Q. Four rods of identical cross-sectional area and made from the same metal form the sides of square. The temperature of two diagonally opposite points are $T$ and $\sqrt{2}\ T$ respectively in the steady state. Assuming that only heat conduction takes place, what will be the temperature difference between other two points
A) $\frac{\sqrt{2} + 1}{2}T$
B) $\frac{2}{\sqrt{2} + 1}T$
C) 0
D) None of these
Explanation
(c) $\frac{Q}{t} = \frac{KA\Delta\theta}{l} = \frac{mL}{t} = \frac{K(\pi r^{2})\Delta\theta}{l}$
$\Rightarrow$ Rate of melting of ice $\left( \frac{m}{t} \right) \propto \frac{Kr^{2}}{l}$
Since for second rod $K$ becomes $\frac{1}{4}th\ r$ becomes double and length becomes half, so
rate of melting will to twice $i.e.\ \left( \frac{m}{t} \right)_{2} = 2\left( \frac{m}{t} \right)_{1} = 2 \times 0.1 = 0.2g/s$
Temperature difference between $C$ and $D$ is zero
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.