Question

Two drops of the same radius are falling through air with a steady velocity of 5 cm/s. If the two drops coalesce, the terminal velocity would be

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

Answer 1

Answer: (C)

$5(4{)}^{1/3}cm/s$

terminal velocity $\left(V\right)\alpha r^2$

$=K{r}^2$

Now the volume of each drop =$\left(\frac{4}{3}\pi r^3\right)$

The volume of coalesced = $2\times \frac{4}{3}\pi R^3$

Radius of coalesce drop = $(2{)}^{\frac{1}{3}}r$

Hence, ter,inal velocity of coalesced drop

$v^1=k\left[\right(2{)}^{\frac{1}{3}}r]$

$=(2{)}^{\frac{2}{3}}\times 5$

$V^{\prime }=(4{)}^{\frac{1}{3}}\times 5$

Hence option (C) is correct