# From a solid sphere of mass M and radius R, a spherical portion of radius  is removed, as shown in the figure. Takinggravitational potential at  , the potential at the centre of the cavity thus formed is :(G5 gravitational constant) Option 1) Option 2) Option 3) Option 4)

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

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

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

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

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)

This option is incorrect.

Option 2)

This option is correct.

Option 3)

This option is incorrect.

Option 4)

This option is incorrect.

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