A circle disc has a tiny hole in it, at a distance `z` from its center. Its mass is `M` and radius `E(Rgtz)`. A horizontal shaft is passed through the hole and held fixed so that the disc can freely swing in the vertical plane. For small distrubance, the disc performs `SHM` whose time period is minimum for `z=`

Explanation

The time period w.r.t the axis T `= 2pi sqrt((I)/(Mgd))`
where I = moment of inertia w.r.t the axis O
d = distance between C.O.M and O
`implies I = (MR^(2))/(2) + Mz^(2)`
d=z
`implies T = 2pi sqrt((((MR^(2))/(2) +Mz^(2)))/(Mgz)) = 2pisqrt((R^(2))/(2gz)+(z)/(g))`
the time period will be minimum when
`(R^(2))/(2z)+z` = minimum
Let say f `= (R^(2))/(2z)+z`
f will be minimum when `(df)/(dz)` = 0
`implies -(R^(2))/(2z^(2))+1=0 implies z = (R)/(sqrt(2))`