The correct option is B 1:64
Excess pressure inside a liquid drop (ΔP)=2TR
Let the excess pressure in drop 1 is ΔP1 and drop 2 is ΔP2 and their radii are R1 and R2 respectively.
Given,
ΔP1=4×ΔP2
⇒2TR1=4×2TR2
⇒R2=4R1
⇒R1R2=14
Mass ratio =mass of droplet 1 (m1)mass of droplet 2 (m2)
Here, m1=43πR31ρ;
m2=43πR32ρ
∴m1m2=R31R32=(R1R2)3=(14)3=164