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 $2R$ from the centre of the sphere. A spherical cavity of radius $R/2$ is now made in the sphere as shown in the figure. The sphere with the cavity now applies a gravitational force $F_2$ on the same particle. The ratio $F_1/F_2$ is

• A. $\frac{1}{2}$
• B. $\frac{3}{4}$
• C. $\frac{7}{8}$
• D. $\frac{9}{7}$

Answer 1

The correct option is D $\frac{9}{7}$

$\text{Let the particle of mass}m\text{be placed on A Then,}$
$F_1=\frac{GMm}{(2R{)}^2}=\frac{GMm}{4R^2}$
$\text{When a spherical part of radius}R/2\text{is taken out from the big sphere,}$
$\text{then the mass of remaining sphere becomes-}$
$=[\frac{4\pi R^3}{3}-\frac{4\pi }{3}(\frac{R}{2}{)}^3]d=\frac{4\pi R^3}{3}\left[\frac{7}{8}\right]d=\frac{7M}{8}[\therefore M=\frac{4\pi R^3}{3}]$
$\text{Now force on m place at A.}{F}_2=\frac{GMm}{4R^2}-\frac{GMm.4}{8\times 9R^2}=\frac{GMm(18-4)}{8\times 9R^2}=\frac{7}{4}\times \frac{GMm}{9R^2}$
$\frac{F_1}{F_2}=\frac{\frac{GMm}{4R^2}}{\frac{7}{4}\times \frac{GMm}{9R^2}}=\frac{9}{7}$