question_answer

From a solid sphere of mass M and radius R, a spherical portion of radius \[\frac{R}{2}\]is removed, as shown in the figure. Taking gravitational potential V = 0 at \[r=\infty ,\] the potential at the centre of cavity thus formed is:

(G = gravitational constant)                                                                                             [JEE MAIN 2015]

A)  \[\frac{-2GM}{3R}\]     

B)       \[\frac{-2GM}{R}\]

C)  \[\frac{-GM}{2R}\]       

D)       \[\frac{-GM}{R}\]

Correct Answer: D

Solution :

[d]