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(D) The power rating of a bulb is given by $P = V^2 / R$,where $V$ is the rated voltage (usually $220 \ V$ or $240 \ V$) and $R$ is the resistance of

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$R_{100} < R_{60} < R_{40}$

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(D) The power rating of a bulb is given by $P = V^2 / R$,where $V$ is the rated voltage (usually $220 \ V$ or $240 \ V$) and $R$ is the resistance of the filament at its operating temperature.Since $V$ is constant for all bulbs,we have $P \propto 1 / R$,which implies $R \propto 1 / P$.Therefore,a bulb with a higher power rating will have a lower resistance at its operating temperature.Although the resistance of the filament increases with temperature,the relative order of resistances remains the same at room temperature as it is at the operating temperature.Given $P_{100} > P_{60} > P_{40}$,it follows that $R_{100} < R_{60} < R_{40}$.

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