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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 $P$ , distance $2 R$ from the centre $O$ 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 same particle placed at $P$ . The ratio $F_{2} / F_{1}$ will be

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$\frac{7}{9}$

$3$

$\frac{1}{2}$

$7$

answer is B.

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Detailed Solution

$C \to$ cavity, $T \to$ Total, $R \to$ remaining

$F_{1} = \frac{G M f m}{(2 R)^{2}} . . . . . (i)$

$F_{K} = F_{2} = F_{r} - F_{C} = F_{1} - F_{C}$

or $F_{2} = \frac{G M m}{(2 R)^{2}} - \frac{G (\frac{M}{8}) m}{(3 R / 2)^{2}}$

or $F_{2} = \frac{14 G M m}{72 R^{2}} . . . . (i i)$

From Eqs. (i) and (ii) we get,

$\frac{F_{2}}{F_{1}} = \frac{7}{9}$

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