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Dependence of Resistance on Temperature

The equivalen...

Question

The equivalent thermal conductivity of the compound bar is

A

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

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B

\frac{{2 K}_{1} K_{2}}{K_{1} + K_{2}}

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C

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

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D

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

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Solution

The correct option is A $\frac{{2 K}_{1} K_{2}}{K_{1} + K_{2}}$
Let K be the thermal conductivity of the compound bar.
Heat current through the compound bar of length $2 L$ is
$H = \frac{K A (T_{1} - T_{2})}{2 L} - - - - - (i)$
At steady state, $H = H_{1} = H_{2}$
$∴ \frac{K A (T_{1} - T_{2})}{2 L} = \frac{K_{1} A (T_{1} - T_{0})}{L}$ $. . . . . . (I I)$
Substituting the value of $T_{0}$ from Eq $(i)$ in $(i i)$ , we get
$\frac{K (T_{1} - T_{2})}{2} = K_{1} [T_{1} - (\frac{K_{1} T_{1} + K_{2} T_{2}}{K_{1} + K_{2}})]$
$= \frac{K_{1} K_{2} (T_{1} - T_{2})}{(K_{1} + K_{2})}$
$\Rightarrow = \frac{2 K_{1} K_{2}}{K_{1} + K_{2}}$

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