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CGPET 2016 Solved Paper

Section: Physics

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Question : 2 of 150

Marks: +1, -0

From a solid sphere of mass M and radius R, aspherical portion of radius $\frac{R}{2}$ is removed, asshown in figure. Taking gravitationalpotential $V = 0$ at $r = a$ . The potential at thecenter of the cavity thus formed is( $G =$ gravitational constant)

$\frac{-2GM}{3R}$
$\frac{-2GM}{R}$
$\frac{-GM}{2R}$
$\frac{-GM}{R}$

Solution:

It is given,

V = 0

at

r = \infty

Gravitational potential at internal point of a solid sphere at a distance r, is given by the following formula

V = -\frac{GM}{R}\left[\frac{3}{2}-\frac{r^2}{2R^2}\right]

Here,

r=\frac{R}{2}

V_1 = -\frac{GM}{R}\left[\frac{3}{2}-\frac{R^2}{8R^2}\right]

= -\frac{GM}{R}\left[\frac{12-1}{8}\right]

= -\frac{11}{8}\frac{GM}{R}

Because of sphere removed-

V_2 = \frac{3}{2} \cdot \frac{\frac{GM}{8}}{\frac{R}{2}} = \frac{3}{2} \times \frac{2GM}{8R}

\frac{3}{8} \frac{GM}{R}

The net potential at the center of the cavity

V = V_1 + V_2

= -\frac{11}{8} \frac{GM}{R} + \frac{3}{8} \frac{GM}{R}

= \frac{GM}{R} \left[\frac{3}{8} - \frac{11}{8}\right]

-\frac{GM}{R}

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