Title 19. Heating and Chemical Effects of current Hard Ans.pdf ✅

1. (a) If suppose resistance of the coil is R so resistance of it’s half will be 2 R . Hence by using R H R V t H 1 2 =    1 2 / 2 = = = R R R R H H Half Full Full Half Note: In general if coil is divided in n equal parts then heat produced by each part will be n times of the heat produced by coil it self i.e. H = Nh 2. (b) By using R V t H 2 = and 2 r l A l R   =  =  ; 2 2 2 l r H l V t r H =     on doubling both r and l heat will be doubled. 3. (b) By using R V t H 2 =  H J 5 2 14.4 10 6 (120 ) 10 60 =    = 4. (c) By using R R A consumed P V V P          = 2  P Watt Consumed 500 456 .6 115 110 2   =      = So % drop in heat output 100 8.6% 500 (500 456 .6) 100  = −  = − = Actual Actual Consumed P P P 5. (d) By using R P R V P i R 1 2 2 = =    (1 4 10 800 ) 500 (1 4 10 200 ) (1 ) (1 ) 4 4 1 2 2 1 2 2 1 +   +    = + + = = − − t P t R R P P    W P 611 1.32 500 1.08 2 =  6. (d) By using . 2 R V t H = Here volume of water is same. So same heat is required in both cases. Resistance is also constant so V 2 t = constant  2 1 V t   20 min 4 1 220 5 110 2 2 2 2 1 2 2 1  =  =       =         = t V t V t t 7. (c) By using R V t H 2 = where l V t A H A l R   2 =  = . Since volume is constant so H is also constant so t  l which gives 1 2 1 2 l l t t =  10 min 3 2 15 2 1 1 2 =  t = l l t 8. (b) Illumination = PConsumed R V 2 = . Initially there were 40 bulbs in series so equivalent resistance was 40 R, finally 39 bulbs are in series so equivalent resistance becomes 39 R. Since resistance decreases so illumination increases with 39 bulbs. 9. (b) In series 1 2 1 2 P P P P PConsumed + =  PConsumed 66W 300 100 200 =  = 10. (d) Comparing the given equation of thermo e.m.f. with E =  t + 2 2 1  t we get  = 40 and 10 1  = − . By using tn tn C o = −  = 400   . 11. (a) As we know thermo electric power dT dE S = . Hence by differentiating the given equation and putting 0 2 1 T = T we get 0 2 1 S = KT . 12. (a) Mass of hydrogen in 11.2 litres of hydrogen M 2 1gm 22.4 11.2 22.4 11.2   =        =      = We know that 1 gm of hydrogen is equal to 1 gm equivalent wt. of hydrogen. It means that 11.2 litre of hydrogen at NTP represents 1 gm equivalent of hydrogen, so for liberation it requires 1 Faraday electricity. 13. (b) Number of gm equivalent 16 / 2 16 gm equivalent weight Given mass = = = 2. Hence 2 Faraday electricity is needed. 14. (c) Mass deposited = density  volume of the metal m =   A  x ...... (i) Hence from Faraday first law m = Zit ......(ii) So from equation (i) and (ii) Zit =   Ax  2.4 10 m 2.4 m 9000 0.05 3.04 10 10 1 36 ...... A Zit x 6 4 3 =  =        =  = − − −
15. (b) Remember mass of the metal deposited on cathode depends on the current through the voltameter and not on the current supplied by the battery. Hence by using m = Zit, we can say Series Parallel Series Parallel i i m m =  1 2.5gm 2 5 mParallel =  = . Hence increase in mass = 2,5 – 1 = 1.5 gm i 2A 5 10 1 = = i 5 A 2 10 2 = = 16. (d) Slope of H vs  graphs give ‘heat capacity’.  