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 the kinetic energy of the big drop formed, the speed of the big drop is

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

Answer 1

Answer: (D)

According to question $\frac{4}{3}\pi a^3$
$n=\frac{4}{3}\pi b^3$
$n={\left(\frac{b}{a}\right)}^3$
$W=T.4\pi \left[n{a}^2-n^2\right]W$
$=\frac{1}{2}m{v}^2=\frac{1}{2}\cdot \frac{4}{3}\pi b^{3}p{v}^2$
$\therefore \frac{1}{2}\times \frac{4}{3}\pi b^{3p{v}^2=}T.4\pi \left[n{a}^2-b^2\right]$
or $v=\sqrt{\frac{6T}{\rho }\left(\frac{n{a}^2}{b^3}-\frac{b^2}{b^3}\right)}$
$\therefore v=\sqrt{\frac{6T}{\rho }\left(\frac{1}{a}-\frac{1}{b}\right)}$