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JEE Mains · Physics · STD 11 - Gravitation
  • JEE Mains 2013
  • Hard
  • MCQ

The gravitational field, due to the 'left over part' of a uniform sphere (from which a part as shown, has been 'removed out'), at a very far off point , \(P\), located as shown, would be (nearly)

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Answer & Solution

Correct Answer

(C) \(\frac{7}{8}\,\frac{{GM}}{{{x^2}}}\)

Step-by-step Solution

Detailed explanation

Let mass of smaller sphere (which has to be removed) is m

\(Radius = \frac{R}{2}\left( {from\,figure} \right)\)

\(\frac{M}{{\frac{4}{3}\pi {R^3}}} = \frac{m}{{\frac{4}{3}\pi {{\left( {\frac{R}{2}} \right)}^3}}}\)

\( \Rightarrow m = \frac{M}{8}\)

Mass of the left over part of the sphere 

\(M' = M - \frac{M}{8} = \frac{7}{8}M\)

Therefore gravitational filed due to the left over part of the sphere

\( = \frac{{GM'}}{{{x^2}}} = \frac{7}{8}\frac{{GM}}{{{x^2}}}\)

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