A light rod of length L can revolve in a vertical circle around point O. The rod carries two equal masses of mass m each such that one mass is connected at the end of the rod and the second mass is fixed at the middle of the rod. u is the velocity imparted to the end P to deflect the rod to the horizontal position. Again mass m in the middle of the rod is removed and mass at end P is doubled. Now v is the velocity imparted to end P to deflect it to the horizontal position. Then v/u is
Text Solution
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Using conservation of energy
In first case, 
= mgL + mg 
or u = 
In second case
(2) v 2 = 2 mgL v = (2gL) 1/2
∴ ∴ 
or v = 
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