One end of a thermally insulated rod is kept at a temperature $T_{1}$ and the other at $T_{2}$ . The rod is composed of two sections of lengths $l_{1}$ and $l_{2}$ and thermal conductivities $k_{1}$ and $k_{2}$ respectively. The temperature at the interface of the two sections is:

(k_{2} l_{2} T_{1} + k_{1} l_{1} T_{2}) / (k_{1} l_{1} + k_{2} l_{2})

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(k_{2} l_{1} T_{1} + k_{1} l_{1} T_{2}) / (k_{2} l_{1} + k_{1} l_{2})

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(k_{2} l_{2} T_{1} + k_{2} l_{1} T_{2}) / (k_{1} l_{2} + k_{2} l_{1})

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(k_{1} l_{1} T_{1} + k_{1} l_{2} T_{2}) / (k_{1} l_{1} + k_{2} l_{2})

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Solution

The correct option is B $(k_{2} l_{2} T_{1} + k_{2} l_{1} T_{2}) / (k_{1} l_{2} + k_{2} l_{1})$
$Solution :$
Let T be the temperature of the interface. Since two section of rods are in series, rate of flow of heat in them are equal.
$\frac{K_{1} A (T_{1} - T)}{l_{1}} = \frac{K_{2} A (T - T_{2})}{l_{2}}$
$∴ T = \frac{K_{1} T_{1} l_{2} + k_{2} T_{2} l_{1}}{K_{2} l_{1} + K_{1} l_{2}} = \frac{K_{1} l_{2} T_{1} + k_{2} l_{1} T_{2}}{K_{1} l_{2} + K_{2} l_{1}}$

$Hence C is the correct option$

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