Question

A water drop of diameter 2 $cm$ is broken into $64$ equal droplets. The surface tension of water is $0.075N/m$. In this process the gain in surface energy will be :

• A. $2.8\times 10^{-4}J$
• B. $1.5\times 10^{-3}J$
• C. $1.9\times 10^{-4}J$
• D. $9.4\times 10^{-5}J$

Answer 1

Answer: (A)

$d=2cm$;
$r=1cm$;
$T=0.075$
$\Delta SE=T\Delta A$
$=0.075\left(A_f-A_1\right)$
$A_i=4\pi r^2$
$A_f=4\pi r_0^2\times 64$
By volume conservation
$\frac{4}{3}\pi r^3=64\cdot \frac{4}{3}\pi r_0^3$
$r_0=\frac{r}{4}$
$A_f=4\pi {\left(\frac{r}{4}\right)}^2\cdot 64=16\pi r^2$
$\Delta SE=0.075\left(16\pi r^2-4\pi r^2\right)$
$=0.075\left(12\pi (0.01{)}^2\right)$
$=2.8\times 10^{-4}J$