Question

Find the viscous force on a steel ball of 2 mm radius ( density 8 g/cc ) acquires a terminal velocity of 4 cm/s in falling freely in the tank of glycerine ( density of glycerine= 1.3 g / cc).

Answer 1

The viscous force is 0.002 N

Explanation:

Given data:

Radius of ball =  2 mm = 2 x 10^-3 m

• The formula for terminal velocity v is given by

V = 2gr^2 (ρsphere−ρmedium) / 9η

"ρsphere" is the density of the still ball = 8 g/cc = 8000 kg/m^3

"ρmedium"  is density of the medium = 1.3 g/cc = 1300 kg/m^3

terminal velocity v is 4 cm/s = 0.04 m/s

η = viscosity

Solution:

η = 2gr^2(ρsphere−ρmedium) / 9v

  = 2 × 9.8 × (2 × 10^−3)^2×(8000−1300) / 9 × 0.04

  = 1.46 pas

Viscous force " f " = 6πηrv = 6π×1.46×2×10−3×0.04

    = 0.002 N

Hence the viscous force is 0.002 N

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