Q.
Two identical thin rings each of radius $R$ are coaxially placed at a distance $R$. If the rings have a uniform mass distribution and each has mass $m_{1}$ and $m_{2}$ respectively, then the work done in moving a mass $m$ from centre of one ring to that of the other is
A) $\frac{Gmm_{1}(\sqrt{2} + 1)}{m_{2}R}$
B) $\frac{Gm(m_{1} - m_{2})(\sqrt{2} + 1)}{\sqrt{2}R}$
C) $\frac{Gm\sqrt{2}(m_{1}{+ m}_{2})}{R}$
D) Zero
Explanation
(b) $V_{A} =$( Potential at $A$ due to $A$) + (Potential at $A$ due to $B$)
$\Rightarrow \ \ V_{A} = - \frac{Gm_{1}}{R} - \frac{Gm_{2}}{\sqrt{2}R}$
Similarly,
$V_{B} =$(Potential at $B$ due to $A$) + (Potential at $B$ due to $B$)
$\Rightarrow \ \ V_{B} = - \frac{Gm_{2}}{R} - \frac{Gm_{1}}{\sqrt{2}R}$
Since, $W_{A \rightarrow B} = m(V_{B} - V_{A}) \Rightarrow W_{A \rightarrow B}$
$= \frac{Gm\left( m_{1} - m_{2} \right)(\sqrt{2} - 1)}{\sqrt{2}R}$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.