P-3 Allens Made Physics Exercise Solution

JEE-Physics   C T   )3 . For (A) Q = 0, W = PdV = 0 so U = 0  T = constant For (B) : (B) by comparing area under curve 1 P  V–2 & PV = µRT  V  (     )   T Since volume increases so temperature decreases. ( For polytropic process PV2 = constant   PV2= ) R R3 R C=C + =C + = RR = v 1x v 12 2 2 Q = µCT = µ  R  T = Negative  2  (C) from 400 K to 500 K, Graph of C vs T become For (C) : asymptotic hence rate of heat absorption become constant (400 K 500 K CT C = C +R  3 R  Q= µCT  )  v 14 2 (D) The rate of heat absorption increases as C is 3 increasing. 3 µ  2 (C        )   = – R  T  Positive  Match the column For (D) : 1 . Process J  K (isochoric) : W = 0, U < 0  Q < 0 P , V , T  Process K  L (isobaric) : W > 0, U > 0  Q > 0 Work = Area under PV curve  positive Process L  M (isochoric) : W = 0, U > 0 Q > 0 U  positive Q  positive (given to system) Process M J , W < 0 , U < 0  Q < 0 Comprehension #1 2 . (A) Bimetallic strip : Works on the thermal expansion 1 . When the piston is pulled out slowly, the pressure drop of solids (different solids expands by different length produced inside the cylinder is almost instantaneously for the same rise of temperature). The energy is neutralised by the air entering from outside into the converted to kinetic energy. cylinder. Therefore the pressure inside is P . 0 ( (    Node6\\E : \\Data\\2014\\Kota\\JEE-Advanced\\SMP\\Phy\\Solution\\Unit 5 & 6\\01.Thermal Physics.p65 )   P) 0 (B) Steam engine ( )  2. Mg = (P – P) R2  P=P – Mg 0 0 Energy is converted (heat–mechanical) R 2 (C) Incandescent lamp ( )  Since the cylinder is thermally conducting, the temperature remains the same. Heat  Light ; radiation from hot body. (D) Electric fuse  ) y Works on melting of fuse wire on heating. P Heat  P.E. of molecules. P0 Mg 39

JEE-Physics ( ) (i) Heat rejected in path CA P (2L × R2) = P (y× R2)  y =  P0  R 2 g  2 L  (C A )  0  P0R 2 M  (Process is isobaric) (  )  3 . Equating pressures ( )   Q =C P T =C ( T – T )  CP  Pf Vf  Pi Vi  CA  R R  P f i + (L ) P = P0 L 0 CP L0  H R  P g –H =  (Pf Vf  Pi Vi ) 0 0 Comprehension #2 Substituting the values ( ) 1 . Force due to the pressure of liquid = The buoyancy 55 force. Q = (P V –2P V ) = – P V 2CA 0 0 00 2 00 (    =)  1 5 Therefore, heat rejected in the process CA is P V . 2 00  P2    P1  2. TP1–=constant  T =T 1 ( CA    5P V ) 2 2 00  P0  g(H  y )  1 3  P0  g(H  2 (ii)Heat absorbed in path AB:  P0  gH 5  P0  gH y)5 =T  =T  (A B)  0 0 ()=Vg  n R T2  (process is isochoric) (  )  P2  3. Buoyancy force =  g  Q = C V T = C ( T – T) AB i V f 2  C  Pf Vf  Pi Vi   CV  R R  R  n R  g T0  P0  g(H  y)5 V (PfVf – PiVi)  g(H   P0 y)  P0  gH   33 =2 (P V – P V )= (3P V –P V ) 3P V = nRgT0 ff ii 2 00 00 0 23 0 (P0  gH )5 (P0  g(H  y )5  Heat absorbed in the process AB is 3P V . Subjective Questions 00 1 . (a) ABCA is a clockwise cyclic process. (AB  3PV  ) (ABCA    )   00 (c) Let Q be the heat absorbed in the process BC BC (BCQ ) BC P Total heat absorbed (  )  B Q=Q +Q +Q 3P0 CA AB BC A C Q    5 P0 V0   (3P0 V0 )  QBC P0  2  V Node6\\E : \\Data\\2014\\Kota\\JEE-Advanced\\SMP\\Phy\\Solution\\Unit 5 & 6\\01.Thermal Physics.p65 V0 2V0  Work done by the gas (   ) Q = Q + P0 V0 W=+Area of triangle ABC (ABC  )  BC 2 11 Change in internal energy U = 0 W = (base) (height) = (2V –V ) (3P –P )=P V 2 2 00 0 0 00 () (b) No. of moles n=1 and gas is monoatomic, therefore Q = W   QBC  P0 V0  P0 V0  QBC  P0 V0 2 2 (  n=1   ) CV = 3 5 R  CV  3 and CP  5  Heat absorbed in the process BC is P0 V0 2 R and CP = 2 R2 R2 2 (BC  )  40

JEE-Physics (d) Maximum temperature of the gas will some where TA=300K between B and C. Line BC is a straight line. Therefore, X P–V equation for the process BC can be written as (B C BC BCPV) P = –mV + c; (y = mx + c) m  2P0 P    2 P0  V  5 P0    dT  =k(T – T)  dT V0  V0   dt  A T  TA =–k.dt Here, and c=5P  0 Multiplying the equation by V T1 dT t1 n  T1  TA  T0 T  TA  T0  TA     k dt    kt1 0 PV = –  2 P0  V 2  5P0 V (PV = RT for n=1)  V0  =–n  350  300  n(2)  kt  400  300   kt = 1 1 RT    2 P0  V2  5P0 V In the IInd part, body X cools by radiation (according  V0  to Newton's law) as well as by conduction (t > t ).  T  1 5P0 V  2 P0 V 2  ...(i) 1 R  V0   (IIX   (t >t)       1 For T to be maximum (T     ) TA T=TA X Y dT  5P – 4 P0 , V=0 V  5 V0 0 0 V0 4 dV i.e., at V= 5 V0 (on line BC), temperature of the gas Therefore, rate of cooling (  ) 4 = (cooling by radiation) + (cooling by conduction) is maximum =       (V=5V 0 BC In conduction () ddQt  KA (T  TA )  C   dT  L  dt  4 From Equation (i) this maximum temperature will be ((i)     )    dT   KA (T  TA )  dt  LC 1   5 V0   2P0  5 V0 2 25 P0 V0 where C = heat capacity of body X T= R 5 P0  4  V0  4   = 8R max   (C= X   ) 2 . In the first part of the question (t < t )    dT   k(T  TA )  KA (T  TA ) ...(ii) 1  dt  CL [   (t <t)]  Node6\\E : \\Data\\2014\\Kota\\JEE-Advanced\\SMP\\Phy\\Solution\\Unit 5 & 6\\01.Thermal Physics.p65 1 At t=0, T =T =400K and at t=t T =T =350K  dT   KA  X0 1X 1   dt    k  CL  (T  TA ) ...(iii) Temperature of atmosphere, T = 300K (constant) A This cools down according to Newton's law of cooling. Let at t = 3t , temperature of X becomes T 12 [  T = 300K  ] Therefore, from Equation (iii) A (t=3t   X  T      (iii)  () 1 2 Therefore, rate of cooling  temperature difference. (  )  T2 dT    k  KA  3 t1 dt  LC  t1 T1 T  TA 41

JEE-Physics  T2  TA   k KA  (2 t1 )  2KA  3  V1  2/ 3   T1  TA   LC     2  V2   1  n      2k t1 t1  UAB = –W = P1 V1 LC AB   n  T2  300  =–2n(2)– 2KAt1 Process B–C (B-C):  350  300  LC W =0 BC  kt =n2 from Equation (i) ((i) )  UBC= Q = Q (Given) 1 BC 2 KAt1   U To ta =  U A B +  U B C  3 P1 V1  V1 2/3   Q 2  V2   1  This equations gives T2  300  12.5e CL kelvin l  3 . The P–V diagram for the complete process will be (iii) Final temperature of the gas as follows (  P-V ) (   )  U Tot a l = nCVT = 2   R 1  ( TC  TA )     3 P1 V1  V1 2/3   Q  5 2R  TC  PAVA  2  V2   1 /3 1  2R   Process A  B is adiabatic compression and  3 P1 V1  V1 2/3   Q  3R  TC  P1 V1  Process B  C is isochoric. 2  V2  1  2R   (AB B C      T = Q  P1 V1  V1 2/3 = T (b) (i) Total work done by the gas process A–B : C 3R 2R  V2 final  (  A–B    )  4 . (i) Number of moles ( ) WAB  PA VA  PB VB  Pi Vi  Pf Vf  P1 V1  P2 V2 n=2, T =300K  1  1 5 1 1 3 During the process A  B (A  B  )  PT = constant or P2V =constant=K(say)  V1    P1 V1  P2 V2  P  K  V2   P1  V1   V P1 V1  P1 V2  V2     P2    2/3    VB VB K dV V  W = P.dV  AB VA VA  3 P1 V1    V1   1   3 P1V1  V1  5 / 31  2 1  V2    V2  1     2 K [ VB  VA ]  2[ KVB  KVA ] 2    3 P1 V1  V1 2/3   2  (PB2 VB )VB  (PA2 VA )VA  (K = P2V) Node6\\E : \\Data\\2014\\Kota\\JEE-Advanced\\SMP\\Phy\\Solution\\Unit 5 & 6\\01.Thermal Physics.p65 2  V2  1  = 2[PBVB – PAVA] = 2[nRTB – nRTA]   = 2nR[T – 2T ] = (2)(2)(R) [300–600] Process B–C : W = 0 (V= constant) 11 BC = –1200R 3  V1 2/ 3  2  V2   1  Work done on the gas in the process AB is 1200R.  W =W + W   P1 V1 Total AB BC  (AB    ) (ii) Total change in internal energy (ii) Heat absorbed/released in different processes. (    ) ()  Process A–B (A-B): Q = 0 (Process is adiabatic) (  )  Since, the gas is monoatomic. AB ()  42

JEE-Physics 3 55 (AB   ) Therefore, C = R and C = R and = V2 P2 3 we get m=0.495 kg. Process A–B :  3   3  6 . Given :  2   2  U = nCVT = (2 ) R (T –T )  (2) R No. of moles, n=2 V D BA 4V0 A 35 (300–600) = –900R CV  2 R &CP  2 R 2V0 C (Monoatomic) V0 QAB = WAB + U = (–1200R) – (900 R) T = 27°C = 300 K VVBA=2 and Q = –2100R(Released) B VVAD=4 A AB Process B–C : Process is isobaric (  )  Let V = V A0  Q = nCPT TA TB T BC then V = 2V B0  5   5   (2 )  2 R  ( TC  TB )  2  2 R  (2 T1  T1 ) and V = V = 4V 0 DC (i) Process A  B : = (5R) (600–300) Q = 1500 R(absorbed) ()  V T TB  VB BC  TA VA Process C–A : Process is isothermal (  )  PC   VB   U =0  PA   VA  and QCA = WC A =nRT  n  TB  TA = (300) (2) = 600K C = nR( 2 T ) n  2 P1  =(2)(R)(600)n(2) TB = 600 K  P1  (ii) Process A B : 1 V  T  P = constant QCA = 831.6 R (absorbed) ()  5 . Let m be the mass of the container. Initial temperature QAB = nC dT = nC (T – T) of container, Ti = (227 + 273) = 500 K and final P PB A temperature of container,  (2)  5 R  (60 0  300) T = (27 + 273) = 300 K  2 f QAB = 1500R (absorbed) Now, heat gained by the ice cube = heat lost by the Process B  C : container i.e., (mass of ice) (latent heat of fusion of ice) + (mass of ice) (specific heat of water) T = constant  dU =0 (m  QBC = W = nRT n  VC  = (2) (R) (600) n  4 V0  BC B  VB   2 V0  Ti = (227 + 273) = 500 K T = (27 + 273) = 300 K = (1200 R)n(2) = (1200 R) (0.693) f  =  QBC  831.6 R (absorbed) Process C  D V = constant Node6\\E : \\Data\\2014\\Kota\\JEE-Advanced\\SMP\\Phy\\Solution\\Unit 5 & 6\\01.Thermal Physics.p65    Q = nC dT = nC (T – T) CD V V D C Tf (300 K – 273 K) = m  S.dT  n  3 R  (T – T) (T = T and T = T)  2 A B D A C B Ti Substituting the values, we have ( )  (2)  3 R  (300 – 600)  2  (0.1) (8 × 104) + (0.1) (103) (27) 300  BT2 300  QCD = –900 R (released) 10700  m AT  2 500 Process D  A : –m (A  BT)dT of T = constant  dU = 0 500 = After substituting the values of A and B and the proper limits 43

JEE-Physics Q D = W = nRT  n  VA  1 (100)2 (10)2 (5) D  VD  U  2 (3.14)(2  103 )2 (2.1  1011 ) J = 0.9478 J A DA = (2) (R) (300)  n  V0  = 600R n  1  When the bob gets snapped, this energy is utilised  4 V0   4  in raising the temperature of the wire. Q  –831.6 R (Released) ( DA     )  (iii) In the complete cycle: dU =0 So, U = ms Therefore, from conservation of energy ( dU=0   )   U 0.9478 ms 0.494(420) Wnet = QAB + QBC + QCD + QDA     C or K W = 1500R + 831.6R – 900 R – 831.6 R   = 4.568 × 10–3°C net  Wnet = W = 600 R total 7 . Given ( )  8 . Volume of the box (  )=1m3 Pressure of the gas (  )=100N/m2  M=100kg Let T be the temperature of the gas. Length of the wire (  ) =5m (T   )  Radius of the wire (  )r=2 × 10–3m Then () Density of wire (  )= 7860 kg/m3 Young's, modulus ( )  (i) Time between two consecutive collisions with one Y= 2.1 × 1011 N/m2 and specific heat wall (    )   (  ) 1 = s. S = 420 J/kg–K Mass of wire, m = (density) (volume) 500 [  m=]   = () (r2) = (7860)()(2 × 10–3)2(5) kg=0.494 kg 2 Elastic potential energy stored in the wire, This time should be equal to vrms ()   2 (  vrms  )  where  is the side of the cube. (  )   2 1  =1m) 1  v rms 500   v = 1 0 0 0 m / s (as U  (stress) (strain)× (volume) rms 2 U  1 () ()× () 3RT Node6\\E : \\Data\\2014\\Kota\\JEE-Advanced\\SMP\\Phy\\Solution\\Unit 5 & 6\\01.Thermal Physics.p65  =1000 2 M T  (1000)2 M (10)6 (4  10 3 )   160K  U  1  Mg     ( r 2  )  1 (Mg). 3R 3(25 / 3) 2  r 2     2  F  (ii) Average kinetic energy per atom  3 kT  AY  2    1 (Mg) 1 M2g2 ( 3 kT )  2 (Mg) (r2 )Y 2  r 2 Y 3 2  (1.38 × 10–23)(160) J = 3.312 × 10–21J Substituting the values, we have ( ) 2 44

