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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. The initial mass of the solid sphere was M. lf a particle of mass m is placed at a distance 2.5R from the centre of the cavity, then what is the gravitational attraction on the mass m?

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

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

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

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

answer is D.

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Detailed Solution

Let mass of the cavity $= M'$

Density of the sphere $=M/ (4 / 3 π R^{3})$

Mass of the cavity cut out $= M' = \frac{4}{3} π \frac{R^{3}}{8} \times \frac{M}{\frac{4}{3} {πR}^{3}}$

$\begin{array}{l} ∴ M' = \frac{M}{8} \Rightarrow F_{net} = F_{Mm} - F_{M' m} \\ = \frac{GMm}{4 R^{2}} - \frac{GM' m}{{(\frac{5}{2} R)}^{2}} = \frac{GMm}{4 R^{2}} - \frac{GMm}{50 R^{2}} \\ F_{net} = \frac{23}{100} \frac{GMm}{R^{2}} \end{array}$

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