A bubble of air of diameter 0.2 cm rises up uniformly in water with velocity of 200m/s if density of water is 1gcm3 find find the coefficient of viscosity of water
Answers
Given info : A bubble of air of diameter 0.2 cm rises up uniformly in water with velocity of 200m/s if density of water is 1 g/cm³.
To find : The coefficient of viscosity of water is ...
solution : radius of bubble air, r = 0.1 cm = 10¯³ m
bubble rises up uniformly in water with velocity, v = 200 m/s
density of water, ρ = 1 g/cm³ = 1000 kg/m³
at equilibrium,
viscous force = buoyancy force
⇒6πηrv = 4/3 πr³ρg
⇒η = 2/9 r²ρg/v
= 2/9 × (10¯³ m)² × (1000 kg/m³) × (10m/s²)/(200 m/s)²
= 2/9 × 10¯⁶ × 10³ × 10/4 × 10⁴
= 2/9 × 10¯²/4 × 10⁴
= 1/18 × 10¯⁶ Ns/m²
Therefore the coefficient of viscosity of water would be 1/18 × 10¯⁶ Ns/m²
Answer :-
Given info : A bubble of air of diameter 0.2 cm rises up uniformly in water with velocity of 200m/s if density of water is 1 g/cm³.
To find : The coefficient of viscosity of water is ...
solution : radius of bubble air, r = 0.1 cm = 10¯³ m
bubble rises up uniformly in water with velocity, v = 200 m/s
density of water, ρ = 1 g/cm³ = 1000 kg/m³
at equilibrium,
viscous force = buoyancy force
⇒6πηrv = 4/3 πr³ρg
⇒η = 2/9 r²ρg/v
= 2/9 × (10¯³ m)² × (1000 kg/m³) × (10m/s²)/(200 m/s)²
= 2/9 × 10¯⁶ × 10³ × 10/4 × 10⁴
= 2/9 × 10¯²/4 × 10⁴
= 1/18 × 10¯⁶ Ns/m²
Therefore the coefficient of viscosity of water would be 1/18 × 10¯⁶ Ns/m²