JEE-Physics m 1 1 . At constant pressure (  ) (iii) From PV = nRT= RT VT V2  T2 Ah2  T2 M V1 T1  Ah1 T1 We get mass of helium gas in the box, (   ) mPVM h2  h1  T2   (1 .0 )  4 00  m  4 m  T1   3 00  3 RT As there is no heat loss, process is adiabatic. Substituting the values we get ( ) ( m  (100)(1)(4)  0.3g )  (25 / 3)(160) 9 . Decrease in kinetic energy = increase in internal For adiabatic process (   ) energy of the gas Tf Vf 1  Ti Vi 1 (   =   )   Vi   1  h  1.4 1  4  0.4  Vf   h   3  1 2 m 3  M v 2  Tf  Ti  ( 4 0 0 ) i  4 0 0 2 0  M   2  0 f m v  nC V T  R T  T  3R 1 2 . When the temperature is increased, volume of the 1 0 . (i) Rate of heat loss per unit area due to radiation cube will increase while density of liquid will decrease. The depth upto which the cube is submerged in the () liquidremains the same, hence the upthrust will not I = e(T4 – T 4) change. 0 ( Here. T = 127 + 273 = 400 K and T = 27 + 273 = 300 K  0 )  17 F = F' I = 0.6 × × 10–8[(400)4 – (300)4]= 595 W/m2  ViLg =Vi''Lg 3 (Vi = volume immersed ( )) (ii) Let  be the temperature of the oil. Then, rate of heat flow through conduction =rate of heat loss due to radiation ( Ah ) (L)(g ) = A(1 + 2s T) (h )  1  L T  g i i    (            =     ) Solving this equation, we get ()  temperature difference  ( 127)  (595)A  = 2  thermal resis tan ce =(595) A   ls  KA  1 3 . Rate of heat conduction through rod= rate of the heat Here, A = area of disc; K = thermal conductivity and lost from right end of the rod. (  Node6\\E : \\Data\\2014\\Kota\\JEE-Advanced\\SMP\\Phy\\Solution\\Unit 5 & 6\\01.Thermal Physics.p65  = thickness (or length) of disc (A =   =       )  ; K =  =   KA (T1  T2 ) =eA (T 4 – T 4) ...(i) )  L 2 s Given that T = T +  T 2 s K  T24  ( Ts  T )4  Ts4 1  T  4  ( 127)   595 Ts     595  10 2 Using binomial expansion, we have (  )  = 595  K  +127  0.167 +127 = 162.6°C T24  Ts4 1  4 T  (as T << T) Ts  s 45

JEE-Physics  T24  Ts4  4(T)(Ts3 ) 15 . 0.05 kg steam at Substituting in Eq.(i), we have ((i)  )   373 K Q1  0.05 kg water at 373 K 0.05 kg water at K (T1  Ts  T ) = 4eTs3.T 373 K Q2  0.05kg water at 273 K L 0.45 kg ice at  K (T1  Ts )   4 e Ts3  K T 253 K Q3  0.45 kg ice at 273 K L  L  0.45 kg ice at K (T1  Ts ) 273 K Q4  0.45 kg water at 273 K  T = (4eLTs3  K ) Q = (50) (540) = 27,000 cal = 27 kcal 1 Q = (50) (1) (100) = 5000 cal = 5 kcal Comparing with the given relation, proportionality 2 Q = (450) (0.5) (20) = 4500 cal = 4.5 kcal constant () Q3= (450) (80) = 36000 cal = 36 kcal 4 K Now since Q + Q > Q but Q + Q < Q + Q  4eLTs3  K 12 3 1234 ice will come to 273K from 253 K, but whole ice will not melt. Therefore, temperature of the mixture is 273K. 1 4 . (a) From Q = msT ( Q + Q > Q Q + Q < Q + Q 12 3 1234 T = Q  20000  50C 273K 253 K ms 1  400     273K ) 1 16. (b) V = VT =  9000  (8 × 10–5) (50) 400°C Ice 0°C = 5 × 10–7 m3 Water  W = P.V = (105) (5 × 10–7) = 0.05 J 100°C (c) U = Q –W =(20000 – 0.05) J A .P B = 19999.95 J L 10x–L k 400  0 A = m (80) ...(i) L kA 400 100 10x  L  = m (540) ...(ii) Divide (i) by (ii) 1080 x = 120 L  L = x  = 9 Node6\\E : \\Data\\2014\\Kota\\JEE-Advanced\\SMP\\Phy\\Solution\\Unit 5 & 6\\01.Thermal Physics.p65 46

JEE-Physics UNIT # 07 (PART - II) CAPACITOR EXERCISE  –I 6. C  =  0 A  t  d ,K   7. dt t  2 1.   k  =  f  5000 =  2,5000  N/m K x 0.2 1 kx2       =  0 A  2 0 A  2C0 2 dd d U SPR 1 CV2 25000  0.2  0.2 d  U CAP   10  106  108 1 2 2K 2 Q1q q q Q2+q  Q3 +Q2+q U 1 CV2 1  40  10 6  9 106 ab cd ef 2 2 2.   P     90kW t t 2  10 3 3 . V0 (C+CV) = CV + (2C) (2V)             C V0  =  V  (Final  pot.  diff.) –+  Here Q1  q  Q2 Q2  Q3  q  q  Q1  2  Q3 1 3CV2 2V Charge  on  a=  Charge  on  f   Ufinal  =  2 (C+  2C)  V2  =  2 +– 2C (a= f) 4 .   (4+2)  V  =  (4  ×  50)  +  (2  ×  100) 4F+ 50V Q  Q1  Q2  Q3 2 2   Q1–q  =  400 200 + 8. 2Q Q V  =  =  V 63 100V F Uinitial  =  1  4  (50)2  1  2  (10 0 )2   106  2 2  + Q +32Q  3Q + Q P 2 2 2                 =  (5000  +  10000)  ×  10–6  =  1.5  ×  10–2  J Ufinal  =  1 4  2  10 6  200  200 2 3 3           = 1.33  × 10–2  J Force  on  either  plate  (   )  Ui  Q12  Q 2 3Q / 22 9Q2 2C1 2 5. Before  sharing =  2 A 0 =  8 A 0 2C2 Force  on  point  'P'  due  to  capacitor  =  0 ()  (P)  After  sharing Uf   Q1  Q2 2 2 C1  C2  Potential  diff.  between  the  plates ()  Node-6\\E:\\Data\\2014\\Kota\\JEE-Advanced\\SMP\\Phy\\Solution\\Unit-07 & 08\\02-Capacitance.p65 ()  = 3Q Q1  Q2 2 2 C1  C2 2C    U  Q12  Q 2 Energy  stored  in  electric  field  between  the  plates 2C1 2  Uf  Ui =  ()  2C2                       Q1C2  Q2 C1 2 C1  C2  2C1C2 –ve  sign  indicates  there  is  decrease  in  energy (  ) 2 But  Q1C2–  Q2C1   0  Q1  C2    Q2C1 =  1 C   3 Q   9Q2 2  2 C  8C   Q140R2    Q240R1  Q1R2   Q2R1 29

JEE-Physics 1 A B C 3 +q 5 13. q 2 5F 15F 4 q +ve +q -ve 5(VA  –  VB)  =  15  (VB–VC)    5(2000–VB)=15(VB–0)   2000  –  VB  =  3VB      VB  =  500V 9. +q q C C CC q 14. Ceff  =  C  +  +  +  + +.... +q 16 2 4 8 Therefore  ( ) C         = 1  1 / 2  = 2C  =  2F q2  =  –  2q,  q3  =  +  2q, q4  =  –  2q  and  q5  =  +  q 10.   1 5 .   For  'n'  plates;  effective  C  will  be  (n–1)C. (n(n–1)C) 3V 1 6 . CV  +  2CV  =  KCV'  +  2CV'    V'  =  K  2 U  1 C1 V 2  1 C2V2  1  C2 )V2 2 2 2 (C1 1 7 . C  =  0 A =  9pF 1  8.85  1012  0.1  102 d 2  0.885  103 2         =      =  10–1J C'  =  0 A = 0 A d d 2d d d  t1  t1  t2  t2 d    1 1 . Each  capacitor  has  potential  difference  'V'  and K2 K2 39 3 9 energy  1 CV2 .  After  reconnecting  total  energy         =  9 0 A  81 pF =  40.5pF 2 2d 2 remains  constant  and  total  voltage  becomes  NV. (V 1 CV2  2        NV k3=6 6C 6C 4C ) 18.   2C P 12. 2C 2C 2C 2C C 2C 2C 2C where  C  =  0 A C C Q C d 2C 6C  2C 6C  4C Ceq  =    3.9 C 8C 10C 2C Node-6\\E:\\Data\\2014\\Kota\\JEE-Advanced\\SMP\\Phy\\Solution\\Unit-07 & 08\\02-Capacitance.p65 2C 19. 2C x a-x 2C C C C 2C 3C C  =  0 ax  K 0 a  x  a dd C  K 0 a2  0 a K  1 x   where  x=  vt dd   C–  t  graph  is  linear  with  negative  slope.     (C–t ) 30

JEE-Physics 1 2msT 30F 10F 2 0 . CV2  =  msT    V  =  2C 2 1 .   C  4 0 a 26. 5F C'= 4 0 ab =  4 0 a  4 0 a  n 4 0 aC 6F 2F ba 1a 1   n  1  b  n  The  system  is  a  balanced  Wheatstone  bridge. 8F 16F () 2 2 .                          Breaking  voltage              20  V                            80  V  C eff   10  30  6  2   9F  10  30 6  2  ()  27. R R Safe  Voltage                          20  V                            10  V R 2R ()   Charge  on  each  capacitor  =  20  ×  8  =  160  C ()  1F 2F 2 3 .                            To  find  the  time  constant  of  a  RC  circuit,  Short circuit  the  battery Breaking  voltage                    6  kV                            4  kV (RC         ()  )  Safe  Voltage                                6  kV                            3  kV 7R 7RC Reff =  4        =  4 ()   Total  voltage    ( ) =  9  kV 2 4 . Capacitance  between  1  and  3  and  between  2  and 2 8 .   There  is  no  closed  path  for  flow  of  current.  Hence 4  are  symmetrical. no  current  flows.  Hence  heat  developed  is  zero. (1324)  (  35F 7F ) 25. 13F q 2 9 . VA  =  3  C  =  3  ×  2.5  =  7.5  volt 10F 2F Node-6\\E:\\Data\\2014\\Kota\\JEE-Advanced\\SMP\\Phy\\Solution\\Unit-07 & 08\\02-Capacitance.p65 The  system  is  a  balanced  Wheatstone  bridge. ()  C eff   35 7  10 2  15 F  35 7 10  2  2 31

JEE-Physics EXERCISE  –II V0 V0 At t=0;    i1 =  R1 ;       i2 =  R 2 1. E  V0  EF  E D  Also  d  R1 = R2           i1 = i2 As    is  less  for  C1  and  hence  it  looses  charges  faster x dx than  C2. 2. (C  C1  C2  1 dy d dy 8 .    Ceff = 1/4 F  dC   K 0 A   y   2d  0  0 A sec C net 3 C eff 1/4  Total no. of rows of capacitor  =   =12 C  =  0 A   ()  2d   Total  no.  of  capacitors  needed  =  12  ×  4  =  48 3. Both  A  and  B  are  always  in  parallel.    ()  (AB) 4 . V  =  V0.e–t/RC 9. dV =  V0 et / RC   =  slope  () +   ++  + +   ++  + dt RC +   ++  + At  t  =  0,  for  R  =  RA;slope  is  least  in  curve–3. +   ++  + (t=0  R=RA        +   ++  +  1+   ++  + 23 45 5.  q  q0et /     i  dq  q0 et/  i0et/ dt                      q0  i0 Charge on plate 1  (1  )  =  0 AV Initial  stored  energy  ( )   d 11 Charge on plate  4 4)= –  2 0 AV             =  CV2 =  (CV)V d 22 11 10. 6C 6C 2C 2C             = 2 (i0) (i0R) =  2 i02 R 10V B 0V 3C C 6 . As B is in  parallel with  C and the  potential develops C slowly.  Hence  during  charging  more  heat  is 10V 20 produced  in  A  than  in  B.  In  steady  state,  same A current  passes  through  A  and  B. 2C   VB  =  7.5  V Node-6\\E:\\Data\\2014\\Kota\\JEE-Advanced\\SMP\\Phy\\Solution\\Unit-07 & 08\\02-Capacitance.p65 (B, C A(V A V–A  V– B)V 6B C=  =1 0(V– B7–0.5)   =  2.5  V B11. Force  on plate  (  ) AB=  2A = Q2 =Kx  =  mg 2 0 2A 0 E 1  E  2 CE2  Q =  2mgA 0 2 2  2  8   V  capacitor =           E capacitor  C  7 . q  q0et/ 12. Ceff  =  CEF  =  0 A   Enet  =  1 CV 2   = 0 AV 2  i  =  dq   =  q 0 e t /  = CV0 et /  = V0 e t /  d 2 2d dt  RC R 32

JEE-Physics 1 3 . i  10e t / RC  2.5  10  e 2/RC 18.   i=  i0  i0et / RC  1  e n 4 / RC 2 2 11 1   RC =  =  n2  & C =  10n2  RC=2   (2+r)  2 =2    r  =  2 For  capacitor  (  ) V0 19.   At  t=0,  VC  =  0   iR3  0 R  10  V0  =  10R  =  100  volt Total  heat  developed  =  Total  initial  energy  stored    10C 5 1  5C 1 1 in capacitor.  (  =    Qmax  =  C   R1R2  R3     R1  R 2   )  =  12 C V 2 = 500  VC 5 n2 R3   IR3 max  5A 1 Thermal  power  in  resistor(   ) Since  R1  and  R2  are  in  parallel  hence  current  ratio P  =  i2R  =  100 Re 2 t / RC of  R1  and  R2  will  remain  same.   Time–constant  ( )=  RC = 1 ( R1    R2      R1   R2   2 2n2 ) 14. 3C 2C 20. q  q0et / RC I  q0 et / RC 1kV 1.5kV RC 6/7 kV 2kV  nI  n  q0   t  n  V0   t 7C 3C  RC  RC  R  RC  As  Imax  does  not  change     R  =  constant Safe  voltages  in  each  arm  are  mentioned. d nI =  0  1   d nI   d nI  RC ()dt  dt I >   dt 2    (1+1.5)  <  (6/7  +2)  Esafe  =  1+1.5  =  2.5  kV   C2  >  C1   C  is  increased 1 5 . Time  constant  ( ) 21. A C B Node-6\\E:\\Data\\2014\\Kota\\JEE-Advanced\\SMP\\Phy\\Solution\\Unit-07 & 08\\02-Capacitance.p65= CReff =100106103s = 50 m/s  2  + Q/2 + Q/216.  i1  = Ve t / RC1,  i2  = Vet / RC2+Q +QQ -Q/2RR + 3Q/2 3Q/2 i  e  e1t/R11tQ Qd + Q/2C2Initial  VAB  C = 0 A C1 2RC2 i2  i1/i2  increases  with  time,  t. Q/2 3Q / 2 Qd  VAB 0 A      (i1/i2  t)  Final  VAB  =  2 0 A +  2 0 A =   d   d  -q q 1 7 . (300q) +  -(360+q) 22.   A -(300+q)  + (360q) C=C0V B C=C0 300  q q 360  q (C0  +  C0V)  V  =  30  C0  2 – 1.5 + 3 =0   q  =  180 V2  +  V  –  30  =  0   V  =  5  volt   q1.5F  =  180  C,  q3F  =  540  C,  q2F  =  480C   VA  =  VB  =  5  volt 33 QA  =  52C0  =  25  C0;  QB  =  5C0

