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 the cavity thus formed is (G = gravitational constant)

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

B)   \[\frac{-2GM}{R}\]

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

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

Correct Answer: D

Solution :

[d]: Potential at point P (centre of cavity) before  removing   the   spherical portion,                  \[{{V}_{1}}=\frac{-GM}{2{{R}^{3}}}\left( 3{{R}^{2}}-{{\left( \frac{R}{2} \right)}^{2}} \right)\] \[=\frac{-GM}{2{{R}^{3}}}\left( 3{{R}^{2}}-{{\frac{R}{4}}^{2}} \right)=\frac{-11GM}{8R}\] Mass of spherical portion to be removed, \[M'=\frac{MV'}{V}=\frac{M\frac{4\pi }{3}{{\left( \frac{R}{2} \right)}^{3}}}{\frac{4\pi }{3}{{R}^{3}}}=\frac{M}{8}\] Potential at point P due to spherical portion to be removed \[{{V}_{2}}=\frac{-3GM'}{2R'}=\frac{-3G{{(M/8)}^{3}}}{2(R/2)}=\frac{-3GM}{8R}\] \[\therefore \]Potential at the centre of cavity formed \[{{V}_{p}}={{V}_{1}}-{{V}_{2}}=\frac{-11GM}{8R}-\left( \frac{-3GM}{8R} \right)=\frac{-GM}{R}\]