Question

The work done in splitting a drop of water of $1mm$ radius into ${10}^6$ droplets is (surface tension of water $72\times {10}^{-3}N/M$ ) ;

• A. $5.98\times {10}^{-5}J$
• B. $10.98\times {10}^{-5}J$
• C. $16.95\times {10}^{-5}J$
• D. $8.95\times {10}^{-5}J$

Answer 1

Answer: (D)

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

Radius of new droplet if be r then, ${10}^6\times \frac{4}{3}\pi r^3=\frac{4}{3}\pi \times {\left(0.001\right)}^3$
$r^3={10}^{-15}\Rightarrow r={10}^{-5}$
Increase in surface area
$=[4\pi \times {\left({10}^{-5}\right)}^2\times {10}^6]-[4\pi \times {\left({10}^{-3}\right)}^2]$
$=[4\pi \times {10}^{-4}]-[4\pi \times {10}^{-6}]=4\pi {10}^{-6}[100-1]$
$=4\pi \times {10}^{-6}\times 99=4\pi \times {10}^{-6}\times 99$
Work done = surface tension x increase in surface area
$=72\times 4\pi \times 99\times {10}^{-6}\times {10}^{-3}=8.95\times {10}^{-5}J$