1.

In the figure shwon there is non-conducting disc of mass M=2k and radius R=4m. On its upper & lower part of circumference +Q and -Q charge are uniformly attached such that liner charge density is Q/(piR). The disc can freely rotate about an horizontal axis passing through O. There is a uniform electric field barE in the vertical direction such that QE=mg. If the disc is rotated by a small angle it performs S.H.M. its time period is given by T=~~sqrt(( pi)/n) then, P+n, is (Take pi^(2)=g)

Answer»


SOLUTION :`:.tau=Ialphaimpliesomega=sqrt((8QE)/(PIMR))impliesomega=sqrt(2PI)=(2pi)/T`
`:. T=sqrt(2pi)`
`:.P+n` is 3


Discussion

No Comment Found

Related InterviewSolutions