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 `vecv_(B)` and at C is `vecv_(C)` respectively.
Q. If `|vecv_(C)|=2|vecv_(B)|` then the value of `theta `as shown is

Options

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

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))`
Tangential acceleration is `a_(t)=g cos theta`, 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 and `t_(2)` is same.
`:. t_(1) lt t_(2)`