A uniform rod of mass `m` and length `l` is placed over a smooth horizontal surface along the `y`-axis and is at rest as shown in Fig. An impulsive force `F` is applied for a small time `/_\t` along `x`-direction at point `A`. The `x`-coordinate of end `A` of the rod when the rod becomes parallel to `x`-axis for the first time is [initially, the coordinate of centre of mass of the rod is `(0, 0)`]

Options

A. `(pil)/12`
B. `l/2(1+pi/12)`
C. `1/2(1-pi/6)`
D. `l/2(1+pi/6)`

Answer

D

Explanation

As torque `=`change in momentum
`F/_\t=mv`(linear momentum)………i
and `(Fl/2)/_\t=(ml^(2))/12 omega` (angular momentum)..ii
Dividing eqn i and ii we get
`2=(12v)/(omegal)implies omega=(6v)/l`
using `S=ut`
Displacement of `CM` is `pi/2=omegat=((6v)/l)t`
and `=vt`
Dividing we get
`(2x)/pi=l/6impliesx=(pil)/12`
coordinates of A will be `[(pil)/12+l/2, 0]`