JEE-Physics 23. C  =  0 ax +  K 0 a  xa U C 2 6 . Q  = E dd        2     =  K 0 a2  0 a K  1 vt KCC KC Q'  = E  E dd V KC  C K 1 Q QV U Q   Q'–Q  =  KCE  CE  K 1CE V =   and U =        K 1 2 2K 1 C 2 V2 24. A d 2d B This  charge  is  supplied  by  battery. + ++ + () q+ + + + + qB + +(2Q q) + qA + ++ + KC C Q 2 K 1 + ++ + K  1 , C eq 2  Q2 2K ++ +   C eq    + ++ + 27. ++ + + + 2 8 .   At  t=  ,    capacitor  gets  open  circuited (t=   ) qd 2Q  q2d 4Q V  =  0 A = 0 A  q  =  3 Total  charge  on  inner  faces  of  A  and  B  =  –2Q (AB) Rest  charge  will  equally  appear  on  their  outer  faces ()  Q  2Q  3Q 15  I  5  3A   VA  –  3  ×  1  –  3  ×  3  =VB =    =  22 Final  charge  on  plate  A  (A   )  3Q 4Q Q VA–VB = 12 V =  – =  23 6 2 9 .   At  t=0,  V  capacitors =  0   Charge flown  through  wire  (   )  6 Q 5Q   I2  =  I3  =  0  and  I1  =  2   =  3A =  Q  –    =  66 6 2 5 .   Final  charge  distribution  (  ) At  t    ,  I1=  I3  = 2  8   =  0.6  A,  I2  =  0 QQ 3 0 . In steady state ( )   +CV CV 120 Node-6\\E:\\Data\\2014\\Kota\\JEE-Advanced\\SMP\\Phy\\Solution\\Unit-07 & 08\\02-Capacitance.p65 +Q +Q Q +Q Iupper  arm  =  Ilower  arm  =  6   =  20A 22 22 For  the right  most loop  (  )                 q 3I  –  3I  +  C2   =  0    q  =  0 Therefore  potential  difference  across  the  capacitor For  the left most loop  (  )  ()  q 20  ×  1  +  C1 –  20  ×  2  =  0 CV  Q Q                 =  C 2 2C  q  =  (40–20)  C1  =  20C1  =  40  C  V  34

JEE-Physics 3 1 . Charge  on  3F  capacitor  (3F   36 .   Wext  =  –U  =Ui–  Uf                               =  6  ×  7  =  42  C        = 1  2F  400  1  1F  400  200J 22 42 V3F  =  3   =  14  volt 11 V3.9 F  =  14+  6  =  20  volt 3 7 .  Uinitial =  2 CV2;             Ufinal  =  2 CV2     U  =  0 Charge  on  3.9  F  capacitor  =  20  ×  3.9  =  78  C   Heat  =  work  done  by  battery (3.9 F     ( =) Total  charge  ( )= 78  +  42  =  120  C                         =  [CV–(–CV)]V    =  2CV2 1 eV  =  m 2 120  3 8 . v 2  v 2 V  12F = 12 =10V 2 1    =  20  +  10  =  30  V 1 32. Energy  ()=  Q2   =  Q2d  1.6  ×  10–19  ×  20  =  ×  9.11  ×  10–31  ×  (v2–0) 2C 2 0 A 2   v  =  2.65  ×  106  m/s As  d  decreases,  E  decreases 3 9 . V  decreases  continuously  from  left  to  right  except in  conductor  where  it  is  constant. (dE) (V 3 3 .   Q  =  CV  =  0 AV  d E  =  V   V / K  V 40. S2 d d Kd S1 W  =  1 Q2  1  1  CV2 1  1  2  C C ' 2 K  34.   = Q0         Q1  =  Q0;       Q2  =   Q 0  C2 Potential  difference  across  each  capacitor  and  cell C1  C 1  combination  is  zero. Q0 1  Q0  2 Q 2 ( C1 2  C1  0 )  V1=V2  =    =  ;      U1  =  C1    =  2C1 Q 1 C  Q0 2 Q 2 C 2 C 2C C= 2  C1 0  41. 1 2 3  = Node-6\\E:\\Data\\2014\\Kota\\JEE-Advanced\\SMP\\Phy\\Solution\\Unit-07 & 08\\02-Capacitance.p65 U2  =  2 2 C 2 1 3 5 .   S–open  ;  Vinner  =  Vouter Initial  charge  on  1  =  Q  when  C1  &  C2  touches S  –closed  ;  Vinner    =  0 (1=Q C1C2) KQ Kq Q1 C 1  Q1  Q 2Q  3R +  R =0   q  =  –  Q/3  Q2   3 ,Q2  2C 2 3 Cinitial  =  4  0 3R  Now  when  Q2  &  Q3  is  touched ( Q2Q3) Cfinal  =  4 0 3R    +  4 0 3R  R   Q2  C2  2C  0 Q2 = 0 3R  R Q3 C3  Cfinal  >  Cinitial 35

JEE-Physics Again  when  Q1  &  Q2  is  touched EXERCISE  –III (Q1 Q2)  Fill  in  the  blanks Q / 3 Q / 3 Q 1. Net  charge  on  capacitor  is  zero.  Hence  total  flux through  a  closed  surface  enclosing  the  capacitor  is Q2  2 3  Q1  3  zero. 9 (        Similarly  we  can  say  after  N  times  it  becomes  )  (N) C   =  All  capacitors  are  in  parallel Q 2. Maximum Q1  =  3N 4 2 . Q  =  2CV–  (–CV)  =  3CV               WB  =  Q(2V)  =  6CV2                         =  3C  =  18F C   =  All  capacitor  are  in  series U  =  Uf–Ui  =  1 C(2V)2  –  1 CV2 = 3CV2 2 2 2 Minimum               ()                          =  C/3  =  2F   Heat  ( )=  WB  –  U  =  9CV2 3. V   =  V  kQ1  kQ 2   Q1  R1 2 12 R1 R2 Q2 R2 Uf  =  1 C 2V2  2CV2 4 . Charge  holding  capacity  increases,  hence  capacity 2 increases. ( )  Heat 9 5 . Air  capacitor  and  dielectric  capacitors  are  in  series.   4   =  2.25 Uf ()  2 0 A   2K 0 A  d   d  C  C1C2  2KC C1  C2 2 0 A  (1  K ) 1K  d  Match  the  column Node-6\\E:\\Data\\2014\\Kota\\JEE-Advanced\\SMP\\Phy\\Solution\\Unit-07 & 08\\02-Capacitance.p65 1 . Initial  charge  ( ) q =CE 12 Final  charge  ( )   q   =  CE 2 Initial  stored  energy  ( )  1 1 CE2 U   =  C(E/2)2  +  C(E/2)2  12 2 4 Final  stored  shergy  (  )  U   =  CE2 22 Charge supplied by battery  ()  CE CE      Q  =  q  – q   =  CE  –  21 2 2 CE2 Work  done  by  battery    W   =  QE  =  B 2 () Heat developed in the system()   CE2  CE2 CE2  CE2 H  =  W  –  U     4   B 2 2 4 36

JEE-Physics C3=3C 3. ( VR1 )   =  V   –  V   =  12  –  10.4  =  1.6  V 2. C1=C C2=2C (3/7)Q t  =  2 0 capacitor  C4=4C 4. VR 2 =  V =  12V (4/7)Q 0  QQ E Comprehension  -3 1 . V   =     (1  –  e–t/RC) At  C   =  V1  Q   and  U1  Q2 1 C 2C b0 At  C   =  V2  Q   and  U2  Q2   110  =  120  (1  –  e–t/RC) 2 2C 4C   e–t/RC  =  1/12  t/RC  =  n/12  =  2.5  t  =  RC  ×  2.5  =  106  ×  10–6  ×  2.5  =  5/2  sec At  C   =  V3  Q   and  U3  3Q2 2 .    =  10–6  ×  10  =  10  s 3 7C 98C 0 At  C   =  V4  Q   and  U4  4Q2 3 . Flash  duration  ( )= 30  =  30  s 4 7C 98C 4 . Energy in flash    Therefore  V   =  V   and  V   =  V   =  V max 1 min 3 4 11 =  CV2  =  ×  1  ×  10–6  ×  110  ×  110  =  6.1  mJ and  U   =  U   and  U   =  U max 1 min 3 22 Comperehension  -1 Comprehension-4 1 . In steady state ( )   1 . q   =  q 2max V 18 1max I   =     =  2A C   and  C   may  be  different  and  hence  E   and  E 12 12 circuit R1  R2 36 may be different.(C  C    E  12 1 VR2  VC2 =  IR   =  2  ×  6  =  12  V E ) 2 2 Q C2  C2 VC2 =  12  ×  4  =  48  C 2. 2  >  1    RC  >  RC    R1  C2 22 11 R2 C1 2. Q   =  Q C1  Q C2 =  IR C   +  IR C initial 11 22 =  3  ×  2  ×  2  +  3  ×  4  ×  4=  12  +  48  =  60C Comprehension-5 Q   =  V(C   +  C )  =  18(2+4)  =  108  C 1. CA  0 A / d   =1  :  K final 1 2 CB K 0 A / d   Q  =  108  –  60  =  48  C  (through  S ) 1 3. U   =  1 C V 2  +  1 C 2 V22 VA  Q /CA  CB initial 2 11 2 VB Q / CB CA 2.  = K  : 1         = 1  2  62  1  4  122 = 324 J    V E (KC) KE 22 3. VA  ;    VA = 2 final C (K  1) K  1 initial 11 (2+4)182  =  972J U   =  (C   +  C )V2  =  final 2 12 2 (VA )Initial K 1   ( VA )Final 2K Node-6\\E:\\Data\\2014\\Kota\\JEE-Advanced\\SMP\\Phy\\Solution\\Unit-07 & 08\\02-Capacitance.p65 Q  =  Q   –  Q   =  48  C  fI W  =  Q.V  =  48  ×  18  =  864  J Battery   Heat  =  W  –  U  =  864  –(972  –  324)  =  216  J V Q  E(KC) 1  E B (V )   = CB (K  1)  K 1 4. ;  (V )   =  2B Initial B Final KC Comprehension-2 1 . Time  Constant  ( ) ( VB )Initial   (VB )Final =  (K+1)  :  2     =  R C  =  8  ×  6  =  48  s 1 2 . V =  V (1  –  e–t/)  =  12(1–  e–2/) Q2 Q2  UA  t  =  2  0 5. (U )   =  ;  (U ) =      U B  Final =  K:1 A final B final 1 1  2CA 2 C B 7.4                 =  12  =  10.4  V 37

JEE-Physics EXERCISE  –IV(A) (ii) Initial  stored  energy  in  first  capacitor qt ()   1 . CV  =     400  ×  10–6  ×  100  =  100  t 11 t Ui= 2 C1V12= 2   ×  0.1  ×  10–6  ×  102  =  5.0  µJ   t  =  400  s 2 . Equivalent capacity between A and B U f 2.5 1  U i   =  5.0   =  2 (AB) 7. C  0 A   ;  q   0 A  V 9 d  d               C =  3 + 3 = 6µF Slope  ()  =  0 A    C2  C1  C3 (i) Stored charge () d Q  =  C V  =  6  ×  1 0 – 6 ×  4  =  2 4  µ C 8 . By  using  KCL C1  (VA–V0)  +  C2  (VB  –V0)  +  C3  (VC–V0)  =  0    V0 (ii) Stored energy () = C1VA  C2VB  C3 VC 11 C1  C2  C3 U = CV2 = ×  6 × 10–6× 16 = 48 µJ 22 3 . Electric  field  ( )  2x A 2µF (10,000  0) 9 . x  =  2x 1                 (2  10 3 ) E  =  VA  VB =  =5  ×  106  V/m (Let  Ceq  =  x) d x 1µF x Q-q q Qq q 2x  2  x B 4. C2     C1  C2 ...(i) x  =  C1 2x x(2  +  x)  =  3x  +  2  2x  +  x2  =  3x  +  2   x2  –  x  –  2  =  0 Use Q q'      Q  q ...(ii) b  b2  4ac 1  1  8 1  3 C1 C2 C1 C2 x   =  =2 Qq 2a 2 2 and  –1 x  =  2,  Ceq  =  2µF Eq. (i)  (ii) : q' =  Q  q 10. CA   =   K 1 0 A    =  3K1   = 3  ×  3 =  9  d/4  5 . Common potential ( ) C1V1  C2V2 2  200  3  400 CB  K 2 0 A  K2  3d / 4  Vcm= C1  C2 =  =  320V 23 Net  capacity Charge  on  C1Q1  =  C1 Vcm  =  2  ×  320 C  =  640 C C  =  CACB  (9CB )(CB ) 9 Charge  on  C2Q2  =  C2 Vcm =  3  ×  320 C  =  960 C CA  CB 9CB  CB  =  10 CB 6 . (i) On  connecting  with  the  second  capacitor  the charge  distributes  equally     9 K 2 0 A  = 6K 2 0 A =  1.2K 2 0 A Node-6\\E:\\Data\\2014\\Kota\\JEE-Advanced\\SMP\\Phy\\Solution\\Unit-07 & 08\\02-Capacitance.p65 (3d / 4)  5d d ( =10 )  11.   E  =  V d C1V1  C2V2 0.1  10   VCM  =  C1  C2   =  0.1  0.1   =  5V V 103   d  =  E   =  106   =  10–3  m Total  stored energy    (  ) Uf  =  1   C1 VC2M   +    1   C2 VC2M Now    C  =  0 r A 2 2 d 11 Cd 88.5  10 12  10 3      = × 0.1 × 10–6 × (5)2+  × 0.1× 10–6× (5)2  A  =  0 r = 8.85  10 12  10 =10–3  m2 22       =2.5 µJ 38

