Physics - Fluid Question with Solution | TestHub

$n\frac{4}{3}\pi r^3=\frac{4}{3}\pi R^3$

$n=\frac{R^3}{r^3}$

$\mathrm{\Delta }U=n 4\pi r^2 T-4\pi R^2 T$

$=4\pi T \left(n{r}^2-R^2\right)$

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

$\therefore$Energy released,

$U=4\pi T{R}^3\left[\frac{1}{r}-\frac{1}{R}\right]$

If this energy is converted into kinetic energy then

$\frac{1}{2}m{v}^2=4\pi R^{3}T\left[\frac{1}{r}-\frac{1}{R}\right]$

$\frac{1}{2}\times \left[\frac{4}{3}\pi R^3\rho \right]v^2=4\pi R^{3}T\left[\frac{1}{r}-\frac{1}{R}\right]$

$v=\sqrt{\frac{6T}{\rho }\left[\frac{1}{r}-\frac{1}{R}\right]}$