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

At the rated condition, the filament is hot and its resistance is $$R=\frac{V^2}{P}=\frac{220^2}{100}.$$ If resistance stayed constant, reducing the voltage to $0.9$ of its value would make the power $$P'=\left(0.9\right)^2\times 100=81\mathrm{W}.$$ But here the filament resistance decreases when the temperature falls, so after the voltage drop the actual resistance becomes less than the rated hot resistance. Therefore the actual power is greater than $81\mathrm{W}$. It must still be less than $90\mathrm{W}$, because for $90\mathrm{W}$ with $198\mathrm{V}$ the required resistance would be $$R'=\frac{198^2}{90},$$ which is less than the rated hot resistance, consistent with cooling, and the true value lies between the constant-resistance result and the linear-voltage guess. Hence the power lies between $81\mathrm{W}$ and $90\mathrm{W}$. Re-checking the options, this matches option $D$.