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(A) 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 of the bulb.Since both bulbs h

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$X$ will glow brighter

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(A) 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 of the bulb.Since both bulbs have the same voltage rating,$R \propto \frac{1}{P}$.Therefore,the bulb with lower power has higher resistance. Thus,$R_X > R_Y$.When connected in series,the current $I$ flowing through both bulbs is the same.The power dissipated (which determines brightness) is given by $P' = I^2 R$.Since $R_X > R_Y$,the power dissipated in bulb $X$ is greater than that in bulb $Y$ $(P'_X > P'_Y)$.Consequently,bulb $X$ will glow brighter than bulb $Y$.

Two bulbs $X$ and $Y$ having the same voltage rating and power ratings of $40\, W$ and $60\, W$ respectively are connected in series across a potential difference of $300\, V$. Then:

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