Question

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

• 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, $R=1 \,cm =1 \times 10^{-2} \,m$
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
$=35 \times 10^{-2} \times 4 \times 9.9 \times \pi \times 10^{-3}$
$=4.35 \times 10^{-2} \,J$