JEE Mains · Physics · STD 11 - 7. gravitation
A solid sphere of radius \(R\) gravitationally attracts a particle placed at \(3 R\) form its centre with a force \(F _{1}\). Now a spherical cavity of radius \(\left(\frac{ R }{2}\right)\) is made in the sphere (as shown in figure) and the force becomes \(F _{2}\). The value of \(F _{1}: F _{2}\) is
- A \(25: 36\)
- B \(36: 25\)
- C \(50: 41\)
- D \(41: 50\)
Answer & Solution
Correct Answer
(C) \(50: 41\)
Step-by-step Solution
Detailed explanation
Let initial mass of sphere is \(m ^{\prime} .\) Hence mass of removed portion will be \(m ^{\prime} / 8\) \(F _{1}= m . E .=\frac{ m \cdot Gm ^{\prime}}{9 R ^{2}}\) \(F_{2}=m\left[\frac{G \cdot m^{\prime}}{(3 R)^{2}}-\frac{G \cdot m^{\prime} / 8}{(5 R / 2)^{2}}\right]\)…
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