Question

Two identical thin rings each of radius $R$ are co-axially 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 the centre of one ring to that of the other is

• A. $0$
• B. $\frac{Gm(m_1-m_2)(\sqrt{2}-1)}{\sqrt{2}R}$
• C. $\frac{Gm\sqrt{2}(m_1+m_2)}{R}$
• D. $\frac{G{m}_{1}m(\sqrt{2}+1)}{m_{2}R}$

Answer 1

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

$V_1=\frac{-G{m}_1}{R}-\frac{G{m}_2}{\sqrt{2}R}$ and $V_2=\frac{-G{m}_2}{R}-\frac{G{m}_1}{\sqrt{2}R}$
$\Delta V=V_2-V_1=\frac{-G{m}_2}{R}-\frac{G{m}_1}{\sqrt{2}R}+\frac{-G{m}_1}{R}-\frac{G{m}_2}{\sqrt{2}R}$
$\Delta V=G(m_1-m_2)\left(\frac{1}{R}-\frac{1}{\sqrt{2}R}\right)$
Hence, $W=m(\Delta V)=Gm(m_1-m_2)(\frac{1}{R}-\frac{1}{\sqrt{2}R})$