JEE Main · 2022 · 29 Jun 2022 · Shift 2 · Fluid Mechanics

A small spherical ball of radius 0.1 mm and density

1 0^{4} kg m^{- 3}

falls freely under gravity through a a distance

h

before entering a tank of water. If after entering the water the velocity of ball does not change and it continue to fall with same constant velocity inside water, then the value of

h

wil be

____

m.
(Given

g = 10 ms^{- 2}

, viscosity of water

= 1.0 \times 1 0^{- 5} N - sm^{- 2}

Answer: 20

Solution

The velocity after falling through height h must equal the terminal velocity in water.

Terminal velocity:

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

where

r = 0.1

mm =

1 0^{- 4}

ρ = 1 0^{4}

kg/m³,

σ = 1 0^{3}

kg/m³ (water density)

v_{t} = \frac{2 \times ( 1 0 ^{- 4} ) ^{2} \times ( 1 0 ^{4} - 1 0 ^{3} ) \times 10}{9 \times 1 0 ^{- 5}} = \frac{2 \times 1 0 ^{- 8} \times 9 \times 1 0 ^{3} \times 10}{9 \times 1 0 ^{- 5}} = 20

m/s

From free fall:

2 g h = 20

2 \times 10 \times h = 400

h = 20

m

Answer: 20

JEE Main · 2022 · 29 Jun 2022 · Shift 2 · Integer answer

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JEE Main · 2022 · 29 Jun 2022 · Shift 2 · Fluid Mechanics

Solution

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