TS EAMCET · Physics · Gravitation
A uniform sphere A with radius R exerts a force F on a small particle B situated at a distance 2R from the centre of the sphere. A spherical portion of diameter R is cut from the sphere A as shown in the figure. If F′ is the new gravitational force between the remaining part of the sphere A and the particle B then the correct relation between F and F′.
- A \(F^{\prime}=\frac{9}{14} F\)
- B \(F^{\prime}=\frac{14}{9} F\)
- C \(F^{\prime}=\frac{7}{9} F\)
- D \(F^{\prime}=\frac{9}{7} F\)
Answer & Solution
Correct Answer
(C) \(F^{\prime}=\frac{7}{9} F\)
Step-by-step Solution
Detailed explanation
\(F = \frac{GMm}{(2R)^2} = \frac{GMm}{4R^2}\) Mass of removed part \(M_{cut} = M \left(\frac{R/2}{R}\right)^3 = \frac{M}{8}\) Distance of removed part's center from B \(r_{cut} = 2R - R/2 = \frac{3R}{2}\) \(F' = \frac{GMm}{(2R)^2} - \frac{G M_{cut} m}{r_{cut}^2}\)…
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