One end of a light string of length `L` is connected to a ball and the other end is connected to a fixed point `O`. The ball is released from a rest at `t=0` with string horizontal and just taut. The ball then moves invertical circular path as shown. The time taken by ball to go from position `A` to `B` is `t_(1)` and from `B` to lowest position `B` to lowest position `C` is `t_(2)`. Let the velocity of ball at `B` is `vec(v)_(B)` and `C` is `vec(v)_(C)` respectively.

If `|vec(v)_(C)-vec(v)_(B)|=|vec(v)_(B)|`, then the value of `theta` as shown is `:`

Options

A. `cos^(-1)((1)/(4))^(1//3)`
B. `sin^(-1)((1)/(4))^(1//3)`
C. `cos^(-1)((1)/(2))^(1//3)`
D. `sin^(-1)((1)/(2))^(1//3)`

Answer

B

Explanation

`V_(B)=sqrt(2gLsintheta)` and `V_(C)=sqrt(2gL)`
`V_(C)=2V_(B)`
Then `2gL=4(2gL sin theta )`
or `sin theta =(1)/(4) ` or `theta=sin ^(-1)((1)/(4))`
`|vec(v)_(B)-vec(v)_(C)|=sqrt(v_(B)^(2)+v_(C)^(2)-2v_(B)v_(C)sin theta )=v_(B)`
`rArr v_(B)^(2)+v_(C)^(2)-2v_(B)v_(C)sin theta -v_(B)^(2)`
`v_(C)=2v_(B) sin theta`
`rArr sqrt(2gl)=2sqrt(2gl sin theta ) sin theta`

`:. sin ^(3) theta =(1)/(4) rArr sin theta =((1)/(4))^(1//3)`
`theta = sin ^(-1)((1)/(4))^(1//3)`