34 JEE MAIN CHAPTERWISE EXPLORER
1. (a) According to equation of motion,
2. (d): Let k be the spring constant of v2 = v12 + 2 gh = (0.4)2 + 2 ´10 ´ 0.2
spring and it gets extended by length » 2 m s –1
x0 in equilibrium position.
In equilibrium, \ According to equation of continuity
kx 0 + FB = Mg
a 1 v1 = a2 v 2
kx0 + s L Ag = Mg 2
2
´ 0.4 =
( ) ( ) p ´ 2
sLAg 8 ´ 10 -3 p ´ d2
2 2 2 ´ 2
Mg - d2 = 3.6 × 10 –3 m
x0 = k
( ) =Mg
k 1 - s2L MA 8. (d) : Here, surface tension, S = 0.03 N m –1
r1 = 3 cm = 3 × 10 –2 m, r2 = 5 cm = 5 × 10 –2 m
3. (d) : According to Newton’s law of cooling the option (d) Since bubble has two surfaces,
represents the correct graph. Initial surface area of the bubble
= 2 × 4pr 1 2 = 2 × 4p × (3 × 10 –2 ) 2
4. (d) : According to Newton’s law of cooling = 72p × 10 –4 m 2
Final surface area of the bubble
d q = - k (q - q 0 ) or q d q = - kdt = 2 × 4pr 2 2 = 2 × 4p(5 × 10 –2 ) 2 = 200p × 10 –4 m 2
dt - q 0 Increase in surface energy
= 200p × 10 –4 – 72p × 10 –4 = 128p × 10 –4
Integrating both sides, we get ò q d q = ò -kdt \ Work done = S × increase in surface energy
- q 0 = 0.03 × 128 × p × 10 –4 = 3.84p × 10 –4
= 4p × 10 –4 J = 0.4p mJ
log e( q – q0 ) = – kt + C
where C is a constant of integration.
So, the graph between loge (q – q 0) and t is a straight line with
a negative slope. Option (d) represents the correct graph.
5. (c) : Increase in length, DL = LaDT F
\ DL = aD T
L
The thermal stress developed is 9. (d) : U = a - b
x12 x6
T DL
S =Y L = Y aD T ( ) Force, F dU d a b
dx dx x12 x6
or T = SYaDT T T = - = - -
From FBD of one part of the wheel, é -12a 6b ù é12a 6b ù
êë x13 x7 ûú êë x13 x7 ûú
or F = 2T = - + = -
Where, F is the force that one part of the wheel applies on
the other part. At equilibrium F = 0
\ F = 2SYaDT
6. (c) : The force due to the surface tension will balance the \ 12a - 6b = 0 or x 6 = 2 a
x13 x7 b
weight.
F = w ( ) ( ) U at equilibrium =a - b
2 a 2a
2TL = w F = 2T L 2
w b b
2 L
T = ab2 b2 b2 b2 b 2
4 a2 2a 4a 2a 4a
Substituting the given values, we get = - = - = -
T = 1.5 ´ 10-2 N = 0.025 Nm -1 w U(x = ¥) = 0
´ 30 ´ 10-2 m
2 D = [U(x = ¥) – U at equilibrium ]
7. (d) : Here, d 1 = 8 × 10 –3 m ( ) = éëê0 - b2 ù b 2
4a ûú 4 a
v1 = 0.4 m s –1 - =
h = 0 .2 m
Properties of Solids and Liquids 35
10. (c) : As ro il < r water , so oil should be over the water. As 15. (a) : The liquid 1 is over liquid 2.
r > r oil, so the ball will sink in the oil but r < r water so it
will float in the water. Therefore r 1 < r2 . If r3 had been greater r3
than r 2 , it will not be partially inside
Hence option (c) is correct. but anywhere inside liquid 2 if r 3 = r 2 l 2
or it would have sunk totally if r 3 had K 2
11. (d) : For the same material, Young’s modulus is the same been greater than r2 .
and it is given that the volume is the same and the area of \ r 1 < r 3 < r 2 . T 2
crosssection for the wire l1 is A and that of l2 is 3A.
V = V1 = V2 16. (d) : Let T be the l 1
V = A × l1 = 3A × l2 Þ l2 = l1/3 temperature of the
Y = F / A Þ F1 = YA Dl1l 1 interface. K 1
D l / l
Since two section of rod
F2 = Y 3 A Dl2l 2 are in series, rate of flow T 1
Given Dl1 = Dl2 = Dx (for the same extension)
of heat in them will be equal
( ) \ \ K1A [T1 - T ] = K2 A [T – T2 ]
F2 = Y ×3A× Dx = 9× YADx = 9F1 or 9F . l1 l2
l1 / 3 l1
or K1 l2 (T1 – T) = K2 l1 ( T – T 2 )
12. (b) : Heat flow can be compared to charges flowing in a or T(K1 l 2 + K 2l 1) = K 1l 2 T 1 + K2 l 1 T2
conductor.
or T = K1 l2 T1 + K2 l1 T2 .
Current is the same. dQ K1 l2 + K2 l1
The potential difference V1 – V q 1 dt q 2 Force ´ L WL
V2 A´l Al
at any point = I × Resistance = I ´ rAl V 1 current V 17. (b) : Y = = P
L
Potential difference is µ l but negative. \ l = WAYL 2
Due to pulley arrangement, the length L
As l increases, potential decreases q q max, Vmax of wire is L/2 on each side and so the
(temperature decreases) but it is a elongation will be l/2. For both sides, W W
elongation = l.
straight line function. W
Potential difference is proportional q min, Vmin
to resistance (thermal as well as x 18. (d) : Terminal velocity = v
electric). viscous force upwards = weight of sphere downwards
13. (d) : The force acting upwards
2prT ; hpr 2rg, the force acting down or or 6phrv = æèç 43 pr3 öø÷ (r – s) g
For gold and silver spheres falling in viscous liquid,
T µ h without making finer corrections. Soap
reduces the surface tension of water. The
height of liquid supported decreases. But v g = rg – s = 19.5 – 1.5 = 18 = 2
vs rs – s 10.5 – 1.5 9 1
it is also a wetting agent. Therefore the meniscus will not \
be convex as in mercury. Therefore (d). v g
2
14. (b) : The forces acting on the solid Vr 2 g or vs = = 02.2 = 0.1 m/s.
ball when it is falling through a
liquid are mg downwards, thrust viscous force 19. (c) : Energy radiated by sun, according to Stefan's law,
E = sT4 × (area 4pR 2) (time)
by Archimedes principle upwards This energy is spread around sun in space, in a sphere of
radius r. Earth (E) in space receives part of this energy.
and the force due to the force
of friction also acting upwards.
The viscous force rapidly
increases with velocity, attaining R
r
a maximum when the ball reaches
Sun
the terminal velocity. Then the mg = Vr 1 g Earth
acceleration is zero. mg – Vr 2g –
kv 2 = ma where V is volume, v is the terminal velocity.
When the ball is moving with terminal velocity = sT 4 ´ 4pR 2 ´ time
a = 0. Therefore Vr 1g – Vr 2g – kv 2 = 0. Energy
Area of envelope 4 pr 2
Þ v = Vg (r1 - r2 ) . Energy incident per unit area on earth = sT 4R 2 ´ time
k r2
36 JEE MAIN CHAPTERWISE EXPLORER
æ R2sT 4 ö 26. (d) : From first surface, Q1 = KA(T2 - T ) t
ç x
\ Power incident per unit area on earth = è r 2 ÷
ø
2sT 4 (2K ) A(T - T1 ) t
\ Power incident on earth = pr 2 ´ R r 2 From second surface, Q2 = ( 4 x )
0
20. (d) : Energy stored per unit volume At steady state,
= 12 ×stress× strain Q1 = Q2 Þ KA(T2x - T )t = 2KA(T - T1 ) t
4 x
or 2(T2 – T) = (T – T 1 )
= Stress × stress = S 2 . or T = 2T 2 + T1
2Y 2Y 3
21. (b) : In a freely falling elevator g = 0 \ Q1 = KxA éëêT2 - 2T 2 + T1 ù t
Water will rise to the full length i.e., 20 cm to tube. 3 ú
û
22. (a) : For conduction from inner sphere to outer one, or é A(T2 - T1 ) K ù f = KA é T2 3- T1 ùúû ´1
êë x ûú x ëê
dQ = -KA dT ´ (time dt )
dr or f = 13 .
or dQ = -K ´ (4pr 2 ) dT 27. (b) : Initial momentum = E/c
dt dr Final momentum = –E/c
\ Radial rate of flow Q = -4 pKr 2 dT \ E çæè E ÷øö 2 E
dr c c c
Change of momentum = - - =
\ r2 dr = -4 pK T2 \ Momentum transferred to surface = 2 E .
r2 c
Qò ò dT
r1 T1
or Q é r1 - r2 úûù = 4 pK [T2 - T1 ] 28. (d) : According to Stefan's law,
êë r1r2 Radiant energy E = (sT4 ) × area × time
or Q = 4pK (T1 - T2 ) r1r2 \ E 2 = s(2T )4 ´ 4p(2R) 2 ´ t = 16 ´ 4
(r2 - r1 ) E1 sT 4 ´ (4pR) 2 ´ t
æ r1r2 ÷øö . \ E 2 = 64.
çè r2 - r1 E1
\ Q is proportional to
29. (d) : Elastic energy per unit volume
23. (c) : Pressure inside the bubble = P0 + 4T = 21 ´ stress ´strain
r \ Elastic energy
Smaller the radius, greater will be the pressure. Air flows
from higher pressure to lower pressure. Hence air flows = 1 ´ stress ´ strain ´ volume
2
from the smaller bubble to the bigger.
24. (b) : Retarding viscous force = 6phRv = 1 ´ F ´ DL ´ ( AL )
obviously option (b) holds goods. 2 A L
= 1 F DL = 1 ´ 200 ´10–3 = 0.1 J .
2 2
25. (d) : Young's modulus FL ....... (i)
Y = Al ....... (ii) 30. (d) : According to Newton's law of cooling, rate of cooling
is proportional to Dq.
\ F = YLAl \ (Dq)n = (Dq) or n = 1.
or dW = F dl = YAlL(dl ) 31. (d) : Wien's law
32. (b) : v = 2gh = 2 ´10 ´ 20 = 20 m/s.
YA l YAl 2
or ò dW = L dl = 2 L 33. (a) : Energy radiated
ò l E = sT 4 × (area 4pR2 ) × time × e
0
or Workdone = YAl 2 E1 = (4000)4 ´ (1)2 ´1´ 4 pse = 1
2 L E 2 (2000)4 ´ (4)2 ´1´ 4 ps e 1 .
From (i) and (ii)
Workdone = Fl . 34. (a) : A good absorber is a good emitter but black holes do
2 not emit all radiations.
Thermodynamics 37
CHAPTER THERMODYNAMICS
8
1. The above pv diagram represents 6. 100 g of water is heated from 30°C to 50°C. Ignoring the
slight expansion of the water, the change in its internal energy
the thermodynamic cycle of an is (specific heat of water is 4184 J kg –1 K– 1)
(a) 4.2 kJ (b) 8.4 kJ (c) 84 kJ (d) 2.1 kJ
engine, operating with an ideal (2011)
monoatomic gas. The amount of
heat, extracted from the source in a
single cycle is 7. A diatomic ideal gas is used in a Carnot engine as the working
(a) 4p 0 v 0 (b) p 0 v0 substance. If during the adiabatic expansion part of the cycle
( ) (c)13 p0v0 ( ) (d)11 p0v0 (2013) the volume of the gas increases from V to 32V, the efficiency
2 2
of the engine is
2. A Carnot engine, whose efficiency is 40%, takes in heat from (a) 0.25 (b) 0.5 (c) 0.75 (d) 0.99
a source maintained at a temperature of 500 K. It is desired (2010)
to have an engine of efficiency 60%. Then, the intake Directions: Question numbers 8, 9 and 10 are based on the
temperature for the same exhaust (sink) temperature must be following paragraph.
(a) 1200 K (b) 750 K Two moles of helium gas are taken over the cycle ABCDA,
(c) 600 K as shown in the P T diagram.
(d) efficiency of Carnot engine cannot be made larger than 50% 2 × 105 A B
(2012)
3. Helium gas goes through a cycle 2P0 B C P(Pa)
ABCDA (consisting of two isochoric P0 A D 1 × 10 5 D C
and two isobaric lines) as shown in 300 K
figure. Efficiency of this cycle is T
500 K
nearly (Assume the gas to be close
to ideal gas) V0 2V0 8. Assuming the gas to be ideal the work done on the gas in
(a) 9.1% (b) 10.5% (c) 12.5% (d) 15.4% taking it from A to B is
(2012) (a) 200R (b) 300R (c) 400R (d) 500R
4. Three perfect gases at absolute temperatures T 1 ,T 2 and T3 are 9. The work done on the gas in taking it from D to A is
mixed. The masses of molecules are m1 , m 2 and m3 and the (a) – 414R (b) + 414R (c) – 690R (d) + 690R
number of molecules are n1 , n 2 and n3 respectively. Assuming
no loss of energy, the final temperature of the mixture is 10. The net work done on the gas in the cycle ABCDA is
(a) zero (b) 276R (c) 1076R (d) 1904R
(T1 + T2 + T3 ) n1T1 + n2T2 + n3T3 (2009)
3 n1 + n2 + n3
(a) (b) 11. An insulated container of gas has two chambers separated by
(c) n1T12 + n2T22 + n3T3 2 (d) n12T12 + n22T22 + n32T3 2 an insulating partition. One of the chambers has volume V1
n1T1 + n2T2 + n3T3 n1T1 + n2T2 + n3T3 and contains ideal gas at pressure P1 and temperature T1 . The
other chamber has volume V2 and contains ideal gas at pressure
(2011) P2 and temperature T 2. If the partition is removed without
doing any work on the gas, the final equilibrium temperature
5. A Carnot engine operating between temperatures T1 and T2 of the gas in the container will be
has efficiency 1 . When T 2 is lowered by 62 K, its efficiency (a) T1T2 (P1V1 + P2V2 ) (b) T1T2 (P1V1 + P2V2 )
6 P1V1T1 + P2V2T2 P1V1T2 + P2V2T1
increases to 13 . Then T 1 and T2 are, respectively (c) P1V1T1 + P2V2T2 (d) P1V1T2 + P2V2T1
(a) 372 K and 310 K (b) 372 K and 330 K P1V1 + P2V2 P1V1 + P2V2
(2004, 2008)
(c) 330 K and 268 K (d) 310 K and 248 K (2011)
38 JEE MAIN CHAPTERWISE EXPLORER
12. A Carnot engine, having an efficiency of h = 1/10 as heat (c) It is not applicable to any cyclic process
(d) It is a restatement of the principle of conservation of
engine, is used as a refrigerator. If the work done on the
energy
system is 10 J, the amount of energy absorbed from the (2005)
reservoir at lower temperature is
(a) 100 J (b) 99 J (c) 90 J (d) 1 J 18. Which of the following statements is correct for any
thermodynamic system?
(2007) (a) The internal energy changes in all processes.
(b) Internal energy and entropy are state functions.
13. When a system is taken from state a f (c) The change in entropy can never be zero.
i to state f along the path iaf, it is (d) The work done in an adiabatic process is always zero.
b (2004)
found that Q = 50 cal and W = 20
(d) 66 cal 19. A Carnot engine takes 3 × 10 6 cal of heat from a reservoir at
cal. Along the path ibf Q = 36 i (2007)
cal. W along the path ibf is 627°C, and gives it to a sink at 27°C. The work done by the
(a) 14 cal (b) 6 cal (c) 16 cal
engine is
14. The work of 146 kJ is performed in order to compress one (a) 4.2 × 106 J (b) 8.4 × 106 J
kilo mole of gas adiabatically and in this process the (c) 16.8 × 10 6 J (d) zero. (2003)
temperature of the gas increases by 7ºC. The gas is 20. Which of the following parameters does not characterize the
(R = 8.3 J mol– 1 K– 1 ) thermodynamic state of matter?
(a) monoatomic (a) temperature (b) pressure
(b) diatomic (c) triatomic (c) work (d) volume. (2003)
(d) a mixture of monoatomic and diatomic. 21. During an adiabatic process, the pressure of a gas is found to
(2006) be proportional to the cube of its absolute temperature. The
15. A system goes from A to B via ratio CP /CV for the gas is
two processes I and II as shown P
(a) 4/3 (b) 2 (c) 5/3 (d) 3/2.
(2003)
in figure. If DU1 and DU2 are II 22. "Heat cannot by itself flow from a body at lower temperature to
the changes in internal energies A B a body at higher temperature" is a statement or consequence of
in the processes I and II
(a) second law of thermodynamics
respectively, then I (b) conservation of momentum
(a) DU2 > DU 1 V (c) conservation of mass
(b) DU2 < DU2 (d) first law of thermodynamics. (2003)
(c) DU1 = DU 2 23. Even Carnot engine cannot give 100% efficiency because we
(d) relation between DU1 and DU 2 cannot be determined cannot
(2005) (a) prevent radiation (b) find ideal sources
16. The temperatureentropy T (c) reach absolute zero temperature
diagram of a reversible 2T 0
engine cycle is given in the T 0 (d) eliminate friction. (2002)
figure. Its efficiency is
(a) 1/3 S 24. Which statement is incorrect?
(b) 2/3 S 0 2S 0 (a) all reversible cycles have same efficiency
(c) 1/2 (b) reversible cycle has more efficiency than an irreversible
(d) 1/4 (2005) one
(c) Carnot cycle is a reversible one
(d) Carnot cycle has the maximum efficiency in all cycles.
(2002)
17. Which of the following is incorrect regarding the first law of 25. Heat given to a body which raises its temperature by 1°C is
thermodynamics?
(a) It introduces the concept of the internal energy (a) water equivalent (b) thermal capacity
(b) It introduces the concept of entropy
(c) specific heat (d) temperature gradient.
(2002)
Answer Key
1. (c) 2. (b) 3. (d) 4. (b) 5. (a) 6. (b)
7. (c) 8. (c) 9. (b) 10. (b) 11. (b) 12. (c)
13. (b) 14. (b) 15. (c) 16. (a) 17. (b, c) 18. (b)
19. (b) 20. (c) 21. (d) 22. (a) 23. (c) 24. (a)
25. (b)
Thermodynamics 39
1. (c) : Heat is extracted from the \ CV = 3 R and CP = 52 R
source in path DA and AB. 2
A B
Along path DA, volume is constant. D C Along the path AB, heat supplied to the gas at constant volume,
Hence, \ DQAB = nCV DT = n 3 RDT = 3 V0 DP = 23 P0V0
DQ DA = nC vDT = nCv ( TA – TD ) 2 2
According to ideal gas equation
Along the path BC, heat supplied to the gas at constant pressure,
pv
pv = nRT or T = nR \ DQBC = nCP DT = n 5 RDT = 25 (2P0 )DV = 5 P0V0
2
For a monoatomic gas, Cv = 32 R
Along the path CD and DA, heat is rejected by the gas
( ) \ 3 é 2 p0v0 p0v0 úùû = 3 Efficiency, h = Work done by the gas ´ 100
DQDA = n 2 R ëê nR - nR 2 p0v0 Heat supplied to the gas
Along the path AB, pressure is constant. Hence = P0V0 ´ 100 = 21030 = 15.4%
DQ AB = nCp DT = nCp ( TB – TA )
23 P0V0 + 5 P0V0
For monoatomic gas, Cp = 52 R
4. (b) : The final temperature of the mixture is
( ) \ 5 R é 2 p0 2v0 2 p0 v0 ù 10 p0v0 T1n1 + n2T2 + n3T3
DQAB = n 2 êë nR - nR úû = 2 Tm ixture = n1 + n2 + n3
\ The amount of heat extracted from the source in a single
cycle is 5. (a) : The efficiency of Carnot engine,
DQ = DQ DA + DQ AB h = çæè1 - T2 ö
T1 ø÷
= 3 p0v0 + 10 p0v0 = 123 p0v0 \ 1 = èçæ1 - T2 øö÷ ( ) Given, h= 1
2 2 6 T1 6
2. (b) : Efficiency of Carnot engine, T2
T1
h = 1 - T2 = 5 Þ T1 = 6T 2 ...(i)
T1 6 5
where T 1 is the temperature of the source and T 2 is the As per question, when T2 is lowered by 62 K, then its efficiency
temperature of the sink. becomes 1
3
For 1s t case
h = 40%, T1 = 500 K \ 31 = æçè1 - T2 - 62 ö
T1 ÷ø
\ 40 = 1 - T2 Þ T2 = 1 - 40 = 3
100 500 500 100 5 T2 - 62 = 1 - 1
T1 3
T2 = 35 ´ 500 = 300 K
T2 56 - T62 2 = 2
For 2n d case 3 (Using (i))
h = 60%, T 2 = 300 K
\ 16000 = 1 - 3T010 Þ 300 = 1 - 60 = 2 5(T2 - 62) = 2
T1 100 5 6T2 3
T1 = 5 ´ 300 = 750 K 5T2 – 310 = 4T2 Þ T2 = 310 K
2 From equation (i),
3. (d) : In case of a cyclic process, work done is equal to the area 6 ´ 310
5
under the cycle and is taken to be positive if the cycle is T1 = = 372 K
clockwise. 6. (b) : DQ = msDT
Here, m = 100 g = 100 × 10– 3 kg
\ Work done by the gas s = 4184 J kg –1 K– 1 and DT = (50 – 30) = 20°C
\ DQ = 100 × 10– 3 × 4184 × 20 = 8.4 × 103 J
W = Area of the rectangle ABCD = P0 V 0
Helium gas is a monoatomic gas.
40 JEE MAIN CHAPTERWISE EXPLORER
As DQ = DU + DW (. . . DW = 0) 11. (b) : As this is a simple mixing of gas, even if adiabatic
\ Change in internal energy conditions are satisfied, PV = nRT for adiabatic as well as
DU = DQ = 8.4 × 103 J = 8.4 kJ isothermal changes. The total number of molecules is
conserved.
