The sphere at `P` is given a downward velocity `v_(0)` and swings in a vertical plane at the end of a rope of `l=1m` attached to a support at `O`. The rope breaks at angle `30^(@)` from horizontal , knowing that it can withstand a maximum tension equal to three times the weight of the sphere. Then the value of `v_(0)` will be `:`
`(g=pi ^(2) m//s^(2))`

Options

A. `(g)/(2) m//s`
B. `(2g)/(3)m//s`
C. `sqrt((3g)/(2))m//s`
D. will have minimum value `(3)/(8)m`

Answer

C

Explanation

`T=mg sin theta=(mv^(2))/(R)`
`rArr 3mg-mg sin 30^(@)`
`=(m.(u_(0)^(2)+2glsin 30^(@)))/(l)`
`:. u_(0)=sqrt(3g//2)`