JEE Mains · Physics · STD 11 - 7. 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 \(V = 0\) at \(r = \infty\), the potential at the centre of the cavity thus formed is
( \(G =\) gravitational constant)
- A \(\frac{{ - GM}}{R}\)
- B \(\;\frac{{ - 2GM}}{{3R}}\)
- C \(\;\frac{{ - 2GM}}{R}\)
- D \(\;\frac{{ - GM}}{{2R}}\)
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…
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