Question:

An engine pumps up $100 \,kg$ of water through a height of $10\, m$ in $5 s$. Given that the efficiency of engine is $60 \%$. If $g=10\, ms ^{-2}$, the power of the engine is

Updated On: Jun 24, 2024
  • 3.3 kW
  • 0.33 kW
  • 0.033 kW
  • 33 kW
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The Correct Option is A

Approach Solution - 1

Power is the rate at which a force does work.
Efficiency of engine, $\eta=60\%$
Thus, power $=\frac{\text { work } / \text { times }}{\eta}=\frac{100}{60} \times \frac{m g h}{t}$
Given, $m=100\, kg , h=10 \,m , t=5\, s$ and $g=10\, ms ^{-2} .$
Hence, power $=\frac{100}{60} \times \frac{100 \times 10 \times 10}{5} $
$=3.3 \times 10^{3} \,W$
$=3.3 \,k W$
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Approach Solution -2

To solve the given problem, we need to calculate the power of the engine. Here is the step-by-step solution:
Step 1: Calculate the work done to pump the water
The work done (\(W\)) to pump the water can be calculated using the formula for gravitational potential energy:
\[ W = m \cdot g \cdot h \]
where:
- \(m = 100 \, \text{kg}\) is the mass of the water,
- \(g = 10 \, \text{m/s}^2\) is the acceleration due to gravity,
- \(h = 10 \, \text{m}\) is the height.
Substituting the values:
\[ W = 100 \, \text{kg} \times 10 \, \text{m/s}^2 \times 10 \, \text{m} \]
\[ W = 10000 \, \text{J} \]
Step 2: Calculate the total power required
Since the efficiency of the engine is 60%, the actual work output is 60% of the total work input. Let \(P_{\text{input}}\) be the total power input and \(P_{\text{output}}\) be the power output. The relationship between power input and output is:
\[ \text{Efficiency} = \frac{P_{\text{output}}}{P_{\text{input}}} \]
\[ 0.60 = \frac{P_{\text{output}}}{P_{\text{input}}} \]
\[ P_{\text{input}} = \frac{P_{\text{output}}}{0.60} \]
We need to calculate the power output first, which is the work done divided by time:
\[ P_{\text{output}} = \frac{W}{t} \]
where:
- \(W = 10000 \, \text{J}\),
- \(t = 5 \, \text{s}\).
Substituting the values:
\[ P_{\text{output}} = \frac{10000 \, \text{J}}{5 \, \text{s}} \]
\[ P_{\text{output}} = 2000 \, \text{W} \]
 Step 3: Calculate the power of the engine
Using the efficiency formula, we can find the power input:
\[ P_{\text{input}} = \frac{2000 \, \text{W}}{0.60} \]
\[ P_{\text{input}} = \frac{2000}{0.60} \]
\[ P_{\text{input}} = 3333.33 \, \text{W} \]
Conclusion
The power of the engine is \(3333.33 \, \text{W}\).i.e. 3.3Kw.
 So The correct Answer is option A
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Concepts Used:

Work, Energy and Power

Work:

  • Work is correlated to force and the displacement over which it acts. When an object is replaced parallel to the force's line of action, it is thought to be doing work. It is a force-driven action that includes movement in the force's direction.
  • The work done by the force is described to be the product of the elements of the force in the direction of the displacement and the magnitude of this displacement.

Energy:

  • A body's energy is its potential to do tasks. Anything that has the capability to work is said to have energy. The unit of energy is the same as the unit of work, i.e., the Joule.
  • There are two types of mechanical energy such as; Kinetic and potential energy.

Read More: Work and Energy

Power:

  • Power is the rate at which energy is transferred, conveyed, or converted or the rate of doing work. Technologically, it is the amount of work done per unit of time. The SI unit of power is Watt (W) which is joules per second (J/s). Sometimes the power of motor vehicles and other machines is demonstrated in terms of Horsepower (hp), which is roughly equal to 745.7 watts.
  • Power is a scalar quantity, which gives us a quantity or amount of energy consumed per unit of time but with no manifestation of direction.