Question

A non-homogenous sphere of radius $R$ has the following density variation.

$\rho =\left\{\begin{array}{cc}{\rho }_0& r\le \frac{R}{3}\\ \frac{{\rho }_0}{2}& \frac{R}{3}\le r\le \frac{3R}{4}\\ \frac{{\rho }_0}{8}& \frac{3R}{4}\le r\le R\end{array}\right.$

The gravitational field at a distance $2R$ from the centre of the sphere is

• A. $0.1\pi GR{\rho }_0$
• B. $0.2\pi GR{\rho }_0$
• C. $0.3\pi GR{\rho }_0$
• D. $0.4\pi GR{\rho }_0$

Answer 1

The correct option is A

$0.1\pi GR{\rho }_0$

Step 1: Given data

1. Radius of the non-homogeneous sphere $R$
2. Density variation.

$\rho =\left\{\begin{array}{cc}{\rho }_0& r\le \frac{R}{3}\\ \frac{{\rho }_0}{2}& \frac{R}{3}\le r\le \frac{3R}{4}\\ \frac{{\rho }_0}{8}& \frac{3R}{4}\le r\le R\end{array}\right.$

Step 2: Formula used

1. $M=\underset{}{\overset{}{\frac{4\pi }{3}\sum }}{R}_i{\rho }_i$ where $M$ is the mass, $R_i$ is the radius ${\rho }_i$ is the density
2. $G\text{'}=\frac{GM}{{\left(R\right)}^2}$ where $G\text{'}$ is the gravitational field acted on a mass $M$ at a distance $R$

Step 3: Find the mass of the non-homogeneous sphere

Density variation.

$\rho =\left\{\begin{array}{cc}{\rho }_0& r\le \frac{R}{3}\\ \frac{{\rho }_0}{2}& \frac{R}{3}\le r\le \frac{3R}{4}\\ \frac{{\rho }_0}{8}& \frac{3R}{4}\le r\le R\end{array}\right.$

$$\begin{gathered} \therefore M=\underset{}{\overset{}{\frac{4\pi }{3}\sum }}{\left(R_i\right)}^3{\rho }_i \\ \Rightarrow M=\frac{4\pi }{3}\left[\left\{{\left(\frac{R}{3}\right)}^3-0\right\}{\rho }_0+\left\{{\left(\frac{3R}{4}\right)}^3-{\left(\frac{R}{3}\right)}^3\right\}\frac{{\rho }_0}{2}+\left\{{\left(R\right)}^3-{\left(\frac{3R}{4}\right)}^3\right\}\frac{{\rho }_0}{8}\right] \\ =\frac{4\pi R^3{\rho }_0}{3}\left[\frac{1}{27}+\left(\frac{27}{64}-\frac{1}{27}\right)\frac{1}{2}+\left(1-\frac{27}{64}\right)\frac{1}{8}\right] \\ =\frac{4\pi R^3{\rho }_0}{3}\left[\frac{4171}{13824}\right] \\ =0.402\pi R^3{\rho }_0 \end{gathered}$$

Step 4: Find the gravitational field at a distance $2R$

Distance$=2R$

Mass$=0.402\pi R^3{\rho }_0$

$$\begin{gathered} \therefore G\text{'}=\frac{GM}{{\left(R\right)}^2} \\ =\frac{0.402\pi G{R}^3{\rho }_0}{{\left(2R\right)}^2} \\ =0.1\pi GR{\rho }_0 \end{gathered}$$

Hence, option A is correct.