A pump of 200 W power is lifting 2 kg water per second from an average depth of 10m. The velocity with which water comes out of the pump (`g=9.8ms^(-2)`)

Options

A. `2 ms^(-1)`
B. `2.5 ms^(-1)`
C. `4 ms^(-1)`
D. `3 ms^(-1)`

Answer

To solve the problem, we need to find the velocity with which water comes out of the pump. We will use the concepts of potential energy and kinetic energy, along with the power equation.

### Step-by-Step Solution:

1. **Identify the Given Values:**
- Power of the pump, \( P = 200 \, \text{W} \)
- Mass of water lifted per second, \( m = 2 \, \text{kg} \)
- Height from which water is lifted, \( h = 10 \, \text{m} \)
- Acceleration due to gravity, \( g = 9.8 \, \text{m/s}^2 \)

2. **Calculate the Potential Energy (PE) for lifting the water:**
\[ \text{Potential Energy} (PE) = mgh \]
Substituting the values:
\[ PE = 2 \, \text{kg} \times 9.8 \, \text{m/s}^2 \times 10 \, \text{m} = 196 \, \text{J} \]

3. **Set Up the Power Equation:**
Power is defined as the rate of doing work or the rate of energy transfer. The total power used by the pump is the sum of the potential energy and the kinetic energy per unit time:
\[ P = \frac{PE + KE}{t} \]
Since we are lifting water at a rate of 2 kg per second, we can express this as:
\[ 200 = \frac{mgh + \frac{1}{2}mv^2}{1} \]

4. **Substituting the Values:**
\[ 200 = 196 + \frac{1}{2} \times 2 \times v^2 \]

5. **Simplifying the Equation:**
\[ 200 = 196 + v^2 \]
\[ v^2 = 200 - 196 \]
\[ v^2 = 4 \]

6. **Finding the Velocity (v):**
Taking the square root of both sides:
\[ v = \sqrt{4} = 2 \, \text{m/s} \]

### Final Answer:
The velocity with which water comes out of the pump is \( v = 2 \, \text{m/s} \).

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