Question

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

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

Answer 1

Answer: (C)

$5 \times (4)^{1/3}cm$persec

Answer 2

If two drops of same radius r coalesce then radius of new

drop is given by R

$\frac{4}{3}\pi R^{3} = \frac{4}{3}\pi r^{3} + \frac{4}{3}\pi r^{3}$ ⇒ $R^{3} = 2r^{3} \Rightarrow R = 2^{1/3}r$

If drop of radius r is falling in viscous medium then it acquire

a critical velocity v and $v \propto r^{2}$

$\frac{v_{2}}{v_{1}} = \left( \frac{R}{r} \right)^{2} = \left( \frac{2^{1/3}r}{r} \right)^{2}$

⇒ $v_{2} = 2^{2/3} \times v_{1} = 2^{2/3} \times (5) = 5 \times (4)^{1/3}m/s$