An electric bulb is rated $220\mathrm{V}-100\mathrm{W}$. The power consumed by it when operated on 110 V will be

$\mathrm{R}=\frac{{\mathrm{V}}_{\mathrm{R}}^2}{{\mathrm{P}}_{\mathrm{R}}}$

We are using same bulb so R is constant.

$\therefore P_R\propto V^2$

Thus, when voltage drop becomes half , power output becomes one fourth.