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The gravitational force acting on a particle due to a solid sphere uniform density and radius R, at a distance of 3R from the center of the sphere id $F_{1}$ . A spherical hole of radius $\frac{R}{2}$ is now made in the sphere as shown. The sphere with hole now exerts a force $F_{2}$ on the same particle. Ratio of $F_{1}$ and $F_{2}$ is

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$\frac{41}{50}$

$\frac{41}{42}$

$\frac{25}{41}$

$\frac{50}{41}$

answer is A.

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Detailed Solution

$F_{1} = \frac{G M m}{{(3 R)}^{2}} = \frac{G M m}{9 R^{2}}$
$F_{2} = \frac{G M m}{9 R^{2}} = \frac{G \frac{M}{8} m}{{(5 \frac{R}{2})}^{2}} = \frac{G M m}{R^{2}} [\frac{1}{9} - \frac{4}{8 \times 25}]$
$\frac{F_{1}}{F_{2}} = \frac{50}{41}$

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