A Spherical ball of radius 1mm 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 N. The value of is (Given, g = 9.8 m/s and )
Show answer and solution
Answer: 7
Solution
At terminal velocity, viscous force = net gravitational force (weight - buoyancy).
F = (4/3)πr³(ρ_ball - ρ_fluid)g
r = 1mm = 10⁻³ m, ρ_ball = 10500 kg/m³, ρ_fluid = 1500 kg/m³, g = 9.8 m/s²
F = (4/3) × (22/7) × (10⁻³)³ × (10500 - 1500) × 9.8
= (4/3) × (22/7) × 10⁻⁹ × 9000 × 9.8
= (88/21) × 10⁻⁹ × 88200
= (88 × 88200)/21 × 10⁻⁹
= (7,761,600/21) × 10⁻⁹
= 369,600 × 10⁻⁹
= 3696 × 10⁻⁷ N
So x = 7.
F = (4/3)πr³(ρ_ball - ρ_fluid)g
r = 1mm = 10⁻³ m, ρ_ball = 10500 kg/m³, ρ_fluid = 1500 kg/m³, g = 9.8 m/s²
F = (4/3) × (22/7) × (10⁻³)³ × (10500 - 1500) × 9.8
= (4/3) × (22/7) × 10⁻⁹ × 9000 × 9.8
= (88/21) × 10⁻⁹ × 88200
= (88 × 88200)/21 × 10⁻⁹
= (7,761,600/21) × 10⁻⁹
= 369,600 × 10⁻⁹
= 3696 × 10⁻⁷ N
So x = 7.
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