A solid sphere of uniform density and radius R applies a gravitational force of attraction equal to F1 on a particle placed at P , distance 2 R from the centre O of the sphere. A spherical cavity of the radius R/2 is now made in the sphere as shown in the figure. The sphere with the cavity now applies a gravitational force F2 on the same particle placed at P . The ratio F2/F1 will be

Question

A solid sphere of uniform density and radius R applies a gravitational force of attraction equal to F1 on a particle placed at P , distance 2 R from the centre O of the sphere. A spherical cavity of the radius R/2 is now made in the sphere as shown in the figure. The sphere with the cavity now applies a gravitational force F2 on the same particle placed at P . The ratio F2/F1 will be <img class="img-fluid question-image" alt="Question" src="https://cdn.tardigrade.in/q/nta/p-rlcsmtrayrrw.png" /> (A) (1/2) (B) (7/9) (C) 3 (D) 7

Tardigrade Admin · Accepted Answer

Gravitational force due to solid sphere,

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

where M and m are mass of the solid sphere and particle respectively and R is the radius of the sphere.
The gravitational force on the particle due to sphere with cavity = force due to a solid sphere - force due to sphere creating a cavity, assumed to be present above at that position.
i.e.,

F_{2} = \frac{GM m}{4 R ^{2}} - \frac{G ( M /8 ) m}{( 3 R /2 ) ^{2}} = \frac{7}{36} \frac{GM m}{R ^{2}}

So

\frac{F _{2}}{F _{1}} = \frac{\frac{7 GM m}{36 R ^{2}}}{( \frac{GM m}{4 R ^{2}} )} = \frac{7}{9}

$\frac{1}{2}$

$\frac{7}{9}$

$3$

$7$

21​

97​

3

7

$\frac{1}{2}$

$\frac{7}{9}$

$3$

$7$