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Q.

A solid sphere of uniform density and radius R applies a gravitational force of attraction $F_{1}$ 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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a

(1/2)

b

(3/4)

c

(7/8)

d

(7/9)

answer is D.

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Detailed Solution

$F_{1} = \frac{G M m}{{(2 R)}^{2}} = \frac{x}{4} - - - (1), w h e r e \frac{G M m}{R^{2}} = x F' = \frac{G M' m}{{(\frac{3}{2} R)}^{2}} = \frac{G (M / 8) m}{(9 / 4) R^{2}} = \frac{x}{18} (a s M ’ i s t h e m a s s o f t h e c a v i t y) T h u s F_{2} = F_{1} - F' = \frac{x}{4} - \frac{x}{18} = \frac{7 x}{36} - - - (2) f r o m (1) a n d (2) w e g e t o r \frac{F_{2}}{F_{1}} = \frac{7 x}{36} \times \frac{4}{x} = \frac{7}{9}$

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A solid sphere of uniform density and radius R applies a gravitational force of attraction F1 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 F2 on the same particle. The ratio F2/F1 is: