Question

Two equal drops are falling through air with a steady velocity of $5cm{s}^{-1}$. If the drop coalesces, the new terminal velocity will become:

• A. $5\times 2cm{s}^{-1}$
• B. $5\times \sqrt{2}cm{s}^{-1}$
• C. $5\times (4{)}^{\frac{1}{3}}cm{s}^{-1}$
• D. $\frac{5}{\sqrt{2}}cm{s}^{-1}$

Answer 1

The correct option is C $5\times (4{)}^{\frac{1}{3}}cm{s}^{-1}$

when the two equal drops coalesces, volume of water remains same and a bigger drop is formed. Lets say $r$ is radius of drop before these combine and $R$ is radius after these combine. we have
$2\times \frac{4}{3}\pi r^3=\frac{4}{3}\pi R^3\Rightarrow R=2^{1/3}r$
and we know that terminal velocity $V_T\propto$ radius${}^2$, so we have
$\frac{V_T^{\prime }}{V_T}=\frac{R^2}{r^2}\Rightarrow V_T^{\prime }={\left(\frac{R}{r}\right)}^2{V}_T={\left(2^{1/3}\right)}^2\times 5cm{s}^{-1}=5\times 4^{\frac{1}{3}}cm{s}^{-1}$