = 2 is silvered, the incident ray retraces its initial path. What is the angle of incident? (A) 90 (B) 180 (C) 45 (D) 55 Solution: (C) As in AED 30 + 90 + D = 180
15.
A 30
i
D
E
r
2
C
B
D = 60 Now as by construction D + r = 90 r = 90 – 60 = 30 Applying Snell’s law at surface AC 1 1 sin i = ( 2 )sin30 = 2 = 2 2 1 i = sin–1 = 45 2
16.
Two plane mirrors are inclined to each other at angle . A ray of light is reflected first at one mirror and then at the other. Find the total deviation of the ray (A) 360 – 2 (B) 360 + 2 (C) 180 – 2 (D) 180 + 2
Solution (A) A ray AB is incident on mirror OM1 at angle and is reflected along BC suffering a deviation 1 = FBC A
M1 E d a
B F
d1
a
N1 N2 b
q O
b d2
C
M2 G
The ray BC falls on mirror OM2 at an angle of incidence and is reflected along CD suffering another deviation 2 = GCD The total deviation is = 1 + 2 It is clear from the diagram that 1 = 180 - 2 and 2 = 180 - 2 = 1 + 2 = 360 –2( + ) Now, in triangle OBC, OBC + BCO + BOC = 180 or (90 – ) + (90 – ) + = 180 or += Hence = 360 – 2 Which is independent of the angle of incidence at the first mirror. 17.
A layer of oil 3cm thick is floating on a layer of coloured water 5cm thick. Refractive index of coloured water is 5/3 and the apparent depth of the two liquids appears to be 36/7cm. Find the refractive index of oil. (A) 1.6 (B) 1.4 (C) 1.9 (D) 0.9
Solution (B)
24 cm
I1
A
I2
O
10 cm x
Apparent depth (A) =
\ \ \
B 6 cm
m
24 - x
t1 t 2 + m1 m2
36 5 3 = + 7 5 / 3 m2 3 36 15 = - 3= m2 7 7 7 m2 = = 1.4 5
18.
A glass sphere ( = 1.5) of radius 8 cm is placed in sunlight. Where is the image of the sun formed by the light passing through the sphere? When refraction by first surface of sphere (A) 24 cm (B) 30 cm (C) 34 cm (D) 50 cm
19.
A glass sphere ( = 1.5) of radius 8 cm is placed in sunlight. Where is the image of the sun formed by the light passing through the sphere? When refraction by second surface of sphere (A) (C)
4 cm 15 cm
(B) (D)
6 cm 50 cm
Solution (18(A); 19(A)) As the sun is very far away, the incident ray can be assumed as coming from infinity. For refraction at first surface: normal normal
I2
I1
m1 m2 m2 - m1 + = u v R 1 1.5 1.5 - 1 Þ + = v1 ¥ +8 Þ v1 = 24 cm. . . . [18 (A)] The rays try to converge at 1 and hence for the second surface it is a virtual object. For second surface: m1 m2 m2 - m1 + = u v R u = –(24 – 16) = –8 cm 1.5 1 1 - 1.5 Þ + = - 8 v2 -8
Using
[19 (A)] Þ v 2 = 4 cm. Hence the final image is formed at 2 which is 4 cm from the other surface of the sphere. 20.
If the critical angle is 4842’ when the media concerned are air and water and 3647’ when they are in air and glass, what is it when they are in water and glass. (A) 5252' (B) 6252' (C) 6262' (D) 7070' Solution (A) We know that 1 sinC = mw 1 1 mw = = sinC sin48°42' 1 Similarly mg = sin36°47' Now for glass and water media sin C = refractive index of water w.r.t. glass mw 1/ sin 48°42' sin36°47' = = or C = 52°52' mg sin36°47' sin 48°42' 21.
Find the size of the image formed in the situation shown in figure. m= 1 1 cm
m= 1.33
C
20 cm
O 40 cm
(A) (C)
0.9 cm 1.2 cm
(B) (D)
0.6 cm 1.4 cm
Solution (B) Here u = –40 cm, R = –20 cm = 1, 2 = 1.33 m= 1 1 cm
C
20 cm
O 40 cm
We have, m2 m1 m2 - m1 = v u R 1.33 1 0.33 =v 40 20 v = –32 cm. The magnification is m = h2 32 =Þ 1 1.33 (- 40) The image is erect.
h2 m1v = h1 m2u h2 = 0.6 cm
m= 1.33
22.
A convergent lens of 6 diopters is combined with a diverging lens of -2 diopters. Find the power of combination? (A) 4 diopter (B) 6 diopter (C) 8 diopter (D) 10 diopter
Solution (A) Here P1 = 6 diopters, P2 = -2 diopters Using the formula P = P1 + P2 = 6 – 2 = 4 diopters 23.
Select a graph between ‘v’ and ‘u’ for a concave mirror. v
v
(A)
(B) O
O
u
u
v
v
(C)
(D) O
O
u
u
Solution (A) For a concave mirror 1 1 1 u v f Curve would be a hyperbola for spherical mirror. v
O
24.
u
A prism is made up of flint glass whose dispersive power is 0.053. Find the angle of dispersion if the mean refractive index of flint glass is 1.68 and the refracting angle of prism is 3. (A) 20.08 (B) 10.08 (C) 0.208 (D) 0.108
Solution (D) As the refracting angle is small deviation = ( – 1) A angle of dispersion = Dv - Dr
= (mv - 1) A - (mr - 1) A = (mv - mr ) A æ m - mr ö =ç v A (mmean - 1) è mmean - 1÷ ø
= w A (mmean - 1) = (0.053) ´ 3 ´ (1.68 - 1) = 0.108°
CMP:
The following figure shows a simple version of a zoom f1 r0
Q
f2=-|f2| r'0
d f
I'
s'2
The converging lens has focal length f1’ and the diverging lens has focal length f2 = –|f2|. The two lenses are separated by a variable distance d that is always less than f1’ also the magnitude of the focal length of the diverging lens satisfies the inequality |f2| > (f1 –d). If the rays that emerge from the diverging lens and reach the final image point are extended backward to the left to the diverging lens, they will eventually expand to the original radius r0 at the same point Q. To determine the effective focal length of the combination lens, consider a bundle of parallel rays of radius r0 entering the converging lens. 25.
At the point where rays enters the diverging lens, the radius of the ray decreases to: f -d f -d (B) r0 = 1 (A) r’0 = 1 r0 r’0 f1 f1 f -f (C) r’0 = 1 2 r0 (D) r’0 = r0 f1 Solution: (A)
bundle
A r0
r'0 B
d f
D C E
From the similar triangles ABC and DEC r0 r' (f d)r0 0 so r’0 1 f1 f1 d f1 26.
To the right of the diverging lens the final image I’ is formed at a distance s’2. (f1 - f2 )d (f1 - d) | f2 | (A) (B) f2 - f1 + d | f2 | -f1 + d f1 - f2 + d (d - f1 )f2 (C) (D) f1 - f' 2 f1 - f2 - d Solution: (B) The image at focal point of the first lens, a distance f 1 to the right of first lens, serves as the object for the second lens. The image is at distance f 1–d to the second lens so s2 f2 (f d)f2 s2 = –(f1 –d) = d – f1 so s’2 = 1 s2 f2 | f2 | f1 d 27. Find the effective focal length? f1 | f2 | f1f2 (B) (A) | f2 | -f1 + d f1 - f2 - d
(C) Solution:
f1 f2 - f1 + d (A)
(D)
r'0 r'0
f2 f1 + f2 + d
I' s'2
From the similar triangles in the sketch r0 r '0 f1(f2 ) so f = | f2 | f1 d f s'2 If f1 = 12 cm, f2 = – 18 cm find the maximum focal length of the combination? (A) 36 cm (B) 34 cm (C) 32 cm (D) 30 cm Solution: (A) Put the numerical values in the expression in the earlier question and we get 36 cm.
28.
29.
In continuation of fourth questions determine the minimum focal length of the combination (A) 21.6 cm (B) 24 cm (C) 30 cm (D) 35 cm Solution: (A) Put the numerical values in the above expression which we found and we will get 21.6 cm. 30.
What value of d gives f = 30 cm? (A) 12 cm (B) 10 cm 216 Solution: (C) f = 30 cm then 30 = 6d 6 + d = 7.2 hence d = 1.2 cm.
