JEE Mains · Physics · STD 11 - Gravitation
- JEE Mains 2015
- Hard
- MCQ
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 \(V = 0\) at \(r = \infty\), the potential at the centre of the cavity thus formed is
( \(G =\) gravitational constant)

Pick an option to check your answer.
Answer & Solution
Correct Answer
(A) \(\frac{{ - GM}}{R}\)
Step-by-step Solution
Detailed explanation
Due to complete solid sphere, potential at point \(p\)
\({V_{sphere}} = \frac{{ - GM}}{{2{R^3}}}\left[ {3{R^2} - {{\left( {\frac{R}{2}} \right)}^2}} \right]\)
\( = \frac{{ - GM}}{{2{R^3}}}\left( {\frac{{11{R^2}}}{4}} \right) = - 11\frac{{GM}}{{8R}}\)
Due to cavity part potential at point \(P\)
\({V_{cavity}} = - \frac{3}{2}\frac{{\frac{{GM}}{8}}}{{\frac{R}{2}}} = - \frac{{3GM}}{{8R}}\)
So potential at the center of carvity
\( = {V_{sphere}} - {V_{cavity}} = - \frac{{11GM}}{{8R}} - \left( { - \frac{3}{8}\frac{{GM}}{R}} \right)\)
\( = \frac{{ - GM}}{R}\)
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