Question

Two identical thin rings each of radius $R$ are coaxially placed at a separation of $R$. If the rings have a uniform mass distribution and masses $m_1$ and $m_2$ respectively, then the work done in moving a mass $m$ from centre of the ring having mass $m_1$ to center of the other ring is

• A. $\frac{Gm(2-\sqrt{2})(m_2-m_1)}{2R}$
• B. $\frac{Gm(2-\sqrt{2})(m_1-m_2)}{2R}$
• C. $0$
• D. $\frac{Gm(2-\sqrt{2})(m_1+m_2)}{2R}$
  

Answer 1

The correct option is B $\frac{Gm(2-\sqrt{2})(m_1-m_2)}{2R}$

Work done will be equal to the change in the potential energy of the two rings and one mass system due to rearrangement.

When mass is at the centre of ring $m_1$,
Initial potential energy,

$U_i=U\left(\text{between two rings}\right)+U\left(\text{between ring of mass}{m}_1\text{and mass m}\right)$
$+U\left(\text{between ring of mass}{m}_2\text{and mass m}\right)$

$\Rightarrow U_i=-\frac{G{m}_1{m}_2}{\sqrt{R^2+R^2}}-\frac{G{m}_{1}m}{R}-\frac{G{m}_{2}m}{\sqrt{R^2+R^2}}$

$\Rightarrow U_i=-\frac{G{m}_1{m}_2}{\sqrt{2}R}-\frac{G{m}_{1}m}{R}-\frac{G{m}_{2}m}{\sqrt{2}R}$

When mass is at the centre of ring $m_2$,
Final potential energy,

$U_f=U\left(\text{between two rings}\right)+U\left(\text{between ring of mass}{m}_1\text{and mass m}\right)$
$+U\left(\text{between ring of mass}{m}_2\text{and mass m}\right)$

$\Rightarrow U_f=-\frac{G{m}_1{m}_2}{\sqrt{R^2+R^2}}-\frac{G{m}_{1}m}{\sqrt{R^2+R^2}}-\frac{G{m}_{2}m}{R}$

$\Rightarrow U_f=-\frac{G{m}_1{m}_2}{\sqrt{2}R}-\frac{G{m}_{1}m}{\sqrt{2}R}-\frac{G{m}_{2}m}{R}$

So, work done,
$W=\Delta U=U_f-U_i=\frac{G{m}_{1}m(2-\sqrt{2})}{2R}-\frac{G{m}_{2}m(2-\sqrt{2})}{2R}$

$\Rightarrow W=\frac{Gm(2-\sqrt{2})(m_1-m_2)}{2R}$

Hence, option (b) is the correct answer.

​​​​

Why this question? The gravitational potential of a ring of radius $R$ and mass $M$ at a distance $r$ from its centre on the axis is given by $V=-\frac{GM}{\sqrt{R^2+r^2}};0\le r\le \infty$