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

see full answer

Start JEE / NEET / Foundation preparation at rupees 99/day !!

21% of IItians & 23% of AIIMS delhi doctors are from Sri Chaitanya institute !!

An Intiative by Sri Chaitanya

$\frac{- G M}{R}$

$\frac{- 2 G M}{3 R}$

$\frac{- 2 G M}{R}$

$\frac{- G M}{2 R}$

answer is A.

(Unlock A.I Detailed Solution for FREE)

Ready to Test Your Skills?

Check your Performance Today with our Free Mock Test used by Toppers!

Take Free Test

Detailed Solution

The correct option is $B \frac{- G M}{R}$
Gravitational potential at any inside point is given as $V = - \frac{G M}{2 R^{3}} (3 R^{2} - r^{2})$ ….(i)
for $r = \frac{R}{2}, V = - \frac{11 G M}{8 R}$
Subtracting potential due to cavity of mass $M_{c} = \frac{M}{8}$ and $R_{c} = \frac{R}{2}$
Gravitational potential at center is obtained by substituting $r = 0$ in equation (i)
$= - \frac{3 G M_{c}}{2 R_{c}}$

$V = - \frac{11 G M}{8 R} - (- \frac{3 G M_{c}}{2 R_{c}}) = - \frac{11 G M}{8 R} + \frac{3 G \frac{M}{8}}{2 \frac{R}{2}} \Rightarrow V = - \frac{G M}{R}$