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)|=2|vec(v)_(B)|` then `:`

Options

A. `t_(1)gtt_(2)`
B. `t_(1)ltt_(2)`
C. `t_(1)=t_(2)`
D. information insufficient

Answer

B

Explanation

Tangential acceleration is `a_(t)=gcostheta`
which decreases with time.
Hence the plot of `a_(t)` versus time may be as shown in graph. Area under graph in time interval `t_(1)=v_(B)-0=v_(B)`
Area under graph in time interval `t_(2)=v_(C)-v_(B)=v_(B)`
Hence area under graph in time `t_(1)` and `t_(2)` is same.
`becauset_(1)gtt_(2)`