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 $F_{1}$. 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 $F_{1}$ and $F_{2}$ is

• A. $\frac{50}{41}$
• B. $\frac{41}{50}$
• C. $\frac{41}{42}$
• D. $\frac{25}{41}$

Answer 1

Answer: (A)

Gravitational force due to solid sphere is
$F_{1}=\frac{G M m}{(3 R)^{2}}=\frac{G M m}{9 R^{2}}$

where, $M$ and $m$ are mass of solid sphere and particle respectively. Gravitational force on particle due to sphere with cavity
$F_{2} =\frac{G M m}{9 R^{2}}-\frac{G\left(\frac{M}{8}\right) m}{(5 R / 2)^{2}}$
$=\frac{G M m}{R^{2}}\left[\frac{1}{9}-\frac{4}{8 \times 25}\right]$
$=\frac{G M m}{R^{2}}\left[\frac{41}{50 \times 9}\right]$
$\therefore \frac{F_{1}}{F_{2}}= \frac{50}{41}$