Title 14. Transmission of Heat Hard Ans.pdf ✅

1. (d) Rate of flow of heat will be equal in both vest and shirt  l K A t l K A t vest  vest shirt  shirt = .  vest shirt shirt vest K K     =  5 3 30 25 25 22 = − − = shirt vest K K . 2. (b) .t l Q K A  =  A l t  [As Q, K and  are constant]                  =  = 1 1 1 1 1 2 2 1 2 1 2 / 2 A A l l A A l l t t 4 2 1 = t t  3 4 12 4 1 2 = = = t t 3. (b) Let the conductivity of each rod is K. By considering the rods B and C are in parallel, effective thermal conductivity of B and C will be 2K. Now with the help of given formula Temperature of interface 1 2 1 1 2 2 K K K K + + =    C K K K K o 60 3 180 2 0 2 90 = = +  +   = . 4. (c) Rate of flow of heat along PQ l K A dt dQ PQ   =      3 ....(i) Rate of flow of heat along PRQ l K A dt dQ s PRQ 2   =      Effective conductivity for series combination of two rods of same length 1 2 1 2 2 K K K K Ks + = So l A K K K K l A K K K K dt dQ PRQ   + =  +  =      . 2 . 2 1 2 1 2 1 2 1 2 Equating (i) and (ii) 1 2 1 2 3 K K K K K + = 5. (a) K1 = 9K2, l1 = 18 cm, l2 = 6 cm, 1 = 100oC, 2 = 0oC Temperature of the junction 2 2 1 1 2 2 2 1 1 1 l K l K l K l K + + =     C K K K K o 75 8 / 12 50 0 18 6 9 0 6 100 18 9 2 2 2 2 = + = + +   = 6. (c) We can consider this arrangement as a parallel combination of two materials having different thermal conductivities K1 and K2 For parallel combination 1 2 1 1 2 2 A A K A K A K + + = A1 = Area of cross-section of internal cylinder = R 2 , A2 = Area of cross-section of outer cylinder =  (2R)2 – (R)2 = 3R 2  4 3 3 . 3 1 2 2 2 2 2 2 1 K K R R K R K R K + = + + =     7. (b) According to Ingen Hausz, 2 K  l  3 10 9 10 2 1 2 1 = = = K K l l . 8. (d) Quantity of heat transferred through wall will be utilized in melting of ice. mL x KA t Q =   =   Amount of ice melted x L KA t m   =   m 1.552 kg 5 10 334 10 0.01 1 (30 0) 86400 2 3 =      −  = − or 1552g 9. (d) Time required in increment of thickness from y1 to y2 ( ) 2 2 1 2 2 y y K L t = −   In first condition y1 = 0, y2 = 1 cm then t1  (12 – 0 2 ) In second condition y1 = 1 cm, y2 = 2 cm then t2  (22 – 1 2 )  3 1 2 1 =   t t   t2 = 3  t1 = 3  7 = 21 hrs. 10. (a) x KA t mL   =   5 0.0075 75 (40 0) 500 80   − t  =  t = 8.9  103 sec = 2.47 hr. R 2R K2 K1 0 oC 90  oC K 2K K K K K
11. (c) ( ) [(10.1) (10) ] 2926 sec 48.77 min . 2 0.005 5 0.91 80 y y 2k l t 2 2 2 1 2 2 − = =    − =   = 12. (c) From the graph it is clear that initially both the bodies are at same temperature but after that at any instant temperature of body x is less then the temperature of body y. It means body x emits more heat i.e. emissivity of body x is more than body y  x y e  e and according to Kirchoff's law good emitter are also good absorber so . x y a  a 13. (d) If a body emits wavelength 1, 2, 3 and 4 at a high temperature then at a lower temperature it will absorbs the radiation of same wavelength. This is in accordance with Kirchoff's law. 14. (a) Black bulb absorbs more heat in comparison with painted bulb. So air in black bulb expands more. Hence the level of alcohol in limb X falls while that in limb Y rises. 15. (b) According to Wien's law T m 1   and from the figure 1 3 2 ( m )  ( m )  ( m ) therefore T1 > T3 > T2. 16. (d) Area under curve represents the emissive power of the body 1 16 2000 2000 = = A A E ET T (given) ....