A water drop of radius $1 \mu \mathrm{~m}$ falls in a situation where the effect of buoyant force is negligible. Coefficient of viscosity of air is $1.8 \times 10^{-5} \mathrm{Nsm}^{-2}$ and its density is negligible as compared to that of water $10^{6} \mathrm{gm}^{-3}$. Terminal velocity of the water drop is: (Take acceleration due to gravity $=10 \mathrm{~ms}^{-2}$)

(D) $123.4 \times 10^{-6} \mathrm{~ms}^{-1}$

(A) $145.4 \times 10^{-6} \mathrm{~ms}^{-1}$

(B) $118.0 \times 10^{-6} \mathrm{~ms}^{-1}$

(C) $132.6 \times 10^{-6} \mathrm{~ms}^{-1}$

JEE Main · 2022 · 28 Jun 2022 · Shift 2 · Fluid Mechanics

A water drop of radius

1 μ m

falls in a situation where the effect of buoyant force is negligible. Coefficient of viscosity of air is

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

and its density is negligible as compared to that of water

1 0^{6} gm^{- 3}

. Terminal velocity of the water drop is:
(Take acceleration due to gravity

= 10 ms^{- 2}

)

(A) $145.4 \times 1 0^{- 6} ms^{- 1}$
(B) $118.0 \times 1 0^{- 6} ms^{- 1}$
(C) $132.6 \times 1 0^{- 6} ms^{- 1}$
(D) $123.4 \times 1 0^{- 6} ms^{- 1}$ ✓ Correct

Answer: (D)

Solution

At terminal velocity,

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

. Solving:

v_{t} = \frac{2 r ^{2} ρ g}{9 η} = \frac{2 \times ( 1 0 ^{- 6} ) ^{2} \times 1 0 ^{3} \times 10}{9 \times 1.8 \times 1 0 ^{- 5}} = 123.4 \times 1 0^{- 6}

m/s.

JEE Main · 2022 · 28 Jun 2022 · Shift 2 · Single correct option

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