Question

Two identical thin rings each of radius R are co-axially 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. $\frac{Gm}{m_{2}R}(\sqrt{2}+1)m$
• B. $\frac{Gm(m_1-m_2)}{\sqrt{2}R}(\sqrt{2}-1)$
• C. $\sqrt{Gm\sqrt{2}}R(m_1+m_2)$
• D. zero

Answer 1

The correct option is B $\frac{Gm(m_1-m_2)}{\sqrt{2}R}(\sqrt{2}-1)$

Let m₁ and m₂ be the masses of two rings having A and B as their centers. The potential of first ring is V_A and that of second ring is V_B

Gravitational potential at A $V_A=\frac{-G{m}_1}{R}-\frac{G{m}_2}{\sqrt{2}R}$

Gravitational potential at B $V_B=\frac{-G{m}_1}{\sqrt{2}R}-\frac{G{m}_2}{R}$

Potential difference is

$V_B-V_A=\frac{-G{m}_1}{\sqrt{2}R}-\frac{G{m}_2}{R}+\frac{G{m}_1}{R}+\frac{G{m}_2}{\sqrt{2}R}$

$=\frac{G}{R}(m_1-m_2)\left(\frac{1-\sqrt{2}}{\sqrt{2}}\right)$

Work done in moving a mass m is

$w=\frac{G}{R}m(m_2-m_1)\left(\frac{1-\sqrt{2}}{\sqrt{2}}\right)$

$=\frac{Gm(m_1-m_2)}{\sqrt{2}R}(\sqrt{2}-1)$