A Solid sphere of radius R carries a charge Q + q distributed uniformly over its volume. A very small point like piece of it of mass m gets detached from the bottom of the sphere and falls down vertically under gravity . This piece carries charge q. if it acquires a speed v when it has fallen through a vertical height y (see I figure ) then : ( assume the remaining portion to be spherical )

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$v^{2} = y [\frac{q Q}{4 π \in_{0} R^{2} y m} + g]$

$v^{2} = 2 y [\frac{Q q R}{4 π \in_{0} {(R + y)}^{3} m} + g]$

$v^{2} = y [\frac{q Q}{4 π \in_{0} R (R + y) m} + g]$

$v^{2} = 2 y [\frac{q Q}{4 π \in_{0} R (R + y) m} + g]$

answer is C.

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Detailed Solution

$\begin{array}{l} G a i n i n k i n e t i c e n e r g y = l o s s i n p o t e n t i a l e n e r g y \\ \Rightarrow KE= Δ U_{g r a v i t a t i o n a l} {+ ΔU}_{e l e c t r i c} \\ \Rightarrow \frac{1}{2} {mv}^{2} = mgy + [\frac{k Q q}{R} - \frac{k Q q}{(R+y)}] \\ \Rightarrow \frac{1}{2} {mv}^{2} = mgy + kQq [\frac{1}{R} - \frac{1}{(R+y)}], w h e r e K = \frac{1}{4 π \in_{0}} \\ \Rightarrow v^{2} = 2 y [g + \frac{Q q}{4 π \in_{0} R (R + y) m}] \end{array}$

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