The sphere at `A` is given a downward velocity `v_(0)` of magnitude `5m//s` and swings in a vertical plane at the end of a rope of length `l=2m` attached to a support at `O` . Determine the angle `theta` at which the rope will breack, knowing that it can withstant a maximum tension equal to twich the weight of the sphere.

Answer

A, D

Explanation

`v^(2)=v_(0)^(2)+2gh=v_(0)^(2)+2gRsintheta`

`=(5)^(2)+2xx10xx2sintheta`
`=(25+40sintheta)`
Now, `2mg-mgsintheta=(mv^(2))/(R)`
or `2g-gsintheta=(v^(2))/(R)`
or `2xx10-10sintheta=25+40sintheta=(25+40sintheta)/(2)`
Solving this equation we get,
`theta=sin^(-1)((1)/(4))`