Q 12-03-079JEE MainJEE Main 2024 (4 Apr, Shift 2)Easy
An electric bulb rated $50\ \text{W}$ – $200\ \text{V}$ is connected across a $100\ \text{V}$ supply. The power dissipated by the bulb is
Answer: (B) $12.5\ \text{W}$
The resistance is fixed, so $P \propto V^2$:
$$P = 50\left(\frac{100}{200}\right)^2 = 12.5\ \text{W}$$
Solution by Sreeraj P, M.Sc Physics