
A washing machine rated 300 W is operated one and half an hour/day. If the cost of units is $\text{Rs}\text{.3}\text{.50}$, find the cost of energy to operate a washing machine for the month of September.
A). $\text{Rs}\text{.27}\text{.90}$
B). $\text{Rs}\text{.35}\text{.25}$
C). $\text{Rs}\text{.47}\text{.25}$
D). $\text{Rs}\text{.55}\text{.90}$
Answer
536.1k+ views
Hint: We know the commercial or bigger unit of electrical energy is Kilowatt-hour. It is the electric energy spent by an electric appliance of power 1 KW used for 1 hour. With this in mind, we can easily solve the given question.
Complete step by step solution:
In the following problem, first we need to convert one and a half hours to a complete hour. Therefore, $\dfrac{90\text{ mins}}{60}\text{ }=\text{ }1.5\text{ hrs}$. We will get the monthly bill by multiplying $\text{Rs}\text{.3}\text{.50}$ with the monthly unit.
We know,
$\text{1 Watthour = 1 watt}\ \times \text{ 1hour = 1 watt }\times \text{ 3600 sec = 3600 Joule}$.
Whereas,
$\text{1 kilowatt-hour = 1KW }\times \text{ 1hour = 1000 Joule/sec }\times \text{ 3600 sec = 3}\text{.6 }\times \text{ 1}{{\text{0}}^{\text{6}}}\text{ Joule = 3}\text{.6 MJ}$.
Now, where the formula is 1 unit$\text{ }\!\!~\!\!\text{ = 1 KWh = }\dfrac{\text{watt }\times \text{ hour}}{1000}\text{ = }\dfrac{\text{Volt }\times \text{ Ampere }\times \text{ hour}}{1000}$
The energy used by the washing machine is 300 Watt used for 1.5 hours per day.
Total hour used in the month of September $= 30 \times 1.5 = \text{ }45\text{ hours}$.
Therefore, the monthly unit of washing machine $=\text{ }\dfrac{300\text{ }\times \text{ }45}{1000}\text{ }=\text{ }1\text{3}\text{.5 Kilowatt-hour}$.
According to the solution we get, the monthly cost of running the washing machine $\text{= 3}\text{.5 }\times \text{ 13}\text{.5 = Rs}\text{.47}\text{.25}$.
So, from the above solution, we find that Option C is correct.
Note: Kilowatt is the commercial unit for electric power. The Kilowatt-hour unit is the commercial unit of electric energy. In domestic environments, this unit is the standard for the electrical energy measured people living in domestic environments pay the electric bill accordingly. This is useful for us in our daily energy consumption. Keeping this in mind, we can solve questions as such.
Complete step by step solution:
In the following problem, first we need to convert one and a half hours to a complete hour. Therefore, $\dfrac{90\text{ mins}}{60}\text{ }=\text{ }1.5\text{ hrs}$. We will get the monthly bill by multiplying $\text{Rs}\text{.3}\text{.50}$ with the monthly unit.
We know,
$\text{1 Watthour = 1 watt}\ \times \text{ 1hour = 1 watt }\times \text{ 3600 sec = 3600 Joule}$.
Whereas,
$\text{1 kilowatt-hour = 1KW }\times \text{ 1hour = 1000 Joule/sec }\times \text{ 3600 sec = 3}\text{.6 }\times \text{ 1}{{\text{0}}^{\text{6}}}\text{ Joule = 3}\text{.6 MJ}$.
Now, where the formula is 1 unit$\text{ }\!\!~\!\!\text{ = 1 KWh = }\dfrac{\text{watt }\times \text{ hour}}{1000}\text{ = }\dfrac{\text{Volt }\times \text{ Ampere }\times \text{ hour}}{1000}$
The energy used by the washing machine is 300 Watt used for 1.5 hours per day.
Total hour used in the month of September $= 30 \times 1.5 = \text{ }45\text{ hours}$.
Therefore, the monthly unit of washing machine $=\text{ }\dfrac{300\text{ }\times \text{ }45}{1000}\text{ }=\text{ }1\text{3}\text{.5 Kilowatt-hour}$.
According to the solution we get, the monthly cost of running the washing machine $\text{= 3}\text{.5 }\times \text{ 13}\text{.5 = Rs}\text{.47}\text{.25}$.
So, from the above solution, we find that Option C is correct.
Note: Kilowatt is the commercial unit for electric power. The Kilowatt-hour unit is the commercial unit of electric energy. In domestic environments, this unit is the standard for the electrical energy measured people living in domestic environments pay the electric bill accordingly. This is useful for us in our daily energy consumption. Keeping this in mind, we can solve questions as such.
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