An electric kettle of $2 kW$ works for $2 h$ daily. Calculate the $(a)$ energy consumed in $SI$ and commercial units, and $(b)$ cost of running it in the month of June at the rate of ₹ $3.00$ per unit.

(N/A) Given: Power, $P = 2 kW = 2000 W$, Time, $t = 2 h$ daily.
Energy consumed per day, $E = P \times t = 2 kW \times 2 h = 4 kWh$.
In $SI$ units (Joules): $4 kWh = 4 \times 3.6 \times 10^6 J = 14.4 \times 10^6 J$.
In commercial units: $4 kWh$ (or $4$ units).
$(b)$ Total energy consumed in June ($30$ days):
Total Energy $= 4 kWh/day \times 30 days = 120 kWh$.
Cost of electricity $= 120 \text{ units} \times ₹ 3.00/unit = ₹ 360$.