Solution:
To just complete the vertical loop, the velocity at the bottom is \(\sqrt{5gL}\). By energy conservation, the velocity at the horizontal position is \(v = \sqrt{3gL}\). The tension at this point is \(T = \frac{mv^2}{L} = 3mg\).
A particle of mass \(m\) is tied to a string of length \(L\) and rotated in vertical circle about other end with critical speed so that it is just able to complete the vertical loop. Then tension in string, when string is at horizontal position will be:
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