The power of a water pump is 2 kW. The amount of water (in litres) it can raise in one minute to a height of 10 m will be : (g = 10 m/s 2 )
1200
Calculating the Amount of Water Raised by a Water Pump
This problem involves understanding the concepts of power, work, and potential energy. The power of a water pump tells us how quickly it can do work. In this case, the work done by the pump is used to lift water against gravity, increasing its potential energy.
We are given the power of the pump, the height to which the water is lifted, and the time duration. We need to find the amount of water lifted, which can be determined by first finding the mass of water lifted and then converting it to volume in litres.
Let's list the given values:
- Power of the pump, \( P = 2 \) kW
- Height, \( h = 10 \) m
- Time, \( t = 1 \) minute
- Acceleration due to gravity, \( g = 10 \) m/s\(^2\)
First, we need to convert the power and time into standard SI units.
- Power: \( P = 2 \) kW \( = 2 \times 1000 \) W \( = 2000 \) W
- Time: \( t = 1 \) minute \( = 1 \times 60 \) seconds \( = 60 \) s
The power is defined as the rate at which work is done:
\( P = \frac{W}{t} \)
Where \( W \) is the work done and \( t \) is the time taken. We can calculate the work done by the pump in 60 seconds:
\( W = P \times t \)
\( W = 2000 \) W \( \times 60 \) s
\( W = 120000 \) J
This work done is used to increase the potential energy of the water. The potential energy gained by lifting a mass \( m \) to a height \( h \) is given by:
\( PE = mgh \)
Assuming all the work done by the pump is used to lift the water, the work done \( W \) equals the potential energy gained \( PE \):
\( W = mgh \)
We can now find the mass \( m \) of the water lifted:
\( 120000 \) J \( = m \times (10 \, \text{m/s}^2) \times (10 \, \text{m}) \)
\( 120000 = m \times 100 \)
Solving for \( m \):
\( m = \frac{120000}{100} \)
\( m = 1200 \) kg
Finally, we need to find the volume of water in litres. For water, the density is approximately 1 kg/litre (or 1000 kg/m\(^3\)). This means that 1 kg of water occupies a volume of approximately 1 litre.
Therefore, the volume of 1200 kg of water is:
Volume \( = 1200 \) litres
So, the water pump can raise 1200 litres of water in one minute to a height of 10 m.
Let's summarize the calculation steps:
| Step | Description | Calculation | Result |
|---|---|---|---|
| 1 | Convert Power to Watts | \(2 \, \text{kW} = 2 \times 1000 \, \text{W}\) | \(2000 \, \text{W}\) |
| 2 | Convert Time to Seconds | \(1 \, \text{minute} = 1 \times 60 \, \text{s}\) | \(60 \, \text{s}\) |
| 3 | Calculate Work Done (\(W = P \times t\)) | \(2000 \, \text{W} \times 60 \, \text{s}\) | \(120000 \, \text{J}\) |
| 4 | Calculate Mass (\(m = W / (g \times h)\)) | \(120000 \, \text{J} / (10 \, \text{m/s}^2 \times 10 \, \text{m})\) | \(1200 \, \text{kg}\) |
| 5 | Convert Mass (kg) to Volume (litres) | \(1200 \, \text{kg} \approx 1200 \, \text{litres}\) | \(1200 \, \text{litres}\) |
The amount of water raised is 1200 litres.
Important Questions from Power
Which one of the following is the value of 1 KWh of energy converted into joules?
Which of the following is correct regarding electric power?
An elevator weighing 400 kg is to be lifted up at a constant velocity of 0.20 m/s. What would be the minimum horsepower of the motor to be used? (g = 9.8 m/s 2and there is no frictional loss)
An elevator weighing 500 kg is to be lifted up at a constant velocity of 0.25 m/s. What would be the minimum horsepower of the motor to be used? G = 9.8 m/s2 and there is no frictional loss.
Units of power is:
