The correct option is B 1/3 rd of the pressure difference for smaller droplet.
Let the radius of bigger droplet is R and radius of smaller droplets be r.
Applying volume conservation for bigger and smaller droplets,
⇒43πR3=27×43πr3
⇒(Rr)3=27
⇒Rr=3
⇒r=R3
The pressure difference between the inner and outer surface of bigger water droplet( consists of single surface) is given as,
ΔPB=2TR
similarly for smaller water droplet,
ΔPS=2Tr=2T(R3)=6TR
Hence,
ΔPB=ΔPS3
∴Option (b) is correct.