7. (c) : For an adiabatic process
TV g–1 = constant \ n1 = P1V1 , n2 = P2V2
R1T1 RT2
\ T1V1g -1 = T2V2 g -1
æ VV12 ö÷ø ( ) g -1 g - 1 Final state = (n 1 + n 2) RT
èç
T1 = T2 = T2 32V = T2 (32) g - 1 P1V1 P2V2 = T2P1VR1T+1TT21 P2V2
V RT1 RT2
(n1 + n2 ) = +
For diatomic gas , g = 7 T = T1n1 + T2n2 , T = T1T2(P1V1 + P2V2 )
5
\ T1 = T2 (32) 7 - 1 = T2 (32) 2/5 = T 2( 25 ) 2/5 = 4T 2 n1 + n2 T2P1V1 + T1P2V2
5
( ) Efficiency of the engine, = 1 T2 = 1 12. (c) : For Carnot engine efficiency h = QH - Q L
h - T1 1 - 4 QL
h = 3 = 0.75 Coefficient of performance of a refrigerator b = 1 -h
4 h
P(in pascal) B 1 - 1
b = 10 = 9
8. (c) : 2 × 10 5 A
1/10
Also b = Q L (where W is the work done)
W
or Q L = b × W = 9 × 10 = 90 J.
1 × 105 D C 13. (b) : According to first law of thermodynamics for the path
300 K
T iaf,
500 K
Q iaf = DU iaf + Wi af
Path AB, P is the same, DT is 200 K. or DU iaf = Q iaf – Wi af
PV = nRT for all process
\ PDV = nRDT = 2R 200 = 400R. = 50 – 20 = 30 cal
Work done on the gas from A to B = 400R.
For the path ibf, a f
9. (b) : D to A, temperature remains the same. b
Q ibf = DU ibf + Wi bf
\ Work done by the gas = W = nRT ln VV12
= nRT ln P1 Since DU iaf = DU ibf , change in
P2
internal energy are path i
Þ W = –600R (0.693) = –415.8R.
This is the work done by the gas independent.
\ Work done on the gas = +415.8R.
Nearest to (b). Q ibf = DU iaf + Wi bf
\ Wi bf = Q ibf – DU iaf = 36 – 30 = 6 cal.
10. (b) : Total work done on the gas when taking from A to B = 400R,
from C to D is equal and opposite. 14. (b) : According to first law of thermodynamics
They cancel each other. DQ = DU + DW
For taking from D to A, work done on the gas = +414R.
Work done on the gas in taking it from B to C, pressure is For an adiabatic process, DQ = 0
decreas ed, temperature remain the same, volume increas es. \ 0 = DU + DW
Þ WB C + WD A = 2 ln 2(500R – 300R).
Þ WB C + DA = (2 ln 2) × (200R) or DU = – DW
= 400R × 0.693 = 277R. or nC V DT = – DW
\ Work done along AB and CD cancel each other because
pressure changes but temperature is the same. or CV = - DW = -(-146) ´10 3
Net work done on the gas of 2 moles of helium through the nDT (1´103 ) ´ 7
whole network = 277R per cycle or nearest to the answer (b).
= 20.8 J mol –1 K– 1
For diatomic gas,
CV = 5 R = 52 ´ 8.3 = 20.8 J mol-1 K -1
2
Hence the gas is diatomic.
15. (c) : DU1 = DU 2, because the change in internal energy
depends only upon the initial and final states A and B.
Thermodynamics 41
16. (a) : Efficiency h = 1 - Q2 20. (c) : The work does not characterize the thermodynamic
Q1 state of matter.
Q2 = T0 (2S0 – S0 ) = T 0 S 0 21. (d) : In an adiabatic process, T g = (constant) Pg -1
Q1 = T0 S0 + T0S0 = 23 T0 S 0 or Tg/g–1 = (constant) P
2
Given T3 = (constant) P
h = 1- T0S 0 ´ 2 = 1- 2 = 13 . \ g g = 3Þ 3g - 3 = g
3T0 S 0 3 -1
\
or 2g = 3 Þ g = 3/2
17. (b, c) : Statements (b) and (c) are incorrect regarding the N.B For monoatomic gas, g = 53 = 1.67
first law of thermodynamics.
For diatomic gas, g = 7 = 1.4
18. (b) : Internal energy and entropy are state functions. 5
when g = 1.5, the gas must be a suitable mixture of monoatomic
19. (b) : Efficiency = 1- T2 = 1- 300 = 1 - 1 = 2 and diatomic gases
T1 900 3 3
\ g = 3/2.
Heat energy = 3 × 10 6 cal = 3 × 10 6 × 4.2 J 22. (a) : Second law of thermodynamics.
\ Workdone by engine = (Heat energy) × (efficiency)
23. (c) : We cannot reach absolute zero temperature.
= (3´106 ´ 4.2) ´ 23 J
= 8.4 × 10 6 J. 24. (a) : All reversible cycles do not have same efficiency.
25. (b) : Thermal capacity.
42 JEE MAIN CHAPTERWISE EXPLORER
CHAPTER KINETIC THEORY OF GASES
9
1. A thermally insulated vessel contains an ideal gas of molecular (c) Tf = 73 T0 (d) Tf = 32 T0 . (2006)
mass M and ratio of specific heats g. It is moving with speed
v and is suddenly brought to rest. Assuming no heat is lost 5. A gaseous mixture consists of 16 g of helium and 16 g of
to the surroundings, its temperature increases by oxygen. The ratio CP /CV of the mixture is
(a) (g - 1) Mv 2 K (b) (g2 g- R1) Mv 2 K (a) 1.4 (b) 1.54 (c) 1.59 (d) 1.62
2(g + 1) R
(2005)
gM v 2
(c) 2 R K (d) (g2- R 1) Mv 2 K (2011) 6. One mole of ideal monoatomic gas (g = 5/3) is mixed with
one mole of diatomic gas (g = 7/5). What is g for the mixture?
2. One kg of a diatomic gas is at a pressure of 8 × 104 N/m2 . g denotes the ratio of specific heat at constant pressure, to
that at constant volume.
The density of the gas is 4 kg/m3 . What is the energy of the (a) 3/2 (b) 23/15 (c) 35/23 (d) 4/3.
(2004)
gas due to its thermal motion?
(a) 3 × 10 4 J (b) 5 × 104 J
(c) 6 × 104 J (d) 7 × 104 J (2009)
7. 1 mole of a gas with g = 7/5 is mixed with 1 mole of a gas
3. If CP and CV denote the specific heats of nitrogen per unit with g = 5/3, then the value of g for the resulting mixture is
mass at constant pressure and constant volume respectively,
(a) 7/5 (b) 2/5 (c) 24/16 (d) 12/7.
then (2002)
(a) CP – CV = 28R (b) C P – CV = R/28 8. At what temperature is the r.m.s. velocity of a hydrogen
(c) CP – C V = R/14
(d) CP – C V = R (2007) molecule equal to that of an oxygen molecule at 47°C?
4. Two rigid boxes containing different ideal gases are placed on (a) 80 K (b) –73 K (c) 3 K (d) 20 K.
a table. Box A contains one mole of nitrogen at temperature T0 , (2002)
while Box B contains one mole of helium at temperature
9. Cooking gas containers are kept in a lorry moving with
(7/3) T0 . The boxes are then put into thermal contact with each uniform speed. The temperature of the gas molecules inside
other and heat flows between them until the gases reach a will
(a) increase
common final temperature. (Ignore the heat capacity of boxes). (b) decrease
(c) remain same
Then, the final temperature of the gases, Tf , in terms of T0 is (d) decrease for some, while increase for others.
(2002)
(a) Tf = 52 T0 (b) Tf = 73 T0
Answer Key
1. (d) 2. (b) 3. (b) 4. (d) 5. (d) 6. (a)
7. (c) 8. (d) 9. (c)
Kinetic Theory of Gases 43
1. (d) : Kinetic energy of vessel = 12 mv2 CV = n1CV1 + n2 CV 2 where CV = 2f R
Increas e in internal energy n1 + n2
DU = nCV D T C P = n1CP1 + n2 C P2 where CP = æ f + 1 ÷øö R
where n is the number of moles of the gas in vessel. n1 + n2 çè 2
As the vessel is stopped suddenly, its kientic energy is used to
increase the temperature of the gas For helium, f = 3, n1 = 4
\ 12 mv2 = D U For oxygen, f = 5, n 2 = 1/2
(( )) (( )) \C P =4 ´ 5 R + 1 ´ 27 R 47 = 1.62.
2 2
1 mv2 =
2 = nCV DT CV 3 1 ´ 52 R 29
4 ´ 2 R + 2
12 mv2 = m CV DT ( ) Q n = m 6. (a) : For mixture of gases,
M M
DT = Mv 2 n1 + n2 = n1 + n2 or 1 +1 = 1 + 1
2C V gm - 1 g1 - g2 - 1 gm -1 5 -1 7 - 1
1
Mv 2 (g ( ) Q
or DT = 2R - 1) K CV = R 3 5
- 1)
(g
2. (b) : The thermal energy or internal energy is U = 52 m RT for or 2 = 3 + 5 = 4Þ g m - 1 = 0.5
diatomic gases. (5 is the degrees of freedom as the gas is gm -1 2 2
\ gm = 1.5 = 3/2.
diatomic) 7. (c) : For mixture of gases
But PV = mRT
V = mass = 4 1 kg = 14 m 3 n1 + n2 = n1 + n2
density kg/m 3 g -1 g -1 g -1
m 1 2
P = 8 × 10 4 N/m2 . ( ) ( ) 1+1 = 1 + 1
\ U= 5 ´ 8 ´ 104 ´ 1 = 5 ´ 104 J g -1 7 -1 5 - 1
2 4
m 5 3
3. (b) : Molar heat capacity = Molar mass × specific 2 = 5 + 3
- 1 2 2
heat capacity g m
So, the molar heat capacities at constant pressure and constant 2 8
- 1 2
volume will be 28CP and 28CV respectively or g =
\ 28CP – 28CV = R or CP - CV = 2R8 . m
or 8gm – 8 = 4
4. (d) : DU = 0 or 8g m = 12
\ 1´ èæç 5 R öø÷ (T f - T0 ) + 1´ 3 R èçæ T f - 73 T0 ÷øö = 0 or g = 12 = 1264
2 2 8
m
or 5Tf – 5T 0 + 3T f – 7T 0 = 0 8. (d) : vr ms = RT
M
or 8Tf = 12T 0
or Tf = 23 T0 . \ (vrm s )O2 = (vrms ) H 2
or 273 + 47 = T2 Þ T = 20 K .
32
5. (d) : For 16 g of helium, n 1 = 16 = 4
4
For 16 g of oxygen, n2 = 16 1 9. (c) : It is the relative velocities between molecules that is
32 = important. Root mean square velocities are different from
lateral translation.
2
For mixture of gases,
44 JEE MAIN CHAPTERWISE EXPLORER
CHAPTER OSCILLATIONS AND WAVES
10
1. The amplitude of a damped oscillator decreases to 0.9 times 6. Two particles are executing simple harmonic motion of the
same amplitude A and frequency w along the xaxis. Their
its original magnitude in 5 s. In another 10 s it will decrease
to a times its original magnitude where a equals mean position is separated by distance X 0 (X 0 > A). If the
maximum separation between them is (X0 + A), the phase
(a) 0.6 (b) 0.7 (c) 0.81 (d) 0.729 difference between their motion is
(2013)
2. An ideal gas enclosed in a vertical cylindrical container (a) p (b) p (c) p (d) p
supports a freely moving piston of mass M. The piston and 2 3 4 6
the cylinder have equal cross sectional area A. When the
piston is in equilibrium, the volume of the gas is V 0 and its (2011)
pressure is P0 . The piston is slightly displaced from the
equilibrium position and released. Assuming that the system 7. A mass M, attached to a horizontal spring, executes SHM
is completely isolated from its surrounding, the piston executes
a simple harmonic motion with frequency with a amplitude A1 . When the mass M passes through its
mean position then a smaller mass m is placed over it and
both of them move together with amplitude A 2 . The ratio of
æ A1 ö is
çè A2 ø÷
(a) 1 MV0 (b) 1 AgP0 (a) M (b) M + m
2 p Ag P0 2 p V0 M M + m M
M 1/ 2 ( ) (d) M + m 1/2
M + m M
(c) 1 V0MP0 (d) 1 A2 gP0 (2013) ( ) (c) (2011)
2 p A2 g 2 p MV0
3. A sonometer wire of length 1.5 m is made of steel. The 8. The transverse displacement y(x,t) of a wave on a string is
given by
tension in it produces an elastic strain of 1%. What is the y( x,t ) = e-(ax2 +bt2 +2 abxt )
This represents a
fundamental frequency of steel if density and elasticity of
steel are 7.7 × 103 kg/m3 and 2.2 × 101 1 N/m2 respectively? a
b
(a) 770 Hz (b) 188.5 Hz (a) wave moving in +xdirection with speed
(c) 178.2 Hz (d) 200.5 Hz (2013) (b) wave moving in –xdirection with speed b
a
4. If a simple pendulum has significant amplitude (up to a factor
(c) standing wave of frequency b
of 1/e of original) only in the period between t = 0 s to 1 (2011)
t = t s, then t may be called the average life of the pendulum.
(d) standing wave of frequency b
When the spherical bob of the pendulum suffers a retardation
(due to viscous drag) proportional to its velocity, with b as 9. The equation of a wave on a string of linear mass density
the constant of proportionality, the average life time of the 0.04 kg m– 1 is given by
( ) y = 0.02 (m) sin éêë2p
pendulum is (as suming damping is small) in seconds t - x úùû .
0.04(s) 0.50(m)
1 2 0.693
(a) b (b) b (c) b (d) b The tension in the string is
(2012) (a) 6.25 N (b) 4.0 N (c) 12.5 N (d) 0.5 N
5. A cylindrical tube, open at both ends, has a fundamental (2010)
frequency, f in air. The tube is dipped vertically in water so 10. If x, v and a denote the displacement, the velocity and the
that half of it is in water. The fundamental frequency of the acceleration of a particle executing simple harmonic motion
aircolumn is now of time period T, then, which of the following does not change
f 3 f (c) 2f (d) f with time?
(a) 2 (b) 4 (2012)
(a) a2 T 2 + 4p2 v 2 (b) aT/x
(c) aT + 2pv (d) aT/v (2009)
Oscillations and Waves 45
11. Three sound waves of equal amplitudes have frequencies 19. A sound absorber attenuates the sound level by
20 dB. The intensity decreases by a factor of
(u – 1), u, (u + 1). They superpose to give beats. The number (a) 100 (b) 1000 (c) 10000 (d) 10
(2007)
of beats produced per second will be
(a) 4 (b) 3 (c) 2 (d) 1
(2009) 20. A coin is placed on a horizontal platform which undergoes
vertical simple harmonic motion of angular frequency w. The
12. A motor cycle starts from rest and accelerates along a straight
path at 2 m/s 2. At the starting point of the motor cycle there amplitude of oscillation is gradually increased. The coin will
is a stationary electric siren. How far has the motor cycle
gone when the driver hears the frequency of the siren at 94% leave contact with the platform for the first time
of its value when the motor cycle was at rest?
(Speed of sound = 330 ms– 1) . (a) at the highest position of the platform
(a) 49 m (b) 98 m (c) 147 m (d) 196 m
(2009) (b) at the mean position of the platform
g
13. A wave travelling along the xaxis is described by the equation (2006)
(c) for an amplitude of w 2
y(x, t) = 0.005 cos(ax – bt). If the wavelength and the time g 2
(d) for an amplitude of w 2 .
period of the wave are 0.08 m and 2.0 s, respectively, then 21. The maximum velocity of a particle, executing simple
a and b in appropriate units are harmonic motion with an amplitude 7 mm, is 4.4 m/s. The
(a) a = 12.50p, b = p (b) a = 25.00p, b = p period of oscillation is
2.0
(d) a = 0.04 , b = 1.0 (a) 100 s (b) 0.01 s (c) 10 s (d) 0.1 s.
(c) a = 0.08 , b = 2.0 p p
p p (2008) (2006)
22. Starting from the origin, a body oscillates simple harmonically
with a period of 2 s. After what time will its kinetic energy
14. The speed of sound in oxygen (O 2) at a certain temperature by 75% of the total energy?
is 460 ms –1. The speed of sound in helium (He) at the same
(a) 1 s (b) 1 s (c) 1 s (d) 1 s .
temperature will be (as sume both gas es to be ideal) 12 6 4 3
(a) 330 ms –1 (b) 460 ms– 1 (2006)
(c) 500 ms– 1 (d) 650 ms –1. (2008) 23. A string is stretched between fixed points separated by 75 cm.
15. A point mass oscillates along the xaxis according to the law It is observed to have resonant frequencies of 420 Hz and
x = x0 cos (wt – p/4). If the acceleration of the particle is 315 Hz. There are no other resonant frequencies between
written as a = Acos(wt + d), then
these two. Then, the lowest resonant frequency for this string is
(a) A = x 0w2 , d = 3p/4 (b) A = x 0, d = –p/4
(c) A = x 0w 2 , d = p/4 (d) A = x 0w2 , d = –p/4 (a) 10.5 Hz (b) 105 Hz
(2007) (c) 1.05 Hz (d) 1050 Hz. (2006)
16. The displacement of an object attached to a spring and 24. A whistle producing sound waves of frequencies 9500 Hz
executing simple harmonic motion is given by
x = 2 × 10– 2 cos pt metre. The time at which the maximum and above is approaching a stationary person with speed
speed first occurs is
(a) 0.25 s (b) 0.5 s (c) 0.75 s (d) 0.125 s v ms –1 . The velocity of sound in air is 300 ms –1. If the person
(2007)
can hear frequencies upto a maximum of 10000 Hz, the
maximum value of v upto which he can hear the whistle is
(a) 30 ms– 1 (b) 15 2 ms-1
17. A particle of mass m executes simple harmonic motion with (c) 15 / 2 ms- 1 (d) 15 ms– 1. (2006)
amplitude a and frequency u. The average kinetic energy during
25. The bob of a simple pendulum is a spherical hollow ball
its motion from the position of equilibrium to the end is filled with water. A plugged hole near the bottom of the
oscillating bob gets suddenly unplugged. During observation,
(a) 2p2 m a 2 u2 (b) p 2m a2 u 2 till water is coming out, the time period of oscillation would
(a) remain unchanged
(c) 1 m a2u 2 (d) 4p2 m a2 u2 (2007) (b) increase towards a saturation value
4 (c) first increase and then decrease to the original value
(d) first decrease and then increase to the original value
18. Two springs, of force k 1 m k 2 (2005)
constants k 1 and k 2 are
connected to a mass m as
shown. The frequency of
oscillation of the mass is f. If both k1 and k2 are made four 26. If a simple harmonic motion is represented by d 2 x ax = 0 ,
times their original values, the frequency of oscillation dt 2 +
becomes its time period is
(a) 2f (b) f/2 (c) f/4 (d) 4f (a) 2pa (b) 2p a (c) 2p/a (d) 2p / a
(2007)
(2005)
46 JEE MAIN CHAPTERWISE EXPLORER
27. Two simple harmonic motions are represented another spring is t2 . If the period of oscillation with the two
springs in series is T, then
çæè100 pt pö
by the equations y1 = 0.1sin + 3 ÷ø and y 2 = 0.1cosp t. (a) T = t1 + t 2 (b) T 2 = t 1 2 + t 2 2
(c) T –1 = t1 – 1 + t 2 –1 (d) T –2 = t1 –2 + t2 –2. (2004)
The phase difference of the velocity of particle 1 with respect
to the velocity of particle 2 is 35. The displacement y of a particle in a medium can be expressed
(a) –p/3 (b) p/6 (c) –p/6 (d) p/3. as:
(2005) y = 10 –6s in(100t + 20x + p/4) m, where t is in second and x
28. The function sin2 ( wt) represents in meter. The speed of the wave is
(a) a simple harmonic motion with a period 2p/w
(b) a simple harmonic motion with a period p/w (a) 2000 m/s (b) 5 m/s
(c) a periodic, but not simple harmonic motion with a period
2p/w (c) 20 m/s (d) 5p m/s. (2004)
(d) a periodic, but not simple harmonic motion with a period
p/w (2005) 36. The bob of a simple pendulum executes simple harmonic
motion in water with a period t, while the period of oscillation
of the bob is t 0 in air. Neglecting frictional force of water and
given that the density of the bob is (4/3) × 1000 kg/m3 . What
29. An observer moves towards a stationary source of sound, relationship between t and t0 is true?
(a) t = t0 (b) t = t0 / 2 (c) t = 2t 0
with a velocity onefifth of the velocity of sound. What is the (d) t = 4t 0 .
(2003)
percentage increase in the apparent frequency?
(a) 5% (b) 20% (c) zero (d) 0.5% 37. A body executes simple harmonic motion. The potential
(2005) energy (P.E.), the kinetic energy (K.E.) and total energy (T.E.)
30. When two tuning forks (fork 1 and fork 2) are sounded are measured as function of displacement x. Which of the
simultaneously, 4 beats per second are heard. Now, some
tape is attached on the prong of the fork 2. When the tuning following statement is true?
forks are sounded again, 6 beats per second are heard. if the
frequency of fork 1 is 200 Hz, then what was the original (a) K.E. is maximum when x = 0
frequency of fork 2?
