Question

A cavity of radius $\frac{R}{2}$ is made inside a solid sphere of radius R. The centre of the cavity is located at a distance $\frac{R}{2}$ from the centre of sphere . The gravitational force on a particle of mass m at a distance $\frac{R}{2}$ from the centre of the sphere on the line joining both the centres of sphere and cavity is ( opposite to the centre of cavity) [ here g = $\frac{GM}{R^2}$ where M is the mass of the sphere].

• A. $\frac{mg}{2}$
• B. $\frac{3mg}{8}$
• C. $\frac{mg}{16}$
• D. $\frac{mg}{4}$

Answer 1

Answer: (B)

$\frac{3mg}{8}$

Gravitational field at mass $m$ due to full solid sphere

${\overline{E}}_1=\frac{p\overline{r}}{3{\epsilon }_0}=\frac{pR}{6{\epsilon }_0}.......[{\epsilon }_0=\frac{1}{4\pi G}]$

Gravitational field at mass $m$ due to cavity $(-p)$

${\overline{E}}_2=\frac{(-p)(R/2{)}^3\overline{r}}{3{\epsilon }_0{R}^2}.......[usingE=\frac{p{a}^3}{3{\epsilon }_0{r}^2}]$

$-\frac{(-p)R^3}{24{\epsilon }_0{R}^2}=-\frac{-pR}{24{\epsilon }_0}$

Net gravitational field

$\overline{E}={\overline{E}}_1+{\overline{E}}_2=\frac{pR}{6{\epsilon }_0}-\frac{pR}{24{\epsilon }_0}=\frac{pR}{8{\epsilon }_0}$

Net force on $m\to F=m\overline{E}=\frac{mpR}{8{\epsilon }_0}$

Here, $p=\frac{M}{\left(4/3\right)\pi R^3}$ & ${\epsilon }_0=\frac{1}{4\pi G}$

Then, $F=\frac{3mg}{8}$