Question

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 $\frac{3L}{2},$ 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. $75^{o}C$

Answer 1

Answer: (A)

$T_1-T_2=120^{o}C$
We define thermal resistance
$R=\frac{L}{KA}$
Given circuit can be reduced to

Equivalent resistance $R_{eq}=\frac{8R}{5}$
Thermal current $i=\frac{T_1-T_2}{\frac{8R}{5}}$
$T_P-T_Q=\frac{T_1-T_2}{\left(\frac{8R}{5}\right)}\left(\frac{3R}{5}\right)$
$=\frac{3}{8}\times 120=45$