Two identical thin rings A and B each of radius R are co-axially placed at a distance R. If mass of rings are m1, m2 respectively, then the work done in moving a mass m from centre of one ring to that of the other is [UP CPMT 2012J

VB = Potential at B due to A + Potential at A due to BVB=-Gm2R--Gm12R⇒ VA=-Gm1R--Gm22RWA→B=mVB-VA=Gmm1-m22R(2-1)

Two identical thin rings A and B each of radius R are co-axially placed at a distance R. If mass of rings are m1, m2 respectively, then the work done in moving a mass m from centre of one ring to that of the other is [UP CPMT 2012J

VB = Potential at B due to A + Potential at A due to BVB=-Gm2R--Gm12R⇒ VA=-Gm1R--Gm22RWA→B=mVB-VA=Gmm1-m22R(2-1)

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Two identical thin rings A and B each of radius R are co-axially placed at a distance R. If mass of rings are m₁, m₂ respectively, then the work done in moving a mass m from centre of one ring to that of the other is [UP CPMT 2012J

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

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

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

answer is B.

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Detailed Solution

V_B = Potential at B due to A + Potential at A due to B

$V_{B} = \frac{- G m_{2}}{R} - (\frac{- G m_{1}}{\sqrt{2} R})$

$\Rightarrow V_{A} = \frac{- G m_{1}}{R} - (\frac{- G m_{2}}{\sqrt{2 R}})$

$W_{A \to B} = m (V_{B} - V_{A}) = \frac{G m (m_{1} - m_{2})}{\sqrt{2} R} (\sqrt{2} - 1)$

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Two identical thin rings A and B each of radius R are co-axially placed at a distance R. If mass of rings are m1, m2 respectively, then the work done in moving a mass m from centre of one ring to that of the other is [UP CPMT 2012J

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Two identical thin rings A and B each of radius R are co-axially placed at a distance R. If mass of rings are m₁, m₂ respectively, then the work done in moving a mass m from centre of one ring to that of the other is [UP CPMT 2012J