Fluid Mechanics

JEE Main level, single correct, medium

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Question 159 of 192

A spherical liquid drop of radius RR acquires the terminal velocity v1v_1 when falls through a gas of viscosity η\eta. Now the drop is broken into 6464 identical droplets and each droplet acquires terminal velocity v2v_2 falling through the same gas. The ratio of terminal velocities v1/v2v_1/v_2 is ________.

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Answer: (D) 1616

The terminal velocity of a spherical drop of radius rr falling through a viscous medium is given by Stokes' law:

v=29r2(ρ−σ)gηv = \dfrac{2}{9} \dfrac{r^2 (\rho - \sigma) g}{\eta}

From this relation, the terminal velocity is directly proportional to the square of the radius of the drop:

v∝r2v \propto r^2

When the large drop of radius RR is broken into 6464 identical droplets of radius rr, the total volume remains conserved:

43πR3=64×43πr3\dfrac{4}{3} \pi R^3 = 64 \times \dfrac{4}{3} \pi r^3

R3=64r3R^3 = 64 r^3

R=4rR = 4r

The ratio of the terminal velocities of the original drop and the smaller droplet is:

v1v2=R2r2\dfrac{v_1}{v_2} = \dfrac{R^2}{r^2}

Substituting R=4rR = 4r:

v1v2=(4r)2r2=16\dfrac{v_1}{v_2} = \dfrac{(4r)^2}{r^2} = 16

Answer: 1616

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