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NEETPhysicsThermal Properties of Matter

Two cylindrical rods, A and B, of the same cross-sectional area are joined end-to-end. The free end of rod A is maintained at 100^ C and the free end of rod B is maintained at 0^ C . In the steady state, the temperature of the junction is found to be 75^ C . If the length of rod A is twice the length of rod B, the ratio of the thermal conductivities of rod A to rod B ( K_A : K_B ) is:

Options

  1. A6 : 1
  2. B1 : 6
  3. C3 : 1
  4. D1 : 3

Correct answer

A. 6 : 1

Step-by-step solution

In steady state, the rate of heat flow through both rods is equal. Let the length of rod B be L , so the length of rod A is 2L . The rate of heat flow is given by H = KA T l . For rod A: H_A = K_A A (100^ C - 75^ C ) 2L = 25 K_A A 2L For rod B: H_B = K_B A (75^ C - 0^ C ) L = 75 K_B A L Equating H_A and H_B : 25 K_A A 2L = 75 K_B A L K_A 2 = 3 K_B K_A K_B = 6 1 The ratio K_A : K_B is 6 : 1 . Answer: 6 : 1

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