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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)

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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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