JEE Main · 2023 · 24 Jan 2023 · Shift 2 · Fluid Mechanics

A Spherical ball of radius 1 mm and density

10.5

g/cc is dropped in glycerine of coefficient of viscosity 9.8 poise and density

1.5

g/cc. Viscous force on the ball when it attains constant velocity is

3696 \times 1 0^{- x}

N. The value of

x

is
(Given,

g = 9.8

m/s

^{2}

and

π = \frac{22}{7}

)

Answer: 7

Solution

At terminal velocity the viscous drag balances the weight less the buoyancy, so

F_{v} = V g (ρ_{ba l l} - ρ_{l i q u i d})

with

V = \frac{4}{3} π r^{3}

. Taking

r = 1 0^{- 3}

m and

ρ_{ba l l} - ρ_{l i q u i d} = (10.5 - 1.5) \times 1 0^{3} = 9 \times 1 0^{3} kg m^{- 3}

F_{v} = \frac{4}{3} \times \frac{22}{7} \times (1 0^{- 3})^{3} \times 9.8 \times 9 \times 1 0^{3} = 3696 \times 1 0^{- 7} N

. Comparing with

3696 \times 1 0^{- x}

N gives

x = 7

JEE Main · 2023 · 24 Jan 2023 · Shift 2 · Integer answer

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