VITEEE 2009 Paper

Let radius of big drop = R. ∴ $\frac{4}{3}\pi R^{3} = \frac{4}{3} \times \pi r^{3}.8$ R = 2r 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 η is coefficient of viscosity and ρ is density of drop, σ is density of liquid. Terminal speed drop is 6cm s$^{-1}$ ∴ $6 = \frac{2}{9} \times \frac{r^{2}}{\eta(\rho-\sigma)g}$...(i) Let terminal velocity becomes v' after coalesce, then $v' = \frac{2}{9} \times \frac{R^{2}}{\eta(\rho-\sigma)g}$ ...(ii) Dividing Eq. (i) by (ii), we get $\frac{6}{v'} = \frac{ \frac{ \frac{2}{9} r^{2} }{ \eta(\rho-\sigma)g } }{ \frac{ \frac{2}{9} R^{2} }{ \eta(\rho-\sigma)g } } \Rightarrow \frac{6}{v'} = \frac{r^{2}}{(2r)^{2}}$ or v' = 24cm s$^{-1}$