Question

The mass density inside a solid sphere of radius $r$ varies as $\rho (r)={\rho }_0{\left(\frac{r}{R}\right)}^{\beta }$, where ${\rho }_0$ and $\beta$ are constants and $r$ is the distance from the centre. Let $E_1$ and $E_2$ be gravitational fields due to sphere at distance $\frac{R}{2}$ and $2R$ from the centre of sphere. If $\frac{E_2}{E_1}=4$, the value of $\beta$ is

• A. 2
• B. 2.5
• C. 3
• D. 4

Answer 1

Answer: (C)

Mass enclosed in a sphere of radius r is


$M_r=\int \rho dV={\int }_0^r{\rho }_0{\left(\frac{r}{R}\right)}^{\beta }\cdot 4\pi r^{2}dr$
$={\left[\frac{4\pi {\rho }_0}{R^{\beta }}\cdot \frac{r^{\beta +3}}{\beta +3}\right]}_0^r$
So, mass enclosed in a sphere of radius $\frac{R}{2}$ is
$M_1=\frac{4\pi {\rho }_0}{R^{\beta }}\frac{{\left(\frac{R}{2}\right)}^{\beta +3}}{(\beta +3)}=\frac{4\pi {\rho }_0{R}^3}{(\beta +3)}\times \frac{1}{2^{\beta +3}}$
and mass enclosed in radius $R$ is
$M_2=\frac{4\pi {\rho }_0}{R^{\beta }}\cdot \frac{R^{\beta +3}}{\beta +3}=\frac{4\pi {\rho }_0{R}^3}{\beta +3}$
So, gravitational field intensities are
$E_1=\frac{G{M}_1}{{\left(\frac{R}{2}\right)}^2}=\frac{4G}{R^2}\times \frac{4\pi {\rho }_0{R}^3}{(\beta +3)}\times \frac{1}{2^{\beta +3}}$
and $E_2=\frac{G{M}_2}{(2R{)}^2}=\frac{G}{4R^2}\times \frac{4\pi {\rho }_0{R}^3}{\beta +3}$
As $\frac{E_2}{E_1}=4$, we get
$\frac{\frac{G}{4R^2}\times \frac{4\pi {\rho }_0{R}^3}{\beta +3}}{\frac{4G}{R^2}\times \frac{4\pi {\rho }_0{R}^3}{\beta +3}\times \frac{1}{2^{\beta +3}}}=4$
$2^{\beta }.2^3=2^6\Rightarrow 2^{\beta }=8$
$\Rightarrow 2^{\beta }=2^3$
$\Rightarrow \beta =3$