Eight drops of water.each radius 2mm are falling through air at a terminal velocity of 8cm/sec If they coalesce to form a single drop the terminal velocity of the combined drop will be?

1 Answer

3232 cm/sec

Explanation:

For small Reynold number, the viscous force on the small sphere of radius rr moving with a velocity vv in a fluid of dynamic viscosity \muμ is given by Stoke's law as

F_v=6\pi\mu r vFv=6πμrv

For terminal or critical velocity v_cvc to achieve, the viscous force on the small sphere (density \rho_sρs) must be equal to the net weight of body in the fluid of density \rho_fρf hence we have

F_v=mg-\text{Buoyant force}Fv=mgBuoyant force

6\pi\mur v_c={4\pi}/3 r^3(\rho_s-\rho_f)g6πμrvc=4π3r3(ρsρf)g

v_c=\frac{2}{9}\frac{(\rho_s-\rho_f)gr^2}{mu}vc=29(ρsρf)gr2μ

v_c\propr^2vcr2

the above relation shows that the terminal velocity v_cvc of the sphere is directly proportional to the square of radius rr keeping other parameters constant

Now, 8 identical water drops each of radius r_1=2\ mm collapse to form a single big drop of radius r_2

Now, by the conservation of volume, volume of big drop of water (radius r_2) must be equal to the volume of 8 identical water drops (r_1=2mm ) then we have

\frac{4\pi}{3}r_2^2=8\times \frac{4\pi}{3}r_1^2

r_2=2r_1

r_2=2(2)=4\ mm

From, terminal velocity relation (derived above), we know

v_c\propr^2

\frac{(v_c)_2}{(v_c)1}=\frac{r_2^2}{r_1^2}

\frac{(v_c)_2}{8}=\frac{4^2}{2^2}

\frac{(v_c)_2}{8}=4

(v_c)_2=32\ \text{cm/sec}