Question

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

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

Answer 1

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

If $R$ is radius of bigger drop formed, then

$\frac{4}{3}\pi R^3=2\times \frac{4}{3}\pi r^3$ or $R=2^{1/3}r$
As $v_0\propto r^2$

$\therefore \frac{v_{01}}{v_0}=\frac{R^2}{r^2}=\frac{(2^{1/3}{)}^2}{r^2}=2^{2/3}$

or $v_{01}=v_0{2}^{2/3}=5(4{)}^{1/3}$