Question

A \[1\,{\text{h}}{\text{.p}}{\text{.}}\] electric motor operates a pump continuously. The work done performed by the motor in one day is:
(A) \[18\,{\text{kWh}}\]
(B) \[36\,{\text{kWh}}\]
(C) \[54\,{\text{kWh}}\]
(D) \[72\,{\text{kWh}}\]

Answer 1

Hint: First of all, we convert the given horsepower into watts and then convert into kilowatts. We know \[24\] hours make one day. We will substitute the required values in the correct formula and manipulate accordingly to obtain the result.

Complete step by step answer:
In the given question, we are supplied the flowing data:
The power of the electric motor is given as \[1\,{\text{h}}{\text{.p}}{\text{.}}\]
The time duration for which the electric motor is run is the whole day.
We are asked to find the work done by the electric motor in one day.
For this we will find out the energy. Before that we will convert the horse power in its equivalent watts.
We know,
\[1\] horse power (\[1\,{\text{h}}{\text{.p}}{\text{.}}\]) is equal to \[746\] watts.
But to find the work done or energy, we need to first convert the power to kilowatts, as this is the commercial system of measuring electrical energy.
\[746\] watts
\[ = \dfrac{{746}}{{1000}}\] kilowatts
Which is equal to \[0.746\,{\text{kW}}\] .
Again, the time duration is given as the whole day.
We know, in commercial measuring of electrical energy we take the time duration in hours unlike seconds which we use in all numerical.
In one day, there are \[24\] hours.
So, now to find the electrical energy or work done, we use the formula:
\[E = P \times t\] …… (1)
Where,
\[E\] indicates electric energy.
\[P\] indicates electric power in kilowatts.
\[t\] indicates time in hours.
Now, we substitute the required values in the equation (1), and we get:
\[
E = P \times t \\
E = 0.746\,{\text{kW}} \times 24\,{\text{h}} \\
E = 17.9\,{\text{kWh}} \\
E \sim 18\,{\text{kWh}} \\
\]
Hence, the work done by the motor in one day is \[18\,{\text{kWh}}\] .
The correct option A.

Note:While solving this problem do remember that, in the commercial system of measuring electrical energy, we use the time in hours. Even if the time mentioned in seconds or in minutes, it must be converted into hours. Same is the case with power which must be in kilowatts. One kilowatt hour is equivalent to one unit.