Q. A cylindrical rod having temperature $T_{1}$ and $T_{2}$ at its ends. The rate of flow of heat is $Q_{1}\ cal/s$. If all the linear dimensions are doubled keeping temperature constant then rate of flow of heat $Q_{2}$ will be
A) $4Q_{1}$
B) $2Q_{1}$
C) $\frac{Q_{1}}{4}$
D) $\frac{Q_{1}}{2}$
Explanation
(b) Rate of heat flow $\left( \frac{Q}{t} \right) = \frac{k\pi r^{2}(\theta_{1} - \theta_{2})}{L} \propto \frac{r^{2}}{L}$
$\therefore\frac{Q_{1}}{Q_{2}} = \left( \frac{r_{1}}{r_{2}} \right)^{2}\left( \frac{l_{2}}{l_{1}} \right) = \left( \frac{1}{2} \right)^{2} \times \left( \frac{2}{1} \right) = \frac{1}{2} \Rightarrow Q_{2} = 2Q_{1}$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.