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 `:`

Options

A. `2pi(ml)/(P)`
B. `(piP)/(ml)`
C. `(pi)/(12)(ml)/(P)`
D. `(piP)/(ml)`

Answer

C

Explanation

Angular impulse `=Pxx(l)/(2)`
= change in angular moment
`therefore" "(Pl)/(2)=l omega=((ml^(2))/(12))omega rArr omega=(6P)/(ml)`
`"Now, "t =(theta)/(omega)=(pi//2)/(6Plml)=(piml)/(12P)`