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Conduction

Two rods of l...

Question

Two rods of lengths $d_{1}$ and $d_{2}$ and coefficients of thermal conductivities $K_{1}$ and $K_{2}$ are kept in contact with each other end to end. Both have the same area of cross-section. The equivalent thermal conductivity is?

A

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

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B

K_{1} + K_{2}

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C

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

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D

\frac{d_{1} + d_{2}}{(\frac{d_{1}}{K_{1}} + \frac{d_{2}}{k_{2}})}

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Solution

The correct option is B $\frac{d_{1} + d_{2}}{(\frac{d_{1}}{K_{1}} + \frac{d_{2}}{k_{2}})}$
Same quantity of heat Q flows in the two rods as they are in contact. It is an arrangement in series.
$Q = \frac{A (T_{1} - T_{2}) t}{\frac{d_{1}}{K_{1}} + \frac{d_{2}}{K_{2}}} = \frac{A (T_{1} - T_{2}) t}{\frac{(d_{1} + d_{2})}{K}}$
$\Rightarrow$ $\frac{d_{1} + d_{2}}{K} = \frac{d_{1}}{K_{1}} + \frac{d_{2}}{K_{2}}$
$\Rightarrow = \frac{d_{1} + d_{2}}{\frac{d_{1}}{K_{1}} + \frac{d_{2}}{K_{2}}}$ .

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