Question

Two drops of same radius are falling through air with a steady velocity of $5\text{cm/s}$. If the two drops coalesce, the terminal velocity will be

• A. $10\text{cm/s}$
• B. $2.5\text{cm/s}$
• C. $5\times (4{)}^{1/3}\text{cm/s}$
• D. $5\times \sqrt{2}\text{cm/s}$

Answer 1

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

Terminal velocity for a spherical body of radius $R$ and density $\rho$ falling in a liquid of density $\sigma$ and dynamic viscosity $\eta$ is given by,
$V_T=\frac{2}{9}\frac{(\rho -\sigma )R^{2}g}{\eta }$
$\Rightarrow V_T\propto R^2$
Let two identical drops of radius $R_1$ coalesce to form bigger drop of radius $R_2$,
$\begin{array}{}\frac{V_{T_2}}{V_{T_1}}=\frac{R_2^2}{R_1^2}& \text{(1)}\end{array}$
Since the volume remains same before and after the drops coalesce. Therefore,
$V_i=V_f$
$2\times \frac{4}{3}\pi R_1^3=\frac{4}{3}\pi R_2^3$
$\Rightarrow R_2=(2{)}^{1/3}{R}_1$
Substituting this in equation $\left(1\right)$,
$\frac{V_{T_2}}{V_{T_1}}=\frac{(4{)}^{1/3}{R}_1^2}{R_1^2}$
$V_{T_2}=5\times (4{)}^{1/3}cm/s$