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One end of a light string of length L is...

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 rest at `t = 0` with string horizontal and just taut. The ball then moves in vertical 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 `C` is `t_(2)`. Let the velocity of ball at `B` is `vec v_(B)` and at `C` is `vec v_(C)` respectively.

If `|vec v_(C)=2|vecv_(B)|` then the value of `theta` as shown is

A

`cos^(-1)((1)/(4))`

B

`sin^(-1)((1)/(4))`

C

`cos^(-1)((1)/(2))`

D

`sin^(-1)((1)/(2))`

Text Solution

Verified by Experts

The correct Answer is:
B
`v_(B)=sqrt(2gLsintheta)` and `v_(C)=sqrt(2gL)`
if `V_(C)=2v_(B)`
Then `2gL=4(2gLsintheta)`
or `sintheta(1)/(4)orthetasin^(-1)((1)/(4))`
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