JEE Main · 2021 · 27 Jul 2021 · Shift 2 · Fluid Mechanics

A raindrop with radius

R = 0.2 mm

falls from a cloud at a heighth

= 2000 m

above the ground. Assume that the drop is spherical through its fall and the force of buoyance may be neglected, then the terminal speed attained by the raindrop is :
[Density of water

f_{w} = 1000 kg m^{- 3}

and Density of air

f_{a} = 1.2 kg m^{- 3}

g = 10 m / s^{2}

Coefficient of viscosity of air

= 1.8 \times 1 0^{- 5} Nsm^{- 2}

]

(A) $4.94 ms^{- 1}$ ✓ Correct
(B) $14.4 ms^{- 1}$
(C) $43.56 ms^{- 1}$
(D) $250.6 ms^{- 1}$

Answer: (A)

Solution

At the terminal speed the weight of the drop is balanced by the viscous drag given by Stokes' law, and buoyancy is neglected:

\frac{4}{3} π r^{3} ρ g = 6 π η r v_{t} \Rightarrow v_{t} = \frac{2 r ^{2} ρ g}{9 η}

With

r = 2 \times 1 0^{- 4}

ρ = 1000

kg m

^{- 3}

g = 10

m s

^{- 2}

and

η = 1.8 \times 1 0^{- 5}

N s m

^{- 2}

v_{t} = \frac{2 \times ( 2 \times 1 0 ^{- 4} ) ^{2} \times 1000 \times 10}{9 \times 1.8 \times 1 0 ^{- 5}} = \frac{8 \times 1 0 ^{- 4}}{1.62 \times 1 0 ^{- 4}}

The terminal speed attained by the raindrop is

4.94

m s

^{- 1}

JEE Main · 2021 · 27 Jul 2021 · Shift 2 · Single correct option

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JEE Main · 2021 · 27 Jul 2021 · Shift 2 · Fluid Mechanics

Solution

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