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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

a. \(\begin{array}{l}\sqrt{\frac{(1+2\sqrt{2})G}{2}}\end{array} \) b. \(\begin{array}{l}\sqrt{G(1+2\sqrt{2})}\end{array} \) c. \(\begin{array}{l}\sqrt{\frac{G}{2}(2\sqrt{2}-1)}\end{array} \) d. \(\begin{array}{l}\sqrt{\frac{G}{2}(1+2\sqrt{2})}\end{array} \) Answer: (a) ⇒ By resolving force F2, we... View Article