An electric refrigerator rated $500$W operates $6$ hours/day. What is the cost of the energy to operate it for $30$ days at $4.5$Rs. per kWh?

Hint: First, we need to find out the energy consumption per day and then we will calculate what the energy consumption would be for $30$ days. Then we will do the multiplication of the cost of the consumption for the required answer.

Complete step-by-step solution:
The energy of the power rating is multiple with the time of the operation is $E = P \times T$
The cost of the energy consumption multiplied by the cost per unit of energy consumed is $\operatorname{Cos} t = E \times \operatorname{Cos} {t_{unit}}$
Since the electric refrigerator rated $500$W operates $6$ hours/day. Thus, the energy consumption is power rating multiplied we get, $E = P \times T \Rightarrow {E_{1day}} = 500 \times 6$
Further solving we get ${E_{1day}} = 3000Wh \Rightarrow 3kWh$ to convert any quantity into its kilo form, just divide it by the thousand.
Now multiple the energy consumption for one day by $30$ and that will be the total energy consumption is ${E_{net}} = {E_{1day}} \times 30 \Rightarrow 3 \times 30 = 90kWh$
Now the cost is energy consumption multiplied by the cost per unit of the energy consumed.
Hence, we get $\operatorname{Cos} t = E \times \operatorname{Cos} {t_{unit}} \Rightarrow 90 \times 4.5$ ($4.5$ per kWh)
Further solving we have $\operatorname{Cos} t = Rs.405$
Therefore, the cost of the energy to operate it for $30$ days $4.5$ per kWh is $Rs.405$.

Note: The electricity bill that we pay also works on this same method as we used above, as the number of the given units on the meter means that many kWh of the energy is used. This is multiplied by the local electric cost of the electricity bill that we will need to pay per unit.
Since in the above we used units, one day, the net is the representation of the given information, and using the formulas we obtained the required result.