Question

Two drops, of the same radius, are falling through air with a steady velocity of $5\,cm\,s^{-1}$ If the two drops coalesce, the terminal velocity would be

• A. $10\,cm\,s^{-1}$
• B. $2.5\,cm\,s^{-1}$
• C. $5 \times (4)^{1/3}\,cm\,s^{-1}$
• D. $5 \times \sqrt{3} cm\,s^{-1}$

Answer 1

Answer: (C)

If $R$ is radius of bigger drop formed, then
$\frac{4}{3}\pi R^3=2\times \frac{4}{3}\mathrm{\pi }{r}^3\mathrm{or}R=2^{1/3}r$
As, $v_0\propto r^2$
∴ $\frac{{{\left(v_0\right)}_1}_{}}{v_0}=\frac{R^2}{r^2}=\frac{(2^{1/3}r{)}^2}{r^2}=2^{2/3}$
or ${\left(v_0\right)}_1=v_0{2}^{2/3}$
$5\times {\left(4\right)}^{1/3}$
$=\left(5\right)\times {\left(2\right)}^{2/3}$
$=5\times {\left(4\right)}^{1/3}{\text{cm s}}^{\text{-1}}$