JEE Main 2020 (5 Sep, Shift 1) Physics Q11: A solid sphere…

Question

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).

Accepted Answer

Option 3: $v^2 = 2y\left[\frac{qQ}{4\pi\epsilon_0 R(R+y)m} + g\right]$

Answer

$v^2 = y\left[\frac{qQ}{4\pi\epsilon_0 R(R+y)m} + g\right]$

Answer

$v^2 = 2y\left[\frac{QqR}{4\pi\epsilon_0 (R+y)^3 m} + g\right]$

Answer

$v^2 = 2y\left[\frac{qQ}{4\pi\epsilon_0 R(R+y)m} + g\right]$

Answer

$v^2 = y\left[\frac{qQ}{4\pi\epsilon_0 R^2 y m} + g\right]$

JEE Main 5 September 2020, Shift 1, Physics Q11

Same idea in other shifts