A uniform stick of length `l` is mounted so as to rotate about a horizontal axis perpendicular to the stick and at a distance `d` from the centre of mass. The time period of small oscillation has a minimum value when `d//l` is

Options

A. `(l)/(sqrt(4))`
B. `(l)/(sqrt(2))`
C. `(l)/(sqrt(3))`
D. `(l)/(sqrt(12))`

Answer

D

Explanation

Time period of the rod oscillating about a horizotal axis with centre of mass at a distance `x` is
`T = 2pi sqrt((I+Mx^(2))/(Mgx))` For least `T, = (dT)/(dx) =0`
`(dT)/(dx) = 2x [(sqrt(Mgx).(1)/(2)(1+Mx^(2))^((-1)/(2)).2Mx-(I+Mx)^(2))/(Mgx)]=0`
`:. (1)/(2) sqrt((Mgx)/(I +Mx^(2)))2Mx = I +Mx^(2)`
`sqrt(Mgx.Mx) =(I +Mx^(2))^((3)/(2))` Solve to get `x` and substitute back in `T` to get `x_("least") = (l)/(sqrt(12))`