A disc of radius R=10cm, oscillations as a physical pendulum. The pendulum oscillates about an axis perpendicular to the plane of the disc at a distance r from its centre. For `r=R/4`, the approximate period of oscillation is _________`pi` s.(Take `g=10ms^(-2)`)

Answer

`00.30`

Explanation

`T=2pisqrt(I/(mgh))`
`I=(mR^(2))/2+mr^(2)`
`=(mR^(2))/2+m(R/4)^(2)`
`=(mR^(2))/2+(mR^(2))/16`
`=9/16mR^(2)`
Here R=10cm,0.1m,`h=R/4`
`impliesT=2pisqrt(((9mR^(2))/16)/((mgR)/4))=2pisqrt((9R)/(4g))=2pisqrt((9xx0.1)/(4xx10))`
`:.T=2pixx3/2xx1/10=0.3pis`