AP EAMCET · PHYSICS · Gravitation
The gravitational force acting on a particle, due to a solid sphere of uniform density and radius \(R\), at a distance of \(3 R\) from the centre of the sphere is \(F_1\). A spherical hole of radius \((R / 2)\) is now made in the sphere as shown in the figure. The sphere with hole now exerts a force \(F_2\) on the same particle. Ratio of \(F_1\) and \(F_2\) is
- A \(\frac{50}{41}\)
- B \(\frac{41}{50}\)
- C \(\frac{41}{42}\)
- D \(\frac{25}{41}\)
Answer & Solution
Correct Answer
(A) \(\frac{50}{41}\)
Step-by-step Solution
Detailed explanation
Gravitational force due to solid sphere is \[ F_1=\frac{G M m}{(3 R)^2}=\frac{G M m}{9 R^2} \] where, \(M\) and \(m\) are mass of solid sphere and particle respectively. Gravitational force on particle due to sphere with cavity…
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