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. The speed of big drop will be

• 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 $\sqrt{\frac{6T}{\rho }[\frac{1}{a}-\frac{1}{b}]}$

Volume of $n$ drops of radius$a=$ Volume of drop with radius $b$

$\Rightarrow n\times \frac{4}{3}\pi a^3=\frac{4}{3}\pi b^3$

$\Rightarrow n{a}^3=b^3$

Work done $=\text{Surface Tension}\times \text{Change in Area}=T\Delta A$

$\Rightarrow \frac{1}{2}m{v}^2=T[4\pi n{a}^2-4\pi b^2]$

$\Rightarrow \frac{1}{2}m{v}^2=4\pi T[n{a}^2-b^2]$

$\Rightarrow \frac{1}{2}\rho \times \frac{4}{3}\pi b^3\times v^2=4\pi T[n{a}^2-b^2]$

$\Rightarrow v=\sqrt{\frac{6T}{\rho }[\frac{n{a}^2}{b^3}-\frac{b^2}{b^3}]}$

$\Rightarrow v=\sqrt{\frac{6T}{\rho }[\frac{1}{a}-\frac{1}{b}]}$