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Question

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

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√6g7l Applying conservation of mechanical energy Loss in gravitational P.E = Gain in rotational K.E ⇒mg(l4)=12Iω2 .......(1) Here, using parallel theorem I=Icm+md2 ⇒I=ml212+m(l4)2=ml212+ml216 ⇒I=7ml248 Substituting in Eq. (1) mg(l4)=12×7ml248ω2 ⇒ω=√24g7l=2√6g7l So, option (b) is correct. Why this question? Caution : while applying mechanical energy conservation, it should be ensured to use I about the hinged point.

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