(a) 196 Hz (b) 204 Hz (c) 200 Hz (d) 202 Hz (b) T.E. is zero when x = 0
(2005)
(c) K.E. is maximum when x is maximum
(d) P.E. is maximum when x = 0. (2003)
38. The length of a simple pendulum executing simple harmonic
motion is increased by 21%. The percentage increase in the
31. In forced oscillation of a particle the amplitude is maximum time period of the pendulum of increased length is
for a frequency w1 of the force, while the energy is maximum (a) 11% (b) 21% (c) 42% (d) 10%.
for a frequency w 2 of the force, then (2003)
(a) w 1 = w2 (b) w 1 > w 2 39. Two particles A and B of equal masses are suspended from
two massless springs of spring constants k 1 and k 2 ,
(c) w1 < w 2 when damping is small and w1 > w 2 when respectively. If the maximum velocities, during oscillations,
are equal, the ratio of amplitudes of A and B is
damping is large
(d) w 1 < w 2 (2004)
32. A particle of mass m is attached to a spring (of spring constant (a) k1 / k 2 (b) k2 /k1 (c) k2 / k1 (d) k 1 /k2 .
(2003)
k) and has a natural angular frequency w0 . An external force
F (t) proportional to coswt (w ¹ w0 ) is applied to the oscillator. 40. A mass M is suspended from a spring of negligible mass. The
The time displacement of the oscillator will be proportional to
m 1 spring is pulled a little and then released so that the mass
(a) w0 2 - w 2 (b) m(w0 2 - w 2 )
executes SHM of time period T. If the mass is increased by
1 m
(c) m(w0 2 + w2 ) (d) w0 2 + w 2 . m, the time period becomes 5T/3. Then the ratio of m/M is
(2004) (a) 3/5 (b) 25/9 (c) 16/9 (d) 5/3.
33. The total energy of a particle, executing simple harmonic (2003)
motion is 41. A tuning fork of known frequency 256 Hz makes 5 beats per
(a) µ x (b) µ x 2 second with the vibrating string of a piano. The beat frequency
(c) independent of x (d) µ x1 /2 decreases to 2 beats per second when the tension in the piano
where x is the displacement from the mean position. string is slightly increased. The frequency of the piano string
(2004) before increasing the tension was
34. A particle at the end of a spring executes simple harmonic (a) (256 + 2) Hz (b) (256 – 2) Hz
motion with a period t 1, while the corresponding period for
(c) (256 – 5) Hz (d) (256 + 5) Hz. (2003)
Oscillations and Waves 47
42. A metal wire of linear mass density of 9.8 g/m is stretched 47. If a spring has time period T, and is cut into n equal parts,
with a tension of 10 kgwt between two rigid supports then the time period of each part will be
1 metre apart. The wire passes at its middle point between the
poles of a permanent magnet, and it vibrates in resonance (a) T n (b) T / n (c) nT (d) T.
when carrying an alternating current of frequency u. The (2002)
frequency u of the alternating source is
(a) 50 Hz (b) 100 Hz (c) 200 Hz (d) 25 Hz. 48. When temperature increases, the frequency of a tuning fork
(2003) (a) increases (b) decreases
(c) remains same
43. The displacement of a particle varies according to the relation (d) increases or decreases depending on the material.
x = 4(cospt + sinpt). The amplitude of the particle is
(2002)
(a) – 4 (b) 4 (c) 4 2 (d) 8. 49. Length of a string tied to two rigid supports is 40 cm.
(2003) Maximum length (wavelength in cm) of a stationary wave
produced on it is
44. The displacement y of a wave travelling in the x direction is (a) 20 (b) 80 (c) 40 (d) 120.
given by (2002)
( ) y
= 10-4 sin 600t - 2x + p metre, 50. A wave y = a sin(wt – kx) on a string meets with another wave
3
where x is expressed in metre and t in second. The speed of producing a node at x = 0. Then the equation of the unknown
the wavemotion, in ms –1 is wave is
(a) 300 (b) 600 (c) 1200 (d) 200. (a) y = asin(wt + kx) (b) y = –asin(wt + kx)
(2003) (c) y = asin(wt – kx) (d) y = –asin(wt – kx).
45. A child swinging on a swing in sitting position, stands up, (2002)
then the time period of the swing will 51. A tuning fork arrangement (pair) produces 4 beats/sec with
(a) increase one fork of frequency 288 cps. A little wax is placed on the
(b) decrease unknown fork and it then produces 2 beats/sec. The frequency
(c) reamains same of the unknown fork is
(d) increases if the child is long and decreases if the child (a) 286 cps (b) 292 cps
is short. (2002) (c) 294 cps (d) 288 cps. (2002)
46. In a simple harmonic oscillator, at the mean position 52. Tube A has both ends open while tube B has one end closed,
(a) kinetic energy is minimum, potential energy is maximum
(b) both kinetic and potential energies are maximum otherwise they are identical. The ratio of fundamental
(c) kinetic energy is maximum, potential energy is minimum
(d) both kinetic and potential energies are minimum. frequency of tube A and B is
(2002)
(a) 1 : 2 (b) 1 : 4 (c) 2 : 1 (d) 4 : 1.
(2002)
Answer Key
1. (d) 2. (d) 3. (c) 4. (c) 5. (d) 6. (b)
7. (d) 8. (b) 9. (a) 10. (b) 11. (a) 12. (b)
13. (b) 14. (*) 15. (a) 16. (b) 17. (b) 18. (a)
19. (a) 20. (c) 21. (b) 22. (b) 23. (b) 24. (d)
25. (c) 26. (d) 27. (c) 28. (d) 29. (b) 30. (a)
31. (a) 32. (b) 33. (c) 34. (b) 35. (b) 36. (c)
37. (a) 38. (d) 39. (c) 40. (c) 41. (c) 42. (a)
43. (c) 44. (a) 45. (b) 46. (c) 47. (b) 48. (b)
49. (b) 50. (b) 51. (b) 52. (c)
48 JEE MAIN CHAPTERWISE EXPLORER
1. (d) : The amplitude of a damped oscillator at a given instant d 2 x = - w 2 x
dt 2
of time t is given by
A = A0 e – bt/2m We get
where A0 is its amplitude in the absence of damping, b is the w2 = gP0 A2 or w = gP0 A 2
damping constant. MV0 MV0
As per question
After 5 s (i.e. t = 5 s) its amplitude becomes w 1 gP0 A 2
2p 2 p MV0
0.9A0 = A 0e –b(5)/2m = A 0e –5b/2m Frequency, u = =
0.9 = e –5b/2m
...(i)
After 10 more second (i.e. t = 15 s), its amplitude becomes 3. (c) : Fundamental frequency of vibration of wire is
aA 0 = A 0e – b(15)/2m = A0 e –15b/2m (Using (i)) u = 1 T
a = (e– 5b/2m) 3 = (0.9)3 2 L m
= 0.729
2. (d) : where L is the length of the wire, T is the tension in the
FBD of piston at equilibrium wire and m is the mass per length of the wire
As m = rA
where r is the density of the material of the wire and A is
the area of crosssection of the wire.
\ u = 1 T
2 L r A
Here tension is due to elasticity of wire
\ T = YA é DL ù éë As Y = Stress = TL ù
ë L û Strain AD L û
Pa tmA + Mg = P0 A …(i) Hence, u = 1 Y DL
2 L r L
FBD of piston when piston is pushed down a distance x Here, Y = 2.2 × 10 11 N/m 2 , r = 7.7 × 10 3 kg/m 3
DL = 0.01, L = 1.5 m
L
Substituting the given values, we get
u = 1 2.2 ´ 1011 ´ 0.01
2 ´ 1.5 7.7 ´ 103
d 2 x
( P0 + dP) A - (Patm A + Mg ) = M dt 2 …(ii)
As the system is completely isolated from its surrounding = 103 2 Hz = 178.2 Hz
therefore the change is adiabatic. 37
For an adiabatic process
4. (c)
PVg = constant
\ Vg dP + Vg–1P dV = 0 5. (d): When the tube of length l is open at l
both ends,
or dP = - gPdV \ f = v ...(i)
V 2l
\ dP = - gP0 ( Ax ) ( .. . dV = Ax) where v is the speed of sound in air.
V0
…(iii) When the tube is dipped vertically in water
Using (i) and (iii) in (ii), we get and half of it is in water, it behaves closed l
2
pipe length 2l ,
M d 2 x = - gP0 A 2 x or d 2 x = - gP0 A 2 l
dt 2 V0 dt 2 MV0 v v 2
f ¢ = = 2l = f (Using(i))
Comparing it with standard equation of SHM, ( ) \ l
4 2
Oscillations and Waves 49
6. (b) ( )T 2p 2
0.04
0.04 ´
M = 2p 2 = 6.25 N
7. (d) : T1 = 2p k ...(i) ( ) 0.05
When a mass m is placed on mass M, the new system is of 10. (b) : For a simple harmonic motion,
mass = (M + m) attached to the spring. New time period of
oscillation acceleration, a = –w 2x where w is a constant = 2 p .
T
(m + M )
T2 = 2p ...(ii) 4p2 aT - 4Tp 2
k a = - T 2 × x Þ x = .
Consider v1 is the velocity of mass M passing through mean The period of oscillation T is a constant.
position and v2 velocity of mass aT
(m + M) passing through mean position. x
Using, law of conservation of linear momentum \ is a constant.
Mv1 = (m + M)v 2 11. (a) : The given sources of sound produce frequencies,
M(A 1(w .. . 1 ) v=1 =(m A + 1w M1 a)n(Ad 2 wv2 2 ) = A 2w 2)
(u – 1), u and (u + 1).
or A1 = (m + M ) w2 For two sources of frequencies u1 and u 2 ,
A2 M w1 y 1 = A cos 2pu 1 t
y2 = A cos 2pu 2 t
( ) ( ) = Superposing, one gets
m + M T1 Q w1 = 2p and w2 = 2 p ( ) ( ) y = 2Acos2p + u2
M ´ T2 T1 T2 u1 - u2 t cos 2p u1 2 t.
2
The resultant frequency obtained is u1 + u 2 and this wave
2
A1 = m + M (Using (i) and (ii))
A2 M u1 - u 2
is modulated by a wave of frequency 2 (rather the
8. (b) : y(x, t) = e-(ax2 + bt2 + 2 ab xt ) difference of frequencies/2).
The intensity waxes and wanes. For a cosine curve (or sine
= e- ( a x+ bt ) 2 curve), the number of beats = u 1 ~ u 2 .
Comparing equation (i) with standard equation Frequencies Mean Beats
y(x, t) = f(ax + bt) u + 1 and u (u + 0.5) Hz 1
As there is positive sign between x and t terms, hence wave u and u – 1 u – 0.5
travel in –x direction. 1
Wave speed = Coefficient of t = b (u + 1) and (u – 1) u 2
Coefficient of x a
Total number of beats = 4.
9. (a) : Here, linear mass density m = 0.04 kg m –1 One should detect three frequencies, u, u + 0.5 and u – 0.5
The given equation of a wave is and each frequency will show 2 beats, 1 beat and 1 beat per
second, respectively.
( ) y = 0.02sin ëêé2 p t
0.04 - x ù Total number of beats = 4
0.50 úû
Compare it with the standard wave equation 12. (b) : The source is at rest, the observer is moving away from
y = Asin(wt – kx)
the source.
we get, \ f ¢ = f (vsound - vo bs )
vsound
2p rad s-1; 2 p m -1
w= 0.04 k = 0.5 rad f ¢
f
Þ ´ vsound = vsound - vo bs
Wave velocity, v = w = (2p / 0.04) m s -1 ...(i) Þ f ¢ ´ vsound - vsound = - vo bs
k (2p / 0.5) f
Also v = T ...(ii) vsound æ f ¢ -1öø = -vo bs
m è f
where T is the tension in the string and m is the linear mass 330(0.94 – 1) = –vobs
density Þ vobs = 330 × 0.06 = 19.80 ms –1 .
Equating equations (i) and (ii), we get
\ s = v2 - u 2 = (19.80) 2 = 98 m.
2a 2 ´ 2
mw 2
w = T or T = k 2 13. (b) : The wave travelling along the xaxis is given by
k m
y(x, t) = 0.005 cos(ax – bt).
50 JEE MAIN CHAPTERWISE EXPLORER
Therefore a = k = 2 p . As l = 0.08 m. = 1 ma2 w2 sin 2 wt
l 2
\ a = 2 p = p Þ a= p ´100.00 = 25.00 p. Average kinetic energy < K >
0.08 0.04 4
= < 1 ma2w2 sin 2 wt >
w=bÞ 2p = b Þ p 2
2.0
= 1 mw 2 a 2 < sin 2 w t >
\ a = 25.00 p, b = p 2
gP gRT = 1 mw 2 a 2 æ 1 ö éëêQ < sin 2 q > = 1 ù
r M 2 è 2 ø 2 ûú
14. (*) : v = =
g for O 2 = 1 + 2/5 = 1.4; = 14 ma2 (2pu ) 2 [Q w = 2pu]
g for He = 1 + 2/3 = 5/3 = p2 ma 2 u 2.
v2 æ g He 32 ö 18. (a) : In the given figure two k 1 m k 2
v1 ççè 4 gO 2 ÷ø÷ ´ springs are connected in
= ´ 460
5 1 32 ´ 5 parallel. Therefore the effective
3 4 7
= 460 ´ ´ ´ = 1420 m/s. spring constant is given by
* The value of the speed of sound in He should have been k eff = k 1 + k 2
Frequency of oscillation,
965 m/s and that of O 2 , about 320 m/s. The value of the
velocity given for O 2 is quite high. Option not given. f = 1 k eff = 1 k1 + k 2 ... (i)
2p m 2 p m
15. (a) : Given : x = x0 cos çèæ wt - p4 ÷øö ... (i)
Acceleration a = A cos (wt + d) ... (ii) As k 1 and k2 are increased four times
New frequency,
Velocity v = dx f¢= 1 4(k1 + k 2 ) = 2 f (using (i).
dt 2 p m
v = -x0w sin çèæ wt - p4 ÷öø ... (iii) 19. (a) : L1 = 10 log çæè I 1 ö ; L2 = 10 log çæè I 2 ö
... (iv) I0 ÷ø I0 ÷ø
Acceleration a = dv
dt æ I1 ö log èçæ I 2 ö
\ L1 - L2 = 10 log çè I0 ÷ø - 10 I0 ÷ø
æ pö x0 w2 cos éêëp p4 ) ûùú
= - x0w2 cos è wt - 4ø = + (wt -
= x0 w2 cos ëéêwt + 3p ù or D L = 10log çæè I 1 ö or 20 dB = 10 log çæè I1 ö
4 úû I2 ÷ø I2 ø÷
Compare (iv) with (ii), we get or 10 2 = I 1 or I 2 = I 1 .
I2 100
A = x0 w2 , d = 34p .
20. (c) : In vertical simple harmonic motion, maximum
16. (b) : Given : displacement x = 2 × 10– 2 cos pt acceleration (aw 2 ) and so the maximum force (maw 2 ) will
Velocity v = dx = -2 ´10-2 p sin p t be at extreme positions. At highest position, force will be
dt
towards mean position and so it will be downwards. At
For the first time when v = v max , sin pt = 1
lowes t position, force will be towards mea n position and
or sin pt = sin 2p or pt = p so it will be upwards. This is opposite to weight direction
2
of the coin. The coin will leave contact will the platform
or t = 1 s = 0.5 s. for the first time when m(aw 2 ) ³ mg at the lowest position
2
of the platform.
17. (b) : For a particle to execute simple harmonic motion its
displacement at any time t is given by 21. (b) : Maximum velocity v m = aw = a çæè 2 p ÷øö
T
x(t) = a(cos wt + f)
T = 2pa = 2 ´ 22 ´ (7 ´10-3 )
where, a = amplitude, w = angular frequency, \ vm 7 4.4
f = phase constant. = 10 –2 sec = 0.01 sec.
Let us choose f = 0
\ x(t) = acoswt 22. (b) : During simple harmonic motion,
Velocity of a particle v = dx = -a wsin wt Kinetic energy = 1 mv 2 = 12 m (a wcos wt ) 2
dt 2
Kinetic energy of a particle is K = 12 mv2 Total energy E = 21 ma2 w2
Oscillations and Waves 51
Q (Kinetic energy) = 17050 ( E ) = (0.1´ p) cos çèæ pt + p2 ÷øö
or 1 ma2w2 cos 2 wt = 75 ´ 1 ma2 w2 \ D f = p - p = - p6 .
2 100 2 3 2
or cos2 wt = 3 Þ cos wt = 3 = cos p y = sin 2 wt = 1 - cos 2wt = 1 - cos 2 wt
4 2 6 2 2 2
28. (d) :
p
\ wt = 6
or t = p = p ) = 6 2p p = 16 sec. It is a periodic motion but it is not SHM
6w 6(2p /T ´2 \ Angular speed = 2w
23. (b) : Let the successive loops formed be p and \ Period T = 2 p = 2 p = p
angular speed 2w w
(p + 1) for frequencies 315 Hz and 420 Hz Hence option (d) represents the answer.
\ u= p T = pv 29. (b) : By Doppler's effect
2l m 2l
\ pv = 315 Hz and ( p + 1) v = 420 Hz u¢ = vs + v O (where v s is the velocity of sound)
2 l 2 l u vs
or (p + 1) v - pv = 420 - 315 = v + (v / 5) = 6
2l 2 l v 5
or v = 105 Þ 12 ´ lv = 105 Hz \ Fractional increase = u¢ - u = æ u¢ - 1øö = æ 6 - 1öø = 1
2l u è u è 5 5
p = 1 for fundamental mode of vibration of string. \ Percentage increase = 1050 = 20%.
\ Lowest resonant frequency = 105 Hz.
24. (d) : u¢ = vs 30. (a) : Let the two frequencies be u1 and u 2
u vs - v
u2 may be either 204 Hz or 196 Hz.
where v s is the velocity of sound in air.
+4 Hz 204 Hz
10000 300 (u2)
9500 = 300 - v
200 Hz
Þ(300 - v ) = 285 Þ v = 15 m /s . (u 1)
– 4 Hz 196 Hz
25. (c) : For a pendulum, T = 2 p l where l is measured upto (u 2)
g
As mass of second fork increases, u2 decreases.
centre of gravity. The centre of gravity of system is at
If u 2 = 204 Hz, a decrease in u2 decreases beats/sec. But
centre of sphere when hole is plugged. When unplugged, this is not given in question
water drains out. Centre of gravity goes on descending. If u 2 = 196 Hz, a decrease in u 2 increased beats/sec.
When the bob becomes empty, centre of gravity is restored This is given in the question when beats increase to 6
to centre. \ Original frequency of second fork = 196 Hz.
\ Length of pendulum first increases, then 31. (a) : In case of forced oscillations
(i) The amplitude is maximum at resonance
decreases to original value. \ Natural frequency = Frequency of force = w 1
(ii) The energy is maximum at resonance
\ T would first increase and then decrease to \Natural frequency = Frequency of force = w 2
\ From (i) and (ii),
the original value. w1 = w 2 .
26. (d) : Standard differential equation of SHM is 32. (b) : In case of forced oscillations,
d 2 x + w2 x = 0
dt 2 x = asin(wt + f) where a = F0 / m
Given equation is dd2t2 x + ax = 0 w02 - w2
\ w 2 = a
or w = a 1
\ x is proportional to m(w02 - w2 ) .
\ 2p 2 p
T = w = a .
33. (c) : Under simple harmonic motion, total energy
27. (c) : v1 = d ( y1 ) = (0.1´100p) cosçæè100 pt + p3 ÷øö = 21 ma2w2
dt Total energy is independent of x.
v2 = d ( y2 ) = (-0.1´ p) sin p t
dt
52 JEE MAIN CHAPTERWISE EXPLORER
34. (b) : When springs are in series, k = k1k 2 or vm = a k
k1 + k2 m
m Q (v m )A = (v m) B
k1
For first spring, t1 = 2 p
m \ a1 k1 = a 2 k2 Þ a1 = k 2 .
k2 m m a2 k1
For second spring t2 = 2p
t12 + t22 = 4p2m + 4 p2 m = 4 p2 m çæ k1 + k 2 ö 40. (c) : Initially, T = 2p M / k
k1 k2 è k1k2 ÷ø
\ 5T M + m
3 k
Finally, = 2 p
or t12 + t 22 = é m ( kk11k+2 k 2 ) ùûú 2 \ 35 ´ 2p M = 2 p M + m
ê2 p k k
ë
or t12 + t22 = T 2 . or 25 M = M + m
9 k k
35. (b) : Given wave equation :
or 9 m + 9 M = 25 M
y = 10-6 sin çæè100t + 20x + p4 ÷öø m or m = 196 .
Standard equation : y = a sin (wt + kx + f) M
Compare the two
\ w = 100 and k = 20 41. (c) : The possible frequencies of piano are
(256 + 5) Hz and (256 – 5) Hz.
+5 Hz 261 Hz
\ w = 100Þ 2 pn = nl = v = 5 256 Hz
k 20 2p / l
\ v = 5 m/s.
– 5 Hz 251 Hz
36. (c) : t0 = 2p l / g ......... (i) 1 T
2 l m
Due to upthrust of water on the top, its apparent weight For piano string, u =
decreases When tension T increases, u increases
upthrust = weight of liquid displaced (i) If 261 Hz increases, beats/sec increase. This is not given
\ Effective weight = mg – (Vsg) = Vrg – Vsg in the question.
Vrg¢ = Vg(r – s), where s is density of water (ii) If 251 Hz increases due to tension, beats per second
or g ¢ = g æ r - s ö decrease. This is given in the question.
çè r ÷ø
Hence frequency of piano = (256 – 5) Hz.
\ t = 2p l / g ¢ = 2 p l r .......(ii) 42. (a) : At resonance, frequency of vibration of wire become
g(r - s ) equal to frequency of a.c.
t = lr ´ g = r For vibration of wire, u = 1 T
t0 g(r - s) l r - s 2l m
\
( ) = 4 ´1000 /3 = 2 \ u = 1 10 ´ 9.8 = 100 = 50 Hz.
40300 - 1000 2 ´1 9.8 ´ 10-3 2
or t = t0 × 2 = 2t0 .
