Question

The gravitational force acting on a particle, due to a solid sphere of uniform density and radius $R$, at a distance of $3R$ from the centre of the sphere is $F_1$. 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 $F_2$ on the same particle. Ratio of $F_1$ to $F_2$ is :

• A. $\frac{50}{41}$
• B. $\frac{41}{50}$
• C. $\frac{41}{25}$
• D. $\frac{25}{41}$

Answer 1

Answer: (A)

$\frac{50}{41}$

Let the mass of the bigger sphere, smaller sphere and particle be $M,M^{\prime }$ and $m$ respectively.
The sphere behaves as a particle of same mass kept at its center, hence,

$F_1=\frac{GMm}{9R^2}$.

In second case, net force will be $F_1$ minus force due to sphere of radius $\frac{R}{2}$ kept in place of hole.

For a sphere, $M\alpha R^3$,
Hence, mass of smaller sphere is $M^{\prime }=\frac{M}{8}$.

$F_2=\frac{GMm}{{9R}^2}-\frac{4G{M}^{\prime }m}{25R^2}=\frac{GMm}{{9R}^2}-\frac{GMm}{50R^2}=\frac{41GMm}{450R^2}$

Hence, $\frac{F_1}{F_2}=\frac{50}{41}$.