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JEE Mains · Physics · STD 11 - 7. gravitation
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)
- A \(\frac{5}{6}\,\frac{{GM}}{{{x^2}}}\)
- B \(\frac{8}{9}\,\frac{{GM}}{{{x^2}}}\)
- C \(\frac{7}{8}\,\frac{{GM}}{{{x^2}}}\)
- D \(\frac{6}{7}\,\frac{{GM}}{{{x^2}}}\)
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…
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