(c) Resistances of both the bulbs are $R_{1} = \frac{V^{2}}{P_{1}} = \frac{220^{2}}{25}$ $R_{2} = \frac{V^{2}}{P_{2}} = \frac{220^{2}}{100}$ $\therefore R_{2} > R_{2}$ Hence $25\ W$ bulb will fuse
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.