Explanation

(c) For brass rod A
Volume $V_{1}$= $\pi(2r)^{2}\ \times l$ …(i)
For volume expansion
$\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ {V'}_{1} = \ V_{1}((1 + \gamma\mathrm{\Delta}t)$
$\Rightarrow$ ${V'}_{1}\ - \ V_{1\ \ } \propto V_{1}$
Or $\mathrm{\Delta}V_{1\ \ \ } \propto V_{1}$ …(ii)
$_{Similarly, for brass rod B}$
$_{Volume}$ $\ \ \ V_{2} = \pi{(r)}^{2} \times 2l$ …(iii)
$_{and}$ $\ \mathrm{\Delta}V_{2}\ \propto V_{2}$ …(iv)
$_{Dividing Eq. (i) by Eq. (ii), we get}$
$\frac{V_{1}}{V_{2}} = \frac{\pi 4r^{2}l}{\pi r^{2}2l} = \frac{2}{1}$
$_{From eqs. (ii) and (iv),}$
$\ \ \ \ \frac{\mathrm{\Delta}V_{1}}{\mathrm{\Delta}V_{2}} = \frac{2}{1}$