Question

A water drop of radius $1mm$ is sprayed into ${10}^6$ droplets of same size at constant temperature. If surface tension of water is $72\times {10}^{-3}N/m$, then the work done is

• A. $8.95\times {10}^{-5}ergs$
• B. $8.95\times {10}^{-5}J$
• C. $17.9\times {10}^{-5}J$
• D. $17.9\times {10}^{-5}ergs$

Answer 1

Answer: (B)

$8.95\times {10}^{-5}J$

Let $R$ be the radius of big drop and $r$ be the radius of each small drop
Volume of big drop$=$ Volume of $N$ small drop
$\frac{4}{3}\pi R^3=N\times \frac{4}{3}\pi r^3$
$\Rightarrow R=N^{1/3}rorr=R{N}^{-1/3}$
Initial surface area, $A_i=4\pi R^2$
Final surface area $A_f=N\pi r^2=N4\pi {\left(R{N}^{-1/3}\right)}^2=N^{1/3}4\pi R^2$
Increase in surface area $=A_f-A_i=\Delta A=4\pi R^2[N^{1/3}-1]$
so, energy expended $=$ Work done
$T\Delta A=72\times {10}^{-3}\times 1243.44=8.95\times {10}^{-5}J$