CBSE Class 10 Science 2020 · Paper 31/3/3 · Question 30(a)

(a) $R=\frac{V^2}{P}=\frac{(200{)}^2}{100}=400\Omega$. (b) Total power of the 3 bulbs $=3\times 100\text{W}=300\text{W}=0.3\text{kW}$. Energy used in one day $=0.3\text{kW}\times 10\text{h}=3\text{kWh}$, so in the 30 days of November the energy consumed $=30\times 3\text{kWh}=90\text{kWh}=90$ units. (c) Total cost $=90$ units $\times$ ₹6.50 per unit $=$ ₹585.