The gravitational force acting on a particle, due to a solid sphere of uniform density and radius R, at a distance of 3 R from the centre of the sphere is F1. A spherical hole of radius (R / 2) is now made in the sphere as shown in the figure. The sphere with hole now exerts a force F2 on the same particle. Ratio of F1 and F2 is

Question

The gravitational force acting on a particle, due to a solid sphere of uniform density and radius R, at a distance of 3 R from the centre of the sphere is F1. A spherical hole of radius (R / 2) is now made in the sphere as shown in the figure. The sphere with hole now exerts a force F2 on the same particle. Ratio of F1 and F2 is (A) (50/41) (B) (41/50) (C) (41/42) (D) (25/41)

Tardigrade Admin · Accepted Answer

Gravitational force due to solid sphere is

F_{1} = \frac{GM m}{( 3 R ) ^{2}} = \frac{GM m}{9 R ^{2}}

where,

M

and

m

are mass of solid sphere and particle respectively. Gravitational force on particle due to sphere with cavity

F_{2} = \frac{GM m}{9 R ^{2}} - \frac{G ( \frac{M}{8} ) m}{( 5 R /2 ) ^{2}}

= \frac{GM m}{R ^{2}} [\frac{1}{9} - \frac{4}{8 \times 25}]

= \frac{GM m}{R ^{2}} [\frac{41}{50 \times 9}]

∴ \frac{F _{1}}{F _{2}} = \frac{50}{41}

$\frac{50}{41}$

$\frac{41}{50}$

$\frac{41}{42}$

$\frac{25}{41}$

4150​

5041​

4241​

4125​

$\frac{50}{41}$

$\frac{41}{50}$

$\frac{41}{42}$

$\frac{25}{41}$