Question

A drop of liquid of radius $R={10}^{-2}m$ having surface tension $S=\frac{0.1}{4\pi }N{m}^{-1}$ divides itself into $K$ identical drops. In this process the total change in the surface energy $\Delta U={10}^{-3}J$. If $K={10}^{\alpha }$ then the value of $\alpha$ is

Answer 1

As the volume of the drop is constant,
Volume of big drop=$k\times$ volume of small drop
let, the radious of small drop is $r$.
So,
$\frac{4}{3}\pi R^3=K\times \frac{4}{3}\pi r^3$
where, $R=r(k{)}^{\frac{1}{3}}$ ...(1)
Now, change in surface energy ($\Delta U)=$ surface tension $\times$ change in surface area
$\Delta U=4\pi T[R^2-r^2]$
$=4\pi T{R}^2[k^{1/3}-1]$
${10}^{-3}=4\pi \times \frac{0.1}{4\pi }\times {10}^{-4}[k^{1/3}-1]$
$k^{1/3}=101\approx 100$
$k={10}^6$
$\alpha =6$