ExamBro
ExamBro
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).

  1. A \(v^{2}=2 y\left[\frac{q Q}{4 \pi \epsilon_{0} R(R+y) m}+g\right]\)
  2. B \(v^{2}=y\left[\frac{q Q}{4 \pi \epsilon_{0} R^{2} y m}+g\right]\)
  3. C \(v^{2}=2 y\left[\frac{q Q R}{4 \pi \epsilon_{0}(R+y)^{3} m}+g\right]\)
  4. D \(v^{2}=y\left[\frac{q Q}{4 \pi \epsilon_{0} R(R+y) m}+g\right]\)
Verified Solution

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

\(\frac{ kQq }{ R }+ mgy\) \(=\frac{ kQq }{ R + y }+\frac{1}{2} mv ^{2}\) \(v ^{2}=2 gy +\frac{2 kQqy }{ mR ( R + y )}\)
From JEE Mains
Explore more questions on app