37. (a) : Kinetic energy is maximum at x = 0. 43. (c) : x = 4(cospt + sinpt)
= 4 ´ 2 é 1 cos pt + 1 sin p t ù
êë 2 2 ûú
38. (d) : Let the lengths of pendulum be (100l) and (121l) or x=4 2 êéësin p cos pt + cos p4 sin pt úûù
4
\ T ¢ = 121 = 11 = 4 2 sin çæè pt + p4 ÷öø
T 100 10 Hence amplitude = 4 2 .
\ Fractional change = T ¢– T = 11 - 10 = 1
T 10 10
44. (a) : Given wave equation :
\ Percentage change = 10%.
y = 10-4 sin çæè 600t - 2 x + p3 ÷øö m
39. (c) : Maximum velocity under simple harmonic motion Standard wave equation : y = asin(wt – kx + f)
Compare them
v m = aw Angular speed = w = 600 sec –1
Propagation constant = k = 2 m –1
\ vm = 2pa = (2pa ) æçè 1 öø÷ = (2pa ) çæ 1 k ö
T T è 2 p ÷
m ø
Oscillations and Waves 53
w = 2 pu = ul = velocity Y = asin(wt – kx) – asin(wt + kx)
k 2p / l At x = 0, Y = a sinwt – asinwt = zero
This option holds good
\ velocity = w = 600 = 300 m/sec . Option (c) gives Y = 2asin(wt – kx)
k 2 At x = 0, Y is not zero
Option (d) gives Y = 0
45. (b) : Time period will decrease. Hence only option (b) holds good.
When the child stands up, the centre of gravity is shifted
upwards and so length of swing decreases. T = 2p l / g .
46. (c) : In a simple harmonic oscillator, kinetic energy is 51. (b) : The wax decreases the frequency of unknown fork.
maximum and potential energy is minimum at mean position. The possible unknown frequencies are (288 + 4)cps and
(288 – 4) cps.
47. (b) : For a spring, T = 2 p m + 4 cps 292 cps
k
For each piece, spring constant = nk 288 cps
\ T ¢ = 2 p m – 4 cps
nk
284 cps
\ T ¢ = 2p m ´ 1 = T . Wax reduces 284 cps and so beats should increases. It is not
k n n
given in the question. This frequency is ruled out. Wax reduced
48. (b) : When temperature increases, l increases 292 cps and so beats should decrease. It is given that the
Hence frequency decreases.
beats decrease to 2 from 4.
l max Hence unknown fork has frequency 292 cps.
2
49. (b) : = 40 Þ l max = 80 cm. 52. (c) : In tube A, l A = 2l AN AN
In tube B, l B = 4l
50. (b) : Consider option (a)
v v
Stationary wave : \ u A = l A = 2l
Y = asin(wt + kx) + asin(wt – kx)
N l
when x = 0, Y is not zero. The option is not acceptable. uB = v = v
l B 4 l
Consider option (b)
Stationary wave : uA = 2 . AN N
uB 1
\ l A = l l B = l
2 4
54 JEE MAIN CHAPTERWISE EXPLORER
CHAPTER ELECTROSTATICS
11
1. Two capacitors C 1 and C2 are charged to 120 V and 200 V An insulating solid sphere of radius R has a uniformly positive
respectively. It is found that by connecting them together the
charge density r. As a result of this uniform charge distribution
potential on each one can be made zero. Then there is a finite value of electric potential at the centre of the
(a) 9C 1 = 4C 2 (b) 5C1 = 3C 2 sphere, at the surface of the sphere and also at a point outside
(c) 3C 1 = 5C2 (d) 3C1 + 5C2 = 0
(2013) the sphere. The electric potential at infinity is zero.
2. Two charges, each equal to q, are kept at x = – a and x = a Statement 1 : When a charge q is taken from the centre to the
on the xaxis. A particle of mass m and charge q0 = q is surface of the sphere, its potential energy changes by qr .
2 3e 0
placed at the origin. If charge q 0 is given a small displacement Statement 2 : The electric field at a distance r(r < R) from
(y < < a) along the yaxis, the net force acting on the particle the centre of the sphere is rr .
3e 0
is proportional to
(a) Statement 1 is true, Statement 2 is false.
- 1 1
(a) y (b) y (c) –y (d) y (b) Statement 1 is false, Statement 2 is true.
(2013) (c) Statement 1 is true, Statement 2 is true, Statement 2
3. A charge Q is uniformly distributed over a long rod AB of is the correct explanation of Statement 1.
length L as shown in the figure. The electric potential at the
point O lying at a distance L from the end A is (d) Statement 1 is true, Statement 2, is true; Statement 2
is not the correct explanation of Statement 1.
(2012)
6. 25
Potential difference
Q ln 2 Q ln 2 V in volts 20
(a) 4 p Î0 L (b) 8 p Î0 L
(2013) 15
3Q Q 10
(c) 4 p Î0 L (d) 4p Î0 L ln 2
4. A charge Q is uniformly distributed over the surface of non 5
conducting disc of radius R. The disc rotates about an axis
perpendicular to its plane and passing through its centre with 0 50 100 150 200 250 300
an angular velocity w. As a result of this rotation a magnetic Time t in seconds
field of induction B is obtained at the centre of the disc. If
we keep both the amount of charge placed on the disc and its The figure shows an experimental plot for discharging of a
angular velocity to be constant and vary the radius of the disc
then the variation of the magnetic induction at the centre of capacitor in an RC circuit. The time constant t of this circuit
the disc will be represented by the figure
lies between
(a) 0 and 50 sec (b) 50 sec and 100 sec
(c) 100 sec and 150 sec (d) 150 sec and 200 sec
(2012)
(a) B (b) B 7. In a uniformly charged sphere of total charge Q and radius
R, the electric field E is plotted as a function of distance from
R R the centre. The graph which would correspond to the above
will be
(c) B (d) B (2012) E
R R (a) E
5. This question has Statement 1 and Statement 2. Of the four (b)
choices given after the statements, choose the one that best
describes the two statements. R r R r
Electrostatics 55
14. Two points P and Q are maintained at the potentials of 10 V
E E and –4 V respectively. The work done in moving 100 electrons
(c) (d) from P to Q is
(a) –9.60 × 10 –17 J (b) 9.60 × 10 –17 J
(c) –2.24 × 10 –16 J (d) 2.24 × 10 –16 J (2009)
R r R r Q
p R4
(2012) 15. Let P( r ) = r be the charge density distribution for a
8. Two identical charged spheres suspended from a common solid sphere of radius R and total charge Q. For a point ‘p’
point by two massless strings of length l are initially a distance
d(d < < l) apart because of their mutual repulsion. The charge inside the sphere at distance r1 from the centre of the sphere,
begins to leak from both the spheres at a constant rate. As a the magnitude of electric field is
result the charges approach each other with a velocity v. Then
as a function of distance x between them (a) 0 (b) Q (c) Qr1 2 (d) Qr1 2
(a) v µ x –1/2 (b) v µ x– 1 (c) v µ x 1/2 (d) v µ x 4 pe 0r12 4 pe 0 R4 3p e 0 R4
(2011)
(2009)
16. This question contains Statement1 and Statement2. Of the
four choices given after the statements, choose the one that
9. The electrostatic potential inside a charged spherical ball is best describes the two statements.
given by f = ar 2 + b where r is the distance from the centre;
a, b are constants. Then the charge density inside the ball is Statement1: For a charged particle moving from point P to
(a) –24pae0 r (b) –6ae0 r (c) –24pae0 (d) –6ae0
(2011) point Q, the net work done by an electrostatic field on the
particle is independent of the path connecting point P to
point Q.
Statement2: The net work done by a conservative force on
10. Two identical charged spheres are suspended by strings of an object moving along a closed loop is zero.
equal lengths. The strings make an angle of 30° with each (a) Statement1 is true, Statement2 is false
other. When suspended in a liquid of density 0.8 g cm– 3, the (b) Statement1 is true, Statement2 is true; Statement2 is
angle remains the same. If density of the material of the the correct explanation of Statement1.
sphere is 1.6 g cm– 3, the dielectric constant of the liquid is (c) Statement1 is true, Statement2 is true; Statement2 is
(a) 1 (b) 4 (c) 3 (d) 2 not the correct explanation of Statement1.
(2010) (d) Statement1 is false, Statement2 is true. (2009)
11. Let there be a spherically symmetric charge distribution with 17. A thin spherical shell of radius R has charge Q spread
uniformly over its surface. Which of the following graphs
( ) charge density varying as 5 r most closely represents the electric field E(r) produced by
r(r ) = r0 4 - R upto r = R, and the shell in the range 0 £ r < ¥, where r is the distance from
the centre of the shell?
r(r) = 0 for r > R, where r is the distance from the origin. The
electric field at a distance r (r < R) from the origin is given by
r0 r 5 - r 4 pr0 r 5 - r
3e 0 4 R 3e 0 3 R
( ) (a) ( ) (b) E(r ) E(r )
r0 r 5 - r 4r 0 r 5 - r
4e 0 3 R 3e 0 4 R
( ) (c) ( ) (d) (2010) (a) (b)
O R r O R r
12. A thin semicircular ring of ^j
radius r has a positive E(r ) E(r )
charge q distributed
uniformr ly over it. The net O ^i (c) (d) (2008)
field E at the centre O is
(2010) O R r
q j^ q j^ O R r
2 p2e 0 r 2 4 p2e 0 r 2
(a) (b)
(c) - q j^ (d) - q j^ 18. A parallel plate capacitor with air between the plates has a
4 p2e 0 r 2 2 p 2 e 0 r 2
capacitance of 9 pF. The separation between its plates is d. The
13. A charge Q is placed at each of the opposite corners of a space between the plates is now filled with two dielectrics.
square. A charge q is placed at each of the other two corners.
If the net electrical force on Q is zero, then the Q/q equals One of the dielectrics has dielectric constant k1 = 3 and thickness
d/3 while the other one has dielectric constant k 2 = 6 and
- 1 thickness 2d/3. Capacitance of the capacitor is now
2
(a) - 2 2 (b) –1 (c) 1 (d) (a) 20.25 pF (b) 1.8 pF
(2009) (c) 45 pF (d) 40.5 pF. (2008)
56 JEE MAIN CHAPTERWISE EXPLORER
19. A parallel plate condenser with a dielectric of dielectric plates are separated by d = 0.1 m and can be treated as
constant K between the plates has a capacity C and is charged infinitely large. An electron is released from rest on the inner
to a potential V volt. The dielectric slab is slowly removed surface of plate 1. What is its speed when it hits plate 2?
from between the plates and then reinserted. The net work (e = 1.6 × 10 –19 C, m e = 9.11 × 10– 31 kg)
done by the system in this process is (a) 32 × 10 –19 m/s (b) 2.65 × 106 m/s
1 CV 2 (c) 7.02 × 101 2 m/s (d) 1.87 × 106 m/s. (2006)
2
(a) zero (b) (K – 1) 26. A electric dipole is placed at an angle of 30º to a nonuniform
(c) CV 2 (K - 1)
(d) (K – 1) CV2 (2007) electric field. The dipole will experience
K
(a) a torque only
20. The potential at a point x (measured in hm) due to some (b) a translational force only in the direction of the field
charges situated on the xaxis is given by (c) a translational force only in a direction normal to the
V(x) = 20/(x2 – 4) volt
direction of the field
The electric field E at x = 4 mm is given by (d) a torque as well as a translational force. (2006)
(a) (10/9) volt/mm and in the +ve x direction 27. A fully charged capacitor has a capacitance C. It is discharged
through a small coil of resistance wire embedded in a thermally
(b) (5/3) volt/mm and in the –ve x direction insulated block of specific heat capacity s and mass m. If the
temperature of the block is raised by DT, the potential
(c) (5/3) volt/mm and in the +ve x direction difference V across the capacitance is
(d) (10/9) volt/mm in the –ve x direction (2007)
21. Charges are placed orn the vertices of a q q msDT 2msD T
square as shown. Let E be the electric field A B (a) C (b) C
and V the potential at the centre. If the
charges on A and B are interchanged with D C 2mCDT mCDT (2005)
thoser o n D and C respectively, then – q – q (c) s (d) s
(a) E changes, V remains unchanged
(2007) 28. A parallel plate capacitor is made by stacking n equally spaced
r
(b) E remr ains unchanged, V changes plates connected alternatively. If the capacitance between
(c) bro th E and V change
(d) E and V remain unchanged any two adjacent plates is C then the resultant capacitance is
(a) C (b) nC
22. A battery is used to charge a parallel plate capacitor till the (c) (n – 1)C (d) (n + 1)C (2005)
potential difference between the plates becomes equal to the 29. Two thin wire rings each having a radius R are placed at a
distance d apart with their axes coinciding. The charges on
electromotive force of the battery. The ratio of the energy stored the two rings are +Q and –Q. The potential difference between
the centers of the two rings is
in the capacitor and the work done by the battery will be
(a) 1/2 (b) 1 (c) 2 (d) 1/4
(2007) Q é 1 1 ù
23. An electric charge 10 –3 mC is placed at the origin (a) zero (b) ê- ú
4p e0 êë R R2 + d 2 ûú
(0, 0) of X – Y coordinate system. Two points A and B are QR Q é 1 - 1 ù
(c) 4p e 0 d 2 ê ú
( ) situated at 2, 2 and (2,0) respectively. The potential (d) 2p e0 êë R
R2 + d 2 úû
difference between the points A and B will be
(a) 4.5 volt (b) 9 volt (c) zero (d) 2 volt (2005)
(2007) 30. Two point charges +8q and –2q are located at x = 0 and x = L
24. Two spherical conductors A and B of radii 1 mm and 2 mm respectively. The location of a point on the x axis at which
are separated by a distance of 5 cm and are uniformly charged. the net electric field due to these two point charges is zero is
If the spheres are connected by a conducting wire then in (a) 8L (b) 4L (c) 2L (d) L/4
equilibrium condition, the ratio of the magnitude of the electric (2005)
fields at the surface of spheres A and B is 31. A charged ball B hangs from a silk +
thread S, which makes an angle q with +
(a) 1 : 4 (b) 4 : 1 (c) 1 : 2 (d) 2 : 1.
a large charged conducting sheet P, as +
(2006) P +
shown in the figure. The surface charge
25. Two insulating plates are Y density s of the sheet is proportional to + q S
both uniformaly charged in (a) sinq +
such a way that the potential 0.1 m (b) tanq +
difference between them is 1 2 +
V 2 – V 1 = 20 V. (i.e. plate 2
is at a higher potential). The X (c) cosq +
(d) cotq B
(2005)
Electrostatics 57
32. Four charges equal to –Q are placed at the four corners of a 2Q (b) 2Q - 2 q
square and a charge q is at its centre. If the system is in (a) 4 pe0 R 4pe0R 4p e 0 R
equilibrium the value of q is
(a) - Q4 (1 + 2 2) (b) Q4 (1+ 2 2) (c) 2Q R + q (d) (q4 p+eQ 0 ) R2 . (2003)
4pe0 4 pe 0 R
(c) - Q2 (1 + 2 2) (d) Q2 (1+ 2 2) . (2004) 38. A sheet of aluminium foil of negligible thickness is introduced
between the plates of a capacitor. The capacitance of the
33. A charged particle q is shot towards another charged particle capacitor
Q which is fixed, with a speed v. It approaches Q upto a (a) decreases (b) remains unchanged
closest distance r and then returns. If q were given a speed (c) becomes infinite (d) increases. (2003)
2v, the closest distances of approach would be 39. If the electric flux entering and leaving an enclosed surface
(a) r (b) 2r (c) r/2 (d) r/4. (2004) respectively is f 1 and f2 , the electric charge inside the surface
will be
34. Two spherical conductors B and C having equal radii and (a) (f2 – f 1) e0 (b) (f1 + f2 ) /e0 (2003)
carrying equal charges in them repel each other with a force (c) (f 2 – f1 ) /e 0 (d) (f1 + f 2) e 0 .
F when kept apart at some distance. A third spherical
conductor having same radius as that of B but uncharged is 40. Capacitance (in F) of a spherical conductor with radius 1 m is
brought in contact with B, then brought in contact with C
and finally removed away from both. The new force of (a) 1.1 × 10– 10 (b) 10– 6
repulsion between B and C is
(a) F/4 (b) 3F/4 (c) F/8 (d) 3F/8. (2004) (c) 9 × 10 –9 (d) 10– 3. (2002)
41. If a charge q is placed at the centre of the line joining two
equal charges Q such that the system is in equilibrium then
the value of q is
35. Three charges –q1 , +q2 and –q3 are placed as shown in the (a) Q/2 (b) –Q/2 (c) Q/4 (d) –Q/4.
figure. The x component of the force on –q1 is proportional to
(2002)
(a) q2 - aq32 cos q –q3 y 42. If there are n capacitors in parallel connected to V volt source,
b2 then the energy stored is equal to
(b) q2 + aq3 2 sin q a q (a) CV (b) 1 nCV 2 (c) CV 2 (d) 21 n CV 2 .
b2 2 (2002)
b
(c) q2 + aq3 2 cos q –q1 +q2 x 43. A charged particle q is placed E F
b2
at the centre O of cube of D q
C G
(d) q2 - aq32 sin q (2003) length L (ABCDEFGH). O B L
b2
Another same charge q is q (2002)
placed at a distance L from H
36. The work done in placing a charge of 8 × 10– 18 coulomb on O. Then the electric flux A
a condenser of capacity 100 microfarad is through ABCD is
(a) 16 × 10 –32 joule (b) 3.1 × 10– 26 joule (a) q/4pe 0 L (b) zero
(c) q/2pe 0L (d) q/3pe 0 L.
(c) 4 × 10 –10 joule (d) 32 × 10– 32 joule. (2003)
37. A thin spherical conducting shell of radius R has a charge q. 44. On moving a charge of 20 coulomb by 2 cm, 2 J of work is
Another charge Q is placed at the centre of the shell. The
electrostatic potential at a point P at a distance R/2 from the done, then the potential difference between the points is
centre of the shell is
(a) 0.1 V (b) 8 V (c) 2 V (d) 0.5 V.
(2002)
Answer Key
1. (c) 2. (b) 3. (a) 4. (d) 5. (b) 6. (c)
7. (b) 8. (a) 9. (d) 10. (d) 11. (c) 12. (d)
13. (a) 14. (d) 15. (c) 16. (c) 17. (b) 18. (d)
19. (a) 20. (a) 21. (a) 22. (a) 23. (c) 24. (d)
25. (b) 26. (d) 27. (b) 28. (c) 29. (d) 30. (c)
31. (b) 32. (b) 33. (d) 34. (d) 35. (b) 36. (d)
37. (c) 38. (b) 39. (a) 40. (a) 41. (d) 42. (b)
43. (*) 44. (a)
58 JEE MAIN CHAPTERWISE EXPLORER
1. (c) : For potential to be made zero, after connection Potential at O due to the rod is
120C1 = 200C 2 V = ò dV 2 L 1 Q dx
6C 1 = 10C 2 4 p Î0 Lx
3C 1 = 5C 2 =ò
2. (b): The situation is as shown in the figure.
L
= 1 QL [ln x ] L2 L = Q ln 2
4p Î0 4p Î0 L
4. (d): Consider a elementary ring of
radius r and thicknes s dr of a disc as
shown in figure. O r dr
R
( ) When a particle of mass m and charge q0 = q placed is at Charge on the ring,
2
dq = Q ( 2prdr ) = 2RQ 2r dr
the origin is given a small displacement along the yaxis, p R 2
then the situation is shown in the figure.
Current due to rotation of charge on ring is
I = dqw = Qrwdr
2p p R2
Magnetic field at the centre due to the ring element
dB = m0I = m0Qrwdr = m0Q wdr
2 r 2pR 2r 2 p R2
Magnetic field at the centre due to the whole disc
B = ò dB = m 0Qw R m 0QwR = m 0 Qw
2pR2 2 p R 2 2p R
ò dr =
0
By symmetry, the components of forces on the particle of Since, Q and w are constants
charge q0 due to charges at A and B along xaxis will cancel \ B µ 1
each other where along yaxis will add up. R
\ The net force acting on the particle is Hence variation of B with R should be a rectangular hyperbola
(Fnet = 2F cos q = 2 4p 1Î 0 qq0 y as represented in option (d).
( y2 + a2 )
) 2 5. (b): Potential at the centre of the sphere,
y 2 + a 2 VC = R 2r
2 e 0
( ) = q q
2 2 y èæQ q0 = q (Given) øö Potential at the surface of the sphere,
4p Î0 ( y2 + a2 ) 2
( y2 + a 2 )2 1 R 2 r
3 e 0
1 q2 y VS =
4p Î0 + a2 ) 3/2
= (y2 When a charge q is taken from the centre to the surface, the
As y < < a change in potential energy is
\ Fn et = 1 q 2 y or F net µ y DU = (VC - VS ) q = æ R2r - 1 R2rö q = 1 R2 rq
4p Î0 a3 çè 2e0 3 e0 ÷ø 6 e 0
Statement 1 is false.
3. (a) : Statement 2 is true.
Consider a small element of length dx at a distance x from O. 6. (c) : During discharging of a capacitor
Charge on the element, dQ = QL dx V = V0 e –t/t
Potential at O due to the element is
where t is the time constant of RC circuit.
At t = t,
V = V0 = 0.37 V0
e
From the graph, t = 0, V 0 = 25 V
\ V = 0.37 × 25 V = 9.25 V
1 dQ 1 Q This voltage occurs at time lies between 100 sec and 500
4p Î0 x 4 p Î0 Lx
dV = = dx sec. Hence, time constant t of this circuit lies between 100
sec and 150 sec.
Electrostatics 59
7. (b): For uniformly charged sphere 10. (d) :
E = 1 Qr (For r < R ) E
4 pe 0 R3
q 30°
E = 1 Q (For r = R ) q T T
4 pe 0 R 2 F+ + F
R r
E = 1 Q (For r > R ) mg mg
4 pe 0 r 2
The variation of E with distance r from the centre is as shown Initially, the forces acting on each ball are
adjacent figure.
