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(A) Let the center of the original sphere be $O$ and the center of the cavity be $P$. The distance $OP = R/2$.The gravitational potential at any point

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$\frac{-GM}{R}$

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(A) Let the center of the original sphere be $O$ and the center of the cavity be $P$. The distance $OP = R/2$.The gravitational potential at any point inside a solid sphere of mass $M$ and radius $R$ at a distance $r$ from its center is given by $V = \frac{-GM}{2R^3}(3R^2 - r^2)$.$1$. Potential at point $P$ due to the original solid sphere:Here,$r = OP = R/2$.$V_{sphere} = \frac{-GM}{2R^3} \left[ 3R^2 - (R/2)^2 ight] = \frac{-GM}{2R^3} \left( 3R^2 - \frac{R^2}{4} ight) = \frac{-GM}{2R^3} \left( \frac{11R^2}{4} ight) = \frac{-11GM}{8R}$.$2$. Potential at point $P$ due to the removed spherical portion (cavity):The mass of the removed portion $M'$ is proportional to the volume: $M' = M 	imes \frac{(4/3)\pi(R/2)^3}{(4/3)\pi R^3} = M/8$.The potential at the center of a solid sphere of mass $M'$ and radius $R' = R/2$ is $V_{cavity} = \frac{-3GM'}{2R'} = \frac{-3G(M/8)}{2(R/2)} = \frac{-3GM}{8R}$.$3$. Potential at the center of the cavity $(P)$:$V_{total} = V_{sphere} - V_{cavity} = \frac{-11GM}{8R} - \left( \frac{-3GM}{8R} ight) = \frac{-11GM + 3GM}{8R} = \frac{-8GM}{8R} = \frac{-GM}{R}$.

From a solid sphere of mass $M$ and radius $R$,a spherical portion of radius $R/2$ 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)

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From a solid sphere of mass $M$ and radius $R$,a spherical portion of radius $R/2$ 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)