Two identical drops of water are falling through air with steady velocity ' $V$ '. If the two drops come…
Two identical drops of water are falling through air with steady velocity ' $V$ '. If the two drops come together to form a single drop. The new velocity of the single drop is
$(2)^{1 / 3} \mathrm{~V}$
$\quad(2)^{3 / 2} \mathrm{~V}$
$\quad(2)^{2 / 3} \mathrm{~V}$
$\quad(2)^{1 / 4} \mathrm{~V}$
Solution
Let the radius of the 2 rain droplets be r each.
They coalesce to form a drop of radius R .
As volume is conserved, $\mathrm{R}^3=2 \mathrm{r}^3$
$\therefore \quad R=2^{\frac{i}{3}} r$
Terminal velocity $\mathrm{V}=\frac{2}{9} \frac{\mathrm{r}^2(\rho-\sigma) g}{\eta}$
$\therefore \quad \mathrm{V} \propto \mathrm{r}^2$
$\therefore \quad \mathrm{V}^{\prime}=\mathrm{V} \frac{\mathrm{R}^2}{\mathrm{r}^2}=\mathrm{V} \times \frac{2^{\frac{2}{3}} \mathrm{r}^2}{\mathrm{r}^2}=\mathrm{V} \cdot 2^{2 / 3}$