The resistance of the filament of an 'incande | vedclass

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(D) The power rating of a bulb is given by $P = \frac{V^2}{R}$,where $V$ is the rated voltage and $R$ is the resistance at operating temperature.Since

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$\frac{1}{R_{100}} > \frac{1}{R_{60}} > \frac{1}{R_{40}}$

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(D) The power rating of a bulb is given by $P = \frac{V^2}{R}$,where $V$ is the rated voltage and $R$ is the resistance at operating temperature.Since the resistance increases with temperature,the resistance at room temperature follows the same trend as the resistance at operating temperature.For a fixed voltage $V$,$R = \frac{V^2}{P}$.Thus,$R_{100} = \frac{V^2}{100}$,$R_{60} = \frac{V^2}{60}$,and $R_{40} = \frac{V^2}{40}$.Comparing these values,we get $R_{40} > R_{60} > R_{100}$.Taking the reciprocal,we get $\frac{1}{R_{100}} > \frac{1}{R_{60}} > \frac{1}{R_{40}}$.

The resistance of the filament of an 'incandescent' bulb increases with an increase in temperature. If $R_{100}$,$R_{60}$,and $R_{40}$ are the resistances of $100 \ W$,$60 \ W$,and $40 \ W$ bulbs at room temperature,respectively,then:

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