From a solid sphere of radius R a point mass ‘m’ is placed at a distance 2R from centre. The force between them is ‘F’. Now solid sphere of radius $\frac{R}{2}$ was hollowed as shown then force on mass 'm' becomes

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$\frac{7}{3} F$

$\frac{2}{9} F$

$\frac{1}{2} F$

$\frac{7}{9} F$

answer is C.

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Detailed Solution

$F_{remaining} = F_{total} - F_{removed}$

$F_{total} = \frac{G M m}{4 R^{2}} \to M \propto R^{3}$

$F_{removed} = \frac{G \frac{M}{8} m}{{(\frac{3 R}{2})}^{2}} (distance = R + \frac{R}{2})$

$F_{remaining} = \frac{G M m}{4 R^{2}} - \frac{G M m}{18 R^{2}} = \frac{7 G M m}{9 (4 R^{2})} = \frac{7 F}{9}$

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