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. `"2 mg"/v`
B. `"mv"/v`
C. `"mg"/v`
D. none of these

Answer

C

Explanation

The impact force,
`F = (Deltap)/(Deltat)`
Where `DeltaP =` change of momentum of water of
mass `Delta m` striking the pendulum with speed v
during time `Deltat`
since water falls dead after collision with the
ping-pong ball.
`Δp = Δmν`
or `F=v(Deltam)/(Deltat)`
Where `(Deltam)/(Deltat)=` rate of flow of water in the nozzle since, the ball is in equilibrium,
`F - mg = 0 or F = mg`
`:. v(Deltam)/(Deltat)=mg`
`(Deltam)/(Delta t)= (mg)/upsilon`