(C)
1.2 cm
(D)
14 cm
For test 1.
An achromatic combination of two prisms - one of quartz and other of flint glass is to be formed. If the refracting angle of the quartz prism is 4, find the refracting angle of the flint glass prism. Also find the net mean deviation produced by the combination. Given that for the flint glass : v = 1.694, r = 1.65 and for the quartz glass : v = 1.557, r = 1.542. (A) –1.282 (B) +1.282 (C) –5.282 (D) –8.282
Solution (A) For an achromatic combination, net dispersion = 0 dispersion by one prism = dispersion by other prism. Þ Þ m1v - m1r A1 = (m2v - m2r ) A 2
(
)
Þ
(1.694 - 1.65) A1 = (1.557 - 1.542) ´
Þ
A1 = 1.364°
4
Dm1 - Dm2 = (m1m - 1) A1 - (m2m - 1) A2 æm + m1r ö æ1.557 + 1.542 ö = ç 1v - 1÷ ´ 1.364 - ç - 1÷ ´ 4 è ø è 2 ø 2
= - 1.641° ´ 2 = - 1.282 Net mean deviation = Dm1 - Dm2 = - 1.282 2.
Find the dispersion produced by a thin prism of 18 having refractive index for red light = 1.56 and for violet light = 1.68. (A) 4.16 (B) 6.16 (C) 2.16 (D) 8.16 Solution (C) We know that dispersion produced by a thin prism q = (mv - mR ) A Here mv = 1.68, mR = 1.56 and A = 18°. q = (1.68 - 1.56) ´ 18° = 2.16°.
3.
Calculate the dispersive power for crown glass μv = 1.523, and μR = 1.5145 (A) 0.1639 (B) 0.01639 (C) 1.639 (D) 2.639 Solution (B) Here mv = 1.523 and mR = 1.5145 Mean refractive index m=
from
the
given
data
1.523 - 1.5145 = 0.01639 (1.51875 - 1)
Dispersive power is given by, m - mR 1.523 - 1.5145 w= v = = 0.01639 (m- 1) (1.51875 - 1) 4.
Two plane mirrors are inclined to each other such that a ray of light incident on the first mirror and parallel to the second is reflected from the second mirror parallel to the first mirror. Determine the angle between the two mirrors. (A) 60 (B) 30 (C) 90 (D) 180 Solution: (A) Let be the angle between the two mirrors OM1 and OM2. The incident ray AB is parallel to mirror OM2 and strikes the mirror OM1 at an angle of incidence equal to . From figure we have M1BA = OBC = M1OM2 = M1 D
E
q B
A
a q
d
N1
a
N2 q
q
q M2
O
C
Similarly for reflection at mirror OM2, we have M2CD = BCO = M2OM1 = Now in triangle OBC, 3 = 180, therefore = 60
5.
A small filament is at the centre of a hollow glass sphere of inner and outer radii 8 cm and 9 cm respectively. The refractive index of glass is 1.50. Calculate the position of the image of the filament when viewed from outside the sphere. (A) 9 cm (B) – 9 cm (C) – 19 cm (D) +19 cm Solution: (B) For refraction at the first surface, u = –8 cm, R1 = –8 cm, 1 = 1, 2 = 1.5 m2 m1 m2 - m1 = v u R1 1.5 1 0.5 + = v' 8 - 8 v' = –8 cm OA = 8 cm OB = 9 cm
m= 1.5
Glass
O
A
B Air
It means due to the first surface the image is formed at the centre. For the second surface u = –9 cm, 1 = 1.5, R2 = –9 cm m2 m1 m2 - m1 = v u R2 1 1.9 1 - 1.5 + = -9 v 5 v = –9 cm Thus, the final image is formed at the centre of the sphere. 6.
A thin rod of length f/3 is placed along the optical axis of a concave mirror of focal length f. Such that its image which is real and elongated just touches the rod. Calculate the magnification. (A) 3/2 (B) 5/2 (C) 1/2 (D) 2/3 Solution: (A) Let be the length of the image. 1 mf Þ 1= Then, m = f 3 3
l
rod
image
f/3
u’ u v
Also image of one end coincides with the object, Þ u' = 2f f f 5f u' = u + Þ 2f - = 3 3 3
æ f mf ö v = - çu + + ÷ . Putting in mirror formula, è 3 3 ø 1 1 1 + = f mf u f + u+ 3 3
Þ
1 2 = m + 6 15
Þ
Þ
3 3 1 + = 5f + f + mf 5f f
m=
3 2
7.
An object of length 2.5 cm is placed at a 1.5 f from a concave mirror where f is the magnitude of the focal length of the mirror. The length of the object is perpendicular to the principal axis. Find the length of the image. Is the image erect or inverted? (A) – 8 cm (B) – 10 cm (C) – 5 cm (D) – 12 cm Solution: (C) f O
F 1.5 f
The focal length F = –f and u = –1.5 f, we have, 1 1 1 1 1 1 + = or + = u v F - 1.5f v f 1 1 1 1 = - =v 1.5f f 3f v = –3f 3f Now, m = - v / u = =-2 1.5f h2 or =-2 or h2 = - 2h1 = - 5 cm h1 The image is 5 cm long. The minus sign shows that it is inverted. 8.
Select a graph for concave mirror between 1/v and 1/u. 1/v
1/v
(A)
(B) O
O
1/u
1/v
1/u
1/v
(C)
(D) O
Solution: (C) For a mirror
1/u
O
1/u
1 1 1 u v f 1 1 let y = , x = v u 1 x+y= , f
y = – x + 1/f
This is the equation of a straight line with slope = (–1) and intercept =
1 f
Hence the graph would be 1/v
1/f
O
9.
1/f
1/u
Show with the help of ray diagrams that a right angled isosceles prism can produce a deviation of 90° If the refractive index is greater than 2 . 45 (A) (B) 45 45
45 45 45 45
45
(C)
(D)
None of these.
45
45
Solution: (A) m> 2 1 1 < m 2 Þ sinC < sin45° Þ C < 45° Hence a ray striking a glass-air surface from inside the prism will be reflected back if angle of incidence is not less than 45°. 90° Deviation is produced if the ray strikes on one of the sides normally as shown. Here, 1 = 45° and hence reflection takes place. 10.
Convex lens of 10 cm focal length is combined with a concave lens of 6 cm focal length. Find the focal length of the combination. (A) 15 cm (B) – 15 cm (C) 30 cm (D) – 30 cm Solution: (B) Here f1 = 10 cm, f2 = -6 cm, F = ?
Use the formula \
1 1 1 1 1 1 = + = - =F f1 f2 10 6 15
F = - 15 cm
11.
Focal length of a convex lens of refractive index 1.5 is 2 cm. Focal length of lens when immersed in a liquid of refractive index 1.25 will be (A) 5 cm (B) 7.5 cm (C) 10 cm (D) 2.5 cm Solution (A) ( g 1) fp 5 fl xfa fa g 2 1 l 5 x 2cm = 5 cm. 2
12.
A double convex lens of focal length 20 cm is made of glass of refractive index 3/2. When placed completely in water (a 4 / 3) , its focal length will be (A) 80 cm (B) 15 cm (C) 17.7 cm (D) 22.5 cm Solution (A) (g 1) fw f w 4 fa g fa 1 w fw = 4 x fa = 4 x 20 = 80 cms.
13.
Find the minimum angle of deviation for a prism with angle A = 60° and = 1.5. (A) 37 (B) 40 (C) 55 (D) 65 Solution: (A) Minimum Deviation The angle of minimum deviation occurs when i = e and r1 = r2 and is given by: æA + Dm ö sin ç è 2 ÷ ø m= A sin 2 Aö æ Dm = 2sin- 1 çmsin ÷ - A Þ è 2ø Substituting = 1.5 and A = 60°, Dm = 2sin- 1 (0.75) - 60° = 37°
14.
Find the maximum angle of deviation for a prism with angle A = 60° and = 1.5. (A) 50 (B) 58 (C) 64 (D) 60 Solution: (B) Maximum Deviation
i 90
r1 r2
e
The deviation is maximum when i = 90° or e = 90° that is at grazing incidence or grazing emergence. Let i = 90° Þ r1 = C = sin- 1 (1/ m) Þ
r1 = sin- 1 (2 / 3 ) = 42°
Þ
r2 = A - r1 = 60° - 42° = 18°
sinr2 1 = sin e m Sin e = u sin r2 = 1.5 sin 18° Þ sine = 0.463 Þ e = 28° Deviation = Dmax = i + e – A = 90° + 28° – 60° = 58° u sin g
CMP :
15.
