Question

A large number of liquid drops, each of radius r, coalesce to form a single drop of radius R. The energy released in the process is converted into the kinetic energy of the big drop so formed. The speed of the big drop is : (T $=$ surface tension, $\rho =$ density of liquid)

• A. $\sqrt{\frac{T}{\rho }(\frac{1}{r}-\frac{1}{R})}$
• B. $\sqrt{\frac{2T}{\rho }(\frac{1}{r}-\frac{1}{R})}$
• C. $\sqrt{\frac{4T}{\rho }(\frac{1}{r}-\frac{1}{R})}$
• D. $\sqrt{\frac{6T}{\rho }(\frac{1}{r}-\frac{1}{R})}$

Answer 1

The correct option is D $\sqrt{\frac{6T}{\rho }(\frac{1}{r}-\frac{1}{R})}$

$n\left(\frac{4}{3}\pi r^3\right)=\frac{4}{3}\pi R^3$
$\Rightarrow n=(\frac{R}{r}{)}^3$
Change in energy $=n8\pi r^{2}S-\pi R^{2}S$

$=8\pi S(n{r}^2-R^2)$
$=8\pi S(\frac{R^3}{r}-R^2)$
$\frac{1}{2}m{v}^2=8\pi S(R^2-\frac{R^3}{r})$
$\frac{1}{2}\times \frac{4}{3}\pi R^3\rho \times v^2=8\pi S(R^2-\frac{R^3}{r})$
$\Rightarrow v=\sqrt{\frac{12S}{\rho }(\frac{1}{R}-\frac{1}{r})}$

Surface energy per unit area is Surface Tension.

So, $T=S/\pi R^2$
$=\sqrt{\frac{6T}{\rho }(\frac{1}{r}-\frac{1}{R})}$