JEE-Physics 12. C   =  0A ,   C   =  50A  C   =  5C X d Y d YX (i) C   and  C   are  in  series,  so  charge  on  each EE XY i   =        and  i   =  (C   C  ) XY q  =  C  V  =  CV           VX   =  5 1 R1 2 R2 X X YY VY (ii) At  steady  state  (t=),  capacitor  has  infinite  V  +  V  =  12          6V   =  12 resistance.    ( (t= )  X Y Y   V  =  12   =  2  volt  and    V   =  10  volt   )   Y X 6 E Hence,    i   =  ,  i   =  0 (ii) Energy  stored  in  capacitor  (  )    1 R1 2 FHG IKJ FGH IJKU q2 UX q2 2C Y CY (iii) 2C UY 2C X Final  potential  difference  across  capacitor  is       =  q2   =  C X =5 E.        1 3 .   CV1  =  3CV2 ....(i) C 3C    Final  energy  stored  (  )  V1  +  V2  =  300  ...(ii)     glass paraffin U  =  1 CE2 2  V1  =  75V;    V2  =  225  V (iv) When  switch  is  opened,  capacitor  will  dis- 300V (v) (i)    E1  =  V1   =  75 100 charge  through  two  resistance  as  R   and  R d1 0.5   =  1.5  ×  104  V/m 12 (both  in  series). V2 225 100 (      R  d2 0.5 1             E2  =     =  4.5  ×  104  V/m R     )  2 (ii)  V1  =  75  V;          V2  =  225  V H e nc e ,     c   =   C   ( R   +   R ) 1 2 When  switch  is  closed,  capacitor  will  charged  C1C2  V  3 C, V  3 2 0 A  300 through  resistance  R   .  C1  C2  4 4  d  2 (iii)  Q  =  (     R 2 Q 6  300  8.89  10 12  )  4  0.5  10 2 =8  ×  10–7  C/m2 S o      =   R C A 2 1 4 . When  SW1  is  closed  and  SW2  is  open  then  capacitor 1 6 . (a) In  steady  state  no  current  in  capacitor's B  is  charged  upto  10V. branch. ( ( SW1  SW2 B, 10V  )  )  Now  SW1  is  open  and  SW2  is  closed  then ( SW1  SW2   )  Node-6\\E:\\Data\\2014\\Kota\\JEE-Advanced\\SMP\\Phy\\Solution\\Unit-07 & 08\\02-Capacitance.p65 C1V1  C2V2 3  10  2  0   =  6V Vcommon= C1  C2 = 32 QA  =  2  ×  10–6  Vcm  =  12  µC So  current  ( ) I =  2 =  1A QB  =  3  ×  10–6  Vcm=  18  µC 0.5  1.5 1 5 . (i) At  t  =  0,  capacitor  has  zero  resistance,  i.e., voltage  across  capacitor R   and  R   are  in  parallel. 1 2 (    )  0       (t  =   R    R     )   VC  =3  +1.5×1  =4.5  V 12 The  simple  circuit  is  shown  in  figure   Q  =  CVC  =  2  ×  10–6  ×  4.5             =  9  ×  10–6  C (     )  39

JEE-Physics 1 7 . For  the  circuit  ACDA  and  the  cell  : 22. 1 Q2 =  Q2d ;  Einitial  =  0 Efinal  =  2 C 2 0 A (A CDA)  Q2d 6 – I1(5) – 6 =  0 I1 =0,   I =  0 Heat  =  –  (Einitial  –  Efinal)  =  2 0 A For  the  loop  BCD  (BCD  ):  V2F  =  6V 20 For  the  loop  ABD  (A BD  ):  V7F =  6V 2 3 .   V2   initial =  2 =10V   Q7F  =  6  ×  7  =  42  C 1 8 .   Total  heat  dissipated  (  )  50 V5   initial =  5   =  10V 11 H  =  CV2  =    ×  5  ×  10–6  ×  200  ×  200  =  0.1  J There  is  no  potential  difference. 22 ()  H1  =  Heat  developed  across  R1=  I2R1dt Hence  no  charge  flows.       (R1 )   () H2  =  Heat  developed  across  R 2  I2R 2dt Heat  produce  is  zero.       (R2 )  ( )  C H1  H2 R1 19 32 R1  R2  H  R1 24.   8/9   C  32 =1 C  =  23 F   H1  =   8/3 R1  R2  0.1  500 25. 1F 2F 1F 1F F          =  500  330  = 60 mJ 8/3 F 1 F 2 3 F 1 9 . Reff=  3  2   +  2.8  =  4 2/3 F V6 I  =  R eff  4   =  1.5  A F   I =  I   3   1.5  3 =  0.9  A  2  3  5 20. 4Q 2Q 2 6 .   60V 144C + 2 F 2 3 A 144C + Ceff  =  2  3 =  1.2F +3Q B   Q2  =  Q3  =  +144C +Q 60V 3F +Q 3Q Initial  effective  charge  =  3Q Q1 C 60V 120C + 2 F Node-6\\E:\\Data\\2014\\Kota\\JEE-Advanced\\SMP\\Phy\\Solution\\Unit-07 & 08\\02-Capacitance.p65 Q2 C 180C + () A B   CV  +  CV  =  Q1  +  Q2 60V 3F                               =  3Q  +  0                = 3Q Qtotal  on  the  middle  plates  (  ) 3Q V  =  2C 1  1 1  1  1 1  ....  1  2                     = + 180+ (–120) = + 60 C   C arm C 2 4  C 21. 1 1  This  charge  flows  from  A  to  B. 8 2  C  (A B) 2C   C  effective =  2Carm  =  2 =C 40

JEE-Physics EXERCISE  –IV(B) 2C 5C 5 . Ceq  3  C  3 q q dq 1 . C –  iR  =  0   C  dt R  =  0    q  =  q0e–t/RC O 4 5 O+ 4 3 32 2 1 dq q0 e–t/RC equivalent  circuit      i  =   dt RC ()  4 A V0  2 0 A V0 Q3  3 0 d  &  Q 5 3 d Where  R  =  L ,  C  k 0 A 6. Qtotal  =  C1V  =    C2C3  SA 4 C1 C2  C3  V0   k 0 5  8.85  10 12 5  8.85   RC  =  S  7.4  1012  7.4 C1 (C2  C3 )V   V0  =  C1C2  C2C 3  C 3C1 q0 8.85  103   i  =  RC e–t/RC   5  8.85  e 12 / 6   Charge  on  C1,  7.4  C 2 V (C 2  C3 ) 1         q1  =  C1V0  =  C1C2  C2C3  C3C1 7.4 1             mA  =  0.2mA Charge  on  C2  and  C3 5 7.4 2 . 2A 1A q2  =  q3  =   C2C3  V0  C1C2 C1C2C3V  C2  C 3   C2C3  C3C1 3A 5 7. 1 C V0 ( V0)C 2 2A 1A 2A 1A VC  =  (5  +  1)  ×  3  +  2  ×  1  =  20V Total charge remains constant()  11 156  C  =  (V0)CV0  +  CV0 Ucap  =  2 CV2  =  2   ×  4  ×  202  =  0.8  mJ   V02  +  V0  –  156  =  0  (  =  1)   (V0  +  13)  (V0–  12)  =  0    V0  =  12  volt ,r1 ,r2 7R/4 2 8. Initial  condition 3 .       I =  A C B 7R CC CC r1  r2  4 Pot.  diff.  across  (,r1)  cell  :    –  Ir1  =  0 Q QQ ()  Node-6\\E:\\Data\\2014\\Kota\\JEE-Advanced\\SMP\\Phy\\Solution\\Unit-07 & 08\\02-Capacitance.p65    =  Ir1    =  2 r1   4(r1  r2 ) R E 7R 7 C r1 CE  r2  4 Q 4. 3 B x dx Final  condition CC A CC 1 dx d dx 2 CE 2 CE CE CE 5CE  dC   KS   7 7 7 77 0 0 KS 0 1  sin x  E d  K1S 0  d dx 2d  4 2d  7 C  0 1 x     Charge  flown  from  B  to  A  =  CE  d    sin  41

JEE-Physics 9 . Extra  weight  needed    (  )  V1  440  440  440  35  350V / d1  9 1 44 6 2 0 A 0 A  V 2 0 A  K1 / d2  1 35 =   0  2 2  d 2  K 2    =  E2  ×      5000 2 8.85  1012  100   E1  =  V1  350 100 =  5  ×  104  V/m  5  103 2 100 100 d 0.7  mg       m  =  4.52  ×  10–3  kg E2  =  V2  90 ×  100  =  3  ×  104  V/m d 0.3 1 1 . Qpq  =  2C2  =  6C1  =  Qbp U1  1 C1 V12  35 2 6C1 C1 b C2 U2 1 C2 V12 9 C1 a 2   Vbp  =   6V   Vbq  =  6  +  2  =  8V P 1 4 . Work done by battery  (   )  q Total  charge  flown  into  right  loop =  QV  =  (3CV)V  =  3CV2 C1 C2 ()  1           =  C2Vbq  +  C1Vbp Energy  stored  in  capacitors  =  (3C)V2           =  3C1  ×    8  +  C1  ×  6  =  30C1 ()  2   Vab  =  Q total  30C1  30 volt (i)    Heat  developed  =  WB–U  =  1 (3C)V2 C ab C1        ( ) 2 1 2 . Applying  junction  law  at  A  : (ii)    Work  done  by  external  agent  =  –  (K–1) (A)      ()  A 20 F (iii)  Final  voltage  after  'dielectric  is  removed  =  V' 2F +– 3F +– +– 5F 4 F     () K 2        3CV' = (K+2)CVV' =  V  3  2(VA–5)  +  3(VA  –  20)  +  5(VA  –  10)  +  4(VA  –  20)=0         Wagent =  Ui–  Uf 100                                   =  1  K  22  1 (K+2)CV2   VA  =  7 =  14.28  volt 2 (3C)V2  3  2   Q2F  =  28.56  C,  Q3F  =  42.84  C, Q5F  =  71.40  C,  Q4F  =  22.88  C (K  2)(K  1)CV 2                 =  6 1 3 . V1C1  =  V2C2  and  V1  +  V2  =  440 Q1 Q2 Q1 C1 C1 C2 Q2 C2 0.7cm 0.3cm 15.    ....(i) K1=3 K2=5 Q1  +  Q2  =  2Q  ....(ii) C1  =  0 A   and  C2  =  0 A Node-6\\E:\\Data\\2014\\Kota\\JEE-Advanced\\SMP\\Phy\\Solution\\Unit-07 & 08\\02-Capacitance.p65 dx dx 440V Q d  x Q d  x  Q2  =  d   and  Q1  =  d V1C1 V1 C 1   V2  =  C2  V1  C2 =  440 dQ2 Q  dx  dQ1 Q  dx dt 2d  dt  dt 2d    dt     =  –   &    =  440C2  440 C1  C2   V1  =    C1 1 I  dQ1  dQ2  Q  dx   200  0.001  2A C2 dt dt d  dt  0.1 42

JEE-Physics Q2  ab  EXERCISE-V-A 2C1  b  a  16. U1  =  ,  C1  4 0  b 1. C  n Ceff( Parallel ) 2. Q2 If  connected  across  V  volts  then  energy  stored U2  =  2C 2 ,  C2  =  40b (V)   U  =  U1  –  U2  =  9  J                =  1 nC  V 2 2C  C 2C 2 1 7 . Cinitial  =  2C  C  3   ;  Cfinal  =  C Capacitance  of  an  isolated  sphere  is (i)   Q  =  C  ×  V ()  C=(4pe0)(Radius)  () C  1  1  0.11  109  1.1  10 10 F 9 109 8  10 18 2 100  106 2C CV 2  30  Work  done  = 1Q2 1            =  C  3  V  3 = 3 =20C 4. 2 C   =   2    1 1 2CV2  1 64 1036  2       = 2  104   =  32  ×  10–32  J (ii) H  =  WB  –  U  =  QV  –  CV 2   2 3 5. 1 CV2  =  msT    V  =  2 m sT 2 C         =  600–  (900  –  600)  =  300  J  =  0.3  mJ 6. Two  plates  stacked  together  form  a  single  capacitor (iii) Energy  supplied  by  the  battery of  capacitance  C.  n  plates  stacked  together  form =  QV  =  600  J  =  0.6  mJ (n–1)  number  of  capacitors  of  effective  capacitance ()  (n–1)C. (iv) Initial  charge  on  each  capacitor (C () n  (n–1)C  (n–1)   )  2C 12          = 3 V  = 40C 7. Final  charge  on  right  capacitor  =  60  C Y ()  0.1m Final  charge  on  left  capacitors  =  0 X () Node-6\\E:\\Data\\2014\\Kota\\JEE-Advanced\\SMP\\Phy\\Solution\\Unit-07 & 08\\02-Capacitance.p65 Applying  law  of  conservation  of  energy   Total  charge  from  through  switch,  S  =  60C () (S)  We  get  1 mv2  eV 2 [Here,  v  =  speed  of  electron,  V=V2–V1=potential difference] [v= V=V2=V1 =] 2eV 2 1.6 1019  20 v  m 9  10 31 On  solving,  we  get  v=2.65×  106  m/s 43

JEE-Physics 8. Energy  stored  in  a  capacitor  when  it  is  charged  by R C n 2 q0 2 4 1 t1  ...(i)    and   q 0et2 /RC a  potential  difference  of  V0  volt  = 2 QV0 t2 2n2 t2  =  2RC  n2...(ii) ( V0 RC  ; =12QV0 ) from  equation  (i)  and  (ii)  t1  1 t2 4 Total  work  done  by  battery  in  sending  a  charge  of    1 2 . 5 Q  through  emf  V0=QV0 V  V0 1  e t / RC  120  =  200  1  e RC (V0Q   R  =  2.7  ×  106       =QV0) 1 13 . Parallel Series hence  energy stored in capacitor  2 QV0  1 c cc R R work done by battery QV0 2 c  1  V0  ...(i)  2QQVV0 0  1 V0   2   v0 2 9 . Net  work  done  by  the  system  in  the  process  is  zero,  v0  tp 1  e R  2C as  in  removing  the  dielectric,  work  done  is  equal   ts and  opposite  to  the  work  done  is  re-inserting  the v0 R    C dielectric. 2  v0 1  e 2  ...(2) (  tp  2 ts from (i) and  (ii)    e 2Rc  e Rc )  ts =  t4p  10 =  2.5  sec 4 10. C  A  9Pf ;  C eq  C1C2 1 4 . t  =  0.37%  of  V0 d C1  C2     =  0.37  ×  25  =  9.25  volt where  is  in  between  100  and  150  sec.  3 0 A K 1   3 0 AK 2   d   2d    d1  40.5pF 15. Common  voltage  ( )= C1Cv11 C2 V2 30AK1  30 AK1 F C2 2 d 2d 11. U =  1 CV2 ; U0  1 C V02 e 2 t1 / RC (positive  plate  of  one  capacitor  is  connected  with Node-6\\E:\\Data\\2014\\Kota\\JEE-Advanced\\SMP\\Phy\\Solution\\Unit-07 & 08\\02-Capacitance.p65 2 2 2 negative  plate  of  second  capacitor) 1  e 2 t1 / RC   (U0  =  1 C V02 ) (   2 2 )  2t1  n2   120  C1  =  200  C2    3C1  =  5C2 RC 44