(i) Tension T
8. (a) : Figure shows equilibrium positions of the two sphere. (ii) Weight mg
O (iii) Electrostatic force of repulsion F
For its equilibrium along vertical,
l q q l Tcosq = mg ...(i)
Tcosq T and along horizontal,
q
T Tsinq = F ...(ii)
F A Tsinq Dividing equation (ii) by (i), we get
B
C F tan q = F ...(iii)
x d mg mg
mg 2 When the balls are suspended in a liquid of density s and
\ T cosq = mg dielectric constant K, the electrostatic force will become
q 2 (1/K) times, i.e. F¢ = (F/K) while weight
d 2
and T sin q = F = 1 mg¢ = mg – Upthrust
4 pe 0
= mg – Vsg [As Upthrust = Vsg]
q 2
\ tan q = 1 d 2 mg mg¢ = mg ëéê1 - s ù éëêAs V = m ù
4p e 0 r úû r úû
When charge begins to leak from both the spheres at a constant
rate, then For equilibrium of balls,
tan q = 1 q 2 tan q¢ = mFg¢¢ = F / r)] ...(iv)
4p e0 x2 mg Kmg[1 - (s
x 1q 2 ( ) Q = x According to given problem, q¢ = q
2l = 4p e0 x2 mg tan q 2l From equations (iv) and (iii), we get
x q 2 1
or 2l µ x2 ( ) K =
s
or q2 µ x 3 Þ q µ x3 /2 1 - r
dq µ 23 x1 / 2 dx r 1.6
dt dt - s) (1.6 - 0.8)
K = (r = = 2
( ) or Q dq
v µ x-1/ 2 dt = constant 11. (c) : Consider a thin spherical shell of radius x and thickness
dx as shown in the figure.
9. (d) : f = ar 2 + b
Electric field, E = -d f = - 2 ar ...(i) dx
dr (Using (i)) x
O Gaussian surface
According to Gauss’s theorem, R r
rr
Ñò E × dS = qi nside
e0
or - 2ar 4 pr2 = qi nside Volume of the shell, dV = 4px2 dx
e 0 Let us draw a Gaussian surface of radius r(r < R) as shown
in the figure above.
q inside = – 8e 0a pr 3 Total charge enclosed inside the Gaussian surface is
Charge density inside the ball is
rinside = qi nside
34 p r3
( ) r r 45 - x 4 px 2 dx
-8e 0 apr 3 R
\ rinside = 43 p r 3 Qin = ò rdV = ò r0
0 0
( ) r x 3
R
= 4p r0 ò
0
rinside = -6ae 0 54 x2 - dx
60 JEE MAIN CHAPTERWISE EXPLORER
= 4p r0 é 5 x 3 - x 4 ùr = 4p r0 é 5 r 3 - r 4 ù 14. (d) : +10 V • ............................... • – 4 V
êë 12 4 R úû 0 êë12 4 R úû P Q
= 4 pr0 é 5 r 3 - r4 ù = pr0 é 53 r 3 - r 4 ù Work done in moving 100e – from P to Q,
4 ëê 3 R úû êë R úû
(Work done in moving 100 negative charges from the positive
According to Gauss’s law to the negative potential).
W = (100e– )(V Q – V P)
E4 pr 2 = Qi n
e 0 = (–100 × 1.6 × 10 –19 )(–14 V) = 2.24 × 10 –16 J.
E4 pr 2 pr0 é 35 r 3 r 4 ù 15. (c) : If the charge density, r = Q r ,
e0 êë R ûú p R4
= -
The electric field at the point p distant r 1 from the centre,
pr0 r 3 é 35 - r ù according to Gauss’s theorem is
4 pr 2 e 0 ë R û
E = E∙4pr 12 = charge enclosed/e 0 Q
E = r0 r é 35 - r ù E × 4p r12 = 1 ò rdV
4e 0 ë Rû e 0
q Þ E × 4pr12 = 1 r1 Qr × 4p r 2 dr r1
p r e0 p R4 R
12. (d) : Linear charge density, l = ò
0
Consider a small element AB ^j Þ E = Qr1 2 4 .
of length dl subtending an 4 pe 0 R
angle dq at the centre O as A dl 16. (c) : Work done = potential difference × charge
dqr B
shown in the figure. dEcosq q = (VB – VA) × q,
q O
\ Charge on the element, VA and VB only depend on the initial and final positions and
dE dEsinq
dq = ldl ^i
( ) = Q d q dl not on the path. Electrostatic force is a conservative force.
lrdq = r
If the loop is completed, VA – VA = 0.
The electric field at the centre O due to the charge element is No net work is done as the initial and
1 dq lrd q final potentials are the same.
4p e0 r2 4p e 0 r 2
dE = = Both the statements are true but A B
statement2 is not the reason for
Resolve dE into two rectangular components statement1.
By symmetry, ò dE cosq = 0 17. (b) : The electric field for a E
uniformly charged spherical
The net electric field at O is
r shell is given in the figure.
E= p j^ ) p lrd q j^ )
= 4p e 0 r 2 sin q(- Inside the shell, the field is zero
ò dE sin q(- ò
and it is maximum at the surface
0 0
=- p qr sin qd q ^ ( ) Q q and then decreases O
ò l = pr µ 1/r 2 . A B
4p 2e0 r3 j R C
0 r
= - p q sin qd q j^ = - q [- cos q]0p j^
ò 4 p2e 0 r 2 Q
4 p2e 0 r 2 E = 4p e0 × r outside shell and zero inside.
0
2
q
=- 2 p 2 e 0 r 2 j^ 18. (d) : C = e0d A = 9 ´ 10-12 F
k 1 = 3 k 2 = 6
13. (a) : The force of repulsion by Q is q – Q + With dielectric, C = e0 kA
cancelled by the resultant attracting A d
force due to q– and q– at A and B. C1 = ed0 A/ 3× 3 = 9C ;
C2 = e20d A/ ×36 = 9C
Force of repulsion,
F = 1 Q2 = 1 × Q 2 Q+ B q – d/3 2d /3
4pe0 (a2 + a2) 4p e 2a 2
0
Total force of attraction along the diagonal
(taking cosq components) \ C total = C1C2 as they are in series.
C1 + C2
{ } = 1 ì 2 üý
4 pe 0 í þ
Qq × 1 + Qq = 1 î Qq
a 2 2 a2 2 4 pe a2
0 9C18´C9C = 9 92 ´ 9 ´10-12 F
2
Q2 Q 2 = ´ C or
2a 2 Qq-
Þ = Qq 2 Þ = - 2 2 (a ).
a2
Þ C total = 40.5 pF.
Electrostatics 61
19. (a) : The potential energy of a charged capacitor 24. (d) : When the spherical conductors are connected by a
conducting wire, charge is redistributed and the spheres
U i q2 attain a common potential V.
=
2C
where U i is the initial potential energy. \ Intensity E A = 1 Q A
4 pe 0 RA2
If a dielectric slab is slowly introduced, the energy
q2 or E A = 1 ´ CAV = (4pe0 RA )V = V
= 4pe0 RA2 4 pe 0 RA2 R A
2K C
Once is taken out, again the energy increases to the old value.
Therefore after it is taken out, the potential energy come back Similarly E B = V
RB
to the old value. Total work done = zero.
20. (a) : Given : Potential V ( x ) = 20 \ EA = R B = 2 .
x2 - 4 EB RA 1
Electric field E = -dV = -d æ 20 ÷øö = 40 x 25. (b) : An electron on plate 1 has electrostatic potential energy.
dx dx çè x2 - ( x2 - 4) 2
4 When it moves, potential energy is converted into kinetic
At x = 4 mm energy.
40 ´ 4 160 10 \ Kinetic energy = Electrostatic potential energy
[16 - 4]2 144 9
\ E = = = V/m m. or 1 mv2 = e DV
2
Positive sign indicate E is +ve x direction. q or v = 2e ´ DV = 2 ´1.6 ´10-19 ´ 20
21. (a) : “Unit positive charge” will be repelled q B m 9.11´ 10- 31
by A and B and attracted by – q and – q A C or v = 2.65 × 10 6 m/s.
downwards in the same direction. If they – q
are exchanged, the direction of the field will 26. (d) : In a nonuniform electric field, the dipole will experience
be opposite. In the case of potential, as it is D a torque as well as a translational force.
a scalar, they cancel each other whatever – q
may be their position. 27. (b) : Energy of capacitor = Heat energy of block
\ Field is affected but not the potential.
\ 1 CV 2 = ms D T
2
22. (a) : Let E be emf of the battery or V = 2 ms D T .
Work done by the battery W = CE 2 C
Energy stored in the capacitor U = 21 CE2 28. (c) : n plates connected alternately give rise to
(n – 1) capacitors connected in parallel
12 CE 2
CE 2 \ Resultant capacitance = (n – 1)C.
\ U = = 12 . 29. (d) : V A = 1 Q- 1 Q
W 4pe0 R 4 pe0 R2 + d 2
23. (c) : r = 2iˆ + 2 ˆj V B = 1 (-RQ ) + 1 Q
r1 4pe0 4 pe0 R2 + d 2
( ) ( ) | rr1 |= r1 =
2 2 A B
2 + 2 = 2 R 2 + d 2
rr2 = 2iˆ + 0 ˆj Y R
or | r A
r2 | = r2 = 2 d
r 1 ( 2 , 2 )
Potential at point A is B X + Q – Q
1q (0, 0) (2, 0) 1 ´ Q é 2 ù
4 pe 0r1 r 2 4 pe0 ê
V A = \ VA - V B = êë 2 - ú
R ûú
R2 + d 2
1 10-3 ´ 10 -6 Q é R1 - 1 ù
= 4pe 0
2 = 2 pe0 ê R2 2 ú .
ëê ûú
Potential at point B is + d
VB = 1 q = 1 10–3 ´10- 6 30. (c) : Resultant intensity = 0 B
4 pe 0 r2 4pe0 2
O A
\ VA – VB = 0. + 8q – 2q
L d
62 JEE MAIN CHAPTERWISE EXPLORER
1 8q - 1 2 q = 0 Finally, 1 m (2v ) 2 = kqQ ............(ii)
4pe0 (L + d ) 2 4 pe0 d 2 2 r1
\ From (i) and (ii)
or (L + d) 2 = 4d 2 1 = r1 Þ r1 = r .
or d = L 4 r 4
\ Distance from origin = 2L.
q 2
31. (b) : Tsinq = sq/e 0 34. (d) : Initially, F = 1 d 2 ...........(i)
4 pe 0
T T cosq
when the third equal conductor touches B, the charge of B
q is shared equally between them
T sinq
qs \ Charge on B = 2q = charge on third conductor.
e 0
æ 2q ÷øö touches
Now this third conductor with charge çè C, their
total charge æ q + 2q ÷öø is equally shared between them.
çè
mg \ Charge on
T cosq = mg
\ tan q = sq C = 3q = Charge of third conductor
e0 mg 4
\ New force between B and C
\ s is proportional to tanq.
32. (b) : Consider the four forces F1 , F2 , F 3 and F4 acting on = 1 æ q ´ 34q ÷öø = 38 F .
charge (–Q) placed at A. 4 pe0 d 2 èç 2
F 3 F 2 35. (b) : Force on (–q 1) due to q2 = - q1q2
45° q q 4 pe0 b2
F 1 A –Q a –Q
B \ F1 = 1 2 along (q 1 q2 )
F 4q 4 pe b2
E 0
a a Force on (–q1 ) due to ( - q3 ) = (-q1 )(-q3 )
q q 4 pe 0 a 2
1 3
F2 = 4 pe a2 as shown
D C 0
–Q a –Q F2 makes an angle of (90° – q) with (q 1q 2 )
Distance CA = 2 a y
Distance EA = 2 a = a –q 3
2 2 q
For equilibrium, consider forces along DA and equate the F1 x
resultant to zero –q 1 (90° – q ) q2
Q´Q (QC´AQ) 2 cos q F 2
( DA)2
\ 1 + 1 45°
4pe0 4 pe 0
Resolved part of F2 along q 1 q 2
- 1 Q ´ q cos 45° = 0 = F2 cos (90° –q)
4 pe 0 ( EA) 2
= q1q3 sin q
Q Q 1 - a2q / 2 ´ 1 = 0 4 pe 0 a 2 along (q1 q2 )
a2 2a2 2 2
or + ´
\ Total force on (–q 1 )
or Q éëê1 + 212 ûùú = q 2 = é q1q2 + q1q3 sin q ù along x axis
ê 4pe0b2 4 pe0 a 2 ú
ëê úû
or q = Q é 2 2 2+ 1 ûùú = Q4 (1 + 2 2 ). \ xcomponent of force µ é q2 + q3 sin q ù .
2 êë 2 ëê b2 a2 ûú
33. (d) : Energy is conserved in the phenomenon 36. (d) : Energy of condenser
= 1 Q 2 = 1 ´ (8 ´10-18 ) 2 = 32 ´10-32 J
Initially, 1 mv 2 = kqQ ..........(i) 2 C 2 (100 ´10-6 )
2 r
Electrostatics 63
37. (c) : Potential at any internal point of charged shell = q 41. (d) : When the system of three charges is in equilibrium,
4 pe 0 R
Q ´ q + Q ´ Q = 0
Potential at P due to Q at centre = 1 2Q 4pe0 d 2 4pe 0 (2d ) 2
4 pe0 R
d d
\ Total potential point
= q + 2 Q = 1 (q + 2Q ). Q q Q
4pe0 R 4pe0 R 4pe 0 R
38. (b) : Aluminium is a good conductor. Its sheet introduced or q = - Q4 .
between the plates of a capacitor is of negligible thickness.
The capacity remains unchanged.
C = e0 A 42. (b) : Total capacity = nC
d
With air as dielectric, \ Energy = 1 nCV 2
2
e0 A e0 A
With space partially filled, C¢ = ( d - t ) = d = C . 43. ( * ) : Electric flux through ABCD = zero for the charge
placed outside the box as the charged enclosed is zero. But
39. (a) : According to Gauss theorem,
q
(f2 - f1 ) = Q Þ Q = ( f2 - f1) e0 . for the charge inside the cube, it is e through all the
e0
0
The flux enters the enclosure if one has a negative charge
q
(–q 2) and flux goes out if one has a +ve charge (+q 1 ). As surfaces. For one surface, it is 6 e . (Option not given).
one does not know whether f 1 > f2 , f 2 > f 1, Q = q1 ~ q 2
0
44. (a) : W = QV
40. (a) : C = 4pe0 R = 1 = 1.1 ´ 10-10 F . \ V = W = 220 = 0.1 volt .
´ 109 Q
9
64 JEE MAIN CHAPTERWISE EXPLORER
CHAPTER CURRENT ELECTRICITY
12
1. The supply voltage to a room is 120 V. The resistance of the 55 W R
lead wires is 6 W. A 60 W bulb is already switched on. What
is the decrease of voltage across the bulb, when a 240 W
heater is switched on in parallel to the bulb? G
(a) 10.04 Volt (b) zero Volt 20 cm
(c) 2.9 Volt (d) 13.3 Volt (2013)
2. Two electric bulbs marked 25 W220 V and 100 W220 V are (a) 55 W (b) 13.75 W
connected in series to a 440 V supply. Which of the bulbs (c) 220 W (d) 110 W. (2008)
will fuse? 7. A 5 V battery with internal resistance 2 W and
(a) 100 W (b) 25 W 2 V battery with internal resistance 1 W are connected to a
(c) neither (d) both (2012) 10 W resistor as shown in the figure. The current in the 10
3. If a wire is stretched to make it 0.1% longer, its resistance W resistor is
will P2
(a) increase by 0.05% (b) increase by 0.2%
(c) decrease by 0.2% (d) decrease by 0.05% (2011) 5 V 10 W 2 V
2 W 1 W
4. Two conductors have the same resistance at 0°C but their
temperature coefficients of resistance are a 1 and a2 . The (a) 0.27 A P1 to P2 P1
respective temperature coefficients of their series and parallel (c) 0.03 A P1 to P2
(b) 0.27 A P2 to P1
combinations are nearly (d) 0.03 A P 2 to P1 . (2008)
(a) a1 + a2 , a1 + a 2 (b) a1 + a 2 , a1 + a 2 Directions : Questions 8 and 9 are based on the following paragraph.
2 2 2 Consider a block of conducting material of resistivity r shown in
a1a 2 the figure. Current I enters at A and leaves from D. We apply
(c) a1 + a 2 , a1 + a2 (d) a1 + a 2 , a1 + a 2 (2010) superposition principle to find voltage DV developed between B
2 and C. The calculation is done in the following steps:
(i) Take current I entering from A and assume it to spread over
5. This question contains Statement1 and Statement2. Of the
a hemispherical surface in the block.
four choices given after the statements, choose the one that (ii) Calculate field E(r) at distance r from A by using Ohm’s
best describes the two statements. law E = rj, where j is the current per unit area at r.
(iii) From the r dependence of E(r), obtain the potential V(r) at r.
Statement1: The temperature dependence of resistance is (iv) Repeat (i), (ii) and (iii) for current I leaving D and superpose
usually given as R = R0 (1 + aDt). The resistance of a wire results for A and D.
changes from 100 W to 150 W when its temperature is increased
from 27°C to 227°C. This implies that a = 2.5 × 10– 3/ °C
Statement2: R = R0 (1 + aDt) is valid only when the change DV
in the temperature DT is small and DR = (R – R 0) << R 0 .
(a) Statement1 is true, Statement2 is false
(b) Statement1 is true, Statement2 is true; Statement2 is I I
the correct explanation of Statement1.
(c) Statement1 is true, Statement2 is true; Statement2 is a b a
A B C D
not the correct explanation of Statement1.
(d) Statement1 is false, Statement2 is true. (2009)
6. Shown in the figure below is a meterbridge set up with 8. DV measured between B and C is
null deflection in the galvanometer. The value of the unknown
resistance R is
Current Electricity 65
rI (b) rI - rI (b) conservation of charge, conservation of momentum
(a) 2p(a - b) pa p(a + b) (c) conservation of energy, conservation of charge
(d) conservation of momentum, conservation of charge.
(c) rI - rI (d) rI - rI (2008)
a (a + b) 2pa 2p(a + b) (2006)
9. For current entering at A, the electric field at a distance r 16. An electric bulb is rated 220 volt 100 watt. The power
from A is consumed by it when operated on 110 volt will be
rI rI (a) 50 watt (b) 75 watt
(a) 4p r 2 (b) 8 p r 2
(c) 40 watt (d) 25 watt. (2006)
r I rI
(c) r 2 (d) 2p r 2 (2008) 17. A thermocouple is made from two metals, antimony and
bismuth. If one junction of the couple is kept hot and the
10. The resistance of a wire is 5 ohm at 50°C and 6 ohm at 100°C. other is kept cold then, an electric current will
The resistance of the wire at 0°C will be (a) flow from antimony to bismuth at the cold junction
(a) 3 ohm (b) 2 ohm (b) flow from antimony to bismuth at the hot junction
(c) 1 ohm (d) 4 ohm (2007) (c) flow from bismuth to antimony at the cold junction
11. A material B has twice the specific resistance of A. A circular (d) not flow through the thermocouple. (2006)
wire made of B has twice the diameter of a wire made of A. 18. In a potentiometer experiment the balancing with a cell is
Then for the two wires to have the same resistance, the ratio at length 240 cm. On shunting the cell with a resistance of
lB /l A of their respective lengths must be 2 W, the balancing length becomes 120 cm. The internal
(a) 2 (b) 1 resistance of the cell is
(c) 1/2 (d) 1/4. (2006) (a) 4 W (b) 2 W
12. The resistance of a bulb filament is 100 W at a temperature (c) 1 W (d) 0.5 W (2005)
of 100ºC. If its temperature coefficient of resistance be 0.005 19. Two sources of equal emf are connected to an external
per ºC, its resistance will become 200 W at a temperature of resistance R. The internal resistances of the two sources are
(a) 200ºC (b) 300ºC R 1 and R 2 (R 2 > R1 ) . If the potential difference across the
source having internal resistance R2 is zero, then
(c) 400ºC (d) 500ºC. (2006)
13. The current I drawn from the 5 volt source will be (a) R = R1R2 (b) R = R1R2
R1 + R2 R2 - R1
10 W
(c) R = R2 ( R1 + R2 ) (d) R = R2 – R 1 (2005)
( R2 - R1 )
5 W 10 W 20 W
20. In the circuit, the galvanometer G shows zero deflection. If
I 10 W the batteries A and B have negligible internal resistance, the
value of the resistor R will be
5 V 500 W
G
(a) 0.17 A (b) 0.33 A (2006) 12 V B 2 V
(c) 0.5 A (d) 0.67 A. R A
14. In a Wheatstone’s bridge, three resistance P, Q and R connected (a) 500 W (b) 1000 W
(c) 200 W (d) 100 W
in the three arms and the fourth arm is formed by two resistance (2005)
S 1 and S2 connected in parallel. The condition for bridge to 21. An energy source will supply a constant current into the load
be balanced will be if its internal resistance is
(a) zero
(a) P = R (b) P = 2 R (b) nonzero but less than the resistance of the load
Q S1 + S2 Q S1 + S2 (c) equal to the resistance of the load
(d) very large as compared to the load resistance
(c) P = R(S1 + S 2 ) (d) P = R(S1 + S 2 ) . (2006) (2005)
Q S1S2 Q 2 S1S2
15. The Kirchhoff’s first law (å i = 0) and second law
(åiR = å E), where the symbols have their usual meanings, 22. The resistance of hot tungsten filament is about 10 times the
cold resistance. What will be the resistance of 100 W and
are respectively based on 200 V lamp when not in use?
