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 will follow these steps:

### Step 1: Determine the mass of water being lifted
Given that the volume of water is 75 liters and the density of water is 1 kg/L, we can calculate the mass of water (m) as follows:

\[ m = \text{Volume} \times \text{Density} = 75 \, \text{L} \times 1 \, \text{kg/L} = 75 \, \text{kg} \]

### Step 2: Calculate the gravitational potential energy (GPE) required to lift the water
The gravitational potential energy (GPE) required to lift the mass to a height (h) can be calculated using the formula:

\[ \text{GPE} = mgh \]

Substituting the values:

\[ \text{GPE} = 75 \, \text{kg} \times 10 \, \text{m/s}^2 \times 5 \, \text{m} = 3750 \, \text{J} \]

### Step 3: Determine the power output (P_out)
Since the water is lifted in 1 second, the power output (P_out) can be calculated as:

\[ P_{\text{out}} = \frac{\text{GPE}}{\text{time}} = \frac{3750 \, \text{J}}{1 \, \text{s}} = 3750 \, \text{W} = 3.75 \, \text{kW} \]

### Step 4: Determine the power input (P_in)
The power consumed by the motor is given as:

\[ P_{\text{in}} = 5 \, \text{kW} \]

### Step 5: Calculate the efficiency
Efficiency (η) can be calculated using the formula:

\[ \eta = \frac{P_{\text{out}}}{P_{\text{in}}} \times 100\% \]

Substituting the values:

\[ \eta = \frac{3.75 \, \text{kW}}{5 \, \text{kW}} \times 100\% = 75\% \]

### Conclusion
The efficiency of the installation is **75%**.

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