Question

Two equal drops are falling through air with a steady velocity of 5cm/s .if two drops coalece, then new terminal velocity will be ?????????????????????????????

Answer 1

Two equal drops are falling through air with a steady velocity of $5\mathrm{cm/s}$ if two drops coalesce, then new terminal velocity will be?

Solution:

Let $r$ be the radius of each drop. The (steady) terminal velocity $v$ of a drop of radius $r$ and density $\rho$, falling through air is given by

$$v = \frac{2}{9} \cdot \frac{r^2 (\rho - \sigma) g}{\eta},$$

where $\sigma$ is density and $\eta$ the viscosity of air. Thus

$$v \propto r^2 = k r^2$$

Now, the volume of each drop $\left(\frac{4}{3}\right) \pi r^2$.

The volume of coalesced drop $= 2 \times \left(\frac{4}{3}\right) \pi r^2 = \left(\frac{4}{3}\right) \pi \left(2^{\frac{1}{3}} r\right)^3$

Radius of the coalesced drop $= (2)^{\frac{1}{3}} r$

Hence, the terminal velocity of the coalesced drop is:

$$v' = k \left((2)^{\frac{1}{3}} r\right)^2 = (2)^{\frac{2}{3}} k r^2 = (2)^{\frac{2}{3}} v = (2)^{\frac{2}{3}} \cdot 5 \frac{\mathrm{cm}}{\mathrm{s}} = 7.94 \frac{\mathrm{cm}}{\mathrm{s}}$$

Answer: new terminal velocity will be $v' = 7.94 \frac{\mathrm{cm}}{\mathrm{s}}$.