An electric bulb rated 220 v and 60 W is connected in series with another electric bulb rated 220 v and 40 W. The combination is connected across 220 volt source of e.m.f. Which of the following statement is true?

$ P_1'< P_2'$

The correct answer is option 2: $ P_1'< P_2'$

When two bulbs are connected in series,
→ Same current flows through both.
→ Power dissipated in each bulb P′∝R

Now,
From the bulb ratings:

R= $\frac{V^2}{P}$

So resistance is inversely proportional to rated power.

Compare Resistances

R(60W) ​:R (40W) = 1/ 60 : 1/ 40 = 40 : 60  = 2 : 3

Hence, R(40W) ​:R (60W)

Compare Powers in Series

Since P′∝R  (same current in series),​

P′ (60W) ​: P′ (40W) = R(60W) ​:R (40W) = 2:3

Therefore, P′ (60W) < P′ (40W)

P1′​<P2′​

The 40 W bulb (which has higher resistance) glows more brightly.