A uniform semi circular ring of radius R and mass m is free to oscillate about its one end in its vertical plane as shown in the figure. Find time period of SHM of its centre of mass of the ring (take `pi^(2)=10`)

Options

A. `2pi sqrt((R )/(g))`
B. `2pisqrt((R )/(g)(sqrt((20)/(7))))`
C. `2pi sqrt((2R)/(g))`
D. `2pisqrt((R )/(g)((14)/(5)))`

Explanation

`T=2pi sqrt(((2mR^(2)))/(mgsqrt(R^(2)+((2R)/(n))^(2))))`
`=2pisqrt((2R)/(gsqrt((1+(4)/(pi^(2))))))=2pisqrt((2R)/(gsqrt(14)))`
`=2pi sqrt((R )/(g)sqrt((40)/(14)))=2pisqrt((R )/(g)(sqrt((20)/(7))))`