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 $2\, R$ 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


$M_{r}=\int \rho d V=\int_{0}^{r} \rho_{0}\left(\frac{r}{R}\right)^{\beta} \cdot 4 \pi r^{2} d r$
$=\left[\frac{4 \pi \rho_{0}}{R^{\beta}} \cdot \frac{r^{\beta+3}}{\beta+3}\right]_{0}^{r}$
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
$E_{1}=\frac{G M_{1}}{\left(\frac{R}{2}\right)^{2}}=\frac{4 G}{R^{2}} \times \frac{4 \pi \rho_{0} R^{3}}{(\beta+3)} \times \frac{1}{2^{\beta+3}}$
and $E_{2}=\frac{G M_{2}}{(2 R)^{2}}=\frac{G}{4 R^{2}} \times \frac{4 \pi \rho_{0} R^{3}}{\beta+3}$
As $\frac{E_{2}}{E_{1}}=4$, we get
$\frac{\frac{G}{4 R^{2}} \times \frac{4 \pi \rho_{0} R^{3}}{\beta+3}}{\frac{4 G}{R^{2}} \times \frac{4 \pi \rho_{0} R^{3}}{\beta+3} \times \frac{1}{2^{\beta+3}}}=4$
$2^{\beta} .2^{3}=2^{6} \Rightarrow 2^{\beta}=8$
$\Rightarrow 2^{\beta}=2^{3}$
$\Rightarrow \beta=3$