Inside a substance such as glass or water, light travels more slowly than it does in a vacuum. If c denotes the speed of light in a vacuum and v c denotes its speed through some other substance, then v = n Where n is a constant called the index of refraction. To almost exact approximation, a substance’s index of refraction does not depend the wavelength of light. For instance, when red and blue light waves enter water, they both slow down by about the same amount. More precise measurements, however, reveal that n varies with wavelength. Table 1 presents some indices of refraction of Cutson glass, for different wavelengths of visible light. A nanometer (nm) is 10–9 meters. In a vacuum, light travesl at c = 3.0 108 m/s. Indices of refraction of Cutson glass (Table 1) approximate wavelength in vacuum (nm) n yellow 580 1.500 yellow orange 600 1.498 orange 620 1.496 orange red 640 1.494
Inside Cutson glass (A) Orange light travels faster than yellow light (B) Yellow light travels faster than orange light (C) Orange and yellow light ravels equally fast (D) We cannot determine which colour of light travels faster Solution: (A) The velocity of light is inversely proportional to the index of refraction c v = , So, the lower the n, the higher the v. According to table, Cutson glass has a n slightly lower n for orange light than it has for yellow light. Therefore, inside Cutson glass, orange light travels slightly faster.
16.
For blue green of wavelength 520 nm, the index of refraction of Cutson glass is probably closest to: (A) 1.49 (B) 1.50 (C) 1.51 (D) 1.52 Solution: (C) Table suggests a linear relationship between wavelength and index of refraction. Starting with yellow light, we can extrapolate with linear relationship to blue green light. As compared to yellow light waves, blue green light waves are 60 nm shorter (520 nm vs. 580 nm). By reading table from bottom to top, we see that each 20 nm decreases in wavelength corresponds to a 0.002 increase in n. So, since yellow and blue green differ in wavelength by 60 nm, the corresponding n’s should differ by 0.006. nblue green = nyellow + 0.006 = 1.500 + 0.006 = 1.506 which is closer to 1.51 than it is to 1.50. You can also address this question by extrapolating table. blue green 520 1.506 540 1.504 560 1.502 yellow 580 1.500 yellow orange 600 1.498 orange 620 1.496 orange red 640 1.494 17.
For visible light, which graph bes expresses the index of refraction of glass as a function of the frequency of the light?
frequency
frequency
frequency
Cutson
frequency
(A) (B) (C) (D) Solution: (A) As just discussed, index of refraction decreases linearly with wavelength Graph b show a linear relationship. But the problem wants the relationship between frequency and n, not between wavelength and n. Since higher wavelengths correspond to lower frequencies, and vica versa, n increasing with frequency. To clarify this reasoning, we should review the relationship between wavelength and frequency. For all waves, not just light waves, v = f, where v denotes velocity, denotes wavelength, and f denotes frequency. In a vacuum, all light waves travel at c speed c. So, frequency and wavelength are inversely proportional. f = . Higher wavelengths correspond to lower frequencies, and vice versa. Therefore, since n decreases at higher wavelengths, it increases at higher frequencies. c Again, extending table 1 can help us spot this relationship. Using f = , we can find the frequency of different colors of light. For instance c 3.0 108 m / s fyellow = = 5.17 1014 Hz 580 10 9 m yellow fyellow–orange =
c yellow orange
3.0 108 m / s = 5.0 1014 Hz 9 600 10 m
18.
Which of the following phenomena happens because n varies with wavelength? (A) A lens focuses light (B) A prism brakes sunlight into different colours (C) Total internal reflection ensures that light travels down a fiber optic cable (D) Light rays entering a pond change direction at the pond’s surface Solution: (B) 19.
The time required for light to pass through a glass slab of 2 mm thickness is ( glass 1.5)
10–8 s (B) 10–11 s (D) 3 d d d. 2x10 x1.5 Solution (C) t sec 1011 sec v c/ c 3x108 (A) (C)
10–6 s None of these
20.
A spherical surface of curvature R separates air (=1.5). The centre of curvature is in the glass. A point object P placed in air is found to have a real image Q in the glass. The line PQ cuts the surface at a point O and PO = OQ. The distance PO is equal to (A) 2R (B) 5R (C) 3R (D) 1.5R 2 1 1 Solution (B) As v u R 1.5 1 1.5 1 2.5R x 5R or x x R 0.5
21.
If the tube length of astronomical telescope is 105 cm and magnifying power is 20 for normal setting, calculate the focal length of the objective (A) 100 cm (B) 10 cm (C) 20 cm (D) 25 cm f Solution (A) L = f0 + fe = 105 cm and M = 0 20 fe f0 = 100 cm and fe = 5 cm. 22.
Time of exposure for a photographic print is 10 seconds, when a lamp of 50 cd is placed at 1 m from it. Then another lamp of luminous intensity I is used, and is kept at 2 m from it. If the time of exposure now is 20 s, the value of I (in cd) is (A) 25 (B) 100 (C) 200 (D) 20 8 c 3x10 3x108 Hz Solution (B) v 1 8 c 3x10 v' 3x107 Hz. ' 10
23.
How does refractive index () of a material vary with respect to wavelength () ? A and B are constants B A 2 (A) = A + B2 (B)
(C)
= A + B
Solution (B) Cauchy's formula is A
(D)
A
B
B 2
24.
Time required for making a print at a distance of 40 cm from a 60 watt lamp is 12.8 second. If the distance is decreased to 25 cm, then time required in making the similar print will be (A) 15 sec (B) 10 sec (C) 5 sec (D) remains some. Solution (C) If prints are similar, I I E1t1 E2 t 2 12 t1 22 xt 2 r1 r2 But I1 = I2 r22 t 2 2 xt1 t 2 5 sec. r1 25.
An achromatic convergent doublet has power of +2D. If power of convex lens is + 5D, then ratio of the dispersive powers of a convergent and divergent lenses will be (A) 3:5 (B) 2:5 (C) 2:3 (D) None of these Solution (A) P = +2D P1 5D P2 3D
As
P2 f1 w w 3 3 1 1 P1 f2 w2 5 w2 5
26.
If a convex lens having focal length f and refractive index 1.5 is immersed in water then new focal length will be (w = 4/3) (A) f (B) 1.5 f 4 f (D) 4f (C) 3 1 1 1 Solution (D) (g 1) . . . (1) f R1 R2 1 g w 1 1 . . . (2) fw w R1 R2 fw 4f. 27.
Radii of curvature of a convex lens are 20 cm and 30 cm and its focal length if 24 cm. Refractive index of the lens will be (A) 1.33 (B) 1.5 (C) 1.71 (D) None of these Solution (B) 1 1 1 As ( 1) f R1 R2
1 1 ( 1) 1 ( 1) 12 1.5 24 20 30
28.
Both radii of curvature of a convex lens are 20 cm and refractive index of the material of the lens is 1.5. Rays parallel to the axis of the lens will converge at (A) 10 cm (B) 20 cm (C) 30 cm (D) 40 cm Solution (B) 1 1 1 As ( 1) f R1 R2
1 1 0.5x2 1 1 (1.5 1) f 20 20 20 20 or f=20 cm i.e., rays will converge at 20 cm from the optical centre of the lens.
29.
Eye piece of an astronomical telescope has focal length of 5 cm. If angular magnification in normal adjustment is 10, then dustance between eye piece and objective should be (A) 15 cm (B) 35 cm (C) 55 cm (D) 75 cm Solution (C) In normal adjustment, L = f 0 + fe f0 f0 and M or 10 f0 50 cm fe 5 L = 5 cm + 50 cm = 55 cm. 30.
Power of a convex lens is + 5D (g= 1.5). When this lens is immersed in a liquid of refractive index , it acts like a divergent lens of focal length 100 cm. Then refractive index of the liquid will be 5 5 (B) (A) 3 4 6 (C) (D) none of these 5 Solution (A) g 1 Pa a 5 5 Pl g 100 /100 1 l 5 l 3
Unit 20 - Wave Optics 1.
