Question

A solid sphere of uniform density and radius $R$ applies a gravitational force of attraction equal to $F_1$ on a particle placed at a distance 3R from the centre of the sphere. A spherical cavity of radius $\frac{R}{2}$ is now made in the sphere, as shown in the figure. The sphere with cavity now applies a gravitational force $F_2$ on the same particle. The ratio $\frac{F_2}{F_1}$ is

• A. $\frac{41}{50}$
• B. $\frac{22}{25}$
• C. $\frac{3}{25}$
• D. $\frac{9}{50}$

Answer 1

Answer: (A)

From super position principle ,

$F_1=F_r+F_c$

Here $F_r=$ force due to remaining part $=F_2$

$F_c=$ force due to mass on the cavity

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

$F_c=\frac{G\left(\frac{M}{8}\right)m}{(\frac{5}{2}R{)}^2}=\frac{GMm}{50R^2}$

$=\succ F_2=F_1-F_c=\frac{GMm}{9R^2}-\frac{GMm}{50R^2}=\frac{41GMm}{450R^2}$

$\Rightarrow \frac{F_2}{F_1}=\frac{41}{50}$