(i) But from Stefan's law E  T 4  4 2000 2000       = T E ET ...(ii) From (i) and (ii) 1 16 2000 4  =      T  2 2000 = T  T = 4000K. 17. (c) As the temperature of body increases, frequency corresponding to maximum energy in radiation (vm) increases this is shown in graph (c). 18. (c)  m T = constant  5 1 1000 200 ( ) ( ) 2 1 1 2 = = = T T m m    5 14 5 ( ) ( ) 1 2 m m m    = = = 2.8 m. 19. (a) According to Wien's law wavelength corresponding to maximum energy decreases. When the temperature of black body increases i.e.  m T = constant  3 4 3 0 / 4 0 2 1 1 2 = = =     T T Now according to Stefan's law 81 256 3 4 4 4 1 2 1 2  =      =         = T T E E . 20. (b) From Wien's displacement law  m T = b  K b T m 7 10 3 10 2.93 10 2.93 10 =   = = − −  . 21. (d) According to Wien's displacement law  m T = b  nm nm nm K T b m 1000 2880 2.88 10 - 6 =   = = i.e. energy corresponding to wavelength 1000 nm will be maximum i.e. U2 will be maximum U1 < U2 > U3 Energy distribution graph with wavelength will be as Follows 22. (b) Wien's law T m 1   or  m  T m increases with temperature. So the graph will be straight line. 23. (a) Q =  A t T4  Q1 Q2 = A1 A2 ( T1 T2 ) 4 = πr1 2 πr2 2 ( T1 T2 ) 4 = ( 4 1 ) 2 × ( 2000 4000) 4 = 16 × 1 16 = 1: 1. 24. (d) Emissive power of a body (T) in a surrounding (T0), ( ) 4 0 4 E =  T − T or ( ) 4 0 4 Q  T1 − T  4 4 4 4 4 0 4 2 4 0 4 1 2 1 (673) (300 ) (473) (300 ) ( ) ( ) − − = − − = T T T T Q Q . 25. (c) 4 A T t Q P = =   4 1 2 1 2 2 2 2 1 2 1 2 1  =      = = = r r A A P P [If T = constant]  (nm) E U1 U2 U3 499 500 1000 900 1499 1500
26. (a) P = A  T 4 = 4r 2 T 4  2 4 P  r T or 4 2 1 T r  [As P = constant]  2 1 2 2 1         = T T r r 27. (d) Energy radiated by body per second 4 A T t Q =  or 4 l b T t Q    [Area = l  b]  4 1 2 1 2 1 2 1 2         =   T T b b l l E E 4 2 3 2 1 2 1       =    E E 64 81 2 = 28. (c) ( ) 4 0 4 T T mc eA dt dT = −  ( ) ( ) (6 ) 4 0 4 3 2 T T a c e a −  =    For the same fall in temperature, time dt  a cm cm a a dt dt 1 2 1 2 1 2 = =  dt2 = 2  dt1 = 2  100 sec = 200 sec [As A = 6a2 and m = V   = a3  ] 29. (b) ( ) 4 0 4 T T mc e A dt dT = −   Rate of cooling r r r R 1 3 4 4 3 2        1 2 2 1 r r R R = But according to problem m1 = 3m2           =    3 2 3 1 3 4 3 3 4 r r  3 2 3 r1 = 3r  1 / 3 1 2 3 1       =         r r  Ratio of rate of cooling 1 / 3 2 1 3 1       = R R . 30. (a) (T T ) mc eA dt dT 4 0 4 −  = ( ) 4 0 4 T T V c eA = −    Rate of cooling R  A [As masses are equal then volume of each body must be equal because materiel is same] i.e. rate of cooling depends on the area of cross-section and we know that for a given volume the area of cross- section will be minimum for sphere. It means the rate of cooling will be minimum in case of sphere. So the temperature of sphere drops to room temperature at last. 31. (b) (T T ) mcJ A dt dT 4 0 4 −  = [In the given problem fall in temperature of body dT = (200 −100) = 100 K Temperature of surrounding T0 = 0K, Initial temperature of body T = 200K] (200 0 ) 3 4 100 4 4 4 3 2 = − r c J r dt      s r c 72 7 s ~– r c 80 7 10 48 4.2 . r c 10 s 48 r c J dt 6 6        =    =   = − − [As J = 4.2] 32. (c) Q =  A t (T4 – T0 4 ) If T, T0,  and t are same for both bodies then Qsphere Qcube = Asphere Acube = 4πr 2 6a2 ..