WAVE NATURE OF LIGHT
Huygens’ Wave Theory (i)
Each point on a wavefront acts as a source of new disturbance and emits its own set of spherical waves called secondary wavelets. The secondary wavelets travel in all directions with the velocity of light so long as they move in the same medium.
The envelope or the locus of these wavelets in the forward direction gives the position of new wavefront at any subsequent time. A surface on which the wave disturbance is in the same phase at all points is called a wavefront. Wave optics involves effects that depend on the wave nature of light. In fact, it is the results of interference and diffraction that prove that light behaves as a wave rather than a stream of particles (as Newton believed). Like other waves, light waves are also associated with a disturbance, which one consists of oscillating electric and magnetic field. The electric field associated with a plane wave propagating along the x-direction can be expressed in the form: E = E o[sin(t - kx + o)] where , k and o bearing their usual meanings. (ii)
Points to remember regarding Interference When two waves with amplitude A1 and A2 superimpose at a point, the amplitude of resultant wave is given by A=
A12 A 22 2A1A 2 cos
Where is the phase difference between the two waves at that point. Intensity (I) =
1 E02 . C = speed of light, E0 = electric field amplitude 20C
Intensity (I) = I1 + I2 + 2 I1I2 cos. Hence for I to be constant, must be constant. When changes randomly with time, the intensity = I1 + I2. When does not change with time, we get an intensity pattern and the sources are said to be coherent. Coherent sources have a constant phase relationship i.e. one that does not change with time. The intensity at a point becomes a maximum when = 2n (n = 0, 1, 2, . .) and there is constructive interference. If = (2n 1) there is destructive interference. (Here n is a non-negative integer) Determination of Phase Difference The phase difference between two waves at a point will depend upon (a) the difference in path lengths of the two waves from their respective sources. (b) the refractive index of the medium (c) initial phase difference, between the source, if any. (d) Reflections, if any, in the path followed by waves.
In case of light waves, the phase difference on account of path difference Optical path difference μ(Geometri cal path difference ) 2 = 2 π λ
=
where is the wavelength in free space. In case of reflection, the reflected disturbance differs in phase by with respect to the incident one if the wave is incident on a denser medium from a rarer medium. No such change of phase occurs when the wave is reflected in going from a denser medium to a rarer medium.
2.
YOUNG’S DOUBLE SLIT EXPERIMENT
A train of plane light waves is incident on a P
barrier containing two narrow slits separated by S1
a distance 'd'. The widths of the slits are small
d
compared with wavelength of the light used, so
O
that interference occurs in the region where the
S2 Screen
light from S1 overlaps that from S2. A series of Double slit
alternately bright and dark bands can be
D
observed on a screen placed in this region of overlap. The variation in light intensity along the screen near the centre O shown in the figure Now consider a point P on the screen. The phase difference between the waves at P is , where 2 = Po
O a bright fringe.
a dark fringe.
fringe width
(where Po is optical path difference, Po=Pg; Pg being the geometrical path difference.)
2 S2P S1P
(here = 1 in air)
As P
As, D >> d, S2P - S1P d sin sin tan( = y/D). [for very small ] Thus, =
S1 d
2 dy D
S2
For constructive interference, = 2n (n = 0, 1, 2...)
2 dy 2n D
y
Similarly for destructive interference,
O dsin D
y=n
D d
y = (2n 1)
D 2d
(n = 1, 2, .........)
Fringe Width W It is the separation of two consecutive maxima or two consecutive minima. Near the centre O [where is very small], W = yn+1 - yn [yn gives the position of nth maxima on screen] =
D d
Intensity Variation on Screen. If A and Io represent amplitude of each wave and the associated intensity on screen, then, the resultant intensity at a point on the screen corresponding to the angular position as in above figure, is given by
Two coherent sources
Intensity
Two Incoherent sources
4I0 One source 2I0 I0 -5
I = Io + Io + 2 I o2 cos,
-5
-4
when =
-2
-
0
+
+2
+3
+4
+5
2d sin = 4Io cos2 2
Illustration 1: A beam of light consisting of two wavelengths 6500 oA and 5200 oA is used to obtain interference fringes in YDE. The distance between the slits is 2.0 mm and the distance between the plane of the slits and the screen is 120 cm. (a) Find the distance of the third bright fringe on the screen from the central maxima for the wavelength 6500 oA. (b) What is the least distance from the central maxima where the bright fringes due to both the wavelengths coincide? Solution:
(i)
y 3 n.
D 3x1.2mx 6500x10 10 m 0.12 cm d 2x10 3 m
Let nth maxima of light with wavelength 6500 Å coincides with that of mth maxima of 5200Å. (ii)
mx6500 A o xD nx5200 A o xD m 5200 4 d d n 6500 5 Least distance = y 4 =
4x6500 x10 10 x1.2 4.D(6500 A o ) 0.16cm = d 2x10 3 m
Illustration 2: The intensity of the light coming from one of the slits in a Young's double slit experiment is double the intensity from the other slit. Find the ratio of the maximum intensity to the minimum intensity in the interference fringe pattern observed. Solution :
Imax l1 l2 Imin l1 l2
2
2
As I1 2I2
2 1 I max 34 Imin 2 1
Displacement of Fringes When a film of thickness 't' and refractive index '' is introduced in the path of one of the sources, then fringe shift occurs as the optical path difference changes. Optical path difference at 1)
P = S2P [S1P+ t t] = S2P S1P ( 1)t = y.d/D ( nth fringe is shifted by
Δy
, t
P
S1
S2
D( 1)t w ( 1)t d
Illustration 3: Monochromatic light of wavelength of 600 nm is used in a YDSE. One of the slits is covered by a transparent sheet of thickness 1.8 x 10 -5 m made of a material of refractive index 1.6. How many fringes will shift due to the introduction of the sheet? Solution:
As derived earlier, the total fringe shift =
w ( 1)t .
As each fringe width = w, The number of fringes that will shift =
total fringe shift fringe width
w ( 1)t ( 1)t (1.6 1)x1.8x10 5 m 18 w 600x10 9 m
Illustration 4: In the YDSE conducted with white S1 light (4000Å-7000Å), consider two points P1 d=1cm and P2 on the screen at y1=0.2mm and S2 y2=1.6mm, respectively. Determine the wavelengths which form maxima at these points Solution: The optical path difference at P1 is p1 =
P2 y2 = 1.6mm P1
y1 = 0.2mm
4m
dy1 10 0.2 5 10 4 mm 5000 A D 4000
In the visible range 4000 - 7000Å n1 =
5000 1.25 4000
and
n2
5000 0.714 7000
The only integer between 0.714 and 1.25 is 1 The wavelength which forms maxima at P is For the point P2, Here n1 =
40000 10 4000
= 5000Å
dy 2 10 p2 = 1.6 4 10 3 mm 40000 A D 4000 40000 and n2 5.71 7000
The integers between 5.71 and 10 are 6, 7, 8, 9 and 10
The wavelengths which form maxima at P2 are 1 = 4000Å for n = 10 2 = 4444Å for n=9
3 = 5000Å for n=8 4 = 5714Å for n=7 5 = 6666Å for n=6 Illustration 5: A transparent paper ( = 1.45) of thickness 0.02 mm is pasted on one of the slits of a Young’s double slit experiment which uses monochromatic light of wavelength 620 nm. How many fringes will cross through the centre if the paper is removed? Solution: Due to pasting the fringes shift which will restore its position after removal. Path difference will be S2P S1P t + t
= t ( 1) +
yd D
for a bright fringe x = n t( 1) + yd/D = n. y = (n t( 1)).
D d
Again after removing y = n
D d
y y = t( 1)
D d
No of fringes shifted will be
y y D D t t( 1) ( D) d d N 3 0.02 10 (0.45 )
n=
=
3.
