Question

A solid sphere of radius $R$ gravitationally attracts a particle placed at $3R$ form its centre with a force $F_1$. Now a spherical cavity of radius $\left(\frac{R}{2}\right)$ is made in the sphere (as shown in figure) and the force becomes $F_2$. The value of $F_1:F_2$ is

• A. 25: 36
• B. 36: 25
• C. 50: 41
• D. 41: 50

Answer 1

Answer: (C)

Let initial mass of sphere is $m$ '. Hence mass of removed portion will be $m^{\prime }/8$ $F_1=m.E.=\frac{m.G{m}^{\prime }}{9R^2}$

$F_2=m\left[\frac{G\cdot m^{\prime }}{(3R{)}^2}-\frac{G.m^1/8}{(5R/2{)}^2}\right]$
$=\frac{G{m}^{\prime }}{9R^2}-\frac{G{m}^{\prime }\times 4}{8\times 25}$
$=\left(\frac{1}{9}-\frac{1}{50}\right)\frac{G{m}^{\prime }}{R^2}$
$F_2=\frac{41}{50\times 9}\cdot \frac{G{m}^{\prime }}{R^2}$
$\frac{F_1}{F_2}=\frac{1}{9}\times \frac{50\times 9}{41}=\frac{50}{41}$