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  4. Temperature difference of 120oC 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 (3 L/2), is connected across AB (See figure). In steady state, temperature difference between P and Q will be close to: <img class=img-fluid question-image alt=Question src=https://cdn.tardigrade.in/q/nta/p-eco31qtdehpuecnn.jpg />

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Q. 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 $\frac{3 L}{2},$ is connected across $AB$ (See figure). In steady state, temperature difference between $P$ and $Q$ will be close to:
Question

NTA AbhyasNTA Abhyas 2022

Solution:

$T_{1}-T_{2}=120^{o}C$
We define thermal resistance
$R=\frac{L}{K A}$
Given circuit can be reduced to
Solution
Equivalent resistance $R_{e q}=\frac{8 R}{5}$
Thermal current $i=\frac{T_{1} - T_{2}}{\frac{8 R}{5}}$
$T_{P}-T_{Q}=\frac{T_{1} - T_{2}}{\left(\frac{8 R}{5}\right) \, }\left(\frac{3 R}{5}\right)$
$=\frac{3}{8}\times 120=45$