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] |
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