Question

n drops of liquid each with surface energy E, join to form a single drop. In this process

• A. No energy will be absorbed
• B. Energy absorbed is E(n - n^2/3)
• C. Energy released will be E(n - n^2/3)
• D. Energy released will be E(2^2/3 - 1)

Answer 1

The correct option is C

Energy released will be E(n - n^2/3)

As is clear from the question,
Initial surface energy =$V_i=nE$
Let the radius of each drop be r and radius of bigger drop formed be R.
The volume in the whole process should not change
$\Rightarrow n\times \frac{4}{3}\pi r^3=\frac{4}{3}\pi R^3$
$\Rightarrow R=(n{)}^{1/3}r$
Let the surface tension of liquid be S
then $nE=4\pi r^{2}S$
now final surface energy of the bigger drop
$=4\pi R^{2}S$
=$4\pi r^2(n{)}^{2/3}S$
=$4\pi r^{2}S(n{)}^{2/3}$
$V_f=E(n{)}^{2/3}$
$\Rightarrow Energyreleased=nE-(n{)}^{2/3}E$
=$E(n-n^{2/3})$