AI-generated practice question — model-generated for extra practice, not a previous-year CBSE board question.
(B) The 40 W bulb
Using $R = V^2/P$, the 40 W bulb has resistance $= 220^2/40 = 1210\ \Omega$, while the 100 W bulb has $220^2/100 = 484\ \Omega$. Lower power rating means higher resistance.
Source: Chapter 11 (Electricity), Section 11.8 Electric Power
Key formula: $R = V^2/P$. Since both bulbs operate at the same voltage (220 V), resistance is inversely proportional to power. The lower the wattage, the higher the resistance. Examiners expect you to recall or derive $R = V^2/P$ from $P = V^2/R$.