Answer to Question #169709 in Physics for Pushpo

Question

A mercury drop of radius 1 mm is sprayed into 103 drops of equal sizes. Compute the energy expended. Surface tension of mercury = 3.5 × 10−2 N/m.  

Answer 1

Surface energy of the initial drop:

$$E_i=\sigma A_i=\sigma·4\pi R^2.$$

Radius of 103 equal-sized smaller drops:

$$r=\sqrt[3]{\frac{3(V_i/103)}{4\pi}}=\frac{R}{\sqrt[3]{103}}.$$

Surface energy of 1 smaller drop:

$$E_f=\sigma A_f=\sigma ·4\pi r^2=4\sigma\pi R^2·103^{2/3}.\\ \Delta E=E_f-E_i=4\pi\sigma R^2(103^{2/3}-1)=9.22\space\mu\text{J}$$