The current flowing through the two bulbs marked as 60 W, 240 V each when connected in series with 240 V source is:

0.125 A

The correct answer is Option (1) → 0.125 A

The resistance of a bulb is -

$R=\frac{V^2}{P}$

where,

V (Voltage in circuit) = 240

P (Power of each bulb) = 60

$R=\frac{(240)^2}{60}$

Now,

As bulbs are connected in series than total resistance ($R_{total}$) is

$R_{total}=R+R$

$=2R$

∴ I (Current in circuit) = $\frac{240}{2R}$

$=\frac{240×60}{2×240×240}$

$=0.125A$