The potential drop across 100W bulb is smaller than that across 60W bulb.
First bulb rating is 200 V - 100 W. So, the resistance of the first bulb is,
P1=V21R1 ⇒ R1=V21P1=2002100=40000100⇒ R1=400 ΩP1=V12R1 ⇒ R1=V12P1=2002100=40000100⇒ R1=400 Ω
Second bulb ratings is 200 V - 60 W. Therefore,
P2=V22R2 ⇒ R2=V22P2=200260=4000060⇒ R2=20003 ΩP2=V22R2 ⇒ R2=V22P2=200260=4000060⇒ R2=20003 Ω
Therefore, when two bulbs are connected in series, the effective resistance is,
Reff=R1+R2 ⇒ Reff=400+20003=1200+20003⇒ Reff=32003 ΩReff=R1+R2 ⇒ Reff=400+20003=1200+20003⇒ Reff=32003 Ω
So, current in the circuit is,
i=VReff=20032003/=6003200⇒ i=316 Ai=VReff=20032003=6003200⇒ i=316 A
The potential drop across each bulb is,
across 100 W bulb is,
V1=iR1=316×400=3×25⇒ V1=75 VV1=iR1=316×400=3×25⇒ V1=75 V
across 60 W bulb is,
V2=iR2=316×20003⇒ V2=125 VV2=iR2=316×20003⇒ V2=125 V
So, the potential drop across 100 W bulb is less than that of the potential drop across 60 W.