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Q.

Two rods of length $d_{1}$ and $d_{2}$ and coefficients of thermal conductivities $K_{1}$ and $K_{2}$ are kept touching each other. Both have the same area of cross-section. The equivalent of thermal conductivity is

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a

$\frac{d_{1} + d_{2}}{(d_{1} / K_{1} + d_{2} / K_{2})}$

b

$\frac{d_{1} K_{1} + d_{2} K_{2}}{d_{1} + d_{2}}$

c

$K_{1} d_{1} + K_{2} d_{2}$

d

$K_{1} + K_{2}$

answer is D.

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Detailed Solution

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$\begin{array}{l} When two rods are connected in series \\ Q = \frac{A (T_{1} - T_{2}) t}{\frac{d_{1}}{K_{1}} + \frac{d_{2}}{K_{1}}} = \frac{A (T_{1} - T_{2}) t}{(d_{1} + d_{2}) / K} \\ \Rightarrow \frac{d_{1} + d_{2}}{K} = \frac{d_{1}}{K_{1}} + \frac{d_{2}}{K_{2}} \\ \Rightarrow K = \frac{(d_{1} + d_{2})}{\frac{d_{1}}{K_{1}} + \frac{d_{2}}{K_{2}}} \end{array}$

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