Question

A drop of mercury of radius $2mm$ is split into $8$ identical drops. Find the increase in surface energy. Surface tension of mercury $=0.465J{m}^{-2}$

Answer 1

$R=2mm$ $\Rightarrow$ $2\times {10}^{3}m$

The volume of drop is given as:

$\frac{4}{3}\pi R^3=8\times \frac{4}{3}\times r^3$

$\therefore$ $R^3=8r^3$

$\frac{R^3}{8}=r^3$ $\Rightarrow$ ${\left(\frac{R}{2}\right)}^3=r^3$

$\therefore$ $r=1mm={10}^{-3}m$

$A_1=4\pi R^2$

$\Rightarrow$ $4\pi {(2\times {10}^{-2})}^2=4\pi \times 4\times {10}^{-6}=16\pi \times {10}^{-6}$

$A_2=8\times 4\pi r^2=32\pi {\left({10}^{-2}\right)}^3=32\pi {10}^{-6}$

$\Delta A=A_1-A_2$

$(32\pi -16\pi ){10}^{-6}=16\pi \times {10}^{-6}$

the increase in surface energy will be:

$E_s=T\Delta A=23.376$