JEE-Physics EXERCISE  –V-B 5 . From Y to X charge flows to plates a and b. (a bYX ) 1 . In  steady  state  condition,  no  current  will  flow ( q  + q )  =  0,  ( q  + q)  = 2 7 C ab b i a f through  the capacitor  C.  Current  in the  outer  circuit, 3F 6F a (C  +– b X +– +– +– 18C 18C 9C 36C ) 3 6 3 1A 6 1AY   1A R 9V 9V Initial Figure Final Figure V (when switch was open) (when switch is closed) AB    i  2V  V V  iV i=0 2R  R 3R  27C charge flows from Y to X. 2R i    ( YX27C  ) 2V 6 . Time constant  ( )=  RC Potential  difference  between  A  and  B  : (AB)   2d   d    3  Vt   3  Vt V   –  V  +  V  +  iR  =  V 1 1  1 AB Where    CC C 0 2 V V 12 0  V  – V  = iR    3R  R  3 BA 6                         C  =  0 t 2. Charging current  ( )   I E  5d  3Vt R e RC MCQ's Taking log both  sides (log) 1 . Before S  is pressed   (S   ) 33                   log I  E  t CV0 CV0 log  R  RC –CV0 –CV0 When  R  is  doubled,  slope  of  curve  increase.  Also  at After S  is pressed  (S )  33 t=0,  the  current  will  be  less.  Graph  Q  represents the  best. 2V0 CV0 –CV0 V0 –CV0 CV0 ( Rt= 0 Q) Subjective 1 . Let  at  any  time  t  charge  on  capacitor  C  be  Q  and 3 . Given : V  = 3V  = 3(V – V ) CR C currents  are  as  shown.  Since,  charge  Q  will  increase with  time  t. Here, V is the applied potential. (tCQ  (V) Qt)  3 31 AR   VC  V  V(1–e–t/RC) =  V  e–t/RC=  S 4 44 N Node-6\\E:\\Data\\2014\\Kota\\JEE-Advanced\\SMP\\Phy\\Solution\\Unit-07 & 08\\02-Capacitance.p65 Here   = cR = 10s  i1  dQ V R + c     B QC dt M R Substituting this value of c in equation and solving – i T ( )  c We get : t = 13.86 s (i) Applying  Kirchhoff's  second  law  in  loop  MNABM 4 . = CR (M NABM) 1 =  (C  +  C 2 )  (R  + R )  =  1 8  s V=(i  –  i )R  +  iR  V  =  2iR  –  i R                          ...(i) 1 1 2 1 1  C1C2   R1R2  8 2 8 Simillarly,  applying  Kirchhoff's  second  law  in  loop  C1  C2   R1  R2  6 3 s 2     MNSTM 9  C1   R1R2  2 (M NSTM)  R1  R2   (6)  3  3  C 2  4 s 45

JEE-Physics Q Q  = C [PD across capacitor in steady state] we  have    V  =  i R+ +iR          ....(ii) 0 1C               [] Eliminating  i  from  equations  (i)  and  (ii),  we  get       =C[ steady state current through R ] (R ) 2Q 2Q 22 V  3i1R  C  3i1R  V  C [R] (R) 2  2 1  2Q  dQ 1  2Q  V   i1  3R  V  C    dt  3R  V  C         C  R1  R2  R2 dt Q dQ t dt CV R2 1 1  dQ     Q0  R1  R2   is  C   C R net 2Q   3R 0 V  3R 2Q 0 V C C R1 This  equation  gives            Q  CV (1  e 2 t / 3RC ) R2 2 (ii) i1  dQ  V e 2 t / 3RC Here, R  is equivalent resistance across capacitor dt 3R net From  equation  (i) after short circuiting the battery. i  V  i1R V  V e 2 t / 3RC (R  2R 3 net  2R )   Current  through  AB  (AB   ) R net  R1R2 R1  R2 V  V e 2 t / 3RC i   =  i  –  i   =  3  V (As R  and R  are in parallel) e 2 t / 3RC 12 21 2R 3R (RR ) 12 VV V i2   e 2 t / 3RC i  =  as  t     2R 6R 2 2R   1  R1  R2 2 . Q  is the steady state charge stored in the capacitor. C  R 1R 2  C R1R2 0    (Q ) R 1 R 2 0 R1 V C R2 Node-6\\E:\\Data\\2014\\Kota\\JEE-Advanced\\SMP\\Phy\\Solution\\Unit-07 & 08\\02-Capacitance.p65 46

JEE-Physics UNIT # 10 ELECTROMAGNETIC INDUCTION & ALTERNATING CURRENT EXERCISE –I t=  W  N2B2A2  0.1mJ 1 . Total change in flux = Total charge flown through v Rt the coil × resistance 1 Br2 2 1  4  0.1 × 13. I=    B r 2  2 1 = Resistance= 0.2 × 10 = 2 Webers R1 R1 2R1 2. N 1 4 . IB > mg  IB < mg S 1 5 .  = NBA cost LN  e= – d =NBA sint  e = NBA max dt e   d  Na2 dB  5volt 3. dt dt e  L I  L  e  8  8  0.1H t I / t (4 / 0.05) 80 16. 4. e   d  NA dB  (100)(40  1 0 4 )  6  1  17. dt dt  2  = 0.8 volt  1volt 5. q   = NBA qR I increases B1 increases B  RR NA So from Lenz's law d dr Current in A is clockwise  (2rB ) 6.  = r2B e  dt dt   7 . W   QE.d  Q E.d = QV 8. d    –A=   I same and flux linked with A increases dt E .d r E.dr So from Lenz's law current in A clockwise 9. P– +Q e = B (2R) v = 2Brv 18. I d a v b ×××××× 10. According to Lenzs law  × × × × × × A db 0I 0 Ia  d  b Node-6\\E:\\Data\\2014\\Kota\\JEE-Advanced\\SMP\\Phy\\Solution\\Unit-9 & 12\\02.EMI & AC.p65 B d 2 x 2  d  × × × × × ×   =   (adx)  n MI but ×××××× ××××××  Plate B will become positively charged.  M  0a n  d  b  M  a 2  d  1 B2 1 9 . Work done = LI 2 = (0.04) (5)2 =1.0 J 11. I  2 0 R 2 0 . According to Lenz law current in loop as shown All spokes are in parallel in figure due to this current, a magnetic force F F N(IB)  N Bv  2 B  N2B24 m  R  tR is acted on bar magnetic if  is the acceleration of the bar magnet then 12. W = = = N 22

JEE-Physics 3 0 . Impedance = R 2  (L )2 At low frequency   0 so impedance  R At high frequency    so impedance  L V 2 (10)2  5 ma = mg – F a = g – F/m  a < g 31. R   P 20 2 1 . 100 = V + V But V = (4000) (15 × 10–3) =60volt In AC circuit RL R V2  V2  R =  z  z  V = 40 volt P = VI cos  cos  =   L z   22. L L  0.5  103 (10)2 (5) R1 =2 × 10–3 & R1  R2 10   L = 5 (5)2  (L )2 90  R1  =4  R = 30, L = 60mH R1 1 5  5 L 2 2L =  f = E 5 2 3 . Here I = = constant  2  3.14  10  103 = 80 Hz 0R  We can't change E or R. L 11 For curve 2 time constant R is more 32. X =  I  I0 (1  e t /  )  I0  I0 (1  e t /  ) c c 2fc 2 24.  t  n2  L n2   300  103  (0.693) = 0.1s 33.   tan 1  XC  R  2   R  25 . Average value = 0 34. Z  R2  X2 rms value = V 0 1 If X is capacitive X= C 26. 35.   tan 1  x   tan 1  L   R   R   tan 1  200 1 = tan–1(1) = 45°  200   3 6 .    T   / 4  T  T = T E reaches at maximum value at phase before 2 I reaches at its peak value at phase  2 T 2 T 8  11 So E is leads I by 2 So E is leads I by 2 but T = 50 s  T = s= 2.5 ms Node-6\\E:\\Data\\2014\\Kota\\JEE-Advanced\\SMP\\Phy\\Solution\\Unit-9 & 12\\02.EMI & AC.p65 400 4  I2dt   4tdt (2 t2 )24 R  V 100  4  dt ( t )24  27. I= 2   12  2 3A 37. I 25 rms 4  dt R 2  (L )2  V  100  5  L = 3 I 20 2 38 . Here V = V 2 8 . Ac ammeter reads rms value LC dc ammeter reads average value  Resonance condition  Voltage across LC combination remains same 29.  Irms   I0  I1 sin t 2    I02  I12sin2t  2I0I1 sin t   I20  I12  I20  0.5 I12 3 9 . tan45° = X C  X L  R  1 – 2fL 2 R 2fc 23

JEE-Physics I 50. L di  q   di   q0 45° dt c  dt  max LC V  R  2fL  11 C  VC 2 fC 2f(2fL  R ) I VL VL 51. I 40. V – V = 50 – 50 = 0 LC VC 1 4 1 . Resonant frequency   LC 52. For LC circuit q = q cos t, 0 42. V = I X = V ( L )   R   2  L L i =q  cos  t  0   V     1  V  100  105 According to given condition  R   2C  R C 103  2 200 = 250 volt q2  1 Li2  q  Q 2C 2 2 43 . V4 = VL + VC = 0 44. V = 100 V, I = VR 100  2A 5 4 .  = (R2)B = B(R0+t)2  d R R R 50  dt = 2B(R0+t) anticlockwise 4 5 . Circuit will be capacitive if X > X CL I  Bv  (0.1)(0.1)(1)  5  1 A Circuit will be inductive if X > X 1  6 / 5 11 / 5 1100 220 LC 55. 4 6 . Reading of voltmeter=0 as V = V LC Reading of ammeter= V 240 =8 A R  Bv Bv 30 2 1 3  5/6 1 11 47. Here   2000rad / s LC 5  103  50  106  1 di di LC 5 6 . e  L ,  is more for 1 so resonance condition dt dt 5 7 . Dimension based : Check yourself.  I  V  20 / 2 2 = 1.4A R 64  1 Node-6\\E:\\Data\\2014\\Kota\\JEE-Advanced\\SMP\\Phy\\Solution\\Unit-9 & 12\\02.EMI & AC.p65 LP and reading of voltmeter= (I) (4) = 4 2 = 5.6V 58. I2R = P  R = P  2  LP L 2U I2 1 LI2 2U    E0   I0    2   2  2 R P 48. P = VI cos  = c o s   0 2   49.   tan 1  L   tan 1  2  50  0.7   tan 1 (1)  45 5 9 . VR  102  82 = 6volt  R   220  wattless current = I sin   tan 1  L   tan 1  VL   tan 1 8  tan 1  4 rms  R   VR   6   3  I  V 1  220 1  1 1 A  1 A Z 2 220 2 2 2 2 2 24

JEE-Physics 6 0 . i = 2 sin 100t + 2 sin(100t) cos 30° EXERCISE –II + 2cos(100t) sin 30° d   d = (2  3 ) sin 100t + cos(100t) 1. e=  [where  = B.A ] e  (BA cos ) dt dt either ,A or B should be changed for induced e.m.f.  I = 21 1 2.   i2   2  3   1   7.5  4 3 Total charge transferred= rms 2 2 R It is independent of time  d  2a 0i adx a 2x    x R1  = B.A 6 1 . Old power factor =  =–(–)=2 R2  (3R )2 10    0ia n2  q  0ian(2) 2 r R1 e b 0i New power factor =  0 de a 2 x vdx  e  R2  (2R )2 5 3. = Bv   e  0iv n(b / a)  i = 0iv n(b / a ) 2 2 R i 62. P b 0i Q a 2x  Force ( )= Bi = b  df  i1dx R 0i  b  0 IV  b   2 V 2  a   a   R f  n i1   2 n   4.  = BA = 0ir2 R 2 2(R2  x2 )3 /2 IP & IQ  clockwise IR = 0 e  d   30ir2R 2x dt (R2  x2 )5/2 For e to be maximum 63. d(e)  0  x  R 5. dx 2 6.  2BR 2   I LI =   i = i = LL Node-6\\E:\\Data\\2014\\Kota\\JEE-Advanced\\SMP\\Phy\\Solution\\Unit-9 & 12\\02.EMI & AC.p65 B L2 e= 2 effective length of the wire frame is 2R  e = 2BR2 64. q= + (Bv) C = (4) (1) (20) (10×10–6) =800C 7.  1 11 A 8. 2 LC  = 6 LC q = –q = –800C B A E E L2  L1 tan 60° = XC Also tan 60° = XL is X = X = 3R R1  R2 R1 R R CL 65.  R1  R2,  L2  L1 R2 series L–C–R is in resonating condition V I = = 2A R P = I2R = 4 × 100 = 400W 25

JEE-Physics V2 V2 1 9 . P   R   100 C = 45002  0.0494 = 10–6 F RP with AC source 0 (e / T )  d 0 e r 2  2R  dt 2RT2 13. = r 2   P   V 2 R  Z  200  Z   0 e r 2 Z2  R2  e  0 e r 2   4R  2R (2 ) Z2  R 2  2L2  L  3H  10. i1  1 W1  1 ; V1 4 AdB i2 4  W2 4 V2 14. e  dt 1 1 . Minimum value of impedance is for 10  10  102 2 4  0   t = 20 ms t 11  0ieet /  b dx X = X  C  L  L  2C = 0.36 mH   B.A 2 x L C 1 5 .   1 2 . In DC (act as open circuit)    0i0et /  b n  a  d  250 2  d  So R = 1 =250  e = d  0i0et / b n  a  d dt 2   d  At resonance L  1 16.  = at (T – t) d  aT  2at C  dt 1 i  d  (aT  2at)  4500L ...(i) Rdt 4500C V=V –V =V T CL R Heat =  (aT  2at)2 Rdt I 1  L  =V    I R = V 2  C  0 1   2 T 2 t  4a2 t3  4a2T t2 T R a 3 2  I12  I22 I2 V2 V2  0 Here I =    R2  2  1   C  L   2 3 4a2T3 2 3  a2T3 a 3  3R  T   2 a T R Heat = 5 / 42   1 1  2502  2502  1 / C  L 2 1 7 . Area in the magnetic field is given by A = x2; = Bx2 1 4 here =2250 rad/sec d dx   L   250 dt dt  C 3  2B x  90°  = 2Bvx x So, 1  4  250  2250L ...(ii) 1 d 2Bv2t Node-6\\E:\\Data\\2014\\Kota\\JEE-Advanced\\SMP\\Phy\\Solution\\Unit-9 & 12\\02.EMI & AC.p65 2250C 3 i=  , it R dt R From (i) & (ii) : 4 1 8 . No induced current in the loop therefore No force acts  250  2250L over the loop. Work done on the loop will be zero 2= 3 1 9 . Equivalent circuit diagram is as shown 4500L 9000L  4  250  2250L  i 4e  e  Bv r 3 4r r r r L  4  250 = 0.0494 H i  Bv r r 3 6750 r 26

JEE-Physics 20. e = Bv E2 where all the three should be perpendicular to each Heat ( )  L2R 2dt  2 R 1 other R   3j  4k [(3i  4j)  2i] OR B( v)  e   e  Heat produced = energy stored in inductor  (3j  4k )(8k )   e=32 volt 1  E  2 LE2 2  R1  = L  2 R 2 1 2 1 . As capacitor blocks the current there will be no current in the circuit HKDE 2 7 . u  1 LI2  u = 32Joule 2 B 2 22. e  L= ?, I = 4A L = 4H, P = i2R  R = 20 ohm 2 effective length for the given diagram is l2 = 2 + L2 e  B(2  L2 ) Time ( L 0.2 sec 2 )= = eff R 2 3 . By flemings left hand rule equivalent circuit diagram 2 9 . As current in the circuit is given by V0 R is as shown current will be P to Q In the disc X 100 × 1  L = X =L =  =100 C PQ 24. e  L di 2  4 5  0  t=10sec cos= R  1 dt Z 2 t 2 5 . In loop CDEF current is independent of the time 30. V = 2V cos  t   as voltage across inductor is 'V ' 0  6  as current will remain all the time more circuit behaves like an inductive circuit is R current lags voltage by an amount 30°. 3V 3V cos  t    30°  6  I=I 0 2V  Circuit equivalent can be considered as V0 Rt 3 1 . For (L – R) circuit current at any time  R [1 e L ] 34 cos  = 5  = 53°  X = R L 3 I 2V For (C – R) circuit 1 cos  =   = 60°  X = 3 R R 2C 2 3V Node-6\\E:\\Data\\2014\\Kota\\JEE-Advanced\\SMP\\Phy\\Solution\\Unit-9 & 12\\02.EMI & AC.p65 In L–C–R power factor is i i.e. reactance is zero By super imposing the two current the net current in it will be 4 R2  4   R1 33 i = V0  V0 e Rt / L  V0   V0 e Rt / L X = X R = 3R2 L C 3 1 RR RR 3 2 . I0  I12  I22  2I1I2 cos 30 2 6 . Decay of current in L–R circuit is given by  Ldi  iR  0 = 3 dt 448 2 i   L di i  i0 (1 R2t I0 = 2 2 3 I= I0  2 3 R dt eff 2 2 e L ) 27