(a) conservation of charge, conservation of energy
66 JEE MAIN CHAPTERWISE EXPLORER
(a) 400 W (b) 200 W 3.3 × 10– 7 kg per coulomb. The mass of the metal liberated
(c) 40 W (d) 20 W
(2005) at the cathode when a 3 A current is passed for 2 second will
23. Two voltameters, one of copper and another of silver, are be
joined in parallel. When a total charge q flows through the (a) 19.8 × 10 –7 kg (b) 9.9 × 10 –7 kg
voltmeters, equal amount of metals are deposited. If the (c) 6.6 × 10 –7 kg (d) 1.1 × 10 –7 kg. (2004)
electrochemical equivalents of copper and silver are z1 and 31. The thermo emf of a thermocouple varies with the temperature
z2 respectively the charge which flows through the silver
voltameter is q of the hot junction as E = aq + bq 2 in volt where the ratio
a/b is 700°C. If the cold junction is kept at 0°C, then the
q z1 q z 2 neutral temperature is
z 2 z1
(a) (b) (a) 700°C (b) 350°C
q q (c) 1400°C
(c) 1 + z1 (d) 1 + z 2
(2005) (d) no neutral temperature is possible for this thermocouple.
z2 z1
(2004)
24. A heater coil is cut into two equal parts and only one part 32. Time taken by a 836 W heater to heat one litre of water from
10°C to 40°C is
is now used in the heater. The heat generated will now be (a) 50 s (b) 100 s (c) 150 s (d) 200 s.
(2004)
(a) one fourth (b) halved
(c) doubled (d) four times (2005)
25. The thermistors are usually made of 33. The length of a given cylindrical wire is increased by 100%.
(a) metals with low temperature coefficient of resistivity Due to the consequent decrease in diameter the change in
(b) metals with high temperature coefficient of resistivity the resistance of the wire will be
(c) metal oxides with high temperature coefficient of (a) 200% (b) 100%
resistivity (c) 50% (d) 300%. (2003)
(d) semiconducting materials having low temperature 34. A 3 volt battery with negligible internal resistance is connected
coefficient of resistivity. (2004) in a circuit as shown in the figure. I
26. In a metre bridge experiment null point is obtained at The current I, in the circuit will
20 cm from one end of the wire when resistance X is balanced be 3 V 3 W 3 W
(a) 1 A 3W
against another resistance Y. If X < Y, then where will be 1.5 A (b)
the new position of the null point from the same end, if (c) 2 A (d) (1/3)
one decides to balance a resistance of 4X against Y? A (2003)
(a) 50 cm (b) 80 cm 35. The length of a wire of a potentiometer is 100 cm, and the
e.m.f. of its standard cell is E volt. It is employed to measure
(c) 40 cm (d) 70 cm. (2004) the e.m.f. of a battery whose internal resistance is 0.5 W. If
the balance point is obtained at l = 30 cm from the positive
27. An electric current is passed through a circuit containing two end, the e.m.f. of the battery is
wires of the same material, connected in parallel. If the lengths
and radii of the wires are in the ratio of 4/3 and 2/3, then
the ratio of the currents passing through the wire will be 30 E 30 E
100.5 100 - 0.5
(a) 3 (b) 1/3 (a) (b)
(c) 8/9 (d) 2. (2004) (c) 3100 E0 - 0.5i , where i is the current in the potentiometer
wire.
28. The resistance of the series combination of two resistances
is S. When they are joined in parallel the total resistance is
P. If S = nP, then the minimum possible value of n is (d) 3100 E0 .
(a) 4 (b) 3 (2003)
(c) 2 (d) 1. (2004) 36. A 220 volt, 1000 watt bulb is connected across a 110 volt
29. The total current supplied to the circuit by the battery is mains supply. The power consumed will be
(a) 1 A (a) 750 watt (b) 500 watt
(b) 2 A 6 V 2 W (c) 250 watt (d) 1000 watt. (2003)
(c) 4 A 6 W
(d) 6 A. (2004) 3 W 37. The negative Zn pole of a Daniell cell, sending a constant
1.5 W current through a circuit, decreases in mass by 0.13 g in 30
30. The electrochemical equivalent minutes. If the electrochemical equivalent of Zn and Cu are
32.5 and 31.5 respectively, the increase in the mass of the
of a metal is
Current Electricity 67
positive Cu pole in this time is 40. If q i is the inversion temperature, q n is the neutral temperature,
q c is the temperature of the cold junction, then
(a) 0.180 g (b) 0.141 g
(c) 0.126 g (d) 0.242 g. (2003) (a) qi + qc = qn (b) qi - qc = 2qn
38. The thermo e.m.f. of a thermocouple is 25 mV/°C at room (c) qi + qc = qn (d) qc - qi = 2qn . (2002)
2
temperature. A galvanometer of 40 ohm resistance, capable
41. A wire when connected to 220 V mains supply has power
of detecting current as low as 10– 5 A, is connected with the
dissipation P1 . Now the wire is cut into two equal pieces
thermocouple. The smallest temperature difference that can which are connected in parallel to the same supply. Power
be detected by this system is dissipation in this case is P2 . Then P2 : P1 is
(a) 16°C (b) 12°C (a) 1 (b) 4 (c) 2 (d) 3. (2002)
(c) 8°C (d) 20°C. (2003) 42. If in the circuit, power R
dissipation is 150 W, then
39. The mass of a product liberated on anode in an electrochemical
cell depends on R is 2 W
(a) 2 W
(a) (It) 1/2 (b) It (c) I/t (d) I 2t .
(where t is the time period for which the current is passed). (b) 6 W 15 V
(2002) (c) 5 W
(d) 4 W.
(2002)
Answer Key
1. (a) 2. (b) 3. (b) 4. (a) 5. (a) 6. (c)
7. (d) 8. (d) 9. (d) 10. (d) 11. (a) 12. (c)
13. (c) 14. (c) 15. (a) 16. (d) 17. (a) 18. (b)
19. (d) 20. (d) 21. (a) 22. (c) 23. (d) 24. (c)
25. (c) 26. (a) 27. (b) 28. (a) 29. (c) 30. (a)
31. (d) 32. (c) 33. (d) 34. (b) 35. (c) 36. (c)
37. (c) 38. (a) 39. (b) 40. (c) 41. (b) 42. (b)
68 JEE MAIN CHAPTERWISE EXPLORER
1. (a) : As P = V 2 = IR1 = 2 ´ (220) 2 = 352 V
R 11 25
Here, the supply voltage is taken as rated Potential difference across 100 W bulb
voltage. (220) 2
100
\ Resistance of bulb = IR2 = 2 ´ = 88 V
11
120 V ´ 120 V
RB = 60 W = 240 W
Thus the bulb 25 W will be fused, because it can tolerate
120 V ´ 120 V = only 220 V while the voltage across it is 352 V.
240 W
Resistance of heater, RH = 60 W 3. (b) : Resistance of wire
Voltage across bulb before heater is switched on, R = rl
A
120 V ´ 240 W ...(i)
240 W + 6 W
V1 = = 117.07 V On stretching, volume (V) remains constant.
As bulb and heater are connected in parallel. Their equivalent So V = Al or A = Vl
res istance is
Req = (240 W)(60 W) = 48 W \ R = rl 2 (Using (i))
240 W + 60 W V
\ Voltage across bulb after heater is switched on Taking logarithm on both sides and differentiating we get,
V2 = 120 V × 48 W = 106. 66 V DR = 2 Dl (Q V and r are constants)
48 W + 6 W R l
Decrease in the voltage across the bulb is or DR % = 2 Dl l %
DV = V 1 – V 2 = 10.41 V ; 10.04 V R
Hence, when wire is stretched by 0.1% its res istance will
2. (b) : 25 W‑220 V 100 W‑2 20 V increase by 0.2%.
440 V 4. (a) : Let R0 be the resistance of both conductors at 0°C.
Let R 1 and R 2 be their resistance at t°C. Then
As R = ( Rated voltage) 2 R1 = R 0( 1 + a1 t)
Rated power R2 = R 0 (1 + a2 t)
Let Rs is the resistance of the series combination of two
conductors at t°C. Then
Rs = R1 + R 2
\ Resistance of 25 W220 V bulb is Rs0 (1 + ast) = R 0( 1 + a 1t) + R 0 (1 + a 2t)
(220) 2 where, Rs0 = R0 + R 0 = 2R 0
25
R1 = W \ 2R 0 (1 + ast) = 2R0 + R 0t(a 1 + a 2 )
Resistance of 100 W220 V bulb is 2R 0 + 2R 0ast = 2R 0 + R 0t(a 1 + a 2 )
R2 = (220) 2 W \ as = a1 + a2
100 2
When these two bulbs are connected in series, the total Let Rp is the resistance of the parallel combination of two
res istance is conductors at t°C. Then
Rs = R1 + R2 = (220) 2 é 1 + 1 ûù = (220) 2 W R p = R1R2
ë 25 100 20 R1 + R2
440 20 = 121 A Rp0 (1 + a p t ) = R0 (1 + a1t) R0 (1 + a2t )
(220)2 / R0 (1 + a1t) + R0 (1 + a 2t )
Current, I =
R0R0 = R0
where, R p 0 = R0 + R0 2
Potential difference across 25 W bulb
Current Electricity 69
R0 R02 (1 + a1t)(1 + a2t ) + 2 – 1(x – y) + 10 ∙ y = 0
2 2R0 + R0 (a1 + a 2 ) t
\ (1 + a p t ) = +x – 11y = 2 ...(ii)
R02 (1 + a1t + a 2t + a1a 2t 2 ) Þ 2x – 22y = 4 ...(iii) = (ii) × 2
R0 (2 + (a1 + a 2 )t)
R0 (1 + a p t ) = (i) – (iii) gives 32y = 1
2
Þ y = 312 A = 0.03 A from P2 to P1.
12 (1 + (1 + a1t + a2t + a1a2 t 2 )
a p t ) = (2 + (a1 + a 2 )t) 8. (d) : Current is spread over an area 2pr 2 . The current I is
As a1 and a 2 are small quantities a surface current.
\ a 1a2 is negligible
A B C D
1 + (a1 + a2 ) t 1 + (a1 + a2 ) t a b a
2 + (a1 + a 2 ) t
\ 1 (1 + a p t ) = = 2 éêë1 + (a1 + a2 ) t ù
2 2 ûú
Current density, j = I
2 p r 2
(a1 + a 2 ) t ù-1
= 12 [1 + (a1 + a 2 )t ] éêë1 + 2 úû rl rr
Resistance = area = 2 p r 2
= 12 [1 + (a1 + a 2 )t ] éëê1 - (a1 + a2 ) t ù E = Ir/2pr 2 .
2 úû
a
[By binomial expansion]
VB -VC = DV = ò -Edr
+ a2 ) 2 t 2 ù
2 úû a +b
= 1 éëê1 - (a1 + a2 )t + (a1 + a2 ) t - (a1 -Ir a 1 - I r êéë- 1 ù a
2 2 2p r 2 2p r úû a +
Þ DV = ò dr =
As (a 1 + a 2) 2 is negligible a +b b
\ 1 (1 + a pt ) = 1 éë1 + 1 (a1 + a 2 ) t ù DV = I r é 1 - a 1 úûù .
2 2 2 û 2 p ëê a +b
a pt = (a1 +a 2 ) t (d) : j × r = E. I r
2 2 p r 2
9. \ E =
ap = a1 + a2 10. (d) : Given : R 50 = 5 W, R 100 = 6 W
2
R t = R0 ( 1 + at)
5. (a) : From the statement given, a = 2.5 × 10 –3 /°C.
where Rt = resistance of a wire at t°C, R 0 = resistance of a
The resistance of a wire change from 100 W to 150 W when
wire at 0°C, a = temperature coefficient of resistance.
the temperature is increased from 27°C to 227°C. \ R 50 = R0 [1 + a 50]
and R1 00 = R 0 [1 + a 100]
It is true that a is small. But (150 – 100) W or 50 W is not or R5 0 – R0 = R 0a(50)
R1 00 – R0 = R 0a (100)
very much less than 100 W i.e., R – R0 << R0 is not true. Divide (i) by (ii), we get ... (i)
... (ii)
55 W R
6. (c) : 5 - R0 = 1
6 - R0 2
20 cm 80 cm or R0 = 4 W.
This is a Wheatstone bridge. or 10 – 2R 0 = 6 – R 0
If r l is the resistance per unit length (in cm)
20r 80r or R = 80 ´ 55 = 220 W. 11. (a) : Resistance of a wire R = rl = rl ´ 4
l = l pr2 p D2
55 R 20
A x P 2 x – y B Q RA = RB
5 V y 2 V \ 4r AlA = 4r Bl B
10 W 1 W pDA2 p DB2
7. (d) :
2 W
C x P 1 x – y D or lB = æ rA öæ D B ö2
lA çè rB ÷ø çè DA ÷ø
Applying Kirchhoff’s law for the loops
AP2 P1 C A and P2 BDP 1 P2 , one gets = æ rA öæ 2 D A 2 = 4 = 2
–10y – 2x + 5 = 0 èç 2r A ÷ø çè 2 1
Þ 2x + 10y = 5 ö
DA ø÷
...(i)
70 JEE MAIN CHAPTERWISE EXPLORER
12. (c) : Given : R 100 = 100 W 19. (d) : I = 2E
a = 0.005ºC –1 R1 + R2 + R
R t = 200 W Q E – IR 2 = 0 E R 1 (Given)
\ R 100 = R0 [ 1 + 0.005 (100)] \ E = IR 2
or 100 = R0 [ 1 + 0.005 × 100] E R 2
R t = R0 [ 1 + 0.005t]
200 = R 0 [1 + 0.005t] .......(i) or E = R1 2 ER2
Divide (i) by (ii), we get .......(ii) + R2 + R
or R1 + R 2 + R = 2R 2 I R
or R = R 2 – R 1 .
100 = [1 + 0.005´100] 20. (d) : For zero deflection in galvanometer,
200 [1 + 0.005t ]
I1 = I 2
1 + 0.005t = 2 + 1
12 = 2
or t = 400ºC. or 500 + R R .
13. (c) : The equivalent circuit is a balanced Wheatstone's bridge. Þ 12R = 1000+ 2R Þ R = 100 W
Hence no current flows through arm BD. 21. (a) : If internal resistance is zero, the energy source will
supply a constant current.
AB and BC are in series B
\ RA BC = 5 + 10 = 15 W 5 W 10 W 22. (c) : Resistance of hot tungsten = V 2 = (200) 2 = 400 W
AD and DC are in series P 100
\ RA DC = 10 + 20 = 30 W A 10 W C Resistance when not in use = 41000 = 40 W .
ABC and ADC are in parallel
\ Re q = ( RABC )( R ADC ) 10 W 20 W 23. (d) : The voltameters are joined in parallel.
( RABC + RADC )
I D Mass deposited = z 1 q 1 = z2 q 2
+
or Req = 15 ´ 30 = 154´53 0 =10 W 5 V \ q1 = z2 Þ q1 + q2 = z1 + z 2 Þ q = çèæ1 + z 2 ö
15 + 30 q2 z1 ÷ø
q2 z1 q2 z1
\ Current I = E = 5 = 0.5 A .
Req 10 or q2 = q .
14. (c) : For balanced Wheatstone's bridge, P = R çèæ1 + z2 ö
Q S z1 ÷ø
Q S = S1S2 (Q S 1 and S2 are in parallel) 24. (c) : Resistance of full coil = R
S1 + S2 Resistance of each half piece = R/2
\ P = R( S1 + S 2 ) . \ H 2 = V 2 t ´ R = 2
Q S1S2 H1 R / 2 V 2 t 1
15. (a) : Kirchhoff's first law [S i = 0] is based on conservation \ H 2 = 2H 1
Heat generated will now be doubled.
of charge
Kirchhoff's second law (S i R = S E) is based on conservation 25. (c) : Thermistors are made of metal oxides with high
temperature coefficient of resistivity.
of energy.
16. (d) : Resistance of the bulb 26. (a) : For meter bridge experiment,
V 2 (220) 2 = 484 W R1 = l1 = l1
( R ) = = R2 l2 (100 – l1 )
P 100
Power across 110 volt = (110) 2 In the first case, X = 20 = 20 = 1
484 Y 100 - 20 80 4
\ Power = 1104´84110 = 25 W . In the second case,
17. (a) : AntimonyBismuth couple is ABC couple. It means 4X = l Þ 4 = l Þ l = 50 cm .
that current flows from A to B at cold junction. Y (100 - l) 4 100 - l
27. (b) : Potential difference is same when the wires are put
18. (b) : The internal resistance of a cell is given by in parallel
r = æ l - ö = R çæ l1 - l 2 ö V = I1R1 = I 1 ´ rl1
R ç 1 ÷ l2 ÷ p r12
1 ø è ø
è
l
2
\ r = 2 é 24012-01 20 ûùú = 2 W. Again V = I2 R2 = I 2 ´ rl 2
êë pr22
Current Electricity 71
\ I1 ´ rl1 = I2 ´ rl2 Þ I1 = æ l2 ö æ r1 ö2 \ The new resistance R¢2 = r l ¢ = r 2 l = 4 R
pr12 pr22 I2 çè l1 ÷ø çè r2 ÷ø r¢2
p p ´ r 2
2
I 1 3 2 2 3´4 13 .
or I 2 = èçæ 4 ÷öø çæè 3 = 4´9 = \ Change in resistance = R¢ - R = 3R
ö÷ø
\ % change = 3R R ´100% = 300% .
28. (a) : In series combination, S = (R 1 + R 2 )
In parallel combination, P = R1R2 34. (b) : Equivalent resistance = (3 + 3) ´ 3 = 18 = 2 W
( R1 + R2 ) (3 + 3) + 3 9
Q S = nP V 3
R 2
\ ( R1 + R2 ) = n R1 R2 \ (R1 + R2 ) 2 = nR1 R 2 \ Current I = = = 1.5 A .
( R1 + R2 )
E volt
For minimum value, R 1 = R 2 = R 35. (c) : Potential gradient along wire = 100 cm
\ (R + R) 2 = n(R × R) Þ 4R2 = nR 2
\ K = 1E00 volt
or n = 4. cm
29. (c) : The equivalent circuits are shown below : For battery V = E¢ – ir, where E¢ is emf of battery.
6 W or K × 30 = E¢ – ir, where current i is drawn from battery
1.5 W 1.5 W 1.5 W
1.5 W E ´ 30 = E ¢ + 0.5 i 30 E
100 100
or or E ¢ = - 0.5 i
2 W Þ 3 W Þ 36. (c) : Resistance of bulb = V 2 = ( 220) 2 = 48.4 W .
3 W P 1000
Required power
6 V 6 V 6 V = V 2 = (110) 2 = 110 ´110 = 250 W .
R 48.4 48.4
I = 16.5 = 4 A.
37. (c) : According to Faraday's laws of electrolysis,
30. (a) : m = Z i t
or m = (3.3 × 10 –7 ) × (3) × (2) = 19.8 × 10 –7 kg. mZn = Z Zn when i and t are same
mCu ZCu
31. (d) : E = aq + bq 2 0.13 = 32.5 0.13´ 31.5 =
mCu 31.5 32.5
\ dE = a + 2 bq \ Þ mCu = 0.126 g
d q
At neutral temperature (q n ), dE = 0 38. (a) : Let the smallest temperature be q°C
d q \ Thermo emf = (25 × 10 –6 ) q volt
or 0 = a + 2bq n Potential difference across galvanometer =
or qn = - a = - 12 ´ (700) = -350° C IR = 10 –5 × 40 = 4 × 10 –4 volt
2b \ (25 × 10 –6 )q = 4 × 10 –4
Neutral temperature is calculated to be –350°C \ q = 4 ´10 -4 = 16°C.
Since temperature of cold junction is 0°C, no neutral 25 ´ 10-6
temperature is possible for this thermocouple.
39. (b) : According to Faraday's laws, m µ It.
32. (c) : Electrical energy is converted into heat energy 40. (c) : qc + qi = 2qn Þ qi + qc = qn .
2
\ 836 × t = 1000 × 1 × (40 – 10) × (4.18)
[Q 4.18 J = 1cal] = V 2
R
t = 1000 ´ 30 ´ 4.18 = 150 sec. 41. (b) : P1
836
or when connected in parallel,
33. (d) : Let the length of the wire be l, radius of the wire be r Re q = (R / 2) ´ ( R / 2) = R V2 = 4 V 2
R + R 4 R / 4 R
\ Resistance R = r l r = resistivity of the wire \ P2 = = 4 P1
pr 2
2 2
Now l is increased by 100% \ l¢ = l + 100 l = 2 l \ P2 = 4.
100 P1
As length is increased, its radius is going to be decreased in 42. (b) : Power = V 2
R
such a way that the volume of the cylinder remains constant.
pr2 ´ l = pr¢2 ´ l¢ Þ r ¢2 = r2 ´l = r2 ´l = r 2 \ 150 = (15)2 + (15) 2 = 225 + 222 5 Þ R = 6 W .
l¢ 2l 2 R 2 R
72 JEE MAIN CHAPTERWISE EXPLORER
CHAPTER MAGNETIC EFFECTS OF
13 CURRENT AND MAGNETISM
1. A metallic rod of length ‘l’ is (a) ra = rp < r d (b) ra > r d > r p (2012)
tied to a string of length 2l and (c) ra = r d > r p (d) ra = r p = r d
made to rotate with angular
speed w on a horizontal table 5. A coil is suspended in a uniform magnetic field, with the
with one end of the string fixed. plane of the coil parallel to the magnetic lines of force. When
If there is a vertical magnetic a current is passed through the coil it starts oscillating; it is
field ‘B’ in the region, the e.m.f. induced across the ends of very difficult to stop. But if an aluminium plate is placed
the rod is near to the coil, it stops. This is due to
(a) induction of electrical charge on the plate.
(a) 5 Bw l 2 (b) 2 Bw l 2 (c) 3B w l 2 (d) 4 Bwl 2 (b) shielding of magnetic lines of force as aluminium is a
2 2 2 2 paramagnetic material.
(c) electromagnetic induction in the aluminium plate giving
(2013) rise to electromagnetic damping.
(d) development of air current when the plate is placed.
2. This question has StatementI and StatementII. Of the four (2012)
choices given after the Statements, choose the one that best
describes the two Statements.
