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
$\vec{E}_{1} = \frac{\rho\vec{r}}{3\varepsilon_{0}} = \frac{\rho R}{6\varepsilon_{0}} ...\left[\varepsilon_{0} =\frac{1}{4\pi G} \right]$
Gravitational field at mass ‘$m$’ due to cavity $(-\rho)$

$\vec{E}_{2} = \frac{\left(-\rho\right)\left(R/2\right)^{3}}{3\varepsilon_{0}R^{2}}...\left[\frac{\rho a^{3}}{3\varepsilon_{0}r^{2}}\right] $
$=-\frac{\left(-\rho\right)R^{3}}{24\varepsilon_{0}R^{2}}= -\frac{-\rho R}{24\varepsilon_{0}}$
Net gravitational field
$\vec{E} = \vec{E}_{1} + \vec{E}_{2} $
$= \frac{\rho R}{6\varepsilon_{0}} - \frac{\rho R}{24\varepsilon_{0}} = \frac{\rho R}{8\varepsilon_{0}}$
Net force on $'m' \to F = m\vec{E} = \frac{m\rho R}{8 \varepsilon_0}$
Here $\rho = \frac{M}{4/3 \pi R^3}$ & $\varepsilon_0 = \frac{1}{4\pi G}$
Then $F = \frac{3mg}{8}$