Question

Two identical thin rings each of radius $R$ are coaxially placed at a distance $R$. If mass of rings are $m_{1}$, $m_{2}$ respectively, then the work done in moving a mass $m$ from centre of one ring to that of the other is

• A. $\frac{Gm\left(m_{1}+m_{2}\right)}{R}\left(\sqrt{2}+1\right)$
• B. $\frac{Gm\left(m_{1}-m_{2}\right)}{\sqrt{2}R}\left(\sqrt{2}-1\right)$
• C. $\frac{Gm\sqrt{2}}{R}\left(m_{1}+m_{2}\right)$
• D. zero

Answer 1

Answer: (B)

$\frac{Gm\left(m_{1}-m_{2}\right)}{\sqrt{2}R}\left(\sqrt{2}-1\right)$

Answer 2

The situation is as shown in the figure. Gravitational potential at the centre of the first ring (i.e., at $O_1$) is $V_{1}=\frac{Gm_{1}}{R}-\frac{GM_{2}}{\sqrt{R^{2}+R^{2}}}$ $=-\frac{Gm_{1}}{R}-\frac{Gm_{2}}{\sqrt{2}R}$ Gravitational potential at the centre of the second ring $\left(i.e. at O_{2}\right)$ is $V_{2}=-\frac{Gm_{2}}{R}-\frac{Gm_{1}}{\sqrt{R^{2}+R^{2}}}$ $=-\frac{Gm_{2}}{R}-\frac{Gm_{1}}{\sqrt{2}R}$ Work done in moving a mass $m$ from $O_{1}$ to $O_{2}$ is $W=m\left(V_{2}-V_{1}\right)$ $=m\left[-\frac{Gm_{2}}{R}-\frac{Gm_{1}}{\sqrt{2}R}-\left(-\frac{Gm_{1}}{R}-\frac{Gm_{2}}{\sqrt{2}R}\right)\right]$ $=m\left[-\frac{Gm_{2}}{R}-\frac{Gm_{1}}{\sqrt{2}R}+\frac{Gm_{1}}{R}+\frac{Gm_{2}}{\sqrt{2}R}\right]$ $=\frac{Gm}{R}\left[\frac{-\sqrt{2}m_{2}-m_{1}+\sqrt{2}m_{1}+m_{2}}{\sqrt{2}}\right]$ $=\frac{Gm}{\sqrt{2}R}[\sqrt{2}\left(m_{1}-m_{2}\right)-\left(m_{1}-m_{2}\right)$ $=\frac{Gm\left(m_{1}-m_{2}\right)}{\sqrt{2}R}\left(\sqrt{2}-1\right)$