A steel ball of radius 2 into 10 to the power minus 3 metre is released in an oil of viscosity 2.32 pascal and density 840 kg metre per cube calculate the terminal velocity of the ball take the density of steel as 7800 for cube as as viewed g is equal to 10 metre per second square
Answers
The terminal velocity of the ball is 2.39 x 10^-2 m/s.
Explanation:
Given data:
Radius of ball "r" = 2 x 10^-3
Viscosity of oil "η" = 2.32 pascal
Density of oil = 840 Kg/m^-3
Density of steel = 7800 Kg/m^-3
Terminal velocity Vt = 2/g r^2 (ρ - ρ)g / η
Vt = 2/g x (2 x 10^-3)^2 x (7800 - 840 ) x 9.8 / 2.32
Vt = 2/g x 4 x 10^-6 x 6950 x 9.8 / 2.32
Vt =55600 x 10-6/2.32
Vt = 5.56 x 10^4 x 10-6/2.32
Vt = 5.56 x 10^-2 /2.32 = 5.56 / 100 x 2.32
Vt = 5.56 / 232 = 0.0239 = 2.39 x 10^-2 m/s
Thus the terminal velocity of the ball is 2.39 x 10^-2 m/s.
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