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(C) The power $P$ consumed by a bulb is given by the formula $P = \frac{V^2}{R}$,where $V$ is the voltage across the bulb and $R$ is its resistance.As

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$60\, W$ bulb

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(C) The power $P$ consumed by a bulb is given by the formula $P = \frac{V^2}{R}$,where $V$ is the voltage across the bulb and $R$ is its resistance.Assuming the voltage $V$ is constant for all bulbs,we have $R = \frac{V^2}{P}$.This shows that resistance $R$ is inversely proportional to power $P$ $(R \propto \frac{1}{P})$.Therefore,the bulb with the highest power rating will have the lowest resistance.Comparing the given values ($25\, W$,$40\, W$,and $60\, W$),the $60\, W$ bulb has the highest power.Thus,the $60\, W$ bulb has the lowest resistance.

You are given three bulbs of $25\, W$,$40\, W$,and $60\, W$. Which of them has the lowest resistance?

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