Question

A non-homogeneous sphere of radius $R$ has the following density variation :
$\rho = \begin{cases} \rho_0; r \le R/3\\ \rho_0/2; (R/3) < r \le (3R/4) \\ \rho_0/8;(3R/4) < r \le R \end{cases} $
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,
$M = \frac{4}{3}\pi\left(\frac{R}{3}\right)^{3}\,\rho_{0} + \left\{\frac{4}{3}\pi\left(\frac{3}{4}R\right)^{3}-\frac{4}{3}\pi\left(\frac{R}{3}\right)^{3}\right\} \frac{\rho_{0}}{2}$
$+\left\{\frac{4}{3}\pi R^{3}-\frac{4}{3}\pi\left(\frac{3}{4}R\right)^{3}\right\} \frac{\rho_{0}}{8}$
$= \frac{4}{3}\pi R^{3}\rho_{0}\left\{\frac{1}{27}+\frac{27}{128}-\frac{1}{54}+\frac{1}{8}-\frac{27}{512}\right\}$
$= \frac{4}{3}\pi R^{3}\rho _{0}\left\{\frac{512+2916-256+1728-729}{13824}\right\}$
$= \frac{4}{3}\pi R^{3}\rho _{0}\left\{\frac{5156-985}{13824}\right\}$
$= \frac{4}{3}\pi R^{3}\rho _{0}\left\{\frac{4171}{13824}\right\}$
$= 0.402\pi R^{3}\rho _{0}$
$\therefore E = \frac{G\left(0.402\pi R^{3}\rho _{0}\right)}{\left(2R\right)^{2}}$
$= 0.1\pi GR\rho_{0}$