A 25kg uniform solid with a 20cm radius respectively by a verticle wire such then the point suspended is velocity about the center of the sphere torque of 0.10N-m and then montain the sphere then at angle of 1.0 rad if sphere is then released its period of the oscillation will be

Options

A. `pi ` second
B. `sqrt(2)pi` second
C. `2pi` second
D. `4pi` second

Answer

D

Explanation

`(D) tau -k theta `
`0.1=-k(1.0),` where `k` is torsional constant o fthe wire.
`k=(1)/(10)`
`T=2pisqrt((I)/(k))=2pi sqrt(((2)/(5)xx25xx(.2)^(2))/(1//10))`
`=2pi sqrt(10xx.2xx.2xx10)=4pi ` second Ans.
`i=[(M(Rsqrt(2))^(2))/(12)+M((R)/(sqrt(2)))^(2)]xx4+mR^(2)`
`=20kgm^(2)`.
`(4M+m)g sin theta -F=(4M+m)a.`
`F.R.=I((a)/(R))`
Solving
`a=(7g)/(24)`
`F=20alemu(4M+m)g cos 30`
`muge(5)/(12sqrt(3))`
`:. mu _(m i n)=(5)/(12sqrt(3))`