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A uniform rod AB of mass m and length l ...

A uniform rod `AB` of mass `m` and length `l` at rest on a smooth horizontal surface . An impulse `P` is applied to the end `B`. The time taken by the rod to turn through a right angle is `:`

A

`(2pi m l)/(P)`

B

`(pi m l)/(3P)`

C

`(pi ml)/(12P)`

D

`(2pi ml)/(3P)`

Text Solution

Verified by Experts

The correct Answer is:
C
`(C )` Impulse `=` change in momentum
`:. P. (l)/(2) = (ml^(2))/(12). omega (` about centre of `AB)`
`rArr omega =(6P)/(ml)`
For `theta =(pi)/(2), (pi)/(2)=omegat`
`rArr t=(pi)/(2omega)=(piml)/(2xx6p)`
`rArr t=(piml)/(12p)`
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Knowledge Check

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