Physics - Fluid Question with Solution | TestHub

Let radius of big drop $=R$. $$\begin{array}{l}\therefore \frac{4}{3}\pi R^3=\frac{4}{3}\times \pi r^3\cdot 8\\ R=2r\end{array}$$ Here $r=$ radius of small drops. Now, terminal velocity of drop in liquid $$v_T=\frac{2}{9}\times \frac{r^2}{\eta }(\rho -\sigma )g$$ where $\eta$ iscoefficient of viscosity and $\rho$ is density of drop $\sigma$ is density of liquid. Term inal speed drop is $6\mathrm{cm}{\mathrm{s}}^{-1}$ $$\therefore 6=\frac{2}{9}\times \frac{r^2}{\eta }(\rho -\sigma )g...(i)$$ Let terminal velocity becomes $v^{\prime }$ after coalesce, then $$v^{\prime }=\frac{2}{9}\frac{R^2}{\eta }(\rho -\sigma )g...(ii)$$ Dividing Eq. (i) by (ii), we get $$\frac{6}{v^{\prime }}=\frac{\frac{2}{9}\frac{r^2}{\eta }(\rho -\sigma )g}{\frac{2}{9}\frac{R^2}{\eta }(\rho -\sigma )g}\Rightarrow \frac{6}{v^{\prime }}=\frac{r^2}{(2r{)}^2}$$ $$\text{ or }{v}^{\prime }=24\mathrm{cm}{\mathrm{s}}^{-1}$$