When electric bulbs of same power but different marked voltage are connected in series across the power line, their brightness will be:
A. Proportional to their marked voltage
B. Inversely proportional to their marked voltage
C. Proportional to the square of their marked voltage
D. Inversely proportional to the square of their marked voltage
E. The same for all of them

Hint: In series the current across all the bulbs will be the same and the power dissipated is proportional to voltage squared and inverse of the resistance. Different marked voltages bulbs will have different resistances.

Complete answer:
We know that power dissipated P, across a resistance R with current I passing through the resistance is given by: $P = {I^2}R$. Using Ohm’s law ($V = IR$) we get, $P = \dfrac{{{V^2}}}{R}$. For bulbs connected in series the power dissipated by the bulb, according to the last relation, will be directly proportional to the voltage across the bulb.
Now since the voltage ratings of the bulbs are different, the resistances of the bulbs will be different and according to the choices above we asked to find a relation of the power output to the marked voltage. Consider two bulbs with power rating as $P$ but with different voltage ratings be ${V_1},{V_2}$. Therefore the resistance of each of the bulb will be ${R_1} = \dfrac{{V_1^2}}{P},{R_2} = \dfrac{{V_2^2}}{P}$. Now when these bulbs are connected in series, the current through each bulb will be the same and consider it as $I$. Since we now have terms for resistance of the bulb and the current through them, power dissipated by each of the bulb using the power dissipated expression above, we get,
${P_1} = {I^2}{R_1} = {I^2}\dfrac{{V_1^2}}{P};$
${P_2} = {I^2}{R_2} = {I^2}\dfrac{{V_2^2}}{P}$
Thus the power output will be proportional to the square of the marked voltage of the bulb. As a result, we can say, option C is correct.

Note: Power rating is the power dissipated by the bulb when connected across the marked voltage. But in series like above the potential difference across the bulb will be lesser than marked voltage and thus affecting the power output.