Question

A activity of radius $R/2$ is made inside a solid sphere of radius $R$. The centre of cavity is located at a distance $R/2$ from the centre of the sphere. The gravitational force on a particle of mass ${}^{\prime }{m}^{\prime }$ 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 sphere]

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

Answer 1

The correct option is B $\frac{3mg}{8}$

$E_{1}1=\frac{Pr}{3{ϵ}_0}=\frac{pR}{6{ϵ}_0}...[{ϵ}_0=\frac{1}{4\pi G}$]

$E_2=\frac{-p}{{\frac{R}{2}}^3}\frac{r}{3{ϵ}_0}{R}^2....[usingE=\frac{p{a}^3}{3{ϵ}_0}{r}^2$]

$-\frac{-p}{R^3}24{ϵ}_0{R}^2=-\frac{-pR}{24{ϵ}_0}$

$E=E_1+E_2=\frac{pR}{6{ϵ}_0}-\frac{pR}{24{ϵ}_0}=\frac{pR}{8{ϵ}_0}$

Net force $m-->F=mE=\frac{mpR}{8{ϵ}_0}$

$p=\frac{M}{\frac{4}{3}}\pi R^{3}and{ϵ}_0=\frac{1}{4\pi G}$

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