Question

The amount of energy dissipated when $8$ water drops of $0.6mm$ radius coalesce to form one big drop is _________ $J$ $($given surface tension of water is $0.072N{m}^{-1})$

• A. $1.3\times {10}^{-7}$
• B. $13\times {10}^{-6}$
• C. $13\times {10}^{-7}$
• D. $0.13\times {10}^{-8}$

Answer 1

Answer: (C)

$13\times {10}^{-7}$

Let $r$ is the radius of each single drop before these coalesce and $R$ is the radius of the new drop after these coalesce. Since volume of liquid remains same, we have

$8\times \frac{4}{3}\pi r^3=\frac{4}{3}\pi R^3$

$\Rightarrow R=8^{\frac{1}{3}}r=2r.......................\left(1\right)$

Initial surface energy of single drop $=\left(4\pi r^2\right)\times T$

Initial surface energy of $8$ drops $=8\times \left(4\pi r^2\right)\times T=32\pi r^{2}T$

final surface energy of big single drop $=\left(4\pi R^2\right)\times T=\left(4\pi {\left(2r\right)}^2\right)\times T=16\pi r^{2}T$

$\because$ final surface energy $<$ initial surface energy

that means some energy has been dissipated.

And the amount of energy dissipated $\left(\Delta E\right)$ in this process is given by

Initial surface energy $-$ final surface energy$=32\pi r^{2}T-16\pi r^{2}T=16\pi r^{2}T$

$\Rightarrow \Delta E=16\pi \times (0.6\times {10}^{-3}{)}^2\times 0.072=13\times {10}^{-7}J$