Question

An engine pumps up $100\,kg$ of water through a height of $10\,m$ in $5\sec$.If the efficiency of the engine is $60.$ what is the power of the engine? Take $g = 10\,m{s^{ - 2}}$
A. $33\,kW$
B. $3.3\,kW$
C. $0.33\,kW$
D. $0.033\,kW$

Answer 1

In order to solve this question, we will use the concept of efficiency of engine which is defined as “the ratio of output power to the input power of an engine” is known as its efficiency and power is defined as rate of change of work done by a system.

Complete step by step answer:
Let us first find the Output performance of the engine which means net output power of the system. Given that, engine pumps of water whose mass $m = 100\,kg$ raise to the height of $h = 5\,m$ in $t = 5\,s$. Now, we can find power by using the formula as,
${P_{output}} = \dfrac{{mgh}}{t}$
Putting the values of parameter and we get,
${P_{output}} = \dfrac{{100 \times 10 \times 10}}{5}$
$\Rightarrow {P_{output}} = 2\,kW$

Now, we have also given that the efficiency of the engine is $60$. Using definition of efficiency we have,
$\text{Efficiency} = \dfrac{{{P_{output}}}}{{{P_{input}}}}$
Where, ${P_{input}}$ is the input power of engine which we need to find,
Now, put the values of given parameters we get,
${P_{input}} = \dfrac{2}{{60}} \times 100$
$\therefore {P_{input}} = 3.33\,kW$

Hence, the correct option is B.

Note: Remember, the efficiency of a system or an engine is always given in percentage and we need to convert it in fraction numerical value by dividing from $100$.And it’s important to know that efficiency of an engine always lies between $0$ and $1$ which also implies that a perfect engine’s efficiency will be equals to $1$ which means total input power is converted into output power by the engine.