JEE Main · 2020 · 05 Sep 2020 · Shift 2 · Fluid Mechanics

In an experiment to verify Stokes law, a small spherical ball of radius

r

and density

ρ

falls under gravity through a distance h in air before entering a tank of water. If the terminal velocity of the ball inside water is same as its velocity just before entering the water surface, then the value of

h

is proportional to: (ignore viscosity of air)

(A) $r^{3}$
(B) $r^{4}$ ✓ Correct
(C) $r^{2}$
(D)r

Answer: (B)

Solution

Falling freely through height

h

, the ball reaches the water at speed

v = 2 g h

, and this equals its terminal speed in water. By Stokes' law the terminal speed is

v_{t} = \frac{2 r ^{2} g ( ρ - ρ _{w} )}{9 η}

. Equating

2 g h = \frac{2 r ^{2} g ( ρ - ρ _{w} )}{9 η}

and squaring gives

2 g h = \frac{4 r ^{4} g ^{2} ( ρ - ρ _{w} ) ^{2}}{81 η ^{2}}

, so

h = \frac{2 r ^{4} g ( ρ - ρ _{w} ) ^{2}}{81 η ^{2}}

. Everything except

r

is fixed, so

h \propto r^{4}

JEE Main · 2020 · 05 Sep 2020 · Shift 2 · Single correct option

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JEE Main · 2020 · 05 Sep 2020 · Shift 2 · Fluid Mechanics

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