An engine pumps 373 kg of water upto height of 16 m in 40 seconds. Rated power of engine if its efficiency is 80% is (1 horse power = 746 W and g =10 m/`s^2`)

Options

A. 1 horse power
B. 2 horse power
C. 2.5 horse power
D. 3 horse power

Answer

To solve the problem step by step, we will calculate the power required to pump the water and then adjust for the efficiency of the engine.

### Step 1: Calculate the potential energy (PE) gained by the water.
The potential energy gained by the water when it is pumped to a height \( h \) is given by the formula:
\[ PE = mgh \]
where:
- \( m = 373 \, \text{kg} \) (mass of water),
- \( g = 10 \, \text{m/s}^2 \) (acceleration due to gravity),
- \( h = 16 \, \text{m} \) (height).

Substituting the values:
\[ PE = 373 \, \text{kg} \times 10 \, \text{m/s}^2 \times 16 \, \text{m} = 59680 \, \text{J} \]

### Step 2: Calculate the power (P) required to pump the water.
Power is defined as the rate of doing work or the rate of energy transfer. It can be calculated using:
\[ P = \frac{E}{t} \]
where:
- \( E \) is the energy (in this case, the potential energy),
- \( t = 40 \, \text{s} \) (time taken).

Substituting the values:
\[ P = \frac{59680 \, \text{J}}{40 \, \text{s}} = 1492 \, \text{W} \]

### Step 3: Adjust for the efficiency of the engine.
The rated power of the engine can be calculated using the efficiency formula:
\[ \text{Rated Power} = \frac{P}{\text{Efficiency}} \]
Given that the efficiency is 80%, or 0.8 in decimal form:
\[ \text{Rated Power} = \frac{1492 \, \text{W}}{0.8} = 1865 \, \text{W} \]

### Step 4: Convert the rated power to horsepower.
To convert watts to horsepower, we use the conversion factor \( 1 \, \text{horsepower} = 746 \, \text{W} \):
\[ \text{Rated Power in horsepower} = \frac{1865 \, \text{W}}{746 \, \text{W/hp}} \approx 2.5 \, \text{hp} \]

### Final Answer:
The rated power of the engine is approximately **2.5 horsepower**.

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