Specific heat capacity = mass heat capacity = 2 tan 45o kJ/kg/oC = 0.5 kJ/kg/oC 17. (b) Req 1 = R1 1 + R2 1 L K (9 R ) 2 eq  = L K R 2 1 + L K 8 R 2 2  Keq = 9 K1 + 8K2 18. (c) Rtotal = R for  = 1800 Two sections of resistance R/2 each are in parallel Req = R/4  Rate of heat flow I1 = 1.2 = T/R/4  0.3 = R T  = 900 two sections of resistance R/4 & 3R/4 in parallel Req = R / 4 3R / 4 R / 4.3R / 4 + = 16 3R  I2 = 3R /16 T = 3 16 (0.3) = 1.6 watt 19. (b) 1 t Q       = d KA( – ) 1 2 = 1 cal s–1 2 t Q       = 2d 2k(4A)( – ) 1 2 = d 4kA( – ) 1 2 = 4 cal s–1 20. (d) Heat current i = – KA dX dT idX = – KA dT   = −   0 T T 2 1 i dX A TdT  i = – A  2 (T T ) 2 1 2 2 −  i = A  2 (T T ) 2 2 2 1 − 21. (b) Parallel combination of cross-section area R 2 and [(2R)2 – R 2 ] = 3R 2 Req 1 = R1 1 + R2 1 with R = KA L L K .4 R 2 eq  = L K R 2 1 + L K (3 R ) 2 2   Keq = 4 K1 + 3K2 22. (c) Electrical circuit R R 2 R 2 R A B C D different between A & D = 200 – 20 = 180oC. Resistances for three parts equal  3 180 = 600C Temperature B = 200 – 60 = 1400C 23. (a) 2 1 K K = 2 2 2 1   24. (b) Q1 = k1 A L t . t1 = K2A L t . t2  2 1 k k = 1 2 t t = 4 7 25. (a) Rate of hoss of heat = 2 2 4 1 1 4 A e A e       = 2 2 2 1 R R [ e1 = e2] 26. (b) 2 1 H H = 2 2 1 1 m s m s = 2 2 2 1 1 1 V s V s   3 10V Volta 3 i1 2 10V Volta 2 i2
27. (b) 14 T 0 6 100 T − = −  1400 – 14 T = 6T  1400 = 20 T  T = 70 0C 28. (a) L x T T x T1 T 2 − − = −  T1L – T1x + Tx – TL = Tx – T2x TL = (T2 – T1)x + T1L 29. (a) KS = 1 2 1 2 K K 2K K + , KP = 2 K1 + K2 P S H H = P S K K = 2 1 2 1 2 (K K ) 4K K + 30. (a) The conductivity more in Cu. 31. (b) KP = 1 2 1 1 2 2 K K K A K A + + 32. (d) A polished is good reflector, hence bad absorber and consequently a bad radiator. 33. (b) T1 = 27 + 273 = 300 K T2 = 327 + 273 = 600 K By Stefan's law 2 1 E E = 4 2 4 1 T T = 4 600 300        E2 = 16 E1 34. (d) By using the formula 1 2 1 2 t t t t t p + = (as we discussed in theory)  t min p 8 (10 40) 10 40 = +  = Note : In this question if coils are connected in series then the time taken by the same quantity of water to boil will be ts = t1 + t2 = 10 + 40 = 50 min 35. (b) Ratio of currents 1 2 5 10 2 1 = = i i by using H = i 2 Rt  H cal sec R H R i i H H 2 / 4 5 1 10 2 2 2 2 2 1 2 2 1 2 1    =        =         = 36. (d) Both the wires are of equal length so they will have same resistance and by using R H R V t H 1 2 =    ; s P P s R R H H =  4 1 2 / 2 = = R R H H P s 37. (b) By using 100 70 70 32 32 0 100 0 e = e + e + e  100 200 64 76 70 = + + e  e 60V 100 70 = 38. (d) In the normal condition current flows from X to Y through cold. While after increasing the temperature of hot junction beyond temperature of inversion. The current is reversed i.e. X to Y through hot junction or Y to X through cold junction. 39. (a) H it (2 10 ) 2.5 (2 60) 6 10 J 6 erg 9 7 = =     =  = − −  S 40. (c) By using m = zit VF A = it  m 16 10 60 2.3 gm 1 96500 23    =  = 41. (b) dR = 4 d K dx 2  = 2 2 0 K d (1 ax) 4dx  +  R =  dR 42. (a) 4K(100 – ) = K ( – 0)  400 – 4 =    = 80oC 43. (d) H1 = H2  1 1 1 k A[T – T]  = 2 2 1 k A[T – T ]  44. (a) 4 6 5 i2 i1 i Line (2) Line (1)
Req = 1 2 1 2 R R R R + 45. (b) dx dT A t Q = −    y = 2 x = 2  – A dx dT = constant = a, t Qx   = a  t Qy   = 2a  t Q t Q t Q x y   +   =   = a + 2a = 3a  Fraction = 3 1 Q / t Q / t x =    46. (d) H1 = H2 or 1 1 R  –  = 2 2 R  –    = 1 2 1 2 2 1 R R R R +  +  47. (c) Resistance of inner cylinder R1 = 2 k1R  Resistance of outer cylinder R2 = k (4 R – R ) 2 2 2    As inner & outer cylinder in parallel Req = 1 2 1 2 R R R R + 48. (b) A × dx ×  × L = x TKAdt  H 0 xdx =   t 0 dt A TKA 2R H x dx 2 H 2 = L TKt  t = 2TK H L 2 