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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 conductivitie $K_{1} \: and\: K_{2}$ respectively. The temperature at the interface of the two sections is
Option: 1
$\frac{\left ( K_{1}l_{1} T_{1}+K_{2}l_{2} T_{2}\right )}{\left ( K_{1}l_{1}+K_{2}l_{2}\right )}$

Option: 2
$\frac{\left ( K_{2}l_{2} T_{1}+K_{1}l_{1} T_{2}\right )}{\left ( K_{1}l_{1}+K_{2}l_{2}\right )}$

Option: 3
$\frac{\left ( K_{2}l_{1} T_{1}+K_{1}l_{2} T_{2}\right )}{\left ( K_{2}l_{1}+K_{1}l_{2}\right )}$

Option: 4
$\frac{\left ( K_{1}l_{2} T_{1}+K_{2}l_{1} T_{2}\right )}{\left ( K_{1}l_{2}+K_{2}l_{1}\right )}$

Answers (1)

best_answer

Let's say temperature at interface is T.

Rate of Heat transfer through the rod is $\frac{dQ}{dt}=\frac{\Delta T}{R}=\frac{T_{1}-T_{2}}{R}$

$R=R_{1}+R_{2}=\frac{l_{1}}{K_{1}A}+\frac{l_{2}}{K_{2}A}=\frac{1}{A} \left(\frac{l_{1}K_{2}+l_{2}K_{1}}{K_{1}K_{2}} \right )$

$\frac{dQ}{dt}=\frac{(T_{1}-T_{2}).AK_{1}K_{2}}{l_{1}K_{2}+l_{2}K_{1}}$

Since rate of heat transfer is same through out the rod.

Rate of heat transfer through the first rod $=\frac{(T_{1}-T).K_{1}A}{l_{1}}=\frac{(T_{1}-T_{2})A.K_{1}K_{2}}{l_{1}K_{2}+l_{2}K_{1}}$

$T_{1}-T=\frac{(T_{1}-T_{2}).K_{1}l_{f}}{l_{1}K_{2}+l_{2}K_{1}}$

$T=T_{1}-\frac{(T_{1}-T_{2}).K_{2}l_{f}}{l_{1}K_{2}+l_{2}K_{1}}=\frac{T_{1}l_{1}K_{2}+T_{1}l_{2}K_{1}-T_{1}l_{1}K_{2}+T_{2}K_{2}l_{1}}{l_{1}K_{2}+l_{2}K_{1}}$

$T=\frac{T_{1}K_{1}l_{2}+T_{2}K_{2}l_{1}}{l_{1}K_{2}+l_{2}K_{1}}$

Correct option is 4.

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