Question

Eight drops of equal size are falling through air- with steady velocity of $10\, cm/s$. If the drop; coalesce, what could be its terminal velocity

• A. 40 cm/s
• B. 20 cm/s
• C. 25 cm/s
• D. 50 cm/s

Answer 1

Answer: (A)

Let $R$ be radius of coalesced drop, and $r$ the radius of $8$ small drops. Also,
Mass of coalesced drop $=8 \times$ mass of small drop
$\Rightarrow \frac{4}{3} \pi R^{3} \times \rho$
$=8 \times \frac{4}{3} \pi r^{3} \times \rho$
$\Rightarrow R=2 r$
Also terminal velocity
$v_{T}=\frac{2}{9} \frac{(\rho-\sigma) r^{2} g}{\eta}$ ...(i)
and of coalesced drop is
$v_{T}'=\frac{2}{9} \frac{(\rho-\sigma) R^{2} g}{\eta}$ ...(ii)
From Eqs. (i) and (ii), we get
$\frac{v_{T}'}{v_{T}} =\left(\frac{R}{r}\right)^{2}=(2)^{2}=4$
$\therefore v_{T}'=4 \times 10=40\, cm / s$