Question

A cavity of radius $R/2$ is made inside a solid sphere of radius $R$. The centre of the cavity is located at a distance $R/2$ from the centre of the sphere. The gravitational force on a particle of mass '$m$' at a distance $R/2$ 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 $g=GM/R^2$, where $M$ is the mass of the solid sphere]

• A. $\frac{mg}{2}$
• B. $\frac{3mg}{8}$
• C. $\frac{mg}{16}$
• D. none of these

Answer 1

Answer: (B)

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

${\vec{E}}_1=\frac{\rho \vec{r}}{3{\epsilon }_0}=\frac{\rho R}{6{\epsilon }_0}...\left[{\epsilon }_0=\frac{1}{4\pi G}\right]$

Gravitational field at mass ‘$m$’ due to cavity $(-\rho )$



${\vec{E}}_2=\frac{\left(-\rho \right){\left(R/2\right)}^3}{3{\epsilon }_0{R}^2}...\left[\frac{\rho a^3}{3{\epsilon }_0{r}^2}\right]$

$=-\frac{\left(-\rho \right)R^3}{24{\epsilon }_0{R}^2}=-\frac{-\rho R}{24{\epsilon }_0}$

Net gravitational field

$\vec{E}={\vec{E}}_1+{\vec{E}}_2$

$=\frac{\rho R}{6{\epsilon }_0}-\frac{\rho R}{24{\epsilon }_0}=\frac{\rho R}{8{\epsilon }_0}$

Net force on ${}^{\prime }{m}^{\prime }\to F=m\vec{E}=\frac{m\rho R}{8{\epsilon }_0}$

Here $\rho =\frac{M}{4/3\pi R^3}$ & ${\epsilon }_0=\frac{1}{4\pi G}$

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