An electric bulb is rated 220 V and 100 W

solution:
Resistance of the electric bulbs,

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

$\left(P=\frac{V^{2}}{R}\right)$

or $\mathrm{R}=\frac{(220)^{2}}{100}=484 \Omega$

Power consumed by the bulb when it is operated at 110 V is given by

$\mathrm{P}^{\prime}=\frac{\mathrm{V}^{2}}{\mathrm{R}}=\frac{(110)^{2}}{484}=\frac{110 \times 110}{484}=\mathbf{2 5} \mathbf{~ W}$

Thus, (4) is the correct answer.