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Q.

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 m₁ and m₂ respectively, then the work done in moving a mass ‘m’ from the centre of one ring to that of the other is:

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a

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

b

$\frac{Gm \sqrt{2} (m_{1} + m_{2})}{R}$

c

Zero

d

$\frac{{Gm}_{1} m (\sqrt{2} + 1)}{m_{2} R}$

answer is B.

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Detailed Solution

$\begin{array}{l} V_{1} = \frac{- {Gm}_{1}}{R} - \frac{{Gm}_{2}}{\sqrt{2} R} and V_{2} = \frac{- {Gm}_{2}}{R} - \frac{{Gm}_{1}}{\sqrt{2} R} \\ ΔV = V_{2} - V_{1} = \frac{- {Gm}_{2}}{R} - \frac{{Gm}_{1}}{\sqrt{2} R} + \frac{- {Gm}_{1}}{R} - \frac{{Gm}_{2}}{\sqrt{2} R} \\ = G (m_{1} - m_{2}) (\frac{1}{R} - \frac{1}{\sqrt{2} R}) \\ Hence W = m (ΔV) = \frac{mG (m_{1} - m_{2}) (\sqrt{2} - 1)}{\sqrt{2} R} \end{array}$

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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 m1 and m2 respectively, then the work done in moving a mass ‘m’ from the centre of one ring to that of the other is: