An engine pumps up `100 kg` water through a height of `10 m` in `5 s`. If efficiency of the engine is `60%`. What is the power of the engine? `Take g = 10 ms^(2)`.

Options

A. (a) `33kW`
B. (b) `3.3kW`
C. (c) `0.33kW`
D. (d) `0.033kW`

Explanation

To find the power of the engine that pumps 100 kg of water through a height of 10 m in 5 seconds with an efficiency of 60%, we can follow these steps:

### Step 1: Calculate the Work Done
The work done (W) against gravity can be calculated using the formula:
\[ W = mgh \]
Where:
- \( m = 100 \, \text{kg} \) (mass of the water)
- \( g = 10 \, \text{m/s}^2 \) (acceleration due to gravity)
- \( h = 10 \, \text{m} \) (height)

Substituting the values:
\[ W = 100 \, \text{kg} \times 10 \, \text{m/s}^2 \times 10 \, \text{m} \]
\[ W = 100 \times 10 \times 10 = 10000 \, \text{J} \]

### Step 2: Calculate the Useful Work Done by the Engine
Since the efficiency (η) of the engine is given as 60%, we can express the useful work done by the engine (W_useful) as:
\[ W_{\text{useful}} = \eta \times W \]
Where:
- \( \eta = 0.60 \)

Substituting the values:
\[ W_{\text{useful}} = 0.60 \times 10000 \, \text{J} \]
\[ W_{\text{useful}} = 6000 \, \text{J} \]

### Step 3: Calculate the Power of the Engine
Power (P) is defined as the work done per unit time. The formula for power is:
\[ P = \frac{W_{\text{useful}}}{t} \]
Where:
- \( t = 5 \, \text{s} \)

Substituting the values:
\[ P = \frac{6000 \, \text{J}}{5 \, \text{s}} \]
\[ P = 1200 \, \text{W} \]

### Step 4: Convert Power to Kilowatts
Since power is often expressed in kilowatts (kW), we convert it:
\[ P = \frac{1200 \, \text{W}}{1000} = 1.2 \, \text{kW} \]

### Final Answer
The power of the engine is **1.2 kW**.

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