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(C) The resistance of a bulb is given by $R = \frac{V^2}{P}$,where $V$ is the marked voltage and $P$ is the marked power.Since the power $P$ is the sa

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proportional to the square of their marked voltage

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(C) The resistance of a bulb is given by $R = \frac{V^2}{P}$,where $V$ is the marked voltage and $P$ is the marked power.Since the power $P$ is the same for both bulbs,the resistance $R$ is directly proportional to the square of the marked voltage $(R \propto V^2)$.When bulbs are connected in series,the current $I$ flowing through each bulb is the same.The brightness of a bulb is determined by the power dissipated in it,given by $P_{dissipated} = I^2 R$.Since $I$ is constant,the brightness is directly proportional to the resistance $(Brightness \propto R)$.Substituting $R \propto V^2$,we get $Brightness \propto V^2$.Therefore,the brightness is proportional to the square of their marked voltage.

When electric bulbs of same power,but different marked voltage are connected in series across the power line,their brightness will be:

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