Question

Temperature difference of ${120}^0$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. ${45}^0$C
• B. ${75}^0$C
• C. ${60}^0$C
• D. ${35}^0$C

Answer 1

The correct option is A ${45}^0$C

$\begin{array}{c}Heatcurrent-\\ I_H=\frac{{\theta }_1-{\theta }_2}{R_{th}}\\ wherein\\ {\theta }_1-{\theta }_2=Temperaturedifferencebetweentwoendsoftheconductor.\\ thermalcurrent=\frac{{\theta }_A-{\theta }_B}{R_{eq}}\\ =\frac{{\theta }_A-{\theta }_B}{\frac{8R}{5}}=\frac{5\times 120}{8\times R}=\frac{75}{R}\\ {\theta }_P-{\theta }_Q=\frac{75}{R}\cdot \frac{3R}{5}\\ ={45}^{0}C\end{array}$

Hence,

option $\left(A\right)$ is correct answer.