Question
Physics Question on Surface tension
Two small drops of mercury, each of radius R, coalesce to form a single large drop. The ratio of the total surface energies before and after the change is
1:21/3
21/3:1
2:1
1:2
21/3:1
Solution
Radius of one drop of mercury is R.
∴ The volume of one drop=34πR3
∴; Total volume of the two drops,
V=2×34πR3=38πR3
Let the radius of the large drop formed be R′ The volume of the large drop is also V.
∴34πR′3=38πR3⇒R′3=2R3⇒R′=21/3R.
Now the surface area of the two drops is
S1=2×4πR2=8πR2
and the surface area of the resultant drop is
S1=2×4πR′2=4π22/3R2
Let T be the surface tension of mercury. Therefore the surface energy of the two drops before coalescing is
U1=S1T=8πR2T
and the surface energy after coalescing,
U1=S2T=22/3×4πR2T
∴U2U1=22/3×4πR2T8πR2T=22/32=21/3