Question

Two drops of same radius are falling through air with steady velocity of $vcm/s$. If the two drops coalesce, what would be the terminal velocity?

• A. $4v$
• B. ${\left(4\right)}^{1/3}v$
• C. $2v$
• D. $64v$

Answer 1

The correct option is B ${\left(4\right)}^{1/3}v$

When the two drops coalesce the volume is conserved, thus if r is the radius of the individual drops and R of the coalesced drop, we get
$2\times \frac{4}{3}\pi r^3=\frac{4}{3}\pi R^3$
$R=(2{)}^{1/3}r$
We know that the terminal velocity of each individual drop is given as $v=\frac{\left(\frac{2}{9}\right)r^{2}g(\rho -\sigma )}{\eta }$
Thus the terminal velocity of the coalesced drop is given as $v^{\prime }=\frac{\left(\frac{2}{9}\right)\left(2^{1/3}r{)}^{2}g\right(\rho -\sigma )}{\eta }$
$v^{\prime }=2^{2/3}v=4^{1/3}v$