From a solid sphere of mass M and radius R , a spherical portion of radius $\mathrm{R} / 2$ is removed, as shown in the figure. Taking gravitational potential $\mathrm{V}=0$ at $\mathrm{r}=\infty$, the potential at the centre of the cavity thus formed is ( $\mathrm{G}=$ gravitational constant)

(A) $-\frac{\mathrm{GM}}{\mathrm{R}}$

(B) $-\frac{2 \mathrm{GM}}{3 \mathrm{R}}$

(C) $-\frac{2 \mathrm{GM}}{\mathrm{R}}$

(D) $\frac{-\mathrm{GM}}{2 \mathrm{R}}$

JEE Main · 2015 · 04 Apr 2015 · Offline · Gravitation

From a solid sphere of mass M and radius R , a spherical portion of radius

R /2

is removed, as shown in the figure. Taking gravitational potential

V = 0

r = \infty

, the potential at the centre of the cavity thus formed is (

G =

gravitational constant)

Answer: (A)

Use superposition: the hollowed object is the full sphere minus a small sphere that would fill the cavity. At a point

R /2

from the centre the complete solid sphere has potential

V_{full} = - \frac{GM}{2 R ^{3}} (3 R^{2} - \frac{R ^{2}}{4}) = - \frac{11 GM}{8 R}

. The removed plug has mass

M /8

and radius

R /2

, and we want the potential at its own centre,

V_{plug} = - \frac{3 G ( M /8 )}{2 ( R /2 )} = - \frac{3 GM}{8 R}

. Subtracting the plug's contribution,

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

JEE Main · 2015 · 04 Apr 2015 · Offline · Single correct option