A pump is required to lift `600 kg` of water per minute from a well `25 m` deep and to eject it with a speed of `50 ms^(-1)`. The power required to perform the above task is

Options

A. 10 kW
B. 15 kW
C. 20 kW
D. 25 kW

Explanation

To solve the problem of calculating the power required by the pump to lift water from a well and eject it, we can break it down into a series of steps.

### Step-by-Step Solution:

1. **Identify the Given Data:**
- Mass of water (m) = 600 kg
- Depth of the well (h) = 25 m
- Speed of ejection (v) = 50 m/s
- Gravitational acceleration (g) = 10 m/s² (approximate value)

2. **Calculate the Work Done Against Gravity:**
The work done (W_g) to lift the water against gravity can be calculated using the formula:
\[ W_g = m \cdot g \cdot h \]
Substituting the values:
\[ W_g = 600 \, \text{kg} \cdot 10 \, \text{m/s}^2 \cdot 25 \, \text{m} = 150000 \, \text{J} \]

3. **Calculate the Kinetic Energy Required:**
The kinetic energy (KE) given to the water when it is ejected can be calculated using the formula:
\[ KE = \frac{1}{2} m v^2 \]
Substituting the values:
\[ KE = \frac{1}{2} \cdot 600 \, \text{kg} \cdot (50 \, \text{m/s})^2 = \frac{1}{2} \cdot 600 \cdot 2500 = 750000 \, \text{J} \]

4. **Calculate the Total Work Done:**
The total work done (W_total) is the sum of the work done against gravity and the kinetic energy:
\[ W_{total} = W_g + KE = 150000 \, \text{J} + 750000 \, \text{J} = 900000 \, \text{J} \]

5. **Calculate the Power Required:**
Power (P) is defined as the work done per unit time. Since the work is done in one minute (60 seconds), we can calculate the power as:
\[ P = \frac{W_{total}}{t} = \frac{900000 \, \text{J}}{60 \, \text{s}} = 15000 \, \text{W} \]
Converting watts to kilowatts:
\[ P = 15 \, \text{kW} \]

### Final Answer:
The power required to perform the above task is **15 kW**.

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