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

JEE Main · 2021 · 25 Feb 2021 · Shift 1 · Gravitation

A solid sphere of radius R gravitationally attracts a particle placed at

3 R

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 :

Answer: (B)

The particle sits at

3 R

from the centre of the uncut sphere, so

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

Cutting the cavity is the same as superposing a sphere of negative mass on the full sphere. The removed sphere has radius

\frac{R}{2}

, hence mass

M (\frac{1}{2})^{3} = \frac{M}{8}

, and its centre lies at

\frac{R}{2}

from the centre of the big sphere, that is at

3 R - \frac{R}{2} = \frac{5 R}{2}

from the particle. Both these spheres are complete, so each pulls as though its mass were at its centre:

F_{2} = \frac{GM m}{9 R ^{2}} - \frac{G ( M /8 ) m}{( 5 R /2 ) ^{2}} = \frac{GM m}{9 R ^{2}} - \frac{GM m}{50 R ^{2}} = \frac{41 GM m}{450 R ^{2}}

Dividing,

F_{1} : F_{2} = \frac{1}{9} : \frac{41}{450} = 50 : 41

JEE Main · 2021 · 25 Feb 2021 · Shift 1 · Single correct option