Question 104 of 126

Temperature difference of ${120}^{o}C$ is maintained between two ends of a uniform rod $AB$ of length $2L.$ Another bent rod $PQ,$ of same cross-section as $AB$ and length $\dfrac{3L}{2},$ is connected across $AB$ (See figure). In steady state, temperature difference between $P$ and $Q$ will be close to:

A${45}^{o}C$B${35}^{o}C$C${60}^{o}C$D${75}^{o}C$

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Answer: (A) ${45}^{o}C$

${T}_{1}-{T}_{2}={120}^{o}C$

We define thermal resistance

$R=\dfrac{L}{KA}$

Given circuit can be reduced to


Equivalent resistance ${R}_{eq}=\dfrac{8R}{5}$

Thermal current $i=\dfrac{{T}_{1}-{T}_{2}}{\dfrac{8R}{5}}$

${T}_{P}-{T}_{Q}=\dfrac{{T}_{1}-{T}_{2}}{\left(\dfrac{8R}{5}\right)}\left(\dfrac{3R}{5}\right)$

$=\dfrac{3}{8}\times 120=45$