Question

What is the work done in breaking a drop of water of 4 mm diameter into 1000 drops of same size, surface tension of water is $72\times 10^{-3}N{m}^{-1}?$

• A. $1.125\times 10^{-4}J$
• B. $1.125\times 10^{-5}J$
• C. $3.25\times 10^{-5}J$
• D. $3.25\times 10^{-6}J$

Answer 1

Answer: (C)

Volume remains constant. Let $R$ and $r$ be the radius of bigger and smaller drops respectively.
$\frac{4}{3}\pi R^3=10^3\times \frac{4}{3}\pi r^3$
$R=10\times r$
$\Rightarrow r=\frac{R}{10}=\frac{2\times 10^{-3}}{10}=2\times 10^{-4}m$
Work done in breaking drop into 1000 drops.
$W=S\times \left(10^3\times 4\pi r^2-4\pi R^2\right)=S\times 4\pi \left(10^3\times r^2-R^2\right)$
$W=72\times 10^{-3}\times 4\times 3.14\left[10^3\times {\left(2\times 10^{-4}\right)}^2-{\left(2\times 10^{-3}\right)}^2\right]$
$W=0.904\left(10^3\times 4\times 10^{-8}-4\times 10^{-6}\right)$
$W=0.904\left(4\times 10^{-5}-4\times 10^{-6}\right)$
$W=0.904\times 4\times 10^{-5}(1-0.1)=3.25\times 10^{-5}J$