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 :

(G5 gravitational constant)

 

  • Option 1)

    \frac{-GM}{2R}\;

  • Option 2)

    \frac{-GM}{R}\;

  • Option 3)

    \frac{-2GM}{3R}\;

  • Option 4)

    \frac{-2GM}{R}

 

Answers (1)
A Aadil Khan

As we learnt in

Potential due to Uniform solid sphere -

R\rightarrow Radius of sphere

M\rightarrow Mass of sphere

r\rightarrow distance from centre of sphere

V_{centre}=\frac{3}{2}V_{surface}

- wherein

Outside the surface

V=-\frac{GM}{r}

on the surface

V_{surface}=-\frac{GM}{R}

Inside the surface 

r<R

V=-\frac{GM}{2R}\left[3-\left ( \frac{r}{R} \right )^{2}\right]

V_{1}\:=\frac{-GM}{2R^{3}}\:\left (3R^{2}-\left ( \frac{R}{2} \right )^{2} \right )

=\:\frac{-GM}{2R^{3}}\:(3R^{2}-\frac{R^{2}}{4})\:=\frac{-11GM}{8R}

Mass of spherical portion to be removed

M'\:=\frac{MV'}{V}\:=\:\frac{M\frac{4}{3}\pi (\frac{R}{2})^{3}}{\frac{4}{3}\pi R^{3}}\:=\frac{M}{8}

Potential at point P due to spehrical portion to be removed V_{2}\:=\frac{-3GM'}{2R'}\:=\frac{-3G(\frac{M}{8})}{2(\frac{R}{2})}\:=\frac{-3GM}{8R}

\therefore V_{P}\:=V_{1}-V_{2}\:=\frac{-11GM}{8R}-(\frac{-3GM}{8R})=\frac{-GM}{R}

 


Option 1)

\frac{-GM}{2R}\;

This option is incorrect.

Option 2)

\frac{-GM}{R}\;

This option is correct.

Option 3)

\frac{-2GM}{3R}\;

This option is incorrect.

Option 4)

\frac{-2GM}{R}

This option is incorrect.

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