A uniform thin rod of length l and mass m is hinged at a distance l/4 from one of the end and released from horizontal position as shown in the figure. The angularvelocity of the rod as it passes the vertical position is

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25g7l

26g7l

3g7l

2gl

answer is B.

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Detailed Solution

Loss in potential energy = Gain in Kinetic energy⇒mgl4=12ml212+ml42ω2⇒ω=24g7l=26g7l

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