JEE-Physics 3 3 . I 2 sint Rdt I1 = I 4 2 . Charge on the capacitor Q = CV = C × (Bv ) I= 0 2 0 As all the quantities are constant rms dQ so dQ =0 Hence i   0 and current In the I more than I = I dt 3 12 4 3 . Replace the induced emf's in the rings by cell I3 > I1 = I2 > I4 3 4 . If I =0 and P moves towards Q, then according to 2- 2 + Lenz law a current in opposite as I1 is induced in C1 C2 Q. Same as I =0 and Q moves towards P when + 1 1 I1 0, I2 0 are in opposite direction then the coils - repels each others. 3 5 . No emf will be induced in any direction of its motion    B  e = B(2r) (2V) = 4BrV  1 C1 e = B (4r) (V) = 4BrV 36. 2 V –V =e +e = 8BrV 2 121 L d d ]  d 1  x   d 2 dt dt  dt  B0 a   44. e=–   [BA cos L [ Area of square = d2,  = ] C2 C1+C2  B0d2 dx  B0d2 V0  V0 i) a dt a (V We know q=q cost  m i = qmcos t    , i = i cos  t   L 2   2  1  R eq 0 4 6 . Charging Maximum current i = qm   1 0 C1     L  C2 For determining R Discharging Maximum charge on (C1 + C2) Req = 2R L  q = i0 m  1  2R  i0 L C1  C2 Maximum charge on C 1 C1 L C1  C2 C1  C2   = i0C1  i0 L C1  C2 3 7 . Sudden increase in the e.m.f. cause the spark in 1 1  3 the inductor 2  3R  2 2 3 8 . In absence of L whole emf of B goes on lamp and 4 7 . Energy per unit volume lamp will glow with full brightness instantaneously but in presence of L some emf is induced in L. B2  1  0ir 2 Voltage drop on P decrease and brightness . 20 20  2 a 2 =  2 rdr Node-6\\E:\\Data\\2014\\Kota\\JEE-Advanced\\SMP\\Phy\\Solution\\Unit-9 & 12\\02.EMI & AC.p65  1  2 i 2 r 2 0i2 a 20 0 4 a 4  2rdr r3dr 39. 1  MI2 , 4=2I   I = 2 A   2 4 2a 4 0 2 40. I  V  0 a4  0i2 Z  4a4 4 16 X = 250 × 2 = 200 L X = 2(400) × 2 = 1600 4 8 . Current through C and L would be equal after L a time f when 1 Li2  q2 2 2C Z increases 8 times current decreases by I/8 4 1 . I is the induced current of i , so the nature of i  11 become should becomes opposite of i ie negative.  28

JEE-Physics 5 3 . The electric field force due to variable magnetic 1   1 dB dB  eR i LC  q field= 2  R  d  q 2 d  q  q0 1  et / Rc , i  q0 et / RC  Acceleration = 1 eR dB RC 2m dt  q 0 e t / RC 5 4 . Induced current RC LC  q0 1  et/RC L / C et / RC  1  et / RC I= e  1   nd  n (2  1 ) R R R  dt  5R t 2et / RC  1 et/RC=n2 5 6 . L  0N2A ; N × 2r = L  t  n2 t=RCn2 L RC As wire is fixed 49. I = 1.57 r 0 Instantaneous voltage across capacitor  N'×2× 2 –L=N×2r  N'=2N 1 (2N )2    r 2  2  E = V0sint, V = XI = 2fc  I0 0 0  i= 0× L L' = 2L 1 2  2  50  100  10 6  1.57 V0 = 50  E = 50 sin 100t T/4 T/4  4V0 2  T  V2dt 0 t dt 5 0 . I= I12  I22 58. V = 0  T/4 rms T/4 I2  2002  V2  dt  dt 0 0 1002  X L  X C 2 2 0 0 2  4  4 V02 T3  V0 = 4+ 1002 = 4 + 4 = 8 T2  3 16  4  3 T  I = 2 I = I 2 4 2 2  rms Area 59 . Average value = time =0 51. XC I1 (in 0–T interval) VL I1 V 60. I V I V & Node-6\\E:\\Data\\2014\\Kota\\JEE-Advanced\\SMP\\Phy\\Solution\\Unit-9 & 12\\02.EMI & AC.p65 I2 90° V=VC= VR2+VL2  1 2 R2 1 R tan1XL  2 c  2 R   R2  c    3  XL I2  (2)2 = R2 9  4R2 4  R2 9 So phase difference between I and I R2  2c2 + 2c2  2c2 12 =   tan 1 X L 1 2R  2c2 5 2 . Time constant of L–R circuit is = L 1 Li2 R From option (A) = 2  2 L 3R2 = 5  3R2= 5X 2  3  Xc i2R  c 5R 2c2 R 29

JEE-Physics EXERCISE –III imean = i0 T0 2  i0 [for half cycle] 4 T0 2 Match the column–I Total Area  0 Total time T 100 t  (C) imean = i0 T 4  2 V = 100 sin(100 t) i = 10 sin  imean = 0 i0 T0 2   imean  i0 T0 / 2  i0 ( A ) Phase difference  4  T0 / 2 cos  = R  Z  2R ; tan   X L  XL  R Total Area Z R ( D ) imean = Total time (B) I0 = V0  Z = 10  Z R2  X 2 imean = i0 T0 / 2 Z L T0 R 2  R 2 = 10  R = 5 2  = XL i0 i0T / 2 2 T/2 1 imean = ihalf cycle = = i0 XL = (L) = 5 2 L 2 10 ( C ) Average power = VL cos  Match the column (IV)  E R1 t at t = 0 at t =   100 10  1  250 2 i1 R1 1  e L 22 i1 = 0 E 18 Match the column II I2  R2  3  6A Peak value current in the circuit is given by V i0 = R 0  If R will be less current will be maximum L E 18 18 Slope of i (v/s) t graph gives time constant is R i2 = R2 = 6  3A I1  6  3A ( A ) Graph III and IV denotes for 'R0' and slope is more Match the column V for III therefore R0,L0 represent III and (IV) represent R0,2L0 (A) tan  = XC  1  10 104 1  10 R C R  106 ( B ) Similarly I and II denotes for '2R0' and slopes is more for 'I'  tan  = 1  = 4 Match the column III (B) tan  = X L  as R = 0    R2 (A) Irms = I0 ( C ) tan  = X C as R = 0     2 R2 Imean for full cycle = 0 ( D ) tan  = X C  X L as R=0   =  Node-6\\E:\\Data\\2014\\Kota\\JEE-Advanced\\SMP\\Phy\\Solution\\Unit-9 & 12\\02.EMI & AC.p65 Imean for half cycle = 0 R2 Imean for half cycle = 2I0 ( E ) tan  = X L  200  5      R 1000 4 (B) i = 4i0 t Match the column VI T d 5 1 16 I20 T/2 i0 (A) =5  i =  Anticlockwise T2 3 dt 10 2 ir = t2Rdt irms = 0 d (B) dt =0  i=0 = zero Total Area imean = Total time imean = 0 for full cycle 30

JEE-Physics (C) d 5  1 = Clockwise Comprehension–3 dt = –5  i =   1 . M is same as that of L 10 2 kL is same as that of d (D) =5  Anticlockwise m LC dt Compreshension–1 1L 1 . f = Bi f = B22 v k is same as that of C mk  C Comprehension–4 R 1 . q = q0 sin () ...(1) dq 4  104  64  104  v dt = q0 cos ()  5 = q0  cos (t + )  3.2 × 10–5= By dividing (1) and (2) we get 2 1 41 2  × 103 = v  25 m/s 5 = tan (t + ) tan () = 5 40  2 . e = Bv = 2 × 10–2 × 8 × 10–2 × 25 = 4 × 10–2V 2 3 . V = E – ir V = Bv – Bv R  V = 3.6 × 10–2V from above equation sin (t + ) = 3 R 2 Comprehension–2 Q = Q0 sin (t + ) 4 = Q0  3   Q0 = 6C 1 . By applying K.V.L. 2. (t + ) =   –  t = 2   sin 1 2 For ABCDE t = 2  2  3    18 – 6i di1  0 ...(i) 2   dt Comprehension–5 By applying K.V.L. for ABCFGE 1 . 1 = B(L2 + 2) as current in both direction are  18 – 6i – 3 (i – i1) = 0 18 – 9i + 3i1=0 additive in nature while 2 = B(L2–2) as current in both the loops are in opposite direction. i = 18  3i1 ...(ii) 9 Substituting its value in equation (i)  18  6 1 8  3 i   di1  0 18– 12 – 2i1= di1  9  dt dt Node-6\\E:\\Data\\2014\\Kota\\JEE-Advanced\\SMP\\Phy\\Solution\\Unit-9 & 12\\02.EMI & AC.p65  t dt  i1 di1  2t  n  3  i1   n (1 ) 2 . By lenz law in both the loop are in clock wise 2 3  i1  3  direction therefore it flours from b to a and d to c 2 in both the loop. 0 3–i1= e–2t  i1 = 3 – e–2t 3 . Again by lenz law current in both direction should be clockwise but it is not possible therefore it is 2 . V1 + V2 = V18 – 6e–2t + V2 = 18 V2 = 6e–2t clockwise in bigger loop and in anticlockwise in smaller loop as e.m.f. due to bigger one is greater 3 than smaller one 4. 2 4 . I1 > I2 31

JEE-Physics EXERCISE –IV (A) they produced a magnetic field which is normal to 1. the plane of the paper but point upwards, i.e. to- Due to current in C no change in  of B. So no wards the reader. This requires that current I and induced current in B. But due to I in A,  is chang- 1 ing in B because A is moving towards B.  is changing (increasing) in B. I flow in the directions shown in the figure. So according to Lenz's law direction of induced current in B will be such that it will try to decrease 2 the  in B so current will be opposite in direction in B than A. Since the resistance per unit length is 1 m–1, the 2 . V0 resistance of wires AD, AE, DF and EF are 1  QP each and those of wires EB, BC and FC are 0.5  B 3 each. Applying Kirchoff's loop rule to loop I SO (AEFDA),  IR 1  103 4  e  Bv  IR  v = B = 2 10  102 = 0.02 ms–1  I1× 1  I1  I2  1  I1  1  I1  1  e1  0 or 4I – I = e = 1 volt..........(1) 12 1 Applying Kirchhoff's loop rule to loop I (BCFCB), we have e – I × 1 – I × 0.5 – (I – I ) × 1 – I × 0.5 = 0 22 2 12 2 or 3I – I = e – 0.5 volt ............(2) 21 2 Solving equation (1) and (2), we get 73 I = and I = A. 1 22 2 11 Referring to figure, the current in segment AE = I 1 7 = A in the direction from E to A, the current 22 3 in segment BE = I = A in the direction from B 2 11 3. E I2 B 7 to E and the current in segment EF = I – I = A 1 2 22 I B II 31 = = A in the direction from F to E. 11 22 e1 (I1I2 ) e2  NBA qR R R NA 4. q= = B= D I1 F I2 C F I F I HG JK GH KJ= 100 Refer to figure Electromotive force (emf) is induced 895  5 in circuits I and II due to change of magnetic flux 100  (3  10 2 )2 = 102 T threading the circuits because magnetic field B is changing with time. As the areas enclosed by the 5. dI circuits remain unchanged, the magnitude of the Induced emf in coil e=M dt induced emf is given by 6. 7. Mutual inductance of system M = 0N1N2A  e= d  d BA  AdB where ()n = N1 , N2 =N,A = R2  dt dt dt   M = 0 n N (R2) Area enclosed by circuit I is Node-6\\E:\\Data\\2014\\Kota\\JEE-Advanced\\SMP\\Phy\\Solution\\Unit-9 & 12\\02.EMI & AC.p65 d A = AD × AE = 1m × 1m = 1m2. Therefore e = 0nN(R2) dt (I0sint) 1 Therefore, the emf induced in circuit I is e = 1m2× 1 Ts–1 = 1 Tm2s–1 = 1V  0n N (R2) I0cost 1 (i) Clockwise B,  Area enclosed by circuit II is (ii)Anticlockwise B,  (iii) Anticlockwise A,  A = EB × EF = 0.5 m × 1m = 0.5 m2 (iv) Clockwise A,  2 Induced emf in the loop = B v – B v = (B – B )  Induced emf in circuit II is 12 1 2 MLN OQP FHG b gIJKv e = 0.5 m2 × 1 Ts–1 = 0.5 V 2 Let I and I be the induced currents in circuits I and 12 II respectively. From Lenz's law, the directions of these currents must be such that they oppose the 0I  0I (v) (a) = 0I a2v 2x 2(x  a) 2 x xa increase in currents. In other words, the directions of the current in circuits I and II must be such that 32

JEE-Physics 2  10 7  50  (0.1)2  10 50 1 1 . Here VL1 = VL2 = = µV 0.2  0.3 3 8 . (i) Maximum Current  L1 di1 = L2 × di2 dt dt I = emax = NBA  50  0.3  2.5  60  4.5A max R R 500  L1i1 = L2i2 (ii) Flux is maximum when plane of coil is at F IF IE L2 90° to the magnetic field.Flux is zero when GH KJHG KJBut i1 + i2 = R = i  i1 = L1  L2 E R plane of coil is at 0° to the magnetic field. (iii) Yes it will work because  related to coil F IF IHence continuous in change. GH JKGH KJ(i) Current in L1 = L1  L2 R L2 E 9 Induced emf in primary coil E E= d d (ii) Current in R = p = dt (0 + 4t) = 4 volt R dt 1 2 . (i) At t=0 Induced emf in secondary coil L act as open circuit Es Ns Ep = Np F I F INs 5000 HG JK HG JK = E Np E= 50 (4) = 400 volt s p 1 0 . (i) 10 So i = 2  3  i =2 A (ii) After some time L act as short circuit I1A Power in R i  10  10 4 1 2  6 3 22  i = 2.5A 7 72 63  2  VA 3  2  VB 0 P = I2R =  2 1 3 . (i) After 100 ms wave is repeated so time 2 1 V –V B = 4V P = 24.5 W 1 period is T = 100 ms. f = T = 10 Hz A (ii) Average value = Area/time period (ii) i0  i0 1  et /   3 (1 / 2)  100  10 i0  2  3  0.6A = (100) = 5 volt L 10  103  2  103 sec b g1 4 . Irms =  I1 cos t  I2 sin t 2    = R eq 5 Node-6\\E:\\Data\\2014\\Kota\\JEE-Advanced\\SMP\\Phy\\Solution\\Unit-9 & 12\\02.EMI & AC.p65 (A) i = 0.6 A = I12 cos2 t  I22 sin2 t  2I1I2 sin t cos t 0 (B) i0  i0 1  e t /    1  1  et/ 2 2 GFH KIJ FGH KIJ b g=11 I12  I22  e t /   1  t /   n2 I12 2  I22 2  2I1I2 0 = 2 2 1 5 . (i) 2 2  3.14 1 t=n2  t = 2 × 10–3 × 0.693   = = 100 = 0.01 s  t = 1.38 × 10–3 sec (ii)  628 Energy stored in L : 1 6 . (i) 60° (30°– C– 30°) 1 Li2 1  0.6 2 Impedance 2 2  2  H=   10  10 3 V0 110 Z = I0 = 5 = 22 (ii)  H = 4.5 × 10–4 J 33

