Question

A non-homogeneous sphere of radius R has the following density variation:
$\rho \{\begin{array}{c}{\rho }_0;r\le R/3\\ {\rho }_0/2;\left(R/3\right)<r\le \left(3R/4\right)\\ {\rho }_0/8;\left(R/3\right)<r\le R\end{array}$
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 $0.1\pi GR{\rho }_0$

The gravitational field at a distance $2R$ from the centre of the sphere is $E=\frac{GM}{{\left(2R\right)}^2}$

where $M$ is the mass of the whole sphere. Here,

$M=\frac{4}{3}\pi {\left(\frac{4}{3}\right)}^3{\rho }_0+\left\{\begin{array}{c}\frac{4}{3}\pi {\left(\frac{3}{4}R\right)}^3-\frac{4}{3}\pi {\left(\frac{R}{3}\right)}^3\end{array}\right\}\frac{{\rho }_0}{2}+\left\{\begin{array}{c}\frac{4}{3}\pi R^3-\frac{4}{3}\pi {\left(\frac{3}{4}R\right)}^3\end{array}\right\}\frac{{\rho }_0}{8}$

$=\frac{4}{3}\pi R^3{\rho }_0\left\{\begin{array}{c}\frac{1}{27}+\frac{27}{128}-\frac{1}{54}+\frac{1}{8}-\frac{27}{512}\end{array}\right\}$

$=\frac{4}{3}\pi R^3{\rho }_0\left\{\begin{array}{c}\frac{512+2916-256+1728-729}{13824}\end{array}\right\}$

$=\frac{4}{3}\pi R^3{\rho }_0\left\{\begin{array}{c}\frac{5156-985}{13824}\end{array}\right\}=\frac{4}{3}\pi R^3{\rho }_0\left\{\begin{array}{c}\frac{4171}{13824}\end{array}\right\}$

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

$\therefore E=\frac{G\left(0.402\pi R^3{\rho }_0\right)}{{\left(2GR\right)}^2}=0.1\pi GR{\rho }_0$