Question

A large number of liquid drops each of radius a area 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. ${\left[\frac{6T}{\rho }(\frac{1}{a}-\frac{1}{b})\right]}^{1/2}$
• B. ${\left[\frac{6T}{\rho }(\frac{1}{b}-\frac{1}{a})\right]}^{1/2}$
• C. ${\left[\frac{\rho }{6T}(\frac{1}{a}-\frac{1}{b})\right]}^{1/2}$
• D. ${\left[\frac{\rho }{6T}(\frac{1}{b}-\frac{1}{a})\right]}^{1/2}$

Answer 1

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

Let the number of small liquid drop be $n$.

As the small drops merged to form bigger drop, thus the total mass of the small drops must be equal to the mass of bigger drop.

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

$⟹n{a}^3=b^3$...........(1)

Total surface area of all small drops, $A_1=n\times \pi a^2$

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

$\therefore$ Decrease in surface area, $\Delta A=A_1-A_2=\pi n{a}^2-\pi b^2$

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

Thus amount of energy released, $E=T\Delta A=$ $4\pi b^3[\frac{1}{a}-\frac{1}{b}]T$

Let the speed of the bigger drop be $v$

$\therefore$ $\frac{1}{2}m{v}^2=E$

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

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