Question

A. liquid drop of radius ‘R’ breaks into 64 tiny droplets each of radius ‘r’. If the surface tension of the liquid is ‘T’ then gain in energy is:

• A. $48\pi R^{2}T$
• B. $12\pi r^{2}T$
• C. $96\pi r^{2}T$
• D. $192\pi r^{2}T$

Answer 1

Answer: (D)

$192\pi r^{2}T$

Let the density of the liquid be $\rho$.

Mass of the liquid drops must be equal to the total mass of the droplets.

$\therefore$ $\rho \times \frac{4\pi }{3}{R}^3=64\times \rho \times \frac{4\pi }{3}{r}^3$ $⟹R=4r$

Change in surface area $\Delta A=64\times 4\pi r^2-4\pi R^2$

$\therefore$ $\Delta A=64\times 4\pi \times r^2-4\pi (4r{)}^2=192\pi r^2$

$\therefore$ Gain in energy $E=T\Delta A=192\pi r^{2}T$