If two electric bulbs have 40,W and 60,W rati | vedclass

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(A) The power $P$ of an electric bulb is given by the formula $P = \frac{V^2}{R}$,where $V$ is the voltage and $R$ is the resistance.Since the voltage

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$3:2$

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(A) The power $P$ of an electric bulb is given by the formula $P = \frac{V^2}{R}$,where $V$ is the voltage and $R$ is the resistance.Since the voltage $V$ is the same for both bulbs,we have $R = \frac{V^2}{P}$.Therefore,the resistance is inversely proportional to the power,i.e.,$R \propto \frac{1}{P}$.For the two bulbs with powers $P_1 = 40\,W$ and $P_2 = 60\,W$,the ratio of their resistances is $\frac{R_1}{R_2} = \frac{P_2}{P_1}$.Substituting the values,we get $\frac{R_1}{R_2} = \frac{60}{40} = \frac{3}{2}$.Thus,the ratio of their resistances is $3:2$.

If two electric bulbs have $40\,W$ and $60\,W$ ratings at $220\,V$,then the ratio of their resistances will be

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