Q.

A solid sphere of radius R gravitationally attracts a particle placed at 3R from its centre with a force $F_{1}$ . Now a spherical cavity of radius $(\frac{R}{2})$ is made in the sphere (as shown in figure) and the force becomes $F_{2}$ . The value of $F_{1} : F_{2}$ is:

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50:41

25:36

41:50

36:25

answer is A.

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Detailed Solution

$\begin{array}{l} F_{1} = \frac{G M m}{(3 R)^{2}} = \frac{G M m}{9 R^{2}} \dots \dots \dots . . (1) \\ F_{2} = \frac{G M m}{9 R^{2}} - \frac{G (\frac{M}{8}) m}{{(\frac{5 R}{2})}^{2}} \\ \Rightarrow F_{2} = \frac{G M m}{9 R^{2}} - \frac{G M m}{50 R^{2}} = \frac{G M m}{R^{2}} (\frac{1}{9} - \frac{1}{50}) = \frac{41}{450} \times \frac{G M m}{R^{2}} \dots \dots \dots \dots . . (2) \\ Dividing (1) & (2) \\ \Rightarrow \frac{F_{1}}{F_{2}} = \frac{\frac{G M m}{9 R^{2}}}{\frac{41}{450} \frac{G M m}{R^{2}}} = \frac{50}{41} \end{array}$

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