Question

A large number of liquid drops each of radius '$a$' are merged 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 the big drop is
[$\rho =$ density of liquid, $T=$ surface tension of liquid]

• A. ${\left[\frac{6T}{\rho }(\frac{1}{a}-\frac{1}{b})\right]}^{\frac{1}{2}}$
• B. ${\left[\frac{6T}{\rho }(\frac{1}{b}-\frac{1}{a})\right]}^{\frac{1}{2}}$
• C. ${\left[\frac{\rho }{6T}(\frac{1}{a}-\frac{1}{b})\right]}^{\frac{1}{2}}$
• D. ${\left[\frac{\rho }{6T}(\frac{1}{b}-\frac{1}{a})\right]}^{\frac{1}{2}}$

Answer 1

The correct option is A ${\left[\frac{6T}{\rho }(\frac{1}{a}-\frac{1}{b})\right]}^{\frac{1}{2}}$

Let there are $n$ small drops that merged to form a bigger single drop. Mass of the liquid is conserved before and after the merging.

$\therefore$ $\rho (n\times \frac{4\pi }{3}{a}^3)=\rho \left(\frac{4\pi }{3}{b}^3\right)$

We get $n=\frac{b^3}{a^3}$

Total surface area of the small drops $A_1=n\left(4\pi a^2\right)$

Surface area of the bigger drop $A_2=4\pi b^2$

Change in surface area $\Delta A=A_1-A_2$

Or $\Delta A=4\pi (n{a}^2-b^2)$

Or $\Delta A=4\pi (\frac{b^3}{a^3}{a}^2-b^2)=4\pi b^3[\frac{1}{a}-\frac{1}{b}]$

Energy released $E=T\Delta A=$ $4\pi T{b}^3[\frac{1}{a}-\frac{1}{b}]$

Kinetic energy of the bigger drop $K.E=\frac{1}{2}\left(\rho \frac{4\pi }{3}{b}^3\right)v^2$

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

$⟹$ $v=[\frac{6T}{\rho }(\frac{1}{a}-\frac{1}{b}){]}^{1/2}$