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 $y$, then (assume the remaining portion to be spherical)
Answer: (D) $v^{2} = 2y\left[\dfrac{qQ}{4\pi\varepsilon_0R(R+y)m} + g\right]$
The remaining sphere (charge $Q$) acts like a point charge at its centre. The piece moves from distance $R$ to $R + y$ from the centre.
$$\frac12mv^{2} = mgy + \frac{qQ}{4\pi\varepsilon_0}\left(\frac1R - \frac1{R+y}\right) = mgy + \frac{qQ\,y}{4\pi\varepsilon_0R(R+y)}$$
$$v^{2} = 2y\left[\frac{qQ}{4\pi\varepsilon_0R(R+y)m} + g\right]$$
Solution by Sreeraj P, M.Sc Physics