JEE-Physics Power factor OR F I 1 F I25 5 4 GH KJ= cos = cos 3 = 2 (lagging) HG KJH = Pt = (VI cos ) t =  × 10 = 4000 22 5 1 7 . R = 100  ; f = 1000 Hz,  = 45° GFHUse IJKR tan  = X L J cos   R Z  XL = R tan  = 100 × 1 =100  (ii)Wattless current = Irms sin  100 F I5  XL = 2fL = 100  L = 2  3.14  1000 GH JK= 2 = 0.0159 H = 15.9 mH 3 = 2.12 A sin   X L  X C  3 ×5 Z5 1 8 . (i) X is resistor and Y is a capacitor 2 2 . Impedance of circuit (ii) Since the current in the two devices is the same Z  R2  (X L  X C )2  (45)2  (4  4)2 = 45 (0.5A at 220 volt) When R and C are in series across the same Total current in circuit I  V  90  2A voltage then Z 45 220 Vrms (Reading of ammeter) R = XC = 0.5 = 440  Irms = R2  X 2 Voltmeter connect across L and C C so reading of voltmeter = V – V LC Now X = X  V =V L C L C 220 220 So reading of voltmeter = 0 = = = 0.35A (440)2  (440)2 440 2 2 3 . Power dissipation 1 9 . (i) resistor (ii) inductor = V I cos  V. V . R V2R ....(1) ZZ = Z2 2 0 . (i) At resonance condition X = X  cos   R ,I  V LC Z Z  11 V = 100 V, R = 10 Z = R 2  (X L )2 = 10 2  0L = 0C     0 = LC  X L  L  2   50  1   (ii)  cos  = R = R = 1  X  1 0 10  ZR L    = 0° No, It is always zero. 2 1 . (i) Impedance of the circuit Put all these value in eq (1) Z = R2  (X L  X C )2 = (4)2  (7  4)2 = 5 P= 100 100 10 Node-6\\E:\\Data\\2014\\Kota\\JEE-Advanced\\SMP\\Phy\\Solution\\Unit-9 & 12\\02.EMI & AC.p65 current flow in ckt. loss (10 2 )2  500 watt Vrms 25 / 2 5 2 4 . Mutual inductance Z = 2A Irms = s B(r2 ) 30I (r2 ) = 30r2 5 Ip I a I a  M= =  Heat developed IGFH KJI=2Rt = 2 25. E=   = B 0 k.i  (v x i  v y j) rms B.L  v 5 × 4 × 80 = 4000 joule  B 0 k.v y k 2 = – vB 0 y 34

JEE-Physics 26. R = 3 6  2  4 EXERCISE –IV (B) eff 3  6 1.  dr  6 3 2 de  r Brdr  1 Br2 02  (i) e   BLv 2 1 2  = = 1A I= R eff R eff  Rt   Rt  4 (ii) I  I0 1  e L B r 2 1  e L  F = ILB = 1 × 1 × 2 = 2N    2R  B Lv T  L B mgR Now torque required for power loses  R  L2B 2 27. mg = ILB =  v=  B 2r 42  T  P 4R   I2R  6 6 2 8 . 6 4 4 4 4 P B2r 42 B2r4 t=0 10V t= 1    4R = 4R Torque required to move the rod in circular motion against gravitational field 2  m g   r co s  = 1  2 mgr cos t 10 10 5 I =  1A I=  A 2 1 64 2 62 4 The total torque I1  0.8   1  2 = B2r4 + 1 mgr cos t I2 4R 2  Direction of torque : clockwise E  L  10 L=10H R=2 R  R  I = =5 29. I= [1  e  t /  ]   2. 02  Charge passed through the battery Magnetic energy 1 LI2 2 E E Idt  R 10V R0 The current in the circuit for one forth of magnetic    e t /  0 1  et/ dt  t  E EL energy I = I0 R eR2 2  (   / e)  (0  )   Rt   Rt   1  e L I0 1  e L But I = I0  2 = I0   Node-6\\E:\\Data\\2014\\Kota\\JEE-Advanced\\SMP\\Phy\\Solution\\Unit-9 & 12\\02.EMI & AC.p65 30. 1 L i 2   1 1 L I 2  i= I  et/5  1 t  n2  2  4  2  2  et/5 = 2  25 t = 5n2 = 5 × 0.693 = 3.47 s where i I  Iet /   L t  n2 3. Refer to figure.  2 R 35 L v BIL t  n 2 idt  (Iet /  )dt   Charge flown = T 0  I et /  0n2  I  1  IL y ma  2   mg x 2R z

JEE-Physics Let v be the velocity of the rod along the positive Due to B, the flux through the loop is x–direction at an instant of time and let the magnetic field B act perpendicular to the table along the    BA  B a2 = 0M × x 2  0Ma2 positive y–direction. 2 x3 2x2 The emf induced in the rod is e = BLv. Therefore, the induced current is Induced emf in the loop is d dx d d e=– = =v dt dt dx dx e BLv I=  = 0Ma2v d 1 3 0 M a 2 v 2 dx  x3  = 2 x4 RR  Induced current int he loop is The rod of length L carrying a current I in magnetic field will experience a force F = BIL ...(2) I= e 3 0 M a 2 v along the negative x–direction. Since the rod is R 2 x4R massless, this force will also be equal to the tension T in the string acting along the positive x–direction, i.e. Magnetic moment of the loop is T = F = BIL  M = I × area enclosed by the loop = I a2 Let a be the acceleration of mass m moving in the 0 downward direction, then ma = net force acting on F 3 0 M a 2 v m= mg – T = mg – F or a = g – ...(3) = x4R m 2 Using (1) , (2) and (3), we have B L B  BL2 v B 2L2 v a =g – =g– =g – ....(4) m mR mR (i) The rod will acquire terminal velocity v when Potential energy t  a = 0. U  M  B  MB cos 180  U  M B Putting a =0 and v = v in eq. (4) t we have 0 = g– B2L2 v t or v = mgR = 3 0 Ma2 V  0  M  mR t 2 Rx4  2 x3  B 2L2 (ii) When the velocity of the rod is half the terminal 3 0M2a2 V 1 4 R x7 vt mgR U   2 = 2B 2L2 velocity i.e. when v= then from equation (4), we have F   dU F  21 20 M 2 a 4 V dx 4 Rx8 a=g– B 2L2 v t / 2 = g – B 2L2 × mgR gg This force is caused by the moving magnet. From mR 2mR B 2L2 =g– 2 =2 Lenz's law, this force opposes the motion of the magnet. 4 . Refer to figure y A 10 I1 P  Loop Magnet 10 I2 I3 O 10 Bv 10 5.  10 a x B Bv v 10 z Node-6\\E:\\Data\\2014\\Kota\\JEE-Advanced\\SMP\\Phy\\Solution\\Unit-9 & 12\\02.EMI & AC.p65 I BL2 50  20  0.1  0.1  5V R =  22 The magnetic field at distance x on the axis of a magnetic of length 2 and dipole moment M is given For the circuit I + I + I = 0 123 by  V  5  V  5  V  0  V  10 10 10 10 3 Mx x2  2 2  B = 0 V1 2  I =  A 10 3 3  x >>, we have B = 0M 2 x 3 36

JEE-Physics  (ii) B is decreasing with B so induced current will try to increase (B) (Lenz's law). So direction of current 6 . Initial current through L for switch in position = R1 anticlockwise 2 (i) For upper half area L2 1    Energy stored in inductor = 2 L   L2 R  2 R 2 1 1  Heat developed across R= 2 R 2 e  AdB =–  2  2   0.87 2 1   2  dB r dt 7 .  = E2r = r2  E = k e = 1.74 Volt dt 2 Total emf in circuit  For an electron a= f  eE  erK E = 1.74 + 20 E = 21.74 volts m m 2m 1 4 . e  Bv , de = (dB)vdx 8. = MI    d = M(2kt) dt  Q = T  dt  kMT2 0R R dB R d(0.2t) R R 9 . E2R = R2  E   (0.2)  dt 2 dt 2 10  Torque on coil = qER = q  R  R  I  mR2 induced current will be clockwise  10  2a  Angular velocity attained in AB : 0 0 I vdx  e1  0 Iv n2 e= 2 x 2 1  = T   qER  t = 40 rad/sec 3a 0 I  0Iv n 3  mR2  2a 2 x 2 2 in e2  vdx  e2 CD : 10. For LC circuit q  Ldi 0 0 Iv n 3  i  e  0 Iv n3 c dt So net e=e2–e1 2 4 R 2R 4  1 q = Q sin( t + )    Force F = iB, dF = i (dx) dB)  0 LC  Work done dW = dF.x. At t=0, q = Q0 Q = Q sin    = /2 2a 0 0 0  0 I   t  W = i  dx   2x   x  q  Q 0 sin  LC 2   1 1 .   BA  Kt  C a2   W= i0I 2a dx = i0I 2a  a 2 0 2 Charge q    2  1 W = 0I2 Va n 3 RR 42R 4 Node-6\\E:\\Data\\2014\\Kota\\JEE-Advanced\\SMP\\Phy\\Solution\\Unit-9 & 12\\02.EMI & AC.p65 At t=0; 1  C a2  R3 16. M C A i1 At t  2 =0 B R1 M i2 R1 k C B D  Ca2  C a 2 R2 R3 i  R  R So q  0   q  E R2 i  E  E R1  R3 R1R 3 R1R2  R2R3  R1R3 1 3 . B= 0.042 – 0.87 t R1  R3  R2 20V i  R1  ER1 R1  R3 R1R2  R2R3  R1R3 37

JEE-Physics 17. i Emax, Emax  dx i2 i2 Now dt =v. Using (1), we have i2 L C1+C2 L d I R  2x  2 m vI C1 C2 F = mv   Bd  Bd  = net dt ( I, R, B and d are constants) E max  E max 1 XL  XC     = 2m I2 (R+2x) [Use equation (1)]  B2d2 imax    L    C1  C2  Using this in equation (2), we get  i1 max  C1 i max 2 m I2 C1  C2 F = BId + B2d2 (R+2x)...(3) (ii) Work done in time dt = Fdx. C1 Imax Therefore, work done per second i.e., power is       i1 max  C1  C2  L  1  d Fdx   C1  C2  P = (Fdx) = = Fv dt dt  and i2  C2 i max Using Equation (1), we have C1  C2 max C2E max FI R  2x      i2 max  C1  C2  L  1  P=   C1  C2  Bd Heat produced per second is Q = I2 (R+2x) 1 8 . i = 3 + 5t, R=4, L = 6H Q I2 R  2x Bd IBd  Ratio P = FI R  2x = F ...(4)  E = iR +L di = (3+5t) 4 + 6(5) dt Using (3) in (4), we get E = 42 + 20t Q B3d3 19.(i) Let v be the velocity of the rod MN at an instant of P  B3d3  2mI R  2x  time t when it is at a distance x from R. Then, the 2 0 . induced emf at that instant is e = Bvd. Since  is Let the magnetic field be perpendicular to the the resistance per unit length of each wire, the total plane of rails and inwards  . if V be the terminal resistance in series with R at that instant is velocity of the rails, then potential difference across x + x = 2 x. E and F would be BVL with E at lower potential Thus the total resistance of the circuit at time t and F at higher potential. The equivalent circuit is = R +2 x. shown in figure (2). In figure (2). Hence the current in the circuit is B induced emf Bvd R1 C A I= = = constant (given) total resistance R  2x (1) E F F I R  2x D Node-6\\E:\\Data\\2014\\Kota\\JEE-Advanced\\SMP\\Phy\\Solution\\Unit-9 & 12\\02.EMI & AC.p65 B v  Bd ...(1) R2 The magnetic force acting on the rod is F = BId R1 m i1 i e=BVL F directed to the left. (2) E R2 The net force acting on the rod is i2 F = F – F =F – BId ...(2) net m F is the force experienced by the rod MN at time net t when its velocity is v at a distance x from R. From Newton's law dv dv dx e F = ma = m =m i1 = R1 ...(1) net dt dx dt 38

JEE-Physics e field is zero, the magnetic flux through the element i2 = R 2 ...(2) d  BdA cos   BdA cos 00  BdA The total magnetic flux linked with the loop is Power dissipated in R1 is 0.76 watt ...(3)    BdA B = B0  y  kˆ Therefore ( ) ei1 = 0.76 watt ...(4)  a  Similarly ( ) ei2 = 1.2 watt depends upon y, we have Now the total current in bar EF is i = i1 + i2 (From E to F) ...(5) y a  y  ya B0 Under equilibrium condition, magnetic force (Fm) y  a  2 on bar EF = weight (Fg) of bar EF    B = B0 = ydy = i.e., Fm = Fg or iLB = mg ...(6) 0 ady y y2 y a = B0 [(y+a)2–y2]......(1) y 2 Fm i  Induced emf is E F e = – d   B0 d [(y+a)2–y2] = – B0 Fg dt 2 dt 2 2 y  a dy  2y dy  = – B0 a dy dt  dt From equation (6) dt mg 0.2 9.8 The induced current is I = e  B0a dy ......(2) i = LB 1.00.6 A or i = 3.27 A R R dt Multiplying equation (5) by e, we get Since the magnetic field B points in the positive z–direction, it follows from Lenz's law that the ei = ei1 + ei2= (0.075 + 1.2) watt direction of the induced current will be along the (From equation 3 and 4) = 1.96 watt negative z–axis. Thus the current in the loop will flow in the counterclockwise direction as shown in figure. 1.96 1.96 e= volt = V or e = 0.6 V i 3.27 Ox But since e = BVL EI F y e 0.6 a V = BL = 0.6 1.0 m/s= 1.0 m/s Hence, terminal velocity of bar is 1.0 m/s. (y+a) mg Power in R1 is 0.76 watt G H y e2 e2 0.62 (ii) The Lorentz force acting on a current element I dI  0.76 = R1  R1 = =   0.47 of length d in magnetic field B is given by 0.76 0.76 R1 = 0.47  F = I  d  B Similarly R2 = e2 = 0 .6 2 L 1.2   0.3 Now, the forces acting on sides EG and FH of the 1.2 loop are equal and opposite. Hence they cancel each other. The forces acting on sides EF and GH are in R2 = 0.3  opposite directions but their magnitudes are different since B depends upon y. Thus, force acting on side 2 1 . Refer to figure z Node-6\\E:\\Data\\2014\\Kota\\JEE-Advanced\\SMP\\Phy\\Solution\\Unit-9 & 12\\02.EMI & AC.p65 O x  EF is  aˆi  B0 y kˆ  B F   a    E I dy FEF = I  = IB 0 yˆj H Ga Similarly, force acting on side GH is y  FGH = I  aˆi  B 0 y  a kˆ     a   = – IB0 y  a ˆj (i) Consider a small element of the loop of width dy  and side a. The area of the element dA = ady. Since the area vector and the magnetic field vector point Therefore, the net force acting on the loop is in the same direction, i.e. the angle  between the normal to the plane of the loop and the magnetic  Fnet = FEF + FGH = IB0 yˆj  IB0 y  a ˆj  IB0aˆj Substituting the expression for I from equation (2), 39