620 10 9
= 14.5
INTERFERENCE BY THIN FILM
A ray of light incident on a thin film of thickness ‘t’ gets partially reflected and refracted at A at surface I and thereafter it gets reflected and refracted at B of surface II. The rays after emerging in the first medium interfere. Now the inference will depend upon the path difference between AD and ABC, as beyond CD path difference is zero. x = (AB + BC) – AD = (AE + EB + BL) – AD
G
= AE–AD+(EB+BC) (I)
i
x = AE – AD + EF
C
i
In LBF = BC = BF
A r x
r
E
In ECF EF = CP cos r = 2 t cos r In ADC and AEC
Surface-II B
AE AD sini and sinr AC AC sini AD μ sinr AE
AE = AD (V)
Putting (IV) and (V) in (III) we obtain
Surface-I
r
r
F
x = 2 t cosr IAD as In reflected at a denser medium it suffers an additional path difference /2 Total path difference the taken place is 2 t cosr – λ / 2 For constructive interference 2 t cosr – / 2 = n. 2 t cosr = n + /2 = (2n + 1)/2 maxima For normal incidence r = 0
2 t = (2n + 1) / 2 = n
2 t cosr – / 2 = (2n – 1) / 2 For normal incidence
2 t cosr = n
2t=n
Illustration 6: White light is incident normally on a glass plate of thickness 0.50 -6 x 10 m and index of refraction 1.50. Which wavelengths in the visible region (400 nm - 700 nm) are strongly reflected by the plate? Solution :
The light of wavelength is strongly reflected if the light rays reflected are interfering constructively. As we know the condition for constructive interference
1 2
2t = n . Here 2t = 2 x 1.5 x (0.5 x 10-6)m = 1.5 x 10-6 m.
1 2
Putting = 400 nm, 1.5 x 10-6 = n 400 x 10-9 n = 3.25
Similarly, by putting = 700 nm.
1 2
1.5 x 10-6 = n (700 x 10-9)
n = 1.66 Thus, within 400 nm to 700 nm, for integral values of n = 2 and 3. Now,
4.
4t 600nm and 429 nm are strongly reflected. 2n 1
DIFFRACTION
5. Diffraction is the bending or spreading of waves that encounter an object ( a barrier or an opening) in their path. 6. In Fresnel class of diffraction, the source and/or screen are at a finite distance from the aperture. 7. In Fraunhoffer class of diffraction, the source and screen are at infinite distance from the diffracting aperture. Fraunhoffer is a special case of Fresnel diffraction. Single Slit Fraunhoffer Diffraction In order to find the intensity at point P on the screen as shown in the figure the slit of width 'a' is divided into N parallel strips of width x. Each strip then acts as a radiator of Huygen's wavelets and produces a characteristic wave disturbance at P, whose position on the screen for a particular arrangement of apparatus can be described by the angle .
The amplitudes Eo of the wave disturbances at P from the various strips may be taken as equal if is not too large. The intensity is proportional to the square of the amplitude. If Im represents the intensity at O, its value at P is
P
O
a
2
sin l = lm ;
where
A minimum occurs when, sin = 0 and 0, so = n, n = 1, 2, 3...
f
a sin 2
I
a sin n a sin n
Angular width of central maxima of diffraction pattern = 21 = 2 sin-1(/a) /a 2/a 0 3/a -3/a -2/a -/a [ 1 gives the angular position of first minima] The concept of diffraction is also useful in deciding the resolving power of optical instruments. Illustration 7: Light of wavelength 6 10-5cm falls on a screen at a distance of 100 cm from a narrow slit. Find the width of the slit if the first minima lies 1mm on either side of the central maximum. Solution:
Here n = 1, = 6 10-5 cm. Distance of screen from slit = 100 cm. Distance of first minimum from central maxima = 0.1 cm. sin = 1 =
Dis tan ceof 1st min ima from the centralmax ima Dis tan ceof the screen from the slit
0.1 1 100 1000
We know that asin = n a=
= 0.06 cm. 1
In YDSE if the source consists of two wavelengths 1 = 4000Å and 2 = 4002Å . Find the distance from the centre where the fringes disappear, if d=1cm ; D=1 m . The fringes disappear when the maxima of 1 fall over the minima of 2. That is
Illustration 8:
Solution:
p p 1 1 2 2
or
p=
Where p is the optical path difference at that point. 1 2 2 2 1
Here 1 = 4000Å, 2 = 4002Å
p = 0.04 cm
In YDSE,
p = dy/D
1000 0.4 40mm D y= p d 10
Illustration 9: A beam of light consisting of two wavelengths 6500A°and 5200A° is used to obtain interference fringes in a Young’s double slit experiment (i) Find the distance of the third fringe on the screen from the central maximum for the wavelength 6500A°. (ii) What is the least distance from the central maximum where the bright fringes due to both wavelength coincide? (iii)The distance between the slits is 2mm and the distance between the plane of the slits and screen is 120cm. What is the fringe width for = 6500A°? Solution:
(i)
The width of the fringe
D d
Then distance of the third fringe 3w =
3D 3 120 6500 10 8 d 0.2
= 0.117 cm (ii)
Let mth and nth bright fringe of the wavelength coincide. Now position mth bright fringe is ym = n1 and
D d
yn = n2
D d
m 1 5200 4 n 2 6500 5
Now (iii)
Fringe width w =
D 6500 10 10 1.2 d 2 10 3
= 3900 107m = 0.039 cm
Illustration 10: In YDSE light of two wavelengths of 700 nm and 500 nm. If D/d = 10 3 find the minimum distance from central maxima where the maxima of two wavelength coincide again. n1 1 n 2 2 Solution: n1 7 n2 5 D1 y n1 d y 7 10 3 500 10 9
35 10 4
y= 3.5 mm POLARISATION Polarisation of two interfering wave must be same state of polarisation or two source of light should be unpolarised.
8.
BREWSTER LAW
According to this law when unpolarised light is incident at polarising angle (i) on an interface separating a rarer medium from a denser medium, of refractive index as shown in Fig., below such that = tan i then light reflected in the rarer medium is completely polarised. Reflected and refractive rays are perpendicular to each other. en ay tr r
R e (P flec ol te ar d is ra ed y )
d ci In i
Rarer Medium
R ra ef ed ct ra y
9.
Denser medium of refractive index m
REDUCTION IN INTENSITY
Intensity of polarised light is 50% of that of the unpolarised light, i.e., Ip = Iu / 2 where I p = Intensity of polarised light and
Iu = Intensity of unpolarised light.
10. ASSIGNMENT 1.
In Young's double slit experiment the light emited from source has l = 6.5 × 10 –7 m and the distance between the two slits is 1 mm. Distance between the screen and slit is 1 metre. Distance between third dark and fifth birth fringe will be (a) 3.2 mm (b) 1.63 mm (c) 0.585 mm (d) 2.31 mm (b) D 5x6.5x107 x1 x5 n 32.5x104 m d 103 D 5x6.3x107 x1 x3 (2n 1) 16.25x104 m 3 2d 2x10 x5 - x3 1.63 mm.
Solution-
2.
Light is incident normally on a diffraction grating through which first order diffraction is seen at 32o. The second order diffraction will be seen at (a) 84o (b) 48o o (c) 64 (d) None of these
Solution(d) For second order diffraction,
2 2 sin 1 d = 2 sin 320 > 1 Which is not possible. Hence there is no second order diffraction. sin 2
3.
In Young's double slit experiment, if the widths of the slit are in the ratio 4:9, ratio of intensity of maxima to intensity of minima will be (a) 25:1 (b) 9:4 (c) 3:2 (d) 81:16
Solution(a) As ratio of slit widths = Ratio of intensities I 9 a12 9 a1 3 1 or 2 or I2 4 a2 4 a2 2 amax = a1 + a2 = 3 + 2 =5; amin. = 3 - 2 = 1 Imax. (a1 a2)2 (3 2)2 25 Imin. (a1 a2)2 (3 2)2 1
4.
A light source approaches the observer with velocity 0.5 cm. Doppler shift for light of wavelength 550 Å is (a) 616 Å (b) 1833 Å (c) 5500 Å (d) 6160 Å
Solution-
(b)
cv v' c v .v and c v c ' 0 c 5500xc 5500x2 3667 A or ' c v c 0.5c 3
In case of light, v '
0
0
(5500 3667) A 1833 A
5.
Angular width of a central max. is 30o when the slit is illuminated by light of wavelength 6000 Å. Then width of the slit will be approx. (a) 12 × 10–6 m (b) 12 × 10–7 m (c) 12 × 10–8 m (d) –9 12 × 10 m
Solution(b) Angular width = 2 = 300 sin 300 a 6000x1010 a 12000x1010 m 0 sin 30 1/ 2 –7 = 12 x 10 m 6.
