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Heat and Thermodynamics question

2007 · Shift 0 · Q90
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Heat and Thermodynamics question

2007 · Shift 0 · Q90

JEE MainPhysicsHeat and ThermodynamicsMCQ+4 / −1
One end of a thermally insulated rod is kept at a temperature T1{T_1}T1​ and the other at T2{T_2}T2​. The rod is composed of two sections of length L1{L_1}L1​ and L2{L_2}L2​ and thermal conductivities K1{K_1}K1​ and K2{K_2}K2​ respectively. The temperature at the interface of the two section is AIEEE 2007 Physics - Heat and Thermodynamics Question 401 English
  1. A
    (K1L1T1+K2L2T2)(K1L1+K2L2){{\left( {{K_1}{L_1}{T_1} + {K_2}{L_2}{T_2}} \right)} \over {\left( {{K_1}{L_1} + {K_2}{L_2}} \right)}}(K1​L1​+K2​L2​)(K1​L1​T1​+K2​L2​T2​)​
  2. B
    (K2L2T1+K1L1T2)(K1L1+K2L2){{\left( {{K_2}{L_2}{T_1} + {K_1}{L_1}{T_2}} \right)} \over {\left( {{K_1}{L_1} + {K_2}{L_2}} \right)}}(K1​L1​+K2​L2​)(K2​L2​T1​+K1​L1​T2​)​
  3. C
    (K2L1T1+K1L2T2)(K2L1+K1L2){{\left( {{K_2}{L_1}{T_1} + {K_1}{L_2}{T_2}} \right)} \over {\left( {{K_2}{L_1} + {K_1}{L_2}} \right)}}(K2​L1​+K1​L2​)(K2​L1​T1​+K1​L2​T2​)​
  4. D
    (K1L2T1+K2L1T2)(K1L2+K2L1){{\left( {{K_1}{L_2}{T_1} + {K_2}{L_1}{T_2}} \right)} \over {\left( {{K_1}{L_2} + {K_2}{L_1}} \right)}}(K1​L2​+K2​L1​)(K1​L2​T1​+K2​L1​T2​)​
View written solutionFree

Correct answer: D

  1. Let the temperature at the interface be TTT.

  2. In steady state, the heat current through both sections of the rod must be the same.

For section 111: rac{Q}{t} = K_1 A \frac{T_1 - T}{L_1}

For section 222: rac{Q}{t} = K_2 A \frac{T - T_2}{L_2}

  1. Equate the two expressions: K1AT1−TL1=K2AT−T2L2K_1 A \frac{T_1 - T}{L_1} = K_2 A \frac{T - T_2}{L_2}K1​AL1​T1​−T​=K2​AL2​T−T2​​

Area AAA cancels: K1T1−TL1=K2T−T2L2K_1 \frac{T_1 - T}{L_1} = K_2 \frac{T - T_2}{L_2}K1​L1​T1​−T​=K2​L2​T−T2​​

  1. Cross-multiply: K1L2(T1−T)=K2L1(T−T2)K_1 L_2 (T_1 - T) = K_2 L_1 (T - T_2)K1​L2​(T1​−T)=K2​L1​(T−T2​)

Expand: K1L2T1−K1L2T=K2L1T−K2L1T2K_1 L_2 T_1 - K_1 L_2 T = K_2 L_1 T - K_2 L_1 T_2K1​L2​T1​−K1​L2​T=K2​L1​T−K2​L1​T2​

  1. Collect terms in TTT: K1L2T1+K2L1T2=(K1L2+K2L1)TK_1 L_2 T_1 + K_2 L_1 T_2 = (K_1 L_2 + K_2 L_1)TK1​L2​T1​+K2​L1​T2​=(K1​L2​+K2​L1​)T

Hence, T=K1L2T1+K2L1T2K1L2+K2L1T = \frac{K_1 L_2 T_1 + K_2 L_1 T_2}{K_1 L_2 + K_2 L_1}T=K1​L2​+K2​L1​K1​L2​T1​+K2​L1​T2​​

  1. Compare with the options:
  • Option A: incorrect
  • Option B: incorrect
  • Option C: incorrect
  • Option D: matches exactly

Therefore, the correct answer is: D\boxed{D}D​

  1. Verification with stored answer: Stored correct answer is DDD, which matches the derived answer.
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