A unit positive point charge of mass `m` is projected with a velocity `V` inside the tunnel as shown. The tunnel has been made inside a uniformly charged nonconducting sphere. The minimum velocity with which the point charge should be projected such that it can reach the opposite end of the tunnel is equal to
(##BMS_V03_C03_E01_124_Q01.png" width="80%">

Options

A. `[(rhoR^(2))/(4mepsilon_(0))]^(1//2)`
B. `[(rhoR^(2))/(24mepsilon_(0))]^(1//2)`
C. `[(rhoR^(2))/(6mepsilon_(0))]^(1//2)`
D. zero because the initial and the final points are at same potential

Answer

A

Explanation

`(KQ)/(R)+1/2m u^(2)=(KQ)/(2R^(3))(3R^(2)-R^(2)/4)+0`
`(1)/(4pi in_(0)R)xxrhoxx4/3piR^(3)+1/2 m u^(2)=(11xxrhoxx4/3piR^(3))/(8xx2R^(2))`
`:.u=[(rhoR^(2))/(4m in_(0))]^(1//2)`