Light of wavelength 6000 12 × 10–6 m is incident on a single slit. First minimum is obtained at a distance of 0.4 cm from the centre. If width of the slit is 0.3 mm, then distance between slit and screen will be (a) 1.0 m (b) 1.5 m (c) 2.0 m (d) 2.3 m
Solution-
(c)
Using a sin = n x D or a. n or a n. D x a.x or D n 4 3x10 x4x103 m 2.0m 1x6000x1010 7.
If velocity of a galaxy relative to earth is 1.2 × 106 ms –2 then % increase in wavelength of light from galaxy as compared to the similar source on earth will be (a) 0.3 % (b) 0.4 % (c) 0.5 % (d) 0.6 % (b) v Since c v x c 1.2x106 12 105 x 8 4x103 or 8 3x10 3 10 x100 0.4 i.e., 0.4%
Solution-
8.
Doppler shift for the light of wavelength 6000 Å emitted from the sun is 0.04 Å. If radius of the sun is 7 × 108 m then time period of rotation of the sun will be (a) 30 days (b) 365 days (c) 24 hours (d) 25 days.
Solution-
(d)
Doppler shift, d
v x R c c
2 or R d T c Substituting values for all parameter, we get Rx2 x or T d c T = 25 days
9.
On introducing a thin nuca sgeet if tgucjbess 2 × 10 –6 m and refractive index 1.5 in the path of one of the waves, central bright maxima shifts by n fringes. Wavelength of the wave used is 5000 Å, then n is (a) 1 (b) 2 (c) 5 (d) 10
Solution(d) shift in no. of fringes is given by n = (u-1) t (u 1)t (1.5 1)x2x106 n fringes 2 5000x1010
10.
Two beams of light having intensities I and 4I interfere to produce a fringe patternon the screen. Phase difference between the beams is at point a and at sities at A 2 and B is (a) 3 I (b) 4 I (c) 5 I (d) 6 I
Solution-
(b)
Given I1 = I, I2 = 4I; 1
and 2 2
IA = I1 + I2 + 2 I1I2 cos 1 I1 I2 5I 2 IB I1 I2 2 I1I2 cos 2 I1 I2 2 I1I2 cos
I1 I2 2 I1I2 cos I
IA IB 5I I 4I
11.
White light is used to illuminate the two slits in Young's double slit experiment, separation between the slits is b and the screen is at a distance d(>>b) from the slits. At a point on the screen, directly in front of the slits, certain wavelengths are missing. Some of these missing wavelengths are 2b2 2b2 b2 b2 (b) (c) (d) (a) 3d d 3d d
Solution-
(a,d)
b2 b2 Path difference = d d 2d 2d For missing wavelengths, b2 (2n 1) 2 2d b2 If n = 1, 2 d b2 b2 or If n = 2, 3 x 2 2d 3d 12.
A radar operates at wavelength 50.0 cm. If the beat freqency between the transmitted singal and the singal reflected from aircraft () is 1 kHz, then velocity of the aircraft will be (a) 800 km/hr (b) 900 km/hr (c) 1000 km/hr (d) 1032 km/hr
Solution(b) when source is fixed and observer is moving towards it ca ' . c when source is moving towards observer at rest
a 1 c c c a " v' . c a c a c a 1 c 1
a a 2a c 1 1 1 c c c 2a 2a ' c 0.5x1000 a 250 ms1 2 2 = 900 km/hr
13.
A plane electromagnetic wave of frequency wo falls normally on the surface of a mirror approaching with a relativisitic velocity . Then frequency of the reflected wave will be given c (1 )w 0 1 1 (a) (c) (d) (b) wo (1 )w 0 (1 ) 1 (1 ) (1 )w o
Solution-
(c)
c .w0 c Frequency of waves reflected from mirror and movi9ng towards source, c w" .w ' c c c c x .w0 or w " w0 c c c c 1 c w" w0 c 1 c 1 c wo 1 wo where c 1 1 c
Frequency of Em waves going towards the approaching mirror, w'=
14.
A spectral line of wavelength 0.59 mm is observed in the directions to the opposite edges of the solar disc along its equator. A difference in wavelength equal to () 8 picometer is observed. Period of Sun's revolution around its own axis will be about (Radius of sun = 6.95 × 108 m) (a) 30 days (b) 24 hours (c) 25 days (d) 365 days
Solution(c) or c c
Change in wavelength for two edges = Total change, 2 c 2 or c 2 2R 2 4R and T = 2Rx c c 4x3.14x6.95x108 x0.59x10 6 days 3x108 x8x1012 x86400 24.8 days 25 days
15.
If light with wavelength 0.50 mm falls on a slit of width 10 mm and at an angle o = 30o to its normal. Then angular position of first minima located on right sides of the central Fraunhoffer's diffraction will be at (a) 33.4o (b) 26.8o o (c) 39.8 (d) None of these
Solution(a) For first diffraction minima at angle d(sin - sin 0) = For right of C.M., sin 1 sin 0 d 0.5 = 0.55 i.e., 1 33.370 = 0.5 + 10 16.
Angular width of central maximum in the Fraunhoffer's diffraction pattern is measured. Slit is illuminated by the light of another wavelength, angular width decreases by 30%. Wavelength of light used is (a) 3500 Å (b) 4200 Å (c) 4700 Å (d) 6000 Å
Solution(b) For first diffraction min. d sin = and if angle is small, sin d i.e. Half angular width, d 2 Full angular width w = 2 = d 2 ' Also w' = d ' w' w' or ' w w 0
= 6000 x 0.7 = 4200 A 17.
A parallel beam of white light falls on a thin film whose refractive index is 1.33. If angle of incidence is 52o then thickness of the film for the reflected light to be coloured yellow ( = 6000 Å) most intensively must be (a) 14 (2n + 1) m (b) 1.4 (2n + 1) m
(c) 0.14 (2n + 1) m
(d) 142 (2n + 1) m
(c) sin i sin i 0.788 sin r 0.6 sin r 1.33
Solution-
cos r 1 sin2 r 1 (0.6)2 0.8 For constructive interference on reflection 2t cos r (2n 1) 2 (2n 1) (2n 1)x0.6 t 4 cos r 4x1.33x0.8 =0.14 (22+1) m
18.
A plane monochromatic light falls normally on a diaphragm with two narrow slits separated by a distance d = 2.5 mm. A fringe pattern is formed on the screen placed at D = 100 cm behind the diaphragm. If one of the slits is covered by a glass plate of thickness 10 m, then distance by which these fringes will be shifted will be (a) 2 mm (b) 3 mm (c) 4 mm (d) 5 mm
Solution(a) xd ( 1) tD ( 1)t x D d 3 (1.5 1)x10 x100 0.2 cm or x = 0.25 19.
A two slit Young's experiment is done with monochromatic light of wavelength 6000 Å. Slits are 2 mm apart and fringes are observed on a screen placed 10 cm away from the slits. If a transparent plate of thickness 0.5 mm is placed in front of one of the slit, interference pattern shifts by 5 mm. Then refractive index of transparent plate should be (a) 1.1 (b) 1.2 (c) 1.3 (d) 1.5 (b) xd xd ( 1)t or 1 D Dt 0.5x0.2 1 0.2 1.2 or 1 10x0.05
Solution-
20.
In Young's double slit experiment, using monochromatic light, fringe pattern shifts by a certain distance on the screen when a mica sheet of refractive index 1.6 and thichness 1.964 mm is introduced in the path of one of the two waves. If now mica sheet is removed and distance between slit and screen is doubled, distance between successive max. or min. remains unchanged. The wavelength of the monochromatic light used in the experiment is (a) 4000 Å (b) 5500 Å (c) 5892 Å (d) 6071 Å
Solution-
(c)
xd (u 1) tD (u 1) t or x D d
when distance between screen and slit is doubled, them fringe width As x 2D (u 1) tD d d ( 1) t (1.6 1)x1.964x106 or 2 2
2D d
0
= 0.5892 x 10–6 m = 5892 A
21.
In a two slit experiment with monochromatic light, fringes are obtained on a screen placed at some distance from the slits. If screen is moved by 5 × 10–2 m towards the slits, then change in fringe width is 3 × 10–5 m. If the distance between slits is 10–3 m then wavelength of the light used will be (a) 4000 Å (b) 6000 Å (c) 5890 Å (d) 8000 Å
Solution-
d d
(b)
1 2 or
22.
