A motor pump of power 1 H.P is discharging water with a velocity of `10ms^(-1)`. In order to discharge water with a velocity of `20 ms^(-1)`, the power of the motor should be

Options

A. 2 H.P
B. 4 H.P
C. 8 H.P
D. 16 H.P

Answer

To solve the problem, we need to determine the power required for a motor pump to discharge water at a different velocity. We start with the relationship between power and velocity.

### Step-by-Step Solution:

1. **Understand the relationship between power and velocity**:
The power \( P \) required to discharge water is given by the formula:
\[ P \propto v^3 \]
This means that power is directly proportional to the cube of the velocity of the water being discharged.

2. **Identify the given values**:
- Initial power \( P = 1 \) H.P (Horsepower)
- Initial velocity \( v = 10 \, \text{m/s} \)
- Final velocity \( v' = 20 \, \text{m/s} \)

3. **Set up the proportionality equation**:
Using the relationship \( \frac{P'}{P} = \left(\frac{v'}{v}\right)^3 \), where \( P' \) is the new power we want to find.
\[ \frac{P'}{1 \, \text{H.P}} = \left(\frac{20 \, \text{m/s}}{10 \, \text{m/s}}\right)^3 \]

4. **Calculate the ratio of the velocities**:
\[ \frac{20}{10} = 2 \]

5. **Cube the ratio**:
\[ \left(2\right)^3 = 8 \]

6. **Calculate the new power**:
\[ P' = 1 \, \text{H.P} \times 8 = 8 \, \text{H.P} \]

### Final Answer:
The power of the motor should be **8 H.P**.

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