Question

A uniform sphere of mass $M$ and radius $R$ exerts a force $F$ on a small mass $m$ situated at a distance of $2R$ from the centre $O$ of the sphere. A spherical portion of diameter $R$ is cut from the sphere as shown in the figure. The force of attraction between the remaining part of the sphere and the mass $m$ will be:

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

Answer 1

The correct option is A

$\frac{7F}{9}$

The force of attraction between the complete sphere and mass $m$ is:
$F=\frac{GmM}{(2R{)}^2}=\frac{GmM}{4R^2}$
Mass of complete sphere is: $M=\frac{4\pi }{3}{R}^3\times \rho$
where $\rho$ is the density of the sphere. [Density $=\frac{\text{Mass}}{\text{Volume}}$]
Mass of the cut-out portion is $m_{\circ }=\frac{4\pi }{3}{\left(\frac{R}{2}\right)}^3\times \rho =\frac{M}{8}$
Now, the distance between the centre of the cut out portion and mass m: $2R-\frac{R}{2}=\frac{3R}{2}$
Hence, the force of attraction between the cut out portion and mass m is:
$f=\frac{G{m}_{\circ }m}{{\left(\frac{3R}{2}\right)}^2}=\frac{G\left(\frac{M}{8}\right)m}{\frac{9R^2}{4}}$

$=\frac{GmM}{4R^2}\times \frac{2}{9}=\frac{2F}{9}$
Therefore, the force of attraction between the remaining part of the sphere and mass m: $F-f=F-\frac{2F}{9}=\frac{7F}{9}$