(D1 D2 ) d(1 2 ) or d D1 D2
0 3x105 x103 6000 A 5x102
White light is used to illuminate two slits in Young's double slit experiment. Separation between the slits is b and the screen is at a distance d (>>b) from the slits. Then wavelengths missing at a point on the screen directly in front of one of the slit are b2 b2 b2 b2 b2 b2 b2 b2 , , , , (a) (b) (c) (d) 2d 4d d 4d 2d 3d d bd (a) b 2 .b b2 d 2d
Solution-
For missing wavelength 1 b2 b2 or 1 2 2d 2d 2 3 2 b b2 or 2 2 2d 3d 2 53 b b2 or 3 2 2d 5d
1 32 53 7 4 , , , , etc. 2 2 2 2
23.
Interference fringes from sodium light (1 = 5890 Å) in a double slit experiment have an angular width 0.20o. To increase the fringe width by 10%, wavelength of light used should be (a) 5892 Å (b) 4000 Å (c) 8000 Å (d) 6479 Å (d) D and angular fringe width, D d d 1 1 / d, 2 2 / d
Solution-
1 1 or 2 1. 2 2 2 1 0 0.22 5890 x 6479 A 0.20
24.
In a Young's double slit experiment, angula width of a fringe formed on a distant screen is 0.1o. If wavelength of light used is 6000 Å , then distance between the slits will be (a) 0.241 mm (b) 0.344 mm (c) 0.519 mm (d) 0.413 mm (b) or d d 5 6x10 x180x7 0.344 mm or d 0.1x22
Solution-
25.
In a Young's double slit experiment two narrow slit 0.8 mm apart are illuminated by the same source of yellow light (l = 5893 Å). If distance between slits and screen is 2m then separation between adjacent bright lines will be (a) 14.73 mm (b) 14.73 cm (c) 1.473 mm (d) 147.3 mm
Solution(a) D 5893x108 x200 0.1473 cm d 0.08 26.
Monochromatic light of wavelength 5000 Å is incident on two slits separated by a distance of 5 × 10–4 m. Interference pattern is seen on the screen placed at a distance of 1 m from the slits. A thin glass plate of thickness 1.5 × 10 –6 m and refractive index 1.5 is laced between one of the slits and the screen. If intensity in the absence of plate was Io then new intensity at the centre of the screen will be I 3I (a) I0 (b) 2 I0(c) o (d) o 2 4 Solution(a) xd ( 1) t D
( 1) Dt (1.5 1)x1.5x106 x1 d 5x104 = 1.5 x 10–3 m
or x
27.
In Young's double slit experiment, separation between the slits is 2 × 10 –3 m and distance of the screen from the slit is 2.5 m. Light in the range of 2000–8000 Å is allowed o fall on the slits. Wavelength in the visible region that will be present on the screen at 10–3 m from the central maxima will be (a) 4000 Å (b) 5000 Å (c) 6000 Å (d) 8000 Å (a) 0 xd 103 x2x103 8000 A D 2.5 For bright line 8000 = n11 n22 n32
Solution-
0
If n1 2 1 4000 A i.e., second bright line for visible region is present. 28.
A double slit experiment is immersed in a liquid of refractive index 1.33. Separation between the slits is 1.0 mm and he distance between slit and screen is 1.33 m. If slits are illuminated by a parallel beam of light whose wavelength is 6300 Å, then fringe width will be (a) 6.3 mm (b) 63 mm (c) 0.63 mm (d) None of these (c) air t l D D D , in liquid 1 air ld d d
Solution-
6300x1010 x1.33 0.63 mm 1.33x103
29.
In a Young's interference experimental arrangement incident yellow light is composed of two wavelength 5890 Å and 5895 Å. between the slits is 1 mm and the screen is placed 1 m away. Order upto which fringes can be seen on the screen will be (a) 384 (b) 486 (c) 512 (d) 589
Solution-
(d)
1 n 1 2 2 n 2 n 1 or 2 n 1
1 2 1 5895 5890 5 2n 5890 5890 1 5890 n 589 10 or
30.
Ratio of intensities between a point A and that of central fringe is 0.853. Then path difference between two waves at point A will be (a) (c) (d) (b) 2 4 8
Solution(c) R2 = a2 + b2 + 2ab cos IR 0.853 Imax. IR 0.853 Imax. 0.853x4I IR = I + I0 + 2I cos = 2I (1 + cos) =0.853 x 4I 4 8
For Test 1.
When light is incident on a soap film of thickness 5 × 10–5 cm, wavelength reflected maximum in the visible region is 5320 Å. Refractive index of the film will be (a) 1.22 (b) 1.33 (c) 1.51 (d) 1.83.
Solution-
(b)
2 (2n 1) (2n 1)x5320x1010 or 2t cos r 2 2x5x105 x102 x1 1.33 2t cos r (2n 1)
2.
A ray of unpolarised light is incident on a glass plate of refractive index 1.54 at polarising angle, then angle of refraction is (a) 33o(b) 44o(c) 57o (d) 90o
Solution(a) tanip or tanip 1.54
i.e., ip = tan–1 1.54 = 570, But r + ip = 900 r = 900 - ip = 900 - 57 = 330.
3.
Two waves of same intensity produce interference. If intensity at maximum is 4I, then intensity at the minimum will be (a) 0 (b) 2I (c) 3I (d) 4I.
Solution - (a) I1 = I 2 = I = a 2 Imax. = (a + a)2 + (2a)2 = 4a2 = 4I Imin. = (a - a)2 = 0
4.
First diffraction minima due to a single slit of width 1.0 x 10 –5 cm is at 300. Then wavelength of light used is 0
0
(a) 400 A
(b) 500 A
0
0
(c) 600 A
(d) 700 A
Solution - (b) As a sin = n or
a sin n
or 1.0x10 5 x sin 300 1.0x10 5 x = 0.5 x 10–5 = 5 x 10–6 cm
1 2
0
= 500 A 0
5.
Light of wavelength 6000 A is normally incident on a slit. Angular position of second minimum from central maximum us 300. Width of the slit should be (a) 12 x 10–5 cm (b) 18 x 10–5 cm (c) 24 x 10–5 cm (d) 36 x 10–5 cm
Solution - (c) a sin n
n 2x6000x1010 a sin sin 300 2x6000x1010 4x6000x1010 1/ 2 = 24 x 10–7 m = 24 x 10–5 cm 0
6.
Light of wavelength 6328 A is incident normally on a slit of width 0.2 mm. Angular width of the central maximum on the screen will be (a) 0.90 (b) 0.180 (c) 0.540 (d) 0.360
Solution - (d) a sin n n 6328x1010 x1 a 0.2x103 –6 = 3164 x 10 rad = 0.003164 radian Angular width 2 sin
1800 0.360 = 2 x 0.003164 x 0
7.
Light of wavelength 5000 A is incident normally on a slit. First minimum of diffraction pattern is formed at a distance of 5 mm from the central maximum. If slit width is 0.2 mm, then distance between slit and screen will be (a) 1 m (b) 1.5 m (c) 2.0 m (d) 2.5 m
Solution - (c) a sin n x or a. n D ax 2x104 x5x103 2.0 metre D n 1x5000x1010 8.
Light of wavelength is incident on a slit. First minima of the diffraction pattern is found to lie at a distance of 6 mm from the central maximum on a screen placed at a distance of 2 m from the slit. If slit width is 0.2 mm, then wavelength of the light used will be 0
(a) 4000 A
0
(b) 6000 A
0
0
(c) 7000 A
(d) 7400 A
Solution - (b) a sin n x ax a. n or D nD 4 0 2x10 x6x103 or 6000 A 1x2
9.
A slit 5 cm wide when irradiated by waves of wavelength 10 mm results in the angular spread of the central maxima on either side of incident light by about (a) 1/2 radian (b) 1/4 radian (c) 3 radian (d) 1/5 radian.
Solution - (d) Angular spread on either side is given by 1 radians a 5 10.
In Young's double slit experiment, ten slits separated by a distance of 1 mm are illuminated by a monochromatic light source of wavelength 5 x 10 –7 m. If the distance between slit and screen is 2 metre, then separation of bright lines in the interference pattern will be (a) 0.5 mm (b) 1.0 mm (c) 1.5 mm (d) 1.75 mm
Solution - (b) D 5x107 x2 103 m 3 d 10 = 1 mm. 0
11.
