A motor is pumping water with a velocity of `20ms^(-1)` from a pump of area of cross section `10cm^(2)`. Its average power is

Options

A. 8kW
B. 12kW
C. 16kW
D. 4kW

Answer

To find the average power of the motor pumping water, we can follow these steps:

### Step 1: Convert the area from cm² to m²
The area of cross-section is given as \(10 \, \text{cm}^2\). We need to convert this to square meters.

\[ \text{Area} = 10 \, \text{cm}^2 = 10 \times 10^{-4} \, \text{m}^2 = 10^{-3} \, \text{m}^2 \]

### Step 2: Identify the density of water
The density of water (\(\rho\)) is approximately \(1000 \, \text{kg/m}^3\) or \(10^3 \, \text{kg/m}^3\).

### Step 3: Use the power formula
The formula for power (\(P\)) in terms of area (\(A\)), density (\(\rho\)), and velocity (\(v\)) is:

\[ P = \frac{1}{2} A \rho v^3 \]

### Step 4: Substitute the known values into the formula
Substituting the values we have:

- Area \(A = 10^{-3} \, \text{m}^2\)
- Density \(\rho = 10^3 \, \text{kg/m}^3\)
- Velocity \(v = 20 \, \text{m/s}\)

\[ P = \frac{1}{2} \times (10^{-3}) \times (10^3) \times (20)^3 \]

### Step 5: Calculate \(v^3\)
First, calculate \(20^3\):

\[ 20^3 = 20 \times 20 \times 20 = 8000 \]

### Step 6: Substitute \(v^3\) back into the power equation
Now substituting \(20^3\) back into the power equation:

\[ P = \frac{1}{2} \times (10^{-3}) \times (10^3) \times 8000 \]

### Step 7: Simplify the equation
The \(10^{-3}\) and \(10^3\) will cancel each other out:

\[ P = \frac{1}{2} \times 8000 \]

### Step 8: Final calculation
Now calculate:

\[ P = 4000 \, \text{W} \]

### Step 9: Convert to kilowatts
Since \(1 \, \text{kW} = 1000 \, \text{W}\):

\[ P = 4 \, \text{kW} \]

### Final Answer
The average power of the motor is \(4 \, \text{kW}\).

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