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A simple pendulum of mass m and length l stands in equilibrium in vertical position. The maximum horizontal velocity that should be given to the bob at the bottom so that it completes one revolution is(A) \(\sqrt{lg}\)(B) \(\sqrt{2lg}\)(C) \(\sqrt{3lg}\)(D) \(\sqrt{5lg}\) |
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Answer» Answer is (D) \(\sqrt{5lg}\) |
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