A cylindrical rod is having temperatures $\theta_1$ and $\theta_2$ at its ends. The rate of heat flow is '…
A cylindrical rod is having temperatures $\theta_1$ and $\theta_2$ at its ends. The rate of heat flow is ' $\mathrm{Q}$ ' $\mathrm{J} \mathrm{s}{ }^{-1}$. All the linear dimensions of the rod are doubled by keeping the temperatures constant. What is the new rate of flow of heat?
$\frac{Q}{2}$
$\frac{Q}{4}$
$2 Q$
$\frac{3 Q}{2}$
Solution
$\mathrm{Q}=\frac{\mathrm{kA} \Delta \theta}{\ell}$
If the radius of the cylindrical rod is doubled, then its area of cross-section will become four times.
$\begin{aligned}
& \therefore \mathrm{A}_2=4 \mathrm{~A}_1 \\
& \text { Also, } \ell_2=2 \ell_1 \\
& \therefore \frac{\mathrm{Q}^{\prime}}{\mathrm{Q}}=\frac{\mathrm{A}_2}{\mathrm{~A}_1} \cdot \frac{\mathrm{d}_1}{\mathrm{~d}_2}=4 \times \frac{1}{2}=2 \\
& \therefore \mathrm{Q}^{\prime}=2 \mathrm{Q}
\end{aligned}$