If voltage across a bulb rated $220$ $volt-$ $100$ $watt$ drops by $2.5\%$ of its rated value, the percentage of the rated value by which the power would decrease is ............... $\%$
AIPMT 2012, Medium
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$\text { Power, } P=\frac{V^{2}}{R}$
As the resistance of the bulb is constant
$\therefore \quad \frac{\Delta P}{P}=\frac{2\, \Delta V}{V}$
$ \%$ decrease in power
$=\frac{\Delta P}{P} \times 100=\frac{2 \Delta V}{V} \times 100 $
$=2 \times 2.5 \%=5 \% $
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