Question

A mercury drop of radius 1 cm is broken into $10^6$ droplets of equal size. The work done is (Surface tension $=35\times 10^{-2}N{m}^{-1})$

• A. $4.35\times 10^{-2}J$
• B. $4.35\times 10^{-3}J$
• C. $4.35\times 10^{-6}J$
• D. $4.35\times 10^{-8}J$

Answer 1

Answer: (A)

Here, radius of the drop, $R=1cm=1\times 10^{-2}m$
Let $r$ be the radius of each droplet. Then Volume of $10^6$ droplets $=$ Volume of the drop
or $10^6\times \frac{4}{3}\pi r^3=\frac{4}{3}\pi R^3$
or $r=\frac{10^{-2}}{10^2}=10^{-4}m$
Increase in surface area $=10^6\times 4\pi r^2-4\pi R^2$
$=10^6\times 4\times \pi \times 10^{-8}-4\pi \times 10^{-4}$
$=4\times 9.9\times \pi \times 10^{-3}{m}^2$
$\therefore$ Work done = surface tension $\times$ increase in surface area
$=35\times 10^{-2}\times 4\times 9.9\times \pi \times 10^{-3}$
$=4.35\times 10^{-2}J$