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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 isA. `cos^(-1)((1)/(4))`B. `sin^(-1)((1)/(4))`C. `cos^(-1)((1)/(2))`D. `sin^(-1)((1)/(2))` |
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Answer» Correct Answer - 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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