Question 428: Properties of Matter

Question

A drop of liquid of radius $$R = {10^{ - 2}}\,m$$ having surface tension $$S = {{0.1} \over {4\pi }}N{m^{ - 1}}$$  divides itself into $$K$$ identical drops. In this process the total change in the surface energy $$\Delta U = {10^{ - 3}}\,J.$$  If $$K = {10^\alpha }$$ then the value of $$\alpha $$ is

Accepted Answer

Let a spherical drop of liquid of radius R divides itself into K spherical drops, each of radius r. The density of liquid does not change in this process. Hence, conservation of mass gives
$${4 \over 3}\pi {R^3} = {4 \over 3}\pi {r^3}K$$, i.e., $$r = R{K^{ - 1/3}}$$. ...... (1)
The surface energy of a drop of radius R is $${U_i} = 4\pi {R^2}S$$ and total surface energy of K drops of radius r is $${U_f} = 4\pi {r^2}KS$$. Note that the surface energy increases when a bigger drop is divided into multiple smaller drops. Total change in surface energy is
$$\Delta U = {U_f} - {U_i} = 4\pi S(K{r^2} - {R^2})$$ ...... (2)
Eliminate r from equations (1) and (2) to get
$$K = {\left( {{{\Delta U} \over {4\pi S{R^2}}} + 1} \right)^3}$$.
Substitute $$\Delta$$U = 10
$$-$$3
J, S = $${{0.1} \over {4\pi }}$$ N/m and R = 10
$$-$$2
m to get $$K = {(100 + 1)^3} \approx {10^6}$$.

Properties of Matter

A drop of liquid of radius $R = {10^{ - 2}}\,m having surface tension S = {{0.1} \over {4\pi }}N{m^{ - 1}} divides itself into K identical drops. In this process the total change in the surface energy \Delta U = {10^{ - 3}}\,J. If K = {10^\alpha } then the value of \alpha $ is

Integer / Numeric Answer Type