A unit positive point charge of mass m is projected with a velocity v inside the tummel as shown. The tunnel has been made inside a uniformly charged non conducting sphere. The minimum velocity with which the point charge should be projected such that it can it reach the opposite end of the tunnel, is equal to:

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)`