The average power of the pump which lifts 20kg of water per second from a well 10m deep and projects it with a velocity `10m//s` is `(g=10m//s^(2))`

Options

A. 9 kW
B. 4.5 kW
C. 2.5kW
D. 3kW

Explanation

To find the average power of the pump that lifts 20 kg of water per second from a well 10 meters deep and projects it with a velocity of 10 m/s, we can follow these steps:

### Step 1: Identify the given values
- Mass of water (m) = 20 kg
- Depth of the well (h) = 10 m
- Velocity of projection (v) = 10 m/s
- Acceleration due to gravity (g) = 10 m/s²

### Step 2: Calculate the change in kinetic energy
The change in kinetic energy (ΔKE) is given by the formula:
\[ \Delta KE = \frac{1}{2} m v^2 \]
Substituting the values:
\[ \Delta KE = \frac{1}{2} \times 20 \, \text{kg} \times (10 \, \text{m/s})^2 \]
\[ \Delta KE = \frac{1}{2} \times 20 \times 100 = 1000 \, \text{J} \]

### Step 3: Calculate the work done against gravity
The work done against gravity (Wg) is given by:
\[ W_g = mgh \]
Substituting the values:
\[ W_g = 20 \, \text{kg} \times 10 \, \text{m/s}^2 \times 10 \, \text{m} \]
\[ W_g = 20 \times 10 \times 10 = 2000 \, \text{J} \]

### Step 4: Calculate the work done by the pump
According to the work-energy theorem:
\[ W_p = \Delta KE + W_g \]
Substituting the values:
\[ W_p = 1000 \, \text{J} + 2000 \, \text{J} = 3000 \, \text{J} \]

### Step 5: Calculate the average power
Power (P) is defined as work done per unit time. Since the pump lifts 20 kg of water per second, the time (t) is 1 second:
\[ P = \frac{W_p}{t} \]
Substituting the values:
\[ P = \frac{3000 \, \text{J}}{1 \, \text{s}} = 3000 \, \text{W} \]

### Step 6: Convert power to kilowatts
To convert watts to kilowatts:
\[ P = \frac{3000 \, \text{W}}{1000} = 3 \, \text{kW} \]

### Final Answer
The average power of the pump is **3 kW**.

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