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

Options

A. `2^(2//3) : 1`
B. `sqrt2 : 1`
C. `sqrt(1//3) : 1`
D. `2 : 1`

Answer

C

Explanation

Given that, the two small drops of mercury and radius R of each drop coalesce to form a large drop of radius r, so the net volume remains constant i.e,
initial volume = final volume
`V_("initial")=V_("final")`
`2xx4/3piR^(3)=1dot4/3pir^(3)rArrr=2^(1//3)R` ...(i)
As, surface tension is constant for both the drops.
So, the surface energy of two small drops,
`E_(1)=2xxTA_(1)=2xx4piR^(2)xxT`
Surface energy of one big drop,
`E_(2)=TA_(2)=4pir^(2)xxT=2^(1//3).4piR^(2)T` (using Eq. (i))
Ratio the total surface energy before and after the change is given as,
`E_(1)/E_(2)=(8piR^(2)T)/(2^(2//3)4piR^(2)T)=(2^(1-2/3))/1=2^(1/3):1`
Hence, the ratio of suface energy of the each drop is `2^(1/3) : 1`