A solid sphere of radius R/2 is cut out of a solid sphere of radius ,R such that the spherical cavity so formed touches the surface on one side and the centre of the sphere on the other side as shown in fig. The initial
mass of the solid sphere was M. If a particle of mass m is placed at a distance 2.5 R from the centre of the cavity, then the gravitational potential at the mass ,n will be

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$\frac{GMm}{4 R^{2}}$

$\frac{GMm}{R^{2}}$

$\frac{23}{100} \frac{GMm}{R^{2}}$

$\frac{GMm}{8 R^{2}}$

answer is D.

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Detailed Solution

Let M' be the mass removed. If p be the density, then
$ρ = \frac{M}{(4 / 3 \cdot {πR}^{3})}$
Now $M^{'} = ρ [\frac{4}{3} π {(\frac{R}{2})}^{2}]$
$= \frac{M}{(4 / 3 {πR}^{3})} \times [\frac{4}{3} π {(\frac{R}{2})}^{2}] = \frac{M}{8}$
So, gravitational force =force due to solid sphere of mass M - force due to mass removed to produce the cavity
$\begin{array}{l} ∴ \frac{GMm}{(2 R)^{2}} - \frac{GMm}{8 \times (2 \cdot 5 R)^{2}} \\ = \frac{GmM}{R^{2}} [\frac{1}{4} - \frac{1}{50}] \\ = \frac{23}{100} \times \frac{GMm}{R^{2}} \end{array}$