A disc of radius `R` and mass `M` is pivoted at the rim and it set for small oscillations. If simple pendulum has to have the same period as that of the disc, the length of the simple pendulum should be

Options

A. `5/4` R
B. `2/3` R
C. `3/4` R
D. `3/2` R

Answer

D

Explanation

`T_(D)=2pisqrt((I)/(mgd))`
`=2pi sqrt((((MR^(2))/(2)+MR^(2)))/(MgR))`
`=2pi sqrt((3R)/(2g))`
Since, `T_(P)=2pisqrt((l)/(g)) and T_(P)=T_(D)`
So, `2pisqrt((l)/(g))=2pisqrt((3R)/(2g))`
`therefore l=(3)/(2)R` (on comparing).