The resistance of the filament of a lamp increases with the increase in temperature. A lamp rated $100\, W, 220\, V$ is connected across $220\, V$ power supply. If the voltage drops by $10\%$ then the power of lamp will be
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(d) Let the resistance of the lamp filament be $R$. Then $100 = \frac{{{{(220)}^2}}}{R}$. When then voltage drops, expected power is $P = \frac{{{{(220 \times 0.9)}^2}}}{{R'}}$. Here $R'$ will be less than $R$, because now the rise in temperature will be less. Therefore $P$ is more than $\frac{{{{(220 \times 0.9)}^2}}}{R} = 81W$
But it will not be $90\%$ of earlier value, because fall in temperature is small. Hence $(d)$ is correct.
But it will not be $90\%$ of earlier value, because fall in temperature is small. Hence $(d)$ is correct.
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