A ping-pong ball of mass `m` is floating in air by a jet of water emerging out of a nozzle. If the water strikes the ping-gong ball with a speed `v` and just after collision water falls dead, the rate of flow of water in the nozzle is equal to

Options

A. `(2mg)/V`
B. `(mV)/g`
C. `(mg)/V`
D. none of these

Answer

C

Explanation

The impact force `F=(/_\p)/(/_\t)` where `/_\p=` change of momentum of water of mass `/_\m` striking the pendulum with a speed `v` during the `/_\t`. Since water falls dead after collision with the ping pong ball.
`/_\p=/_\mvimpliesF=v(/_\m)/(/_\t)`
where `(/_\m)/(/_\t)=` rate of flow of wate in the nozzle.
Since the ball is in equlibrium
`F-mg=0impliesF=mg`
` implies(v/_\m)/(/_\t)=mgimplies(/_\m)/(/_\t)=(mg)/v`