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
• Correct answer
  C

  3
• D

  4

Solution

$$ \text { Mass enclosed in a sphere of radius } r \text { is } $$
$$ \begin{aligned} M_p & =\int \rho d V=\int_0^\rho \rho_0\left(\frac{r}{R}\right)^\beta \cdot 4 \pi r^2 d r \\ & =\left[\frac{4 \pi \rho_0}{R^\beta} \cdot \frac{r^{\beta+3}}{\beta+3}\right]_0^r \end{aligned} $$ 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 $$ \begin{array}{cc} & \frac{\frac{G}{4 R^2} \times \frac{4 \pi \rho_0 R^3}{\beta+3}}{\frac{4 G}{R^2} \times \frac{4 \pi \rho_0 R^3}{\beta+3} \times \frac{1}{2^{\beta+3}}}=4 \\ \Rightarrow \quad & \frac{2^{\beta+3}}{16}=4 \Rightarrow 2^{\beta+3}=64 \\ 2^\beta .2^3=2^6 \Rightarrow 2^\beta=8 \Rightarrow 2^\beta=2^3 \Rightarrow \beta=3 \end{array} $$

Hence, the correct answer is option: 'C'.