JEE Mains · Physics · STD 12 - 2. Electric potential and capacitance
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 figure), then : (assume the remaining portion to be spherical).

- A \(v^{2}=2 y\left[\frac{q Q}{4 \pi \epsilon_{0} R(R+y) m}+g\right]\)
- B \(v^{2}=y\left[\frac{q Q}{4 \pi \epsilon_{0} R^{2} y m}+g\right]\)
- C \(v^{2}=2 y\left[\frac{q Q R}{4 \pi \epsilon_{0}(R+y)^{3} m}+g\right]\)
- D \(v^{2}=y\left[\frac{q Q}{4 \pi \epsilon_{0} R(R+y) m}+g\right]\)
Answer & Solution
Correct Answer
(A) \(v^{2}=2 y\left[\frac{q Q}{4 \pi \epsilon_{0} R(R+y) m}+g\right]\)
Step-by-step Solution
Detailed explanation
\(\frac{ kQq }{ R }+ mgy\) \(=\frac{ kQq }{ R + y }+\frac{1}{2} mv ^{2}\) \(v ^{2}=2 gy +\frac{2 kQqy }{ mR ( R + y )}\)
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