Question

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

Answer: (A)

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,
$M=\frac{4}{3}\pi {\left(\frac{R}{3}\right)}^3{\rho }_0+\left\{\frac{4}{3}\pi {\left(\frac{3}{4}R\right)}^3-\frac{4}{3}\pi {\left(\frac{R}{3}\right)}^3\right\}\frac{{\rho }_0}{2}$
$+\left\{\frac{4}{3}\pi R^3-\frac{4}{3}\pi {\left(\frac{3}{4}R\right)}^3\right\}\frac{{\rho }_0}{8}$
$=\frac{4}{3}\pi R^3{\rho }_0\left\{\frac{1}{27}+\frac{27}{128}-\frac{1}{54}+\frac{1}{8}-\frac{27}{512}\right\}$
$=\frac{4}{3}\pi R^3{\rho }_0\left\{\frac{512+2916-256+1728-729}{13824}\right\}$
$=\frac{4}{3}\pi R^3{\rho }_0\left\{\frac{5156-985}{13824}\right\}$
$=\frac{4}{3}\pi R^3{\rho }_0\left\{\frac{4171}{13824}\right\}$
$=0.402\pi R^3{\rho }_0$
$\therefore E=\frac{G\left(0.402\pi R^3{\rho }_0\right)}{{\left(2R\right)}^2}$
$=0.1\pi GR{\rho }_0$