Two electric bulbs marked $25\text{W}-220\text{V}$ and $100\text{W}-220\text{V}$ are connected in series to a $440\text{V}$ supply. Which of the bulbs will fuse?

Resistance $R=\frac{V^2}{P}$.$R_1(25\text{W})=\frac{220^2}{25}=1936\Omega$.$R_2(100\text{W})=\frac{220^2}{100}=484\Omega$.In series, the current is $I=\frac{V_{\text{total}}}{R_1+R_2}=\frac{440}{1936+484}=\frac{440}{2420}=\frac{2}{11}\text{A}$.Voltage across $25\text{W}$ bulb: $V_1=I{R}_1=\frac{2}{11}\times 1936=2\times 176=352\text{V}$.Voltage across $100\text{W}$ bulb: $V_2=I{R}_2=\frac{2}{11}\times 484=2\times 44=88\text{V}$.Since the voltage across the $25\text{W}$ bulb ($352\text{V}$) exceeds its rated voltage ($220\text{V}$), it will fuse. The $100\text{W}$ bulb remains safe.