A star is moving away from earth and shift in spectral line of wavelength 5700 A is 0
1.90 A . Velocity of the star is (a) 50 km s–1
(b) 70 km s–1
(c) 80 km s–1 Solution - (d) c 3x108 x1.9x1010 or 100 km s1 10 5700x10 12.
(d) 100 km s–1
c
Intensity of central bright fringe due to interference of two identical coherent monochromatic sources is I. If one of the source is switched off, then intensity of central bright fringe becomes I I (a) (b) 2 3 I (c) (d) I. 4
Solution - (c) I = (a+a)2 = 4a2, I'=(a)2 = a2 I I' a2 1 I' . 2 , 4 I 4a 4 13.
In two separate set ups of Young's double slit experiment, fringes of equal width are observed when lights of wavelengths in the ratio 1:2 are used. If the ratio of slit separation in two cases is 2:1, then ratio of distances between the plane of slits and the screen in the two set ups is (a) 1 : 2 (b) 2 : 1 (c) 1 : 4 (d) 4 : 1
Solution - (d) D D 1 1 2 2 d1 d2 D d 2 2 1 1 x 2 x 4 :1 D2 d2 1 1 1 14.
In an experiment to demonstrate interference of flight using Young's slits, separation of two slits is doubled. In order to maintain same spacing of fringes, distance D of screen from slits must be changed to D (a) D (b) 2 3D (c) 2D (d) 4 (c) D (2d) d 2d i.e., D' = 2D.
Solution-
15.
In an interference pattern by two identical slits, intensity of central maxima is I. If one slit is closed, intensity of central maxima changes to I0. Then I and I0 are related by (a) I = I0 (b) I = 2I0 (c) I = 3I0 (d) I = 4I0
Solution - (d) I1 = I 2 = a 2 Imax. = (a+a)2 = 4a2 = I If one slit is closed, Intensity, I0 = (a)2 I 4a2 2 4 or I 4I0 I0 a
16.
In a young's double slit experiment, two slits are illuminated by a mixture of two 0
0
wavelengths 12000 A and 10000 A . At 6.0 mm from the common central bright fringe on a screen 2 m away from the slits, a bright fringe of one interference pattern coincides with the bright fringe of other. Distance between the slits should be (a) 1.0 mm (b) 1.5 mm (c) 2.0 mm (d) 2.5 mm Solution - (c) D D x n (n 1) ' d d n x 12000 = (n+1) x 10000 D or n = 5 and x = n d n D 5x12000x1010 x2 d m x 6x103 = 2 x 10–3 m = 2 mm 17.
In a young's double slit experiment, slits are illuminated by a monochromatic source 0
of wavelength 6000 A and fringes are obtained. If screen is moved by a distance of 5 cm towards slits, change in fringe width is 3 x 10–5 m. Then separation between the slits will be (a) 1 mm (b) 1.2 mm (c) 1.5 mm (d) 1.63 mm Solution - (a) D' D and ' d d (D D') ' d (D D') d or ( ')
6000x1010 x5x102 m 103 m 1mm 5 3x10
18.
An unpolarized beam of light is incident on a group of three polarizing sheets which are arranged in such a way that plane of rotation of one make an angle of 40 0 with the adjacent one. The 0% of incident light transmitted by first polarizer will be (a) 33% (b) 16.6% (c) 50% (d) 25%
Solution - (c) First polariser, polarises the light and hence intensity of light reduces by 50%. 19.
80 gm of impure sugar when dissolved in a litre of water gives an optical rotation of 9.90 when placed in a tube of length 20 cm. If concentration of sugar solution is 75 gm/litre then specific rotation of sugar is (a) 440 (b) 550 0 (c) 66 (d) 730
Solution - (c) 9.9 S 660. c or S ic 2x0.075 lS 20.
In Young's double slit experiment, distance between the slits S1 and S2 is d and the distance between slits and screen is D. Then first missing wavelength on the screen in front of S1 is d2 d2 (a) (b) D 2D D (d) None of these (c) 2 d
Solution - (a) d2 2D For first missing wavelength, d2 d2 or 2D 2 D
Path difference, x
21.
In Young's double slit experiment, 12 fringes are obtained to be formed in a certain segment of the screen when light of wavelength 600 nm is used. If wavelength of light is changed to 400 nm, number of fringes observed in the same segment of the screen is given by (a) 18 (b) 24 (c) 30 (d) 36.
Solution - (a) n11 n22
n2
22.
n11 12x600 18 400 2
In Young's double slit experiment, interference pattern is found to have an intensity ratio between bright and dark fringes as 9. Then amplitude ratio will be (a) 1 (b) 2 (c) 9 (d) 3.
Solution - (b) amax. a1 a2 ; amin. a1 a2 Imax. (a1 a2 )2 9 a1 2 and a2 1 Imin. (a1 a2 )2 1
a1 2 2 a2 1
0
23.
Interference fringes from sodium light ( 5890 A) in a double slit experiment have an angular width 0.200, To increase the fringe width by 10%, new wavelength should be 0
0
(a) 5896 A
(b) 7321A
0
0
(c) 6300 A
(d) 6479 A
Solution - (d) 1 1 , 2 2 d d 1 1 2 1x 2 2 2 1 0 0.22 6479 A or 2 5890x 0.20 24.
Sol
25.
Sol
If A is the amplitude of the wave coming from a point source at distance r then (A) (B) A µ r- 2 A µ r- 1 (C) (D) A µ r2 A µ r1 1 (B) For a point source, I µ 2 r 1 1 and A µ I \ Aµ 2 or Aµ r r If A is the amplitude of the wave coming from a line source at a distance r, then Aµr (A) (B) A µ r - 1/ 2 -2 (D) (C) A µ r1/ 2 Aµr 1 (B) For a line source, I µ r \ A µ r - 1/ 2 and A µ I
26.
Phase difference between two waves having same frequency (v) and same amplitude (A) is 2/3. If these waves superimpose each other, then resultant amplitude will be (A) 2A (B) 0 (C) A (D) A2.
Sol
(C) A R = A 2 + A 2 + 2A ×A cos q
æ2p ö æ 1ö = A 2 + A 2 + 2A 2 cos ç ÷ = A 2 + A 2 + 2A 2 ç- ÷ è3 ø è 2ø A2 + A2 - A2 = A2 = A
27.
Sol
28.
Sol
Ratio of amplitudes of the waves coming from two slits having widths in the ratio 4 : 1 will be (A) 1:2 (B) 2:1 (C) 1:4 (D) 4:1 (B) Intensity slit width I1 4 = I2 1
\
a1 I = 1 = 4 = 2 :1 a2 I2
\
a1 = 2 :1 a2
Two slits S1 and S2 illuminated by a white light source give a white central maxima. A transparent sheet of refractive index 1.25 and thickness t 1 is placed in front of S1. Another transparent sheet of refractive index 1.50 and thickness t2 is placed in front of S2. If central maxima is not effected, then ratio of the thickness of the two sheets will be (A) 1:2 (B) 2:1 (C) 1:4 (D) 4:1 (B) Since there is no shift in central maxima. Therefore path difference introduced by the two sheets are equal i.e. (m1 - 1) t1 = (m2 - 1) t 2 where m1 and m2 are refraction index i.e.
29.
æp ö Two coherent waves are represented by y1 = a1 cos t and y2 = a2 cos ç - wt ÷ . è2 ø Their resultant intensity after interference will be a1 - a2 a1 + a2 (A) (B) (C)
Sol
t1 (m2 - 1) (1.5 - 1) 0.5 = = = =2 t 2 (m1 - 1) (1.25 - 1) 0.25
a12 - a22
(D)
æp ö (D) y1 = a1 cos wt ; y 2 = a2 cos ç - wt ÷ è2 ø Since the two wave differ in phase by \
30.
Sol
a12 + a22
(
p 2
)
y = a12 + a22 .
If a thin film of thickness t and refractive index is placed in the path of light coming from a source S, then increase in length of optical path is (A) t (B) /t (C) ( – 1) t (D) None of these. (C) Increase in optical path = mt - t = (m- 1) t
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