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JEE Mains · Physics · STD 11 - 7. gravitation
Two identical particles each of mass \(m\) go round a circle of radius \(a\) under the action of their mutual gravitational attraction. The angular speed of each particle will be
- A \(\sqrt{\frac{G m}{2 a^3}}\)
- B \(\sqrt{\frac{G m}{8 a^3}}\)
- C \(\sqrt{\frac{G m}{4 a^3}}\)
- D \(\sqrt{\frac{G m}{a^3}}\)
Answer & Solution
Correct Answer
(C) \(\sqrt{\frac{G m}{4 a^3}}\)
Step-by-step Solution
Detailed explanation
\(F = m \omega^2 r\) \(\Rightarrow \frac{G m m}{(2 a)^2}=m \omega^2 a\) \(\Rightarrow \omega=\sqrt{\frac{G m}{4 a^3}}\)
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