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A solid sphere of uniform density and radius R applies a gravitational force of attraction equal to $F_{1}$ on a particle placed at a distance 2R from the centre of the sphere. A spherical cavity of radius R/2 is now made in the sphere as shown in figure. The sphere with cavity now applies a gravitational force $F_{2}$ on the same particle. The ratio $(F_{2} / F_{1})$ is

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7/9

3/4

1/2

7/8

answer is D.

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Detailed Solution

We have the following:

$F_{1} = \frac{GMm}{{(2 R)}^{2}} = \frac{GMm}{4 R^{2}}$

and the mass of hollow portion of the sphere $= \frac{m}{8},$

Therefore, force applied by the mass of hollow portion of the sphere,

$F_{3} = \frac{GM (\frac{m}{8})}{{(\frac{3 R}{2})}^{2}} = \frac{GMm \times 4}{8 \times 9 R^{2}} = \frac{GMm}{18 R^{2}}$

$\Rightarrow F_{2} = F_{1} - F_{3} = \frac{GMm}{R^{2}} [\frac{1}{4} - \frac{1}{18}] = \frac{GMm}{R^{2}} [\frac{9 - 2}{36}]$

$\Rightarrow F_{2} = \frac{GMm}{4 R^{2}} \times \frac{7}{9} = \frac{7}{9} F_{1}$

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