Eight equal drops of water are falling through air with a steady speed of $10 \mathrm{~cm} / \mathrm{s}$. If the drops coalesce, the new velocity is:-

(B) $40 \mathrm{~cm} / \mathrm{s}$

(A) $10 \mathrm{~cm} / \mathrm{s}$

(C) $16 \mathrm{~cm} / \mathrm{s}$

(D) $5 \mathrm{~cm} / \mathrm{s}$

JEE Main · 2023 · 11 Apr 2023 · Shift 2 · Fluid Mechanics

Eight equal drops of water are falling through air with a steady speed of

10 cm / s

. If the drops coalesce, the new velocity is:-

(A) $10 cm / s$
(B) $40 cm / s$ ✓ Correct
(C) $16 cm / s$
(D) $5 cm / s$

Answer: (B)

Solution

When drops fall through air at terminal velocity, the viscous drag force equals the weight. For a spherical drop, terminal velocity

v \propto r^{2}

(from Stokes' law). When eight drops coalesce, volume is conserved:

8 \times \frac{4}{3} π r^{3} = \frac{4}{3} π R^{3}

, giving

R = 2 r

. Using the proportionality:

\frac{v _{1}}{v _{2}} = (\frac{r}{R})^{2} = (\frac{1}{2})^{2} = \frac{1}{4}

. Therefore

v_{2} = 4 v_{1} = 40 cm/s

JEE Main · 2023 · 11 Apr 2023 · Shift 2 · Single correct option

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