JEE-Physics we get EXERCISE –V(A)  B 2 a 2  dy ˆj 0 dt Fnet = –  R  1.  The negative sign ˆj indicates that the force acts along the negative y–direction. 11 1 1  (iii) As force Fnet is directed along the negative In parallel L L1 L2 L3 y–axis and the gravitational force mg is along the positive y–direction, the total force on the loop in the downward (positive y) direction is  B 2 a 2  dy 1  1 1  1  3 1 as L=1H 0  L 333 3 F = mg –  R  dt dy 2. A B where = v, the speed with which the loop is v x xxxxxxx dt x xxxxxxx C falling downwards. From newton's second law, the x xxxxxxx equation of motion of the loop is x xxxxxxx x xxxxxxx dv x xxxxxxx F = mass × acceleration or F = m x xxxxxxx x xxxxxxx dt x xxxxxxx x xxxxxxx D  B 2 a 2  The equivalent circuit of the conductor will be  0   m dv v dt =mg – R AB or dv = g –  B 2 a2  v or dv =g –kv ....(3) e=Bvl dt  0 R  dt m B 2 a 2 0 where k = DC mR The core of transformer is laminated so as to reduce dv 4. the energy loss due to eddy currents. 6. Rearranging expression (3), we have g  kv = dt 7. Intergrating, we have I1=2A I2=–2A v dv t 1 g  kv t=0.05, e= 8V  k loge   =t  g  kv  =  dt or –  g  0 0 e  Ldi  L  2A  2 A  dt  0.05  which gives  v = g 1  ekt gmR   B 2 a 2 t   4  k 1  0   0 .0 5  =   e x p   8  L 2 2  B 0 a mR This is the required expression for the speed v of L = 2 × 0.05 = 0.1 H the loop as a function of time t. When the loop Number of turns = n Node-6\\E:\\Data\\2014\\Kota\\JEE-Advanced\\SMP\\Phy\\Solution\\Unit-9 & 12\\02.EMI & AC.p65 Resistance of coil = R acquires terminal velocity v0 no force acts on it. Resistance of galvanometer= 4R Hence its acceleration dv/dt is zero. e Using this in equation (3), we have Induced current = g mgR R 0 = g –kvt or vt = = k B 2 a 2 0 (dA) d   0 I  dx     2  1  1  2 x   t  Resitance of 22. flux =BA; d = (dB) circuit  ab 0I ab 1 dx    n W2  nW1  1 2 a x  t  5R   d = a 0 I nx aa b   0 I n a  b  n  W2  W1  2 2  a   5Rt  40

JEE-Physics  1 2 . =10t2–50t +250 8. d The flux associated with coil = NBAcost dt =20t–50  = B r2 cos t d 2 e = dt =50–20t e(t=3) = 50 – 20 × 3 = – 10V d Br2 sin t e=  1 3 . L = 100 mH; R=100; E=100V dt 2 L Pinst.  e2  Br2 sin2 t R   2  R R   Pav  sin2 tdt Pdt B2r42 dt AB  E dt 4R T sin2 tdt  1 T Long time after, current in the circuit is 02 E I0  R  1A B22r42 1 Hence Pav  4R 2 On short circuiting R t I  Ioe L B 2 2 r 4 2 I  1e 100 1 0 3 Pav  8R 1 0 0 1 0 3  Br2 2 1   1  A  e  Pav  8R 9 . The emf developed across the ends of the pivoted L rod is B2 e 2 rad R 5 , B = 0.2 × 10–4T, L=1m sec e  0.2  104  12  5 =5 × 10–5 = 50V 1 4 . RMS value of electric field = 720 N/C 2 10. Isteady = V  2  1A ;  Rt Peak value of electric field = 2  720 N/C R2 I  I0 1  e L  the average total energy density of electromagnetic R 2 2000 wave = 1 0 E 2 ; u av  1 0 E 2 L  300  103  300 2 0 2 0 Node-6\\E:\\Data\\2014\\Kota\\JEE-Advanced\\SMP\\Phy\\Solution\\Unit-9 & 12\\02.EMI & AC.p65 On solving we get I0  Rt uav = 4.58 × 10–6 J/m3 2  I0 1  e L  1 R t n2 1 5 . Inductance of the coil L = 10 H 1e L t   0.1s Resistance of the coil R=5 2 L/R As R-L is connected across battery, hence nature of current that will flow through the circuit will be 1 1 . The vertical arm of the both tubes will becomes a transient current, i.e., I=I0(1–e–t/) battery of emf Blv. L 10 V5 where   =2s and I0  R  5 =1A Blv Blv R 5 hence I=1(1–e–2/2) ; I = (1–e–1)A The emf induced in the circuit is 2 Blv. 41

JEE-Physics 16. M  0N1N 2 A  2.4   104 H 2 3 . R1=0.2  R2=3×10–3 1 7 . Equivalent resistance across L(by short circuiting battery) = Req = 2 then current through L i = i0(1 – e–Rt/L) 4 E 12 0.15 where R = 2 L = 10 H, i0 = R  2 = 6A  R 2 R 2 i = 6(1 – e–5t) .... (i) 0 1 2 Mutual inductance = 2 ( R 2  X2 )3/2 p.d. across L is 1 di 4 d   0  . 2 2 R 2 R 2 i  4  1 2 so flux through trigger coil VL = L dt = × [6(1 – e–5t)] = 12 e–5t ( R 2  X 2 )3/2 10 1 dt This question contains Statement–1 and Statement– 10 7  (3.14 )2 (2  10 –1 )2 (3  10 –3 )2  2  2 2. Of the four choices given after the statements,  ((0.2)2  (0.15)2 )3/2 choose the one that best describes the two statements. ALTERNATING CURRENT 2 4 . On drawing the impedance triangle; we get 1 8 . At t = 0 inductor behaves as broken wire then V V Z  R2  XL2 i XL=L R2 R2  R at t =  Inductor behaves as conducting wire V V RR The power factor cos   Z  R 2  2L2  i  R1 2 6 . Let Qmax = Qmax R2 R1R2 / R1  R2 1 Q2  1 LI2 ...(i) 2C 2 1 9 . Circuit can be reduced as [Given that energy stored in capacitor= Energy stored in conductor]. i1 R iR According to the conservation of energy we know i2 e 2vB R i  3R / 2 3R e = vB 1 Q2  1 Q2  1 LI2 max that i vB 2 C 2C 2 i1  i2  2  3R 1 Q2 1 Q2 1 Q2 max   [from eq. (i)] 2 0 . e = Bv = (5 × 10–5) (2) (1.50) = 0.15 mV 2 C 2C 2C Q2  2Q2 Node-6\\E:\\Data\\2014\\Kota\\JEE-Advanced\\SMP\\Phy\\Solution\\Unit-9 & 12\\02.EMI & AC.p65 max 2 1 . B = 0.3 × 10–4 wb/m2 CC BH Q2 max Q2  C WE Q  Qmax V 2 e = vB= 5 × 20 × 0.3 × 10–4 = 3mV 2 7 . A DC meter measure the average value and the average value of AC over one full cycle is zero. 2 2 . Due to conducting nature of Al eddy currents are Hence, DC meters can't measure AC. produced 42

JEE-Physics 2 8 . Voltage across LC combination = |VL–VC| 3 5 . Given that Voltage across LC combination = |50–50|=0V E=E0sin(t) and I\\I0sin(t–/2) 1  2 9 . The resonant frequency r  LC The phase difference between E and I is . For 1=2 1 1 2 L1C L22C Power dissipated in an AC circuit P = Erms Irms cos = 0 On squaring both sides, we get So, power dissipated for this situation where phase  1 1 ; L2 1 ; L2  L1  L1  2 difference between voltage and current is will  L1C L2 2C 2 2 be zero. 3 0 . In order to transfer maximum power the generator should work at resonant frequency, i.e., C should 3 6 . tan   X L  X C be such so that R tan 30° = XC R R  XC = 3 f  1  f2  1 XL R 2 LC 4 2 LC R 3 tan 30° =   XL = 11 XL = XC  Condition for resonance  C  42Lf2  4  10  10  2500 So  = 0°  C  106 F  1F P = VI cos0° R V 2 2202 cos   P   242W 31. Power factor= R 200 Z cos   12  4 =0.8 3 7 . Energy is shared equally between L and C at 15 5 t= T , 3T ...... where T  2  2 LC 3 2 . If resistance will be the part of circuit, phase 88  difference between voltage and current can not be so tT 2 LC  LC    . 884 2 Node-6\\E:\\Data\\2014\\Kota\\JEE-Advanced\\SMP\\Phy\\Solution\\Unit-9 & 12\\02.EMI & AC.p65 3 3 . The voltage across L =VL = IXL Voltage At resonance, current = Resistance ; 100V I= 103   0.1 Also at resonance XL=XC 1 1 104 XC = C  200  2  106  4  104 VL  IX L  0.1  4 =250V 3 4 . Magnetic flux associated with rotating coil = NBAcost =NBAcost d dt =–(NBA)sin(t) e = e 0 s i n t Maximum value of emf generated in coil = NBA 43

JEE-Physics EXERCISE –V-B and L  L 8.4  103  1.4  103 s R  1 . A motional emf, e = Bv is induced in the rod. Or we 6 can say a potential difference is induced between the and i = 1A (given) two ends of the rod AB, with A at higher potential and Substituting these values in equation (i), we get B at lower potential. Due to this potential difference, t = 0.97 × 10–3s there is an electric field in the rod.  t = 0.97 ms  t 1ms AB 5.  d dB ××××× dt S  E.d   ××××× dt × × × × v× dB a2 dB ××××× E (2r) = a2 for r > a  E  B dt 2r dt 2 . Magnetic field produced by a current i in a large 1 square loop at its centre.  Induced electric field  r ii dB r dB B  say B =K For r < a; E(2r) = r2 E  dt 2 dt E  r LL  Magnetic flux linked with smaller loop, a dB At r = a, E = = B.S     i  (2 )  K L  2 dt E Therefore, variation of Therefore, the mutual inductance a dB 1 2 dt r E E with r(distance from  2 2 Er centre)will be as follows : M  K M  r iL L r=a 3 . For understanding, let us assume that the two loops The equations of I (t), I (t) and B(t) will take the are lying in the plane of paper as shown. The 6 . 12 current in loop 1 will produce * magnetic field in following forms : loop 2. Therefore, increase in current in loop 1 will I (t) = K (1–e–k2t)  current growth in L–R circuit 1 1 produce an induced current in loop 2 which produce B(t) = K (1–e–k2t)  B(t)  I (t); B =  Ni in case of 3 10  magnetic field passing through it i.e., induced solenoid coil and 0Ni in case of circular coil i.e., current in loop 2 will also be clokwise as shown in 2R the figure. Bi  I (t) = K4e–k2t 2 1 2   e2 and e2  dI1 : e2  M dI1  F I2 (t) R dt dt  F Perpendicular to paper outwards Therefore the product I (t) B(t) = K e5 –k2t (1–e–k2t). Perpendicular to 2 paper inwards The value of this product is zero at t=0 and t=. Node-6\\E:\\Data\\2014\\Kota\\JEE-Advanced\\SMP\\Phy\\Solution\\Unit-9 & 12\\02.EMI & AC.p65 FF Therefore, the product will pass through a maximum value (K : K : K : K and K are positive 1234 5 constants and M is the mutual inductance between The loops will now repel each other as the currents at the coil and the ring). the nearest and farthest points of the two loops flow in the opposite directions. The corresponding graph will be as follows : 4 . The current–time (i–t) equation in L–R circuit is given I2(t) B(t) I2(t)B(t) by [Growth of current in L–R circuit] i = i (1– e–t/L) ...(i) 0 i0 V 12 t tt where   =2A R 6 44

JEE-Physics 7 . Electric field will be induced in both AD and BC. 3 . When resistivity is low current induced will be more; therefore impulsive force on the ring will also be 8 . When current flows in any of the coils, the flux linked more and it jumps to higher levels. [But for this mass with the other coil will be maximum in the first case. should be either less or equal to the other] Therefore, mutual inductance will be maximum in case (a). 9 . When switch S is closed magnetic field lines passing Comprehension#1 through Q increases in the direction from right to 1 . Charge on capacitor at time t is : left. So, according to Lenz's law induced current in Q q = q (1– e )–t/ 0 i.e., I will flow in such a direction, so that the Here, magneQ1tic field lines due to I passes from left to  q = CV and t = 2 right through Q. This is possQi2ble when I flows in 0 anticlockwise direction as seen by E. OppQo1site is the q = CV (1– e )–2/ = CV (1– e–2) case when switch S is opened i.e., I will be 2 . From conservation of energy, Q2 clockwise as seen by E. e2 1 L I 2m ax 1 CV2 C 1 0 . Power P = 2  2  Imax  V e L R  d 3 . Comparing the LC oscillations with normal SHM we Here, e = induced emf =  dt  get,  dB  d2Q  2 Q Here, 2 = 1 d2Q where  = NBA e = –NA  dt  dt2 LC  Q = –LC dt2 1 Subjective R  r2 1 . Magnetic field (B) varies with time (t) as shown in Also, B(T) where R = resistance, r = radius,  = length N2r2 P1 0.8   P2  P 1 figure. 1 1 . As the current i leads the emf e by 0 0.2 0.4 0.6 0.8 t(s)  XC dB 0.8 , it is an R–C circuit. tan  =   4T / s 4R dt 0.2 1 Induced emf in the coil due to change in magnetic or   C  CR = 1 flux passing through it, tan R d dB 4 e   NA As  = 100 rad/s dt dt The product of C–R should be 1 s–1. Here, A = Area of coil = 5 × 10–3 m2 N = Number of turns = 100 100 Substituting the values, we get Node-6\\E:\\Data\\2014\\Kota\\JEE-Advanced\\SMP\\Phy\\Solution\\Unit-9 & 12\\02.EMI & AC.p65 1 2 . Polarity of emf will be opposite in the two case while e = (100) (5 × 10–3) (4) V = 2V entering and while leaving the coil. Only in option (b) polarity is changing. Therefore, current passing through the coil 1 3 . In uniform magnetic field, change in magnetic flux is e zero. i= (R = Resistance of coil = 1.6 ) Therefore, induced current will be zero. R 14 . As area of outer loop is bigger therefore emf 2 induced in outer loop is dominant and therefore  i  1.6  1.25A according to lenz law current in outer loop is Note that from 0 to 0.2 s and from 0.4s to 0.6s, Anticlockwise and inner loop is clockwise magnetic field passing through the coil increases, while during the time 0.2s to 0.4s and from 0.6s to MCQ 0.8s magnetic field passing through the coil decreases. Therefore, direction of current through 1 . Electrostatic and gravitational field do not make closed loops. 45