Problem 27 - Master Resource Book in JEE Main Physics

Question

An engine pumps up \(100 \mathrm{~kg}\) of water through a height of \(10 \mathrm{~m}\) in 5 s. Given that the efficiency of the engine is \(60 \%\). If \(g=10 \mathrm{~m} / \mathrm{s}^{2}\), the power of the engine is (a) \(3.3 \mathrm{~kW}\) (b) \(0.33 \mathrm{~kW}\) (c) \(0.033 \mathrm{~kW}\) (d) \(33 \mathrm{~kW}\)

Step-by-Step Solution

Answer

The power of the engine is approximately 3.3 kW (option a).

Step 1: Calculate the work done

First, we calculate the work done by the engine in lifting the water. The work done \( W \) is given by the formula for gravitational potential energy, \( W = mgh \), where \( m \) is the mass of the water (100 kg), \( g \) is the acceleration due to gravity (10 m/s²), and \( h \) is the height (10 m).Substitute these values into the formula:\[ W = 100 \times 10 \times 10 = 10000 \text{ J} \]

Step 2: Calculate the power without efficiency

Power \( P \) is defined as the work done per unit time. The engine does 10000 J of work in 5 seconds, so the power produced without considering efficiency is:\[ P = \frac{W}{t} = \frac{10000}{5} = 2000 \text{ W} \]Since 1 kW = 1000 W, we convert this to kilowatts:\[ P = 2 \text{ kW} \]

Step 3: Adjust for efficiency

Since the engine is only 60% efficient, we need to adjust the calculated power to reflect this efficiency. The actual power output relative to the input is given by:\[ P' = \frac{P}{\text{Efficiency}} = \frac{2}{0.6} \]Calculate:\[ P' = \frac{2}{0.6} = 3.33 \text{ kW} \]

Step 4: Select the closest answer

After calculating, we find that the power of the engine considering its efficiency is closest to the option (a) 3.3 kW.

Key Concepts

Work Done

Work done is an energy transfer concept that describes how much energy is exerted to move an object. In this example, the engine lifts a mass of water against the gravitational force. The work done can be calculated using the formula for gravitational potential energy:

• Work Done, \( W = mgh \)
• \( m \): Mass of the object (here, 100 kg)
• \( g \): Acceleration due to gravity (10 m/s²)
• \( h \): Height the object is moved (10 m)

Plugging in the values:\[ W = 100 \times 10 \times 10 = 10000 \text{ J} \]
This calculation tells us that 10,000 joules of energy is needed to lift the water to the specified height.

Efficiency

Efficiency measures how effectively the input energy is converted to useful work. It is expressed as a percentage, representing the fraction of energy that accomplishes the intended purpose.

• In our example, the engine's efficiency is given as 60%.
• High efficiency means most energy is used efficiently in achieving the task.

When the efficiency of an engine or any system is known, it helps evaluate and improve performance. Remember, real engines always lose some energy to the surroundings, primarily as heat.

Gravitational Potential Energy

Gravitational potential energy (GPE) is the energy stored in an object due to its position above the ground. This energy is potential because it has the capacity to do work due to the object's position.

• GPE is calculated using \( W = mgh \).
• As the height or mass increases, the gravitational potential energy also increases.

In this scenario, the GPE of the lifted water is 10,000 J, signifying that when raised to a height of 10 meters, the water possesses this much potential energy.
GPE is a critical concept for understanding energy conservation, as it can transform into other energy types when the object moves.

Power Calculation

Power is the rate at which work is done or energy is transferred. It helps understand how quickly a task is performed.
Power is calculated using the formula:

• \( P = \frac{W}{t} \)
• \( W \): Work done, or energy transferred
• \( t \): Time taken to do the work

In our case:\[ P = \frac{10000}{5} = 2000 \text{ W} \]
Convert the power into kilowatts:\[ 2000 \text{ W} = 2 \text{ kW} \]Considering the engine is 60% efficient, adjust the power by dividing by the efficiency:\[ P' = \frac{2}{0.6} = 3.33 \text{ kW} \]Thus, the actual power output considering the engine's efficiency is about 3.3 kW.