JEE PYQ: Gravitation Question 8
Question 8 - 2021 (24 Feb Shift 1)
Four identical particles of equal masses 1 kg made to move along the circumference of a circle of radius 1 m under the action of their own mutual gravitational attraction. The speed of each particle will be:
(1) $\sqrt{\frac{(1+2\sqrt{2})G}{2}}$
(2) $\sqrt{G(1+2\sqrt{2})}$
(3) $\sqrt{\frac{G}{2}(2\sqrt{2}-1)}$
(4) $\sqrt{\frac{G}{2}(1+2\sqrt{2})}$
Show Answer
Answer: (1)
Solution
For 4 particles on a circle of radius $R = 1$ m: net force toward centre provides centripetal force. Computing forces from 3 other particles: $F_1 + 2F_2\cos 45° = \frac{MV^2}{R}$. Solving gives $V = \frac{1}{2}\sqrt{G(1 + 2\sqrt{2})} = \sqrt{\frac{(1+2\sqrt{2})G}{2}}$ (with $R = 1$, $M = 1$).
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