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

Explanation

To solve the problem, we need to determine the rate of flow of water in the nozzle that allows a ping-pong ball of mass \( m \) to float in air when struck by water moving with speed \( v \). The water stops moving after hitting the ball, meaning its velocity becomes zero.

### Step-by-step Solution:

1. **Understand the System**:
- A ping-pong ball of mass \( m \) is floating in air due to the upward force provided by the water jet.
- The water strikes the ball with a speed \( v \) and comes to rest after the collision.

2. **Identify the Momentum Change**:
- The momentum of the water before the collision is given by \( \Delta p = \Delta m \cdot v \), where \( \Delta m \) is the mass of the water that strikes the ball in time \( \Delta t \).

3. **Force Acting on the Water**:
- After the collision, the water comes to a stop, which means the change in momentum must equal the force acting on it over the time interval. The force acting on the water is due to gravity, which is \( F = mg \).

4. **Relate Momentum Change to Force**:
- The change in momentum can also be expressed as \( F \cdot \Delta t \). Thus, we have:
\[ \Delta m \cdot v = mg \cdot \Delta t \]

5. **Rearranging the Equation**:
- We can rearrange this equation to find the mass flow rate:
\[ \Delta m = \frac{mg \cdot \Delta t}{v} \]

6. **Rate of Flow of Water**:
- The rate of flow of water, which is the mass of water flowing per unit time, is given by:
\[ \frac{\Delta m}{\Delta t} = \frac{mg}{v} \]

### Final Expression:
- Therefore, the rate of flow of water in the nozzle is:
\[ \frac{\Delta m}{\Delta t} = \frac{mg}{v} \]

### Conclusion:
- The correct answer for the rate of flow of water in the nozzle is \( \frac{mg}{v} \).