Question

A large number of liquid drops each of radius $a$ coalesce to form a single spherical drop of radius $b$. The energy released in the process is converted into kinetic energy of the big drop formed. What will be the speed of the bigger drop, if it starts from rest?
[Given, $T$ is the surface tension and $\rho$ is the density of liquid]

• A. $\sqrt{\frac{6T}{\rho }[\frac{1}{a}-\frac{1}{b}]}$
• B. $\sqrt{\frac{4T}{\rho }[\frac{1}{a}-\frac{1}{b}]}$
• C. $\sqrt{\frac{8T}{\rho }[\frac{1}{a}-\frac{1}{b}]}$
• D. $\sqrt{\frac{5T}{\rho }[\frac{1}{a}-\frac{1}{b}]}$

Answer 1

The correct option is A $\sqrt{\frac{6T}{\rho }[\frac{1}{a}-\frac{1}{b}]}$

Let $n$ no. of small drops coalesced.

$\Rightarrow n\frac{4}{3}\pi a^3=\frac{4}{3}\pi b^3\Rightarrow b=a.n^{1/3}$
$\therefore n={\left(\frac{b}{a}\right)}^3$

Change in surface area
$\Delta A=4\pi b^2-n.4\pi a^2$ (this is -ve, hence energy is released)

$\therefore \Delta A=4\pi a^2(n^{2/3}-n)$

Now, the energy released is given by, $U=T\times \Delta A$
$\Rightarrow U=4\pi a^{2}T(n-n^{2/3})$
$=4\pi a^{2}T{\left(\frac{b}{a}\right)}^3-{\left(\frac{b}{a}\right)}^2$

This energy converts into $K.E$.

$\Rightarrow \frac{1}{2}m{V}^2=U$

$\frac{1}{2}\left(\rho \frac{4}{3}\pi b^3\right)V^2=4\pi a^{2}T\frac{b^2}{a^2}\left(\frac{b-a}{a}\right)$

$\Rightarrow V=\sqrt{\frac{6T}{\rho }(\frac{1}{a}-\frac{1}{b})}$

Hence, $\left(A\right)$ is the correct answer.