A bulb of 220 V and 300 W is connected across 110 V circuit. The percentage reduction in power is:

75%

P = 300 W, V = 220 V

$R = \frac{V^2}{P}=\frac{220 \times 220}{300}=\frac{484}{3} \Omega$

If this bulb is connected across 110 V, then the power delivered is

$P' = \frac{V^2}{R}=\frac{(110)^2}{484 / 3}=\frac{110 \times 110 \times 3}{484}$ = 75 W

∴ Percentage reduction in power = $\frac{300-75}{300}$ × 100% = 75%