Two uniform brass rods A and B of length ' \(l\) ' and ' \(2 l\) ' and their radii ' \(2 \mathrm{r}\) ' and ' \(\mathrm{r}\) ' respectively are heated to same temperature. The ratio of the increase in the volume of \(\operatorname{rod} \mathrm{A}\) to that of \(\operatorname{rod} \mathrm{B}\) is

Let the original temperature be \(0^{\circ} \mathrm{C}\)
Volume of \(\mathrm{A}=\mathrm{V}_1=l \times \pi(2 \mathrm{r})^2\)
After heating, volume of A will become
\(\begin{aligned}
& \mathrm{V}_1^{\prime}=\mathrm{V}_1(1+\gamma \Delta \mathrm{T}) \\
& \frac{\left(\mathrm{V}_1^{\prime}-\mathrm{V}_1\right)}{\mathrm{V}_1}=\gamma \Delta \mathrm{T} \Rightarrow \mathrm{V}_1^{\prime}-\mathrm{V}_1 \propto \mathrm{V}_1
\end{aligned}\)
Similarly for rod B,
\(\begin{aligned}
& \frac{\left(\mathrm{V}_2^{\prime}-\mathrm{V}_2\right)}{\mathrm{V}_2}=\gamma \Delta \mathrm{T} \Rightarrow \mathrm{V}_2^{\prime}-\mathrm{V}_2 \propto \mathrm{V}_2 \\
\therefore \quad & \frac{\Delta \mathrm{V}_1}{\Delta \mathrm{V}_2}=\frac{l(2 \mathrm{r})^2}{2 l \mathrm{r}^2}=\frac{2}{1}
\end{aligned}\)