Question

A cylindrical rod having temperature $T_{1}$ and $T_{2}$ at its ends. The rate of flow of heat is $Q_{1} cal / \sec$. If all the linear dimensions are doubled keeping temperature constant, then rate of flow of heat $Q_{2}$ will be

• A. $4\, Q_{1}$
• B. $2\, Q_{1}$
• C. $\frac{Q_{1}}{4}$
• D. $\frac{Q_{1}}{2}$

Answer 1

Answer: (B)

Rate of heat flow $\left(\frac{Q}{t}\right)=\frac{k \pi r^{2}\left(\theta_{1}-\theta_{2}\right)}{L} \propto \frac{r^{2}}{L}$
$\therefore \frac{Q_{1}}{Q_{2}}=\left(\frac{r_{1}}{r_{2}}\right)^{2}\left(\frac{l_{2}}{l_{1}}\right)$
$=\left(\frac{1}{2}\right)^{2} \times\left(\frac{2}{1}\right)=\frac{1}{2}$
$\Rightarrow Q_{2}=2 Q_{1}$