Answer:518.4
Explanation:
Let the power of bulb used be [tex]P[/tex] watts.
In [tex]1[/tex] second,[tex]P[/tex] joules of energy is consumed.
In [tex]1[/tex] minute,[tex]P\times 60=60P[/tex] joules of energy is consumed.
In [tex]1[/tex] hour, [tex]60P\times 60=3600P[/tex] joules of energy is consumed.
In [tex]1[/tex] day,[tex]3600P\times 24=86400P[/tex] joules of energy is consumed.
In [tex]1[/tex] month,[tex]30\times 86400P=2592000P[/tex] joules of energy.
As we know that [tex]1KWh=3600000joules[/tex]
So,required energy in [tex]KWh[/tex] is [tex]\frac{2592000P}{3600000}=0.72PKWh[/tex]
Given that the power of the bulb used is [tex]100[/tex] watts.
So,required energy is [tex]0.72\times 100=72KWh[/tex]
Given that cost of one [tex]KWh[/tex] is [tex]7.2[/tex]
So,cost of [tex]72KWh[/tex] is [tex]72\times 7.2=518.4[/tex]
