An installation consists of an electric motor which drives a water pump to lift 75 L of water per second to a height of 5m, where water is disbursed at neglible speed. If the motor consumes a power of 5 kW, then what is the efficiency (%) of the installation?
`[g = 10 ms^(-2)]`

Answer

To find the efficiency of the installation consisting of an electric motor and a water pump, we can follow these steps:

### Step 1: Understand the given data
- Volume of water pumped per second (V) = 75 L/s
- Height to which water is lifted (h) = 5 m
- Power consumed by the motor (P_motor) = 5 kW = 5000 W
- Acceleration due to gravity (g) = 10 m/s²

### Step 2: Convert volume of water to mass
Since 1 L of water has a mass of approximately 1 kg, we can convert the volume of water to mass:
\[ \text{Mass of water (m)} = 75 \, \text{L/s} = 75 \, \text{kg/s} \]

### Step 3: Calculate the work done to lift the water
The work done (W) to lift the water can be calculated using the formula:
\[ W = mgh \]
Substituting the values:
\[ W = 75 \, \text{kg/s} \times 10 \, \text{m/s}^2 \times 5 \, \text{m} \]
\[ W = 75 \times 10 \times 5 = 3750 \, \text{J/s} = 3750 \, \text{W} \]

### Step 4: Calculate the efficiency of the installation
Efficiency (η) is defined as the ratio of useful power output to the total power input, expressed as a percentage:
\[ \eta = \left( \frac{\text{Power output}}{\text{Power input}} \right) \times 100\% \]
Substituting the values:
\[ \eta = \left( \frac{3750 \, \text{W}}{5000 \, \text{W}} \right) \times 100\% \]
\[ \eta = 0.75 \times 100\% = 75\% \]

### Final Answer
The efficiency of the installation is **75%**.

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