StatementI : Higher the range, greater is the resistance of 6. A current I flows in an infinitely long wire with crosssection
ammeter. in the form of a semicircular ring of radius R. The magnitude
StatementII : To increase the range of ammeter, additional of the magnetic induction along its axis is
shunt needs to be used across it. (a) m0 I (b) m 0 I (c) m0 I (d) m0 I
p 2 R 2 p 2 R 2p R 4p R
(a) StatementI is false, StatementII is true.
(b) StatementI is true, StatementII is true, StatementII is (2011)
the correct explanation of StatementI. 7. Two long parallel wires are at a distance 2d apart. They carry
steady equal currents flowing out of the plane of the paper
(c) StatementI is true, StatementII is true, StatementII is as shown. The variation of the magnetic field B along the line
XX¢ is given by
not the correct explanation of StatementI.
B
(d) StatementI is true, StatementII is false. (2013)
3. Two short bar magnets of length 1 cm each have magnetic (a) X X¢
moments 1.20 Am2 and 1.00 Am2 respectively. They are placed
on a horizontal table parallel to each other with their N poles d d
pointing towards the South. They have a common magnetic
equator and are separated by a distance of 20.0 cm. The value (b) X B X¢
of the resultant horizontal magnetic induction at the mid d d
point O of the line joining their centres is close to
(Horizontal component of earth’s magnetic induction is (c) X B X¢
3.6 × 10– 5 Wb/m2 ) d d
(a) 5.80 × 10 –4 Wb/m2 (b) 3.6 × 10– 5 Wb/m2
(c) 2.56 × 10 –4 Wb/m2 (d) 3.50 × 10– 4 Wb/m2
(2013)
4. Proton, deuteron and alpha particle of the same kinetic energy (d) X B X¢ (2010)
are moving in circular trajectories in a constant magnetic d d
field. The radii of proton, deuteron and alpha particle are
respectively r p, r d and ra. Which one of the following relation
is correct?
Magnetic Effects of Current and Magnetism 73
Directions : Question numbers 8 and 9 are based on the following 13. A charged particle moves through a magnetic field
paragraph. perpendicular to its direction. Then
A current loop ABCD is held fixed on (a) kinetic energy changes but the momentum is constant
the plane of the paper as shown in the
figure. The arcs BC (radius = b) and DA B (b) the momentum changes but the kinetic energy is constant
( ra d iu s = a) of t he l oop a re j oi n ed b y I
two straight wires AB and CD. A steady A (c) both momentum and kinetic energy of the particle are
current I is flowing in the loop. Angle a C
made by AB and CD at the origin O is not constant
I 1 O 30°
(d) both, momentum and kinetic energy of the particle are
b D
constant (2007)
30°. Another straight thin wire with steady current I 1 flowing 14. A charged particle with charge q enters a regri on of cro nstant,
out of the plane of the paper is kept at the origin. (2009) ((wvucaeni))lti ohfcoovvrrirut my==t aBBvrrran n´´ypd EEcerr hrm//p a EBeunn22tg udeai cilunlyl am ro art((gtodbhn ))obi tgouvtvorrdh ne= =a EorlEErr rf ´´aidenBiBrlrrdde// cs BEBr t E i22,o anna odnf d c vorB m . w(Te2si0ht he0on 7ua )t
8. The magnitude of the magnetic field (B) due to loop ABCD
at the origin (O) is
(a) zero (b) m0 I (b - a ) 15. A current I flows along the length of an infinitely long,
24ab straight, thin walled pipe. Then
(a) the magnetic field at all points inside the pipe is the same,
(c) m0 I é b - a ù (d) m 0 I ëé2(b - a) + p3 (a + b) ùû but not zero
4p ë ab û 4p (b) the magnetic field is zero only on the axis of the pipe
(c) the magnetic field is different at different points inside
9. Due to the presence of the current I1 at the origin the pipe
(a) the forces on AB and DC are zero (d) the magnetic field at any point inside the pipe is zero
(2007)
(b) the forces on AD and BC are zero
(c) the magnitude of the net force on the loop is given by
I41p I m0 é 2(b - a) + p3 (a + b) ùû 16. A long straight wire of radius a carries a steady current i. The
ë
(d) the magnitude of the net force on the loop is given by current is uniformly distributed across its cross section. The
m 0 II1 (b - a ). ratio of the magnetic field at a/2 and 2a is
24 ab
(a) 1/2 (b) 1/4 (c) 4 (d) 1 (2007)
10. A horizontal overhead powerline is at a height of 4 m from 17. A long solenoid has 200 turns per cm and carries a current
i. The magnetic field at its centre is 6.28 × 10 –2 weber/m2 .
the ground and carries a current of 100 A from east to west. Another long solenoid has 100 turns per cm and it carries
a current i/3. The value of the magnetic field at its centre
The magnetic field directly below it on the ground is is
(a) 1.05 × 10 –4 Wb/m2 (b) 1.05 × 10 –2 Wb/m2
(m 0 = 4p × 10 –7 T m A –1 ) (c) 1.05 × 10 –5 Wb/m2 (d) 1.05 × 10– 3 Wb/m2 . (2006)
(a) 2.5 × 10 –7 T northward
(b) 2.5 × 10 –7 T southward
(c) 5 × 10 –6 T northward
(d) 5 × 10 –6 T southward. (2008) 18. In a region, steady and uniform electric and magnetic fields
11. Relative permittivity and permeability of a material are e r are present. These two fields are parallel to each other. A
and mr , respectively. Which of the following values of these
charged particle is released from rest in this region. The path
quantities are allowed for a diamagnetic mateiral?
of the particle will be a
(a) er = 1.5, m r = 1.5 (b) er = 0.5, mr = 1.5
(c) e r = 1.5, m r = 0.5 (d) er = 0.5, m r = 0.5 (a) circle (b) helix
(2008) (c) straight line (d) ellipse. (2006)
12. Two identical conducting wires AOB and COD are placed at 19. Needles N 1, N 2 and N3 are made of a ferromagnetic, a
paramagnetic and a diamagnetic substance respectively. A
right angles to each other. The wire AOB carries an electric
magnet when brought close to them will
current I1 and COD carries a current I 2. The magnetic field on
a point lying at a distance d from O, in a direction (a) attract all three of them
perpendicular to the plane of the wires AOB and COD, will (b) attract N 1 and N 2 strongly but repel N3
(c) attract N1 strongly, N 2 weakly and repel N3 weakly
be given by (d) attract N1 strongly, but repel N2 and N3 weakly.
1 (2006)
(a) m0 ( I12 + I 2 2 ) (b) m0 æ I1 + I2 ö 2 20. A uniform electric field and a uniform magnetic field are
2 p d 2 p èç d ÷ø acting along the same direction in a certain region. If an
electron is projected along the direction of the fields with
(c) m0 ( I12 + 1 (d) m0 ( I1 + I 2 ) (2007) a certain velocity then
2 p d 2 p d
I 22 ) 2
74 JEE MAIN CHAPTERWISE EXPLORER
(a) it will turn towards right of direction of motion 27. The magnetic field due to a current carrying circular loop of
(b) it will turn towards left of direction of motion radius 3 cm at a point on the axis at a distance of 4 cm from
(c) its velocity will decrease the centre is 54 mT. What will be its value at the centre of
(d) its velocity will increase (2005) the loop?
21. A charged particle of mass m and charge q travels on a (a) 250 mT (b) 150 mT
circular path of radius r that is perpendicular to a magnetic (c) 125 mT (d) 75 mT. (2004)
field B. The time taken by the particle to complete one 28. A long wire carries a steady current. It is bent into a circle
revolution is of one turn and the magnetic field at the centre of the coil
2 p qB 2 pm is B. It is then bent into a circular loop of n turns. The
(a) m (b) qB
magnetic field at the centre of the coil will be
(c) 2p mq (d) 2pmq (2005) (a) nB (b) n 2B
B qB
(c) 2nB (d) 2n 2B . (2004)
22. Two concentric coils each of radius equal to 29. A current i ampere flows along an infinitely long straight thin
2p cm are placed at right angles to each other.
walled tube, then the magnetic induction at any point inside
3 ampere and 4 ampere are the currents flowing in each coil
the tube is
respectively. The magnetic induction in weber/m2 at the center
(a) infinite (b) zero
of the coils will be (m 0 = 4p ´ 10– 7 Wb/Am)
m4 p0 × 2i 2i
(a) 5 ´ 10 –5 (b) 7 ´ 10– 5 (c) r tesla (d) r tesla . (2004)
(c) 12 ´ 10 –5 (d) 10– 5 (2005)
23. A moving coil galvanometer has 150 equal divisions. Its 30. The materials suitable for making electromagnets should have
current sensitivity is 10 divisions per milliampere and voltage (a) high retentivity and high coercivity
sensitivity is 2 divisions per millivolt. In order that each (b) low retentivity and low coercivity
division reads 1 volt, the resistance in ohms needed to be (c) high retentivity and low coercivity
connected in series with the coil will be (d) low retentivity and high coercivity. (2004)
(a) 99995 (b) 9995 31. The length of a magnet is large compared to its width and
(c) 10 3 (d) 105 (2005) breadth. The time period of its oscillation in a vibration
24. Two thin long, parallel wires, separated by a distance d carry magnetometer is 2 s. The magnet is cut along its length into
a current of i A in the same direction. They will
three equal parts and three parts are then placed on each other
m0i 2 with their like poles together. The time period of this
(2p d 2 )
(a) attract each other with a force of combination will be
(a) 2 s 2 (c) (2 3) s (d) æ 2 ö s
(b) 3 s ç 3 ÷
m0i 2 è ø
(2p d 2 )
(b) repel each other with a force of (2004)
(c) attract each other with a force of m0 i 2 32. An ammeter reads upto 1 ampere. Its internal resistance is
(2p d )
0.81 ohm. To increase the range to 10 A the value of the
(d) repel each other with a force of m0 i 2 required shunt is
(2pd )
(2005) (a) 0.03 W (b) 0.3 W
(c) 0.9 W (d) 0.09 W. (2003)
25. A magnetic needle is kept in a nonuniform magnetic field. 33. A particle of charge –16 × 10 –18 coulomb moving with velocity
It experiences 10 ms– 1 along the xaxis enters a region where a magnetic
(a) a force and a torque field of induction B is along the y axis, and an electric field
(b) a force but not a torque of magnitude 10 4 V/m is along the negative z axis. If the
(c) a torque but not a force charged particle continues moving along the xaxis, the
(d) neither a force nor a torque (2005) magnitude of B is
26. Two long conductors, separated by a distance d carry current (a) 10 3 Wb/m2 (b) 10 5 Wb/m2
I 1 and I 2 in the same direction. They exert a force F on each
other. Now the current in one of them is increased to two times (c) 10 16 Wb/m2 (d) 10 –3 Wb/m2 . (2003)
and its direction is reversed. The distance is also increased to
3d. The new value of the force between them is 34. A particle of mass M and charge Q moving with velocity r
(a) –2F (b) F/3 (c) –2F/3 (d) –F/3. v
(2004)
describes a circular path of radius R when subjected to a uniform
transverse magnetic field of induction B. The work done by
the field when the particle completes one full circle is
Magnetic Effects of Current and Magnetism 75
(a) æ Mv 2 ö 2 pR (b) zero (2003)
ç R ÷ (d) BQv 2pR.
è ø 39. The time period of a charged particle undergoing a circular
(c) BQ 2pR (2003) motion in a uniform magnetic field is independent of its
35. A thin rectangular magnet suspended freely has a period of (a) speed (b) mass
oscillation equal to T. Now it is broken into two equal halves (c) charge (d) magnetic induction. (2002)
(each having half of the original length) and one piece is 40. If a current is passed through a spring then the spring will
made to oscillate freely in the same field. If its period of (a) expand (b) compress
T ¢ (c) remains same (d) none of these. (2002)
oscillation is T¢ , the ratio is
41. If an electron and a proton having same momenta enter
T
perpendicular to a magnetic field, then
1 1 1
(a) 2 2 (b) (c) 2 (d) .
2 (a) curved path of electron and proton will be same (ignoring
4
(2003) the sense of revolution)
36. Curie temperature is the temperature above which (b) they will move undeflected
(a) a ferromagnetic material becomes paramagnetic
(b) a paramagnetic material becomes diamagnetic (c) curved path of electron is more curved than that of the
(c) a ferromagnetic material becomes diamagnetic
(d) a paramagnetic material becomes ferromagnetic. (2003) proton
(d) path of proton is more curved. (2002)
42. If in a circular coil A of radius R, current I is flowing and
in another coil B of radius 2R a current 2I is flowing, then
37. The magnetic lines of force inside a bar magnet the ratio of the magnetic fields, B A and B B , produced by them
will be
(a) are from northpole to southpole of the magnet
(b) do not exist (a) 1 (b) 2
(c) depend upon the area of crosssection of the bar magnet (c) 1/2 (d) 4. (2002)
(d) are from southpole to northpole of the magnet. 43. If an ammeter is to be used in place of a voltmeter, the we
(2003) must connect with the ammeter a
38. A magnetic needle lying parallel to a magnetic field requires (a) low resistance in parallel
W units of work to turn it through 60°. The torque needed (b) high resistance in parallel
to maintain the needle in this position will be (c) high resistance in series
( ) (c) 3 (d) low resistance in series. (2002)
2
(a) 3W (b) W W (d) 2W.
Answer Key
1. (a) 2. (a) 3. (c) 4. (a) 5. (c) 6. (a)
7. (b) 8. (b) 9. (b) 10. (d) 11. (c) 12. (c)
13. (b, c) 14. (b) 15. (d) 16. (d) 17. (b) 18. (c)
19. (c) 20. (c) 21. (b) 22. (a) 23. (b) 24. (c)
25. (a) 26. (c) 27. (a) 28. (b) 29. (b) 30. (b)
31. (b) 32. (d) 33. (a) 34. (b) 35. (b) 36. (a)
37. (d) 38. (a) 39. (a) 40. (b) 41. (a) 42. (a)
43. (c)
76 JEE MAIN CHAPTERWISE EXPLORER
1. (a) : = m : 2m : 4 m = 1: 2 :1
e e 2 e
Þ ra = r p < r d
Consider a element of length dx at a distance x from the 5. (c) 6. (a) 7. (b)
fixed end of the string. 8. (b) : O is along the line CD and AB. A B
They do not contribute to the a I
e.m.f. induced in the element is
C
de = B(wx)dx magnetic induction at O. The field I 1 O 30°
due to DA is positive or out of the
Hence, the e.m.f. induced across the ends of the rod is
e = 3l = Bw é x 2 ù 3 l = Bw [(3l )2 - (2l)2 ] paper and that due to BC is into b D
êë 2 úû 2l 2
ò Bwxdx
the paper or negative.
2l
5 Bwl 2
2
= The total magnetic field due to loop ABCD at O is
2. (a) B = BAB + BBC + BCD + BDA
3. (c) : The situation is as Þ B = 0 - m0I ´ p + 0 + m0 I ´ p
shown in the figure. 4pb 6 4pa 6
As the point O lies on Þ B = m0 I (b - a ) , out of the paper or positive.
broadside position with 24 ab
respect to both the
magnets. Therefore, 9. (b) : The straight wire is perpendicular to the segments and
The net magnetic field at point O is
the fields are parallel. There will be no force. Due to parts
B net = B1 + B 2 + B H
AB and CD, their fields are equal and opposite and their
m0 M1 m0 M 2 m0 effects also cancel each other.
4p r3 4p r 3 4 p r 3
Bnet = + + BH = ( M 1 + M 2 ) + BH
Substituting the given values, we get 10.* (d) : i
Bnet = 4p 4p ´ 10 -7 ) 3 [1.2 + 1] + 3.6 ´ 10 -5 r
´ (10 ´ 10-2 By Ampere’s theorem, B × 2p d = m0 i
= 10 -7 ´ 2.2 + 3.6 ´ 10 -5 r m0 i 4p ´ 10-7 ´100 A
10-3 B = 2p´ 4 m
= = 50 ´10-7 T
= 2.2 × 10 –4 + 0.36 × 10– 4 = 2.56 × 10 –4 Wb/m 2 . 2 pd
4. (a) : The radius of the circular path of a charged particle in Þ B = 5 × 10 –6 T southwards.
the magnetic field is given by * It is assumed that this is a direct current. If it is a.c, the
current at the given instant is in the given direction.
r = mv
Bq
11. (c) : The values of relative permeability of diamagnetic
Kinetic energy of a charged particle, materials are slightly less than 1 and e r is quite high.
According to the table given, one takes
K = 1 mv2 or v = 2 K
2 m
er = 1.5 and m r = 0.5. Then the choice (c) is correct.
\ r = m 2K = 2 Km
qB m qB 12. (c) : The field at the same point at the same distance from the
mutually perpendicular wires carrying current will be having
As K and B are constants the same magnitude but in perpendicular directions.
\ r µ m \ B = B12 + B2 2 \ B = m 0 ( I12 + I 22 )1 / 2 .
q 2 p d
rp : rd : ra = m p : qmd d : ma 13. (b, c) : Due to Lorentzian force, F = qv × B,
qp qa When a charged particle enters a field with its velocity
perpendicular to the magnetic field, the motion is circular
Magnetic Effects of Current and Magnetism 77
with qvB = mv 2 . v constantly changes its direction (but 20. (c) : Magnetic field applied parallel to motion of electron
r exerts no force on it as q = 0 and force = Bevsinq = zero
Electric field opposes motion of electron which carries a
not the magnitude). Therefore its tangential momentum negative charge
\ velocity of electron decreases.
changes its direction but its energy remains the same
æ 12 I w 2 = constant øö . Therefore the answer is (b).
è
2p 2 pr
If angular momentum is taken, Iw is a constant. 21. (b) : T = w = v ...........(i)
.........(ii)
As 1 Iw 2 is also constant, (c) is the answer. Q centripetal force = magnetic force
2
\ mv 2 = qvB Þ v = qBr
rm
* The questions could have been more specific, whether by
From (i) and (ii)
“momentum” it is meant tangential momentum or angular 2pr ´ m = 2q pBm .
qBr
momentum. r \ T =
r
14. (b) : When E and B are perpendicular and velocity has no
22. (a) : Magnetic induction at centre of one coil B1 = m0i1
changes then qE = qvB i.e., v = E . The two forces oppose 2 r
r r B vr = rr Similarly B2 = m0i2
E´B i.e., E ´ B 2 r
each other if v is along
r r B2 B2 B12 B2 2 = æ m0i1 ö2 æ m0i2 ö2 m0 2 i12 i2 2 )
çè 2r ÷ø çè 2 r ÷ø 4 r 2
As E anrd Br are perpendicular to each other \ = + + = ( +
E´B = EB sin 90° = E B = m0 4p ´10 -7 (3)2 + (4) 2
B2 B2 B 2 r 2 ´ (2p ´ 10-2 )
\ i12 + i2 2 =
For historic and standard experiments like Thomson’s e/m
value, if v is given only as E/B, it would have been better or B = 5 × 10 –5 Wb/m 2 .
from the pedagogic view, although the answer is numerically
correct. 23. (b) : Vmax = 1520 = 75 m V
15. (d) : Magnetic field is shielded and no current is inside the Imax = 11500 = 15 mA = I g
pipe to apply Ampère’s law. (Compare to electric field inside Resistance of galvanometer G = 75/15 = 5 W
a hollow sphere). For conversion into a voltmeter, a high resistance should
be connected in series with the galvanometer
16. (d) : Uniform current is flowing. Current enclosed in the
1s t ampèrean path is I × pr12 = Ir1 2 = 15 (150+0R0 )
pR2 R2
V = I g (G + R) = 15 (5 + R) Þ1 50
m0 ´ current m0 × Ir12 m 0 Ir1 1000
B = path = 2 pr1 R 2 = 2 p R2
\
or 5 + R = 150 ´151 000 = 10000 \ R = 9995 W.
Magnetic induction at a distance r2 = m0 × I
2 p r2 24. (c) : Force of attraction between wires = m0i 2 L .
2 p d
\ B1 = r1r2 = a2 × 2 a = 1 .
B2 R2 a2 N.B. The options do not mention L, perhaps by slip.
17. (b) : In first case, B 1 = m0 n1 I 1 25. (a) : A force and a torque act on a magnetic needle kept
In second case, B 2 = m 0n 2 I 2 in a nonuniform magnetic field.
\ B2 = n2 ´ I 2 = 100 ´ i / 3 = 1 26. (c) : Initially, F = m0 I1I 2 l
B1 n1 I1 200 i 6 2 p d
\ B2 = B1 = 6.28 ´10 -2 = 1.05 × 10 –2 Wb/m 2 . Finally, F¢ = m0 (-2I1)( I 2 ) l
6 6 2p 3d
18. (c) : Magnetic field exerts a force \ F ¢ = -m0 2 I1I2 l ´ 2pd = - 2
= Bevsinq = Bevsin0 = 0 F 2p 3d m0I1I2 l 3
Electric field exerts force along a straight line. \ F¢ = – 2F/3.
The path of charged particle will be a straight line.
m0i R 2
19. (c) : Magnet will attract N 1 strongly, N 2 weakly and repel 27. (a) : Field along axis of coil B = 2(R2 + x2 ) 3/ 2
N3 weakly.
At the centre of coil, B ¢ = m0i
2 R
78 JEE MAIN CHAPTERWISE EXPLORER
B ¢ m2 R0i ´ 2( R 2 + x2 )3 / 2 (R2 + x 2 ) 3 / 2 Again az = 0 as the particle traverse through the region
B m0 iR 2 R 3 undeflected
\ = =
´(R2 + x 2 ) 3/ 2 54 ´[(3)2 + (4)2 ]3 / 2 \ E z = v x B y or B y = E z = 104 = 103 Wb .
R3 (3)3 vx 10 m 2
B ¢ = B = = 54 ´125
\ 27 34. (b) : Workdone by the field = zero.
or B¢ = 250 mT. I
MB
28. (b) : Initially, r 1 = radius of coil = l/2p 35. (b) : For an oscillating magnet, T = 2 p
B = m0i = 2m 0 i p where I = ml 2/ 12, M = xl, x = pole strength
2r1 2 l
\ When the magnet is divided into 2 equal parts, the magnetic
l dipole moment
2 pn
Finally, r2 = radius of coil = M¢ x´l M
2 2
= Pole strength × length = = .....(i)
\ B ¢ = m0i ´ n = nm0i ´ 2pn = 2 m0 in 2 p I ¢ = Mass× (length) 2
2r2 2l 2 l 12
\ B¢ = 2m 0 in 2 p ´ 2 l = n 2 \ B¢= n2 B. = (m / 2)(l / 2) 2 = ml 2 = I .....(ii)
B 2l 2 m0 ip 12 12 ´ 8 8
29. (b) : Magnetic field will be zero inside the straight thin \ Time period T¢ = 2 p I¢
walled tube according to ampere's theorem. M ¢B
30. (b) : Materials of low retentivity and low coercivity are suitable \ T¢ = I¢ ´ M = I¢ ´ M .........(iii)
for making electromagnets. T M¢ I I M ¢
31. (b) : For a vibrating magnet, T = 2 p I \ T ¢ = 1 ´ 2 = 12 .
MB T 8 1
where I = ml 2 /12, M = xl, x = pole strength of magnet 36. (a) : A ferromagnetic material becomes paramagnetic above
Curie temperature.
I¢ = çæè m ø÷ö èæç l ø÷ö2 ´ 3 = ml 2 = I (For three pieces together)
3 3 12 9 ´12 9
37. (d) : The magnetic lines of force inside a bar magnet are
M ¢ = ( x) æçè 3l øö÷ ´ 3 = xl = M
(For three pieces together) from south pole to north pole of magnet.
38. (a) : W = – MB (cos q2 – cos q1 )
\ T ¢ = 2p I¢ = 2 p I / 9 = 13 ´ 2 p I = T = – MB (cos 60° – cos 0) = MB
M ¢B MB MB 3 2
T 23 sec \ MB = 2W ............(i)
3
\ T ¢ = = . Torque = MB sin 60° = (2W) sin 60°
32. (d) : S = Ig Þ S = I g G = 2W ´ 3 = 3 W .
S +G I I - I g 2
\ S = 1´ 0.81 = 0.98 1 = 0.09 W in parallel. 39. (a) : mRw2 = BqRw Þ w = Bq ÞT = 2 pm
10 - 1 m Bq
33. (a) : Particle travels along xaxis. Hence vy = vz = 0 T is independent of speed.
Field of induction B is along yaxis. Bx = Bz = 0
Electric field is along the negative zaxis. 40. (b) : The spring will compress. It will be
on account of force of attraction between
\ Ex = Ey = 0 r = r + vr ´ r two adjacent turns carrying currents in
Net force on particle F q(E B) the same direction.
Resolve the motion along the three coordinate axis 41. (a) : Bqv = mv 2 Þ r = mv = p
r Bq Bq
\ ax = Fx = q ( E + v y Bz - vz B y )
m m x r will be same for electron and proton as p, B and q are
F y mq ( Ey of same magnitude.
m
ay = = + vz Bx - vx B z ) m0 2 pI = m0 I
4p R 2 R
42. (a) : B =
az = Fz = mq ( Ez + vxBy - v y B x ) BA IA R B æèç 1 ø÷ö çæè 12 ÷öø =
m BB IB RA 2
\ = ´ = 1
Since E x = Ey = 0, v y = v z = 0, B x = B z = 0
\ ax = a y = 0, az = mq (-Ez + vxB y ) 43. (c) : High resistance in series with a galvanometer converts
it into a voltmeter.
Electromagnetic Induction and Alternating Current 79
CHAPTER ELECTROMAGNETIC INDUCTION
14 AND ALTERNATING CURRENTS
1. In an LCR circuit as shown below both 6. A rectangular loop has a P l
switches are open initially. Now switch sliding connector PQ of
S 1 is closed, S 2 kept open. (q is charge length l and resistance R W R W R W v R W
on the capacitor and t = RC is and it is moving with a I 2
I
capacitive time constant). Which of the speed v as shown. The I 1 Q
following statement is correct? setup is placed in a uniform
(a) At t = 2t , q = CV(1 – e –1 ) magnetic field going into the plane of the paper. The three
(b) Work done by the battery is half of the energy dissipated
currents I 1 , I2 and I are
in the resistor (a) I1 = I 2 = Blv , I = Blv (b) I1 = - I 2 = Blv , I = 2 Blv
(c) At t = t, q = CV/2 6R 3 R R R
(d) At t = 2t, q = CV(1 – e– 2)
(2013) (c) I1 = I 2 = Blv , I = 2 Blv (d) I1 = I 2 = I = Blv
3R 3 R R
2. A circular loop of radius 0.3 cm lies parallel to a much bigger (2010)
circular loop of radius 20 cm. The centre of the small loop 7. Let C be the capacitance of a capacitor discharging through
is on the axis of the bigger loop. The distance between their a resistor R. Suppose t1 is the time taken for the energy stored
in the capacitor to reduce to half its initial value and t 2 is the
centres is 15 cm. If a current of 2.0 A flows through the time taken for the charge to reduce to onefourth its initial
smaller loop, then the flux linked with bigger loop is
(a) 6.6 × 10 –9 weber (b) 9.1 × 10– 11 weber value. Then the ratio t1 / t2 will be
(c) 6 × 10 –11 weber (d) 3.3 × 10– 11 weber (a) 2 (b) 1 (c) 1 (d) 1
2 4
(2013)
3. A boat is moving due east in a region where the earth’s (2010)
magnetic field is 5.0 × 10– 5 N A –1m – 1 due north and horizontal.
The boat carries a vertical aerial 2 m long. If the speed of the 8. In the circuit shown below, the V K
boat is 1.50 m s –1, the magnitude of the induced emf in the
wire of aerial is key K is closed at t = 0. The L R 1
(a) 1 mV (b) 0.75 mV (c) 0.50 mV (d) 0.15 mV current through the battery is
(2011)
V (R1 + R2 ) V R2
R1R2 R2
(a) at t = 0 and at t = ¥
4. A fully charged capacitor C with initial charge q0 is connected (b) VR1R2 at t = 0 and V at t = ¥
to a coil of self inductance L at t = 0. The time at which the R12 + R22 R2
energy is stored equally between the electric and the magnetic (c) V at t = 0 and V (R1 + R2 ) at t = ¥
R2 R1R2
fields is
(a) p LC (b) p LC (c) 2 p LC (d) LC V VR1R2 at t = ¥
4 (2011) R2 R12 + R22
(d) at t = 0 and (2010)
5. A resistor R and 2 mF capacitor in series is connected through 9. In a series LCR circuit R = 200 W and the voltage and the
a switch to 200 V direct supply. Across the capacitor is a frequency of the main supply is 220 V and 50 Hz respectively.
neon bulb that lights up at 120 V. Calculate the value of R On taking out the capacitance from the circuit the current lags
to make the bulb light up 5 s after the switch has been closed. behind the voltage by 30°. On taking out the inductor from
(log1 02 .5 = 0.4) the circuit the current leads the voltage by 30°. The power
(a) 1.3 × 10 4 W
(b) 1.7 × 105 W dissipated in the LCR circuit is
(c) 2.7 × 10 6 W (d) 3.3 × 107 W (2011) (a) 242 W (b) 305 W (c) 210 W (d) zero W
(2010)
80 JEE MAIN CHAPTERWISE EXPLORER
10. An inductor of inductance L = 400 18. The phase difference between the alternating current and
emf is p/2. Which of the following cannot be the constituent
mH and resistors of resistances of the circuit?
(a) LC (b) L alone (c) C alone (d) R, L
R1 = 2 W and R2 = 2 W are connected E L (2005)
to a battery of emf 12 V as shown
R 1
in the figure. The internal resistance S R2 19. A circuit has a resistance of 12 ohm and an impedance of 15
of the battery is negligible. The
switch S is closed at t = 0. The ohm. The power factor of the circuit will be
potential drop across L as a function
(a) 1.25 (b) 0.125 (c) 0.8 (d) 0.4
of time is
(2005)
12 e - 3 t V
(a) 6e –5t V (b) t 20. The self inductance of the motor of an electric fan is 10 H.
(c) 6(1 - e-t /0.2 ) V
(d) 12e –5t V (2009) In order to impart maximum power at 50 Hz, it should be
11. Two coaxial solenoids are made by winding thin insulated wire connected to a capacitance of
over a pipe of crosssectional area A = 10 cm2 and length =
(a) 1 mF (b) 2 mF (c) 4 mF (d) 8 mF
20 cm. If one of the solenoids has 300 turns and the other 400 (2005)
turns, their mutual inductance is (m 0 = 4p × 10– 7 T mA– 1 ) 21. A coil of inductance 300 mH and resistance 2 W is connected
to a source of voltage 2 V. The current reaches half of its
(a) 2.4p × 10 –4 H (b) 2.4p × 10– 5 H steady state value in
(a) 0.15 s (b) 0.3 s (c) 0.05 s (d) 0.1 s
(c) 4.8p × 10 –4 H (d) 4.8p × 10– 5 H. (2008) (2005)
12. An ideal coil of 10 H is connected in series with a resistance
of 5 W and a battery of 5 V. 2 second after the connection is
made, the current flowing in ampere in the circuit is 22. One conducting U tube can slide inside another as shown
(a) (1 – e –1) (b) (1 – e) (c) e (d) e –1 in figure, maintaining
(2007) electrical contacts ´ ´ ´ ´ ´ ´
13. In an a.c. circuit the voltage applied is E = E 0 sin wt. The between the tubes. The A B
resulting current in the circuit is I = I0 sin çæè wt - p2 ÷öø . The power magnetic field B is ´ ´ ´ ´ ´ ´
perpendicular to the plane ´ ´ ´ ´ ´´
D C
consumption in the circuit is given by of the figure. If each tube
(a) P = 2 E0I0 (b) P = E0I 0 moves towards the other ´ ´ ´ ´ ´ ´
2
at a constant speed v, then
(d) P = E0I 0 the emf induced in the circuit in terms of B, l and v where
2
(c) P = zero (2007) l is the width of each tube, will be
14. An inductor (L = 100 mH), a resistor L (a) zero (b) 2Blv (c) Blv (d) –Blv
(R = 100 W) and a battery (E = 100
(2005)
V) are initially connected in series as R 23. A metal conductor of length 1 m rotates vertically about
shown in the figure. After a long time one of its ends at angular velocity 5 radian per second. If
the battery is disconnected after short A B the horizontal component of earth’s magnetic field is
circuiting the points A and B. The E 0.2 × 10 –4 T, then the e.m.f. developed between the two
ends of the conductor is
current in the circuit 1 ms after the short circuit is (a) 5 mV (b) 50 mV (c) 5 mV (d) 50 mV.
(2004)
(a) 1 A (b) (1/e) A (c) e A (d) 0.1 A.
(2006)
15. The flux linked with a coil at any instant t is given by 24. In a LCR circuit capacitance is changed from C to 2C. For
f = 10t2 – 50t + 250. The induced emf at t = 3 s is
(a) 190 V (b) –190 V (c) –10 V (d) 10 V. the resonant frequency to remain unchanged, the inductance
(2006)
should be changed from L to
(a) 4L (b) 2L (c) L/2 (d) L/4.
16. In an AC generator, a coil with N turns, all of the same area (2004)
A and total resistance R, rotates with frequency w in a magnetic 25. In a uniform magnetic field of induction B a wire in the form
field B. The maximum value of emf generated in the coil is of a semicircle of radius r rotates about the diameter of the
(a) NABw (b) NABRw (c) NAB (d) NABR circle with angular frequency w. The axis of rotation is
(2006) perpendicular to the field. If the total resistance of the circuit
17. In a series resonant LCR circuit, the voltage across R is 100 is R the mean power generated per period of rotation is
volts and R = 1 kW with C = 2 mF. The resonant frequency (a) Bpr 2 w ( Bpr 2w) 2
2 R 8 R
w is 200 rad/s. At resonance the voltage across L is (b)
(c) (Bpr w) 2
(a) 4 × 10 –3 V (b) 2.5 × 10– 2 V 2 R ( Bpr w2 ) 2
8 R
(c) 40 V (d) 250 V. (2006) (d) . (2004)
Electromagnetic Induction and Alternating Current 81
26. A coil having n turns and resistance R W is connected with (a) 0.2 H (b) 0.4 H (c) 0.8 H (d) 0.1 H.
a galvanometer of resistance 4R W. This combination is (2003)
moved in time t seconds from a magnetic field W 1 weber 32. Two coils are placed close to each other. The mutual
to W2 weber. The induced current in the circuit is
inductance of the pair of coils depends upon
(a) - W52 R-nWt 1 (b) - n(W52 R-tW1 ) (a) the rates at which currents are changing in the two coils
(b) relative position and orientation of the two coils
(c) - (W2R-ntW1 ) (d) - n(W2R-t W1 ) . (c) the materials of the wires of the coils
(2004) (d) the currents in the two coils. (2003)
27. Alternating current cannot be measured by D.C. ammeter 33. A conducting square loop of side + ++ + ++
L and resistance R moves in its + ++ + ++
because plane with a uniform velocity v + ++ + ++
perpendicular to one of its sides.
(a) A.C. cannot pass through D.C. ammeter A magnetic induction B constant + ++ + ++ v
in time and space, pointing + ++ + ++
(b) A.C. changes direction perpendicular and into the plane + ++ + ++
at the loop exists everywhere
(c) average value of current for complete cycle is zero + ++ + ++
+ ++ + ++
(d) D.C. ammeter will get damaged. (2004)
L
28. In an LCR series a.c. circuit, the voltage across each of the with half the loop outside the field, as shown in figure. The
components, L, C and R is 50 V. The voltage across the LC induced emf is
combination will be (a) zero (b) RvB (c) vBL/R (d) vBL.
(a) 50 V (b) 50 2 V (2002)
(c) 100 V (d) 0 V (zero). (2004) 34. In a transformer, number of turns in the primary coil are 140
29. The core of any transformer is laminated so as to and that in the secondary coil are 280. If current in primary
(a) reduce the energy loss due to eddy currents
(b) make it light weight coil is 4 A, then that in the secondary coil is
(c) make it robust & strong
(d) increase the secondary voltage. (a) 4 A (b) 2 A (c) 6 A (d) 10 A.
(2002)
(2003) 35. The power factor of an AC circuit having resistance (R) and
30. In an oscillating LC circuit the maximum charge on the inductance (L) connected in series and an angular velocity
capacitor is Q. The charge on the capacitor when the energy
is stored equally between the electric and magnetic field is w is
(a) Q/2 (b) Q / 3 (c) Q / 2 (d) Q.
(2003) (a) R/wL (b) R/(R2 + w 2 L 2 )1 /2
(c) wL/R (d) R/(R 2 – w 2L 2 ) 1/2. (2002)
36. The inductance between A and D is
31. When the current changes from +2 A to –2 A in 0.05 second, (a) 3.66 H
an e.m.f. of 8 V is induced in a coil. The coefficient of self
induction of the coil is (b) 9 H A 3 H 3 H 3 H D
(c) 0.66 H
(d) 1 H.
(2002)
Answer Key
1. (d) 2. (b) 3. (d) 4. (b) 5. (c) 6. (c)
7. (d) 8. (c) 9. (a) 10. (d) 11. (a) 12. (a)
13. (c) 14. (b) 15. (c) 16. (a) 17. (d) 18. (d)
19. (c) 20. (a) 21. (d) 22. (a) 23. (b) 24. (c)
25. (b) 26. (b) 27. (c) 28. (d) 29. (a) 30. (c)
31. (d) 32. (c) 33. (d) 34. (b) 35. (b) 36. (d)
82 JEE MAIN CHAPTERWISE EXPLORER
4. (b) : Charge on the capacitor at any instant t is
1. (d): q = q 0 coswt ...(i)
Equal sharing of energy means
Energy of a capacitor = 1 Total energy
2
As switch S 1 is closed and switch S 2 is kept open. Now, 1 q 2 = 1 æ 1 q0 2 ö Þ q = q0
capacitor is charging through a resistor R. 2 C 2 è 2 C ø 2
From equation (i)
Charge on a capacitor at any time t is q0
2
q = q 0( 1 – e –t/t ) = q0 cos w t
q = CV(1 – e –t/t )
[As q 0 = CV] cos wt = 1
2
At t = t
2
( ) wt = cos -1 1 = p
q = CV (1 - e-t /2t ) = CV (1 - e-1/2 ) 2 4
At t = t p p ( ) Q w = 1
4w 4 LC
q = CV (1 - e-t/t ) = CV (1 - e-1 ) t = = LC
At t = 2t, R 2 mF
q = CV (1 - e-2t /t ) = CV (1 - e-2 )
5. (c) :
200 V S
2. (b): In case charging of capacitor through the resistance is
As field due to current loop 1 at an axial point V = V0 (1 - e-t / RC )
\ B1 = m 0 I1 R 2 Here, V = 120 V, V 0 = 200 V, R = ?
2(d 2 + R2 ) 3/2 C = 2 mF and t = 5 s.
\ 120 = 200(1 - e-5/R ´2´10- 6 )
Flux linked with smaller loop 2 due to B 1 is or e-5/ R ´ 2 ´ 10 -6 = 80
200
f2 = B1A2 = m0I1R 2 pr 2 Taking the natural logarithm on both sides, we get
2(d 2 + R2 ) 3/2
-5 = ln(0.4) = - 0.916
R ´ 2 ´ 10 -6
The coefficient of mutual inductance between the loops is
f2 m 0 R 2 pr 2 Þ R = 2.7 × 10 6 W
I 1 2(d 2 + R2 ) 3/2
M = = 6. (c) : Emf induced across PQ is e = Blv.
The equivalent circuit diagram is as shown in the figure.
Flux linked with bigger loop 1 is
L P O
f1 = MI 2 = m0R2pr 2 I 2 e = Blv
2(d 2 + R2 ) 3/2
Substituting the given values, we get R R R
f1 = 4p ´ 10-7 ´ (20 ´ 10-2 )2 ´ p ´ (0.3 ´ 10-2 )2 ´ 2
2[(15 ´ 10-2 )2 + (20 ´ 10-2 )2 ]3/2
I I 2
f 1 = 9.1 × 10 –11 weber M I 1 Q N
3. (d) : Here, BH = 5.0 × 10 –5 N A –1 m –1 Applying Kirchhoff’s first law at junction Q, we get
l = 2 m and v = 1.5 m s –1 I = I 1 + I 2 ...(i)
Induced emf, e = BHvl = 5 × 10 –5 × 1.50 × 2 Applying Kirchhoff’s second law for the closed loop PLMQP,
= 15 × 10 –5 V = 0.15 mV
we get
Electromagnetic Induction and Alternating Current 83
–I 1R – IR + e = 0 8. (c) : V t = 0 K
I 1R + IR = Blv ...(ii) L R 1
Again, applying Kirchhoff’s second law for the closed loop R 2
PONQP, we get At time t = 0, the inductor acts as an open circuit. The
corresponding equivalent circuit diagram is as shown in the
–I 2R – IR + e = 0 figure (i).
I 2R + IR = Blv ...(iii) V
Adding equations (ii) and (iii), we get
2IR + I 1R + I 2R = 2Blv
2IR + R(I 1 + I 2 ) = 2Blv I
R2
2IR + IR = 2Blv (Using (i))
3IR = 2Blv (i)
I = 2 Blv ...(iv) The current through battery is I = V
3 R R2
Substituting this value of I in equation (ii), we get I 1 = Blv At time t = ¥, the inductor acts as a short circuit. The
3 R
Substituting the value of I in equa tion (iii), we get I 2 Blv corresponding equivalent circuit diagram is as shown in the
3 R
= figure (ii).
Hence, I1 = I2 = Blv , I = 2 Blv I V
3R 3 R
7. (d) : During discharging of capacitor through a resistor, R 1
q = q 0e –t/RC ...(i) R2
(ii)
The energy stored in the capacitor at any instant of time t is
\ The current through the battery is
U = 1 q 2 = 1 ( q0 e -t / RC ) 2 (Using(i))
2 C 2 C
V V (. . . R 1 and R2 are in parallel)
1 q0 2 I = Req = R1R2
2 C
= e-2t / RC = U0 e -2t / RC ...(ii) R1 + R2
where U 0 = 1 q0 2 , the maximum energy stored in the capacitor. = V ( R1 + R2 )
2 C R1R2
According to given problem
U 0 = U 0e -2t1 / RC (Using (ii)) ...(iii) 9. (a) : Here, R = 200 W, V rms = 220 V, u = 50 Hz
2 When only the capacitance is removed, the phase difference
and q0 = q0 e - t2 / RC (Using (i)) ...(iv) between the current and voltage is
4
X L
From equation (iii), we get tan f = R
1 = e-2t1 / RC tan 30 ° = X L or X L = 1 R
2 R 3
Taking natural logarithms of both sides, we get When only the inductance is removed, the phase difference
ln1 - ln 2 = - 2t 1 or t1 = RC ln 2 (Q ln1 = 0) between current and voltage is
RC 2
X C
From equation (iv), we get tan f¢ = R
1 = e- t2 / RC tan 30° = X C or XC = 1 R
4 R 3
Taking natural logarithms of both sides of the above equation,
we get As XL = XC , therefore the given series LCR is in resonance.
\ Impedance of the circuit is Z = R = 200 W
ln1 - ln 4 = - t 2
RC The power dissipated in the circuit is
(. . . ln4 = 2ln2)
t2 = RC ln4 = 2RC ln2 P = V rmsI rmsc osf
\ t1 = RC ln 2 ´ 1 = 1 Vr m2 s ( ) Q Vr ms
t2 2 2RC ln 2 4 = Z cos f I rms = Z
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