The resistance of the filament of a lamp increases with the increase in temperature. A lamp rated 100 W and 220 V is connected across 220 V power supply. If the voltage drops by 10%, then the power of the lamp will be

Let the resistance of the lamp filament be R. Then $100={\left(200\right)}^2/R.$ When the voltage drops, expected power is  $P={(220\times 0.9)}^2/R\text{'}.$Here $R\text{'}$ will be less than R,  because now the rise in temperature will be less. Therefore, P is more than ${(220\times 0.9)}^2/R=81W.$ But it will not be 90% of the earlier value, because fall in temperature is small. Hence, option (4) is correct.