(i) But according to problem, volume of sphere = Volume of cube  4 3 πr 3 = a 3  a = ( 4 3 π) 1/3 r Substituting the value of a in equation (i) we get Qsphere Qcube = 4πr 2 6a 2 = 4πr 2 6{( 4 3 π) 1/3 r} 2 = 4πr 2 6( 4 3 π) 2/3 r 2 = ( π 6 ) 1/3 : 1 33. (c) According to Newton's law of cooling rate of cooling depends upon the difference of temperature between the body and the surrounding. It means that when the difference of temperature between the body and the surrounding is small then time required for same fall in temperature is more in comparison with the same fall at higher temperature difference between the body and surrounding. So according to problem T1 < T2 < T3. 34. (d) According to Newton's law of cooling       − +  −      2 1 2 1 2 t For first condition       − +  − 30 2 80 60 60 80 60 ..(i) and for second condition       − +  − 30 2 60 50 60 50 t ....(ii) By solving (i) and (ii) we get t = 48 sec ~– 50 sec. 35. (b) According to Newton's law of cooling 1 2 1+ 2 2k k k k
      − +  −      2 1 2 1 2 t For first condition       − +  − 30 2 62 61 62 61 T .....(i) and for second condition       − +  − 30 2 46 45.5 46 45.5 t .(ii) By solving (i) and (ii) we get t = T sec. 36. (d) (T T ) mc A dt dT 4 0 4 −  = . If the liquids put in exactly similar calorimeters and identical surrounding then we can consider T0 and A constant then mc T T dt dT ( ) 4 0 4 −  ......(i) If we consider that equal masses of liquid (m) are taken at the same temperature then dt c dT 1  So for same rate of cooling c should be equal which is not possible because liquids are of different nature. Again from (i) equation mc T T dt dT ( ) 4 0 4 −   V c T T dt dT  ( ) 4 0 4 −  Now if we consider that equal volume of liquid (V) are taken at the same temperature then dt c dT  1  . So for same rate of cooling multiplication of  c for two liquid of different nature can be possible. So option (d) may be correct. 37. (b)       − +  −      2 1 2 1 2 t For first condition       − +  −  2 60 50 10 60 50  1 = K [55 – ]......(i) For second condition       − +  −  2 50 42 10 50 42  0.8 = K (46 –  ) .....(ii) From (i) and (ii) we get  = 10oC 38. (b) Rate of cooling  difference in temperature dT dt    dT dt = K in first case dT = 61− 59 = 2  = 60 − 30 = 30 dt = 4 minute  60 1 30x4 2 dt dT K = =  = For second case dT = 2  = 50 − 30 = 20  6min . x20 2 K dT dt 60 1 = =  = Hence (B) is correct 39. (d) 1 2 1 2 Q d KA( ) or ( ) t d K  −  =  −   Now, 1 0 1 d K  −   1 2 1 d K  −   Adding 1 2 1 2 1 1 d k k    −   +     Also, 1 2 1 (2d) k    −        1 2 1 2 2 1 2 1 1 2k k or k k k k k k = + = + Hence (D) is correct. 40. (d) From wein’s Law m T constant. Where T is temperature of black body and  m is wavelength corresponding to maximum energy of emission. Energy distribution of blackbody radiation is given below : 1. U1 and U3 are not zero because a blackbody emits nearly radiation’s of all wavelengths. 2. Since U1 corresponds lower wavelength and U3 corresponds higher wavelength and U2 corresponds medium wavelength. Hence U2  U1 . Hence (D) is correct. 41. (a) Q Q KA( ) ( ) 1 2 1 2 or t l t l KA  −   −  = = or Thermal resistance 1 2 ( )t Q  −  = So, dimensional formula is [KT] /[M−1L −2T −2 ] Or [M−1L −2T 3K]