Question

A water drop of radius $1cm$ is broken into $729$ equal droplets. If surface tension of water is $75dync/cm$, then the gain in surface energy upto first decimal place will be :
[Given $\pi =3.14]$

• A. $8.5\times 10^{-4}J$
• B. $8.2\times 10^{-4}J$
• C. $7.5\times 10^{-4}J$
• D. $5.3\times 10^{-4}J$

Answer 1

Answer: (C)

Initial surface energy $=TA$
Where $T$ is surface tension and $A$ is surface area
$U_i=\left(\frac{75\times 10^{-5}}{10^{-2}}\frac{N}{m}\right)\times \left[4\pi {\left(1\times 10^{-2}\right)}^2\right]$
$=75\times 10^{-3}\times 4\pi \times 10^{-4}=942\times 10^{-7}J$
To get final radius of drops by volume conservation
$\frac{4}{3}\pi R^3=729\left(\frac{4}{3}\pi r^3\right)$
$R=$ Initial radius
$r=$ final radius
$r=\frac{R}{(729{)}^{1/3}}=\frac{R}{9}=\frac{1}{9}cm$
Final surface energy
$U_f=729[TA]$
$=729\left[\frac{75\times 10^{-5}}{10^{-2}}\frac{N}{m}\right]\times \left[4\pi {\left(\frac{1}{9}\times 10^{-2}\right)}^2\right]$
$=729\left[75\times 10^{-3}\times \frac{4\pi \times 10^{-4}}{81}\right]$
$=9\left[942\times 10^{-7}J\right]$
Gain in surface energy $\Delta U=9\times 942\times 10^{-7}-942\times 10^{-7}$
$=8\times 942\times 10^{-7}J=7536\times 10^{-7}J$
$=7.5\times 10^{-4}J$