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, distant $2R$ from the center of the sphere. A spherical cavity of radius $R/2$ is now made on the sphere with cavity now applies a gravitational force$F_2$. Then $\frac{F_2}{F_1}$ will be

• A. $1/2$
• B. $3/4$
• C. $7/8$
• D. $14/9$

Answer 1

The correct option is C $7/8$

Let, M is the mass of the sphere of uniform density $\rho$ and m is the mass of the particle.

The gravitational force of attraction $F_1$ is given by,

$F_1=\frac{GMm}{(2R{)}^2}$ ........................(1)

The mass of sphere of volume V, is given by

$M=\rho V$

$M=\rho \frac{4}{3}\pi R^3$ ......................(2)

From equations (1) and (2), we get

$F_1=\frac{G\left[\frac{4}{3}\pi R^3\right]m}{(2R{)}^2}$ ...................(3)

When the spherical cavity is made in sphere, the gravitational force $F_2$ will be

$F_2=\frac{G[M-\Delta M]m}{(2R{)}^2}$

$F_2=\frac{G[\frac{4}{3}\pi R^3-\frac{4}{3}\pi (\frac{R}{2}{)}^3]m}{(2R{)}^2}$

$F_2=\frac{G\left[\frac{4}{3}\pi R^3\right(1-\left(\frac{1}{2}{)}^3\right)]m}{(2R{)}^2}$ .......................(4)

Now,

$\frac{F_2}{F_1}=\frac{\frac{G\left[\frac{4}{3}\pi R^3\right(1-\left(\frac{1}{2}{)}^3\right)]m}{(2R{)}^2}}{\frac{G\left[\frac{4}{3}\pi R^3\right]m}{(2R{)}^2}}$ ...............from equations (3) and (4)

$\frac{F_2}{F_1}=1-(\frac{1}{2}{)}^3$

$\frac{F_2}{F_1}=1-\left(\frac{1}{8}\right)$

$\frac{F_2}{F_1}=\frac{7}{8}$