An electric bulb rated for 500 watts at 100 volts is used in a circuit having a 200 volts supply. The resistance R that must be put in series the bulb, so that the bulb delivers 500 watt is …… ohm. 

We know, 

$\text{ Power }\mathrm{P}=\frac{{\mathrm{V}}^2}{\mathrm{R}}\therefore {\mathrm{R}}_{\text{bulb }}=\frac{{\mathrm{V}}^2}{\mathrm{P}}=\frac{100\times 100}{500}=20\mathrm{\Omega }$

This would be possible only when $\mathrm{R}=20\mathrm{\Omega }$ is in series with the bulb resistance $\left({\mathrm{R}}_{\mathrm{bulb}}=20\mathrm{\Omega }\right)$because in that case both resistances will share equal potential difference of 100 V each.