Question

A number of little droplets of a liquid of surface tension T, density $\rho$ , all of same radius r, combine to form a single of radius R. If the total energy released in this process is converted into kinetic energy of the bigger drop, velocity of the bigger drop it

• A. $\sqrt{\frac{3T(R-r)}{Rr\rho }}$
• B. $\sqrt{\frac{6T(R-r)}{Rr\rho }}$
• C. $\sqrt{\frac{T(R-r)}{Rr}}$
• D. $\sqrt{\frac{2T(R-r)\rho }{R}}$

Answer 1

The correct option is B

$\sqrt{\frac{6T(R-r)}{Rr\rho }}$

$R^3=N{r}^3$ so N =$\frac{R^3}{r^3}$

Energy released = $T\times 4\pi (N{r}^2-R^2)=4\pi T(\frac{R^3}{r^3}{r}^2-R^2)=4\pi T{R}^2\left(\frac{R-r}{r}\right)$

So, $\frac{1}{2}m{v}^2=4\pi T{R}^2\left(\frac{R-r}{r}\right)Or\frac{1}{2}\times \frac{4}{3}\pi R^3\rho V^2=4\pi T{R}^2\left(\frac{R-r}{r}\right)orV=\sqrt{\frac{6T(R-r)}{Rr\rho }}$