Fluid Mechanics

JEE Main level, numerical answer, medium

161 of 192

Question 161 of 192

Sixty four rain drops of radius 1 mm each falling down with a terminal velocity of 10 cm/s10 \mathrm{~cm} / \mathrm{s} coalesce to form a bigger drop. The terminal velocity of bigger drop is ____\_\_\_\_ cm/s\mathrm{cm} / \mathrm{s}.

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Answer: 160

At terminal velocity, gravitational force equals Stokes drag: mg=6πηrvmg = 6\pi\eta rv
For a sphere with mass m=43πr3ρm = \frac{4}{3}\pi r^3 \rho, this gives v=2r2ρg9ηv = \frac{2r^2\rho g}{9\eta}
Therefore, v∝r2v \propto r^2
When 64 drops coalesce: 64×43πr13=43πr2364 \times \frac{4}{3}\pi r_1^3 = \frac{4}{3}\pi r_2^3, so r2=4r1r_2 = 4r_1
New terminal velocity: v2v1=(r2r1)2=16\frac{v_2}{v_1} = \left(\frac{r_2}{r_1}\right)^2 = 16
v2=16×10=160v_2 = 16 \times 10 = 160 cm/s

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