The viscous drag acting on a metal sphere of diameter 1 mm, falling through a fluid of viscosity 0.8 Pa s with a velocity of 2 is equal to [2023]

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Solution

Q.
The viscous drag acting on a metal sphere of diameter 1 mm, falling through a fluid of viscosity 0.8 Pa s with a velocity of 2 ${ms}^{- 1}$ is equal to [2023]

1 $15 \times 10^{- 3} N$

2 $30 \times 10^{- 3} N$

3 $1.5 \times 10^{- 3} N$

4 $20 \times 10^{- 3} N$

Ans.
(1)

Stoke's law relates the backward dragging force F acting on a small sphere of radius $r$ moving through a viscous medium of viscosity $η$ with velocity $v$ .

$F = 6 π r η v$

$= (6) (3.14) (\frac{1 \times 10^{- 3}}{2}) (0.8) (2) = 15 \times 10^{- 3}$ $N$

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