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Question

# A uniform thin rod of length l and mass m is hinged at a distance l4 from one of the end and released from horizontal position as shown in the figure. The angular velocity of the rod as it passes the vertical position is : (acceleration due to gravity =g)

A
25g7l
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B
26g7l
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C
3g7l
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D
2gl
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Solution

## The correct option is B 2√6g7lUsing conservation of energy.The cm of the rod moves by a distance l4 when the rod is vertical.Decrease in potential energy is converted into rotational kinetic energy.Δ(PE)=+Krotation=12Iω2=12(ml212+m(l4)2)×ω2=12ml2×748ω2∴mgl4=12ml2×748ω2⇒ω2=24g7l⇒ω=2√6g7l

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