Question

Two uniform brass rods A and B of lengths $l$ and $2l$ and radii $2r \, $ and $r$ respectively are heated to the same temperature. The ratio of the increase in the volumes of A to that of B is

• A. $1:1$
• B. $1:2$
• C. $2:1$
• D. $1:4$

Answer 1

Answer: (C)

For brass rod A
Volume $V_{1}$ = $\pi \left(2 r\right)^{2} \, \times l$ ...(i)
For volume expansion
$ \, \, \text{V}_{1}^{′}= \, V_{1}\left(\right.1+\gamma \Delta t\left.\right)$
$⇒ \, \, \, \text{V}_{1}^{′} \, - \, V_{1 \, \, }∝V_{1}$
Or $\Delta V_{1 \, \, \, }∝V_{1}$ ...(ii)
Similarly, for brass rod B
Volume $ \, \, \, V_{2}=\pi \left(\right. r \left.\right)^{2}\times 2l$ ...(iii)
and $ \, \Delta V_{2} \, ∝V_{2}$ ...(iv)
Dividing Eq. (i) by Eq. (ii), we get
$\frac{V_{1}}{V_{2}}=\frac{\pi 4 r^{2} l}{\pi r^{2} 2 l}=\frac{2}{1}$
From eqs. (ii) and (iv),
$ \, \frac{\Delta V_{1}}{\Delta V_{2}}=\frac{2}{1}$