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 gravitional potential V = 0 at \[r=\infty ,\] the potential at the centre of cavity thus formed is: (G = gravitational constant) [JEE Main Solved Paper-2015 ]

A) \[\frac{-2GM}{3R}\]                       

B) \[\frac{-2GM}{R}\]

C) \[\frac{-GM}{2R}\]                         

D) \[\frac{-GM}{R}\]

Correct Answer: D

Solution :

\[{{V}_{B}}={{V}_{A}}+{{V}_{B}}\] \[{{V}_{B}}={{V}_{M}}-{{V}_{m}}\] \[=\frac{-11GM}{8R}+\frac{3GM}{8R}=\frac{-GM}{R}\]