Question

A water drop of radius $1 \,cm$ is broken into $729$ equal droplets. If surface tension of water is $75 \,dync / cm$, 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 $