Question

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

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

Answer 1

Answer: (D)

$n\frac{4}{3}\pi r^{3}=\frac{4}{3}\pi R^{3}$
$n=\frac{R^{3}}{r^{3}}$
$\Delta U=n \, 4\pi r^{2} \, T-4\pi R^{2} \, T$
$=4\pi T \, \left(n r^{2} - R^{2}\right)$
$=4\pi T \, \left(\frac{R^{3} r^{2}}{r^{3}} - R^{2}\right)$
$∴ \, \, $ Energy released,
$U = 4 \pi T R^{3} \left[\right. \frac{1}{r} - \frac{1}{R} \left]\right.$
If this energy is converted into kinetic energy then
$\frac{1}{2} m v^{2} = 4 \pi R^{3} T \left[\right. \frac{1}{r} - \frac{1}{R} \left]\right.$
$\frac{1}{2} \times \left[\right. \frac{4}{3} \pi R^{3} \rho \left]\right. v^{2} = 4 \pi R^{3} T \left[\right. \frac{1}{r} - \frac{1}{R} \left]\right.$
$v = \sqrt{\frac{6 T}{\rho } \left[\right. \frac{1}{r} - \frac{1}{R} \left]\right.}$