Question

Two drops of same radius are falling through air with steady speed $2^{\frac{1}{3}}m/s$. If the two drops coalesce, what would be the terminal speed in $m/s$ ?

• A. $1$
  
• B. $4$
  
• C. $2$
  
• D. $8$
  

Answer 1

The correct option is C $2$

In equilibrium, the equation for each drop is
$6\pi \eta rv=\frac{4}{3}\pi r^3(\rho -\sigma )g$ .....(1)
When two drops coalesce, volume will be conserved.
Hence,
$2\times \frac{4}{3}\pi r^3=\frac{4}{3}\pi r^{\prime 3}$
$\therefore r^{\prime }=2^{\frac{1}{3}}r$
Hence, at equilibrium,
$6\pi \eta r^{\prime }{v}^{\prime }=\frac{4}{3}\pi r^{\prime 3}(\rho -\sigma )g$ .....(2) where, v' is the terminal speed after both drops coalesce.
Dividing LHS and RHS of (1) and (2) respectively
$\frac{rv}{r^{\prime }{v}^{\prime }}=\frac{r^3}{r^{\prime 3}}$
$\therefore \frac{v^{\prime }}{v}=\frac{r^{\prime 2}}{r^2}$
$\therefore v^{\prime }=2^{\frac{2}{3}}v=2^{\frac{2}{3}}{2}^{\frac{1}{3}}=2^{(\frac{2}{3}+\frac{1}{3})}=2^1=2$