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

Question

Quantrex Academy · Accepted Answer

Correct answer: A. 6 : 1

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 = \dfrac{KA\Delta T}{l}

.

For rod A:

H_A = \dfrac{K_A A (100^{\circ}\mathrm{C} - 75^{\circ}\mathrm{C})}{2L} = \dfrac{25 K_A A}{2L}

For rod B:

H_B = \dfrac{K_B A (75^{\circ}\mathrm{C} - 0^{\circ}\mathrm{C})}{L} = \dfrac{75 K_B A}{L}

Equating

H_A

and

H_B

:

\dfrac{25 K_A A}{2L} = \dfrac{75 K_B A}{L}

\Rightarrow \dfrac{K_A}{2} = 3 K_B

\Rightarrow \dfrac{K_A}{K_B} = \dfrac{6}{1}

The ratio

K_A : K_B

is

6 : 1

.

Answer:

6 : 1

Two cylindrical rods, A and B, of the same cross-sectional area are...

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