A spherical drop of oil falls at A constant speed of 4cm/s in steady air çalculate the radius of the drop the density of the oil is 0.9 g/ cm res to the power 3 density of air is 1.0 g/CMcube and the coefficient of viscosity of air is 1.8×10^-4 poise.(g=980 CM/s square)
Answers
Answer :
The radius of the spherical drop is 0.006 cm
Explanation :
Given,
A spherical drop of oil falls at a constant speed of 4 cm/s in steady air.
i.e., terminal velocity, v = 4 cm/s = 0.04 m/s
The density of the oil is 0.9 g/cm³
⇒ ρ = 0.9 g/cm³
⇒ ρ = 0.9 × 1000 kg/m³
⇒ ρ = 900 kg/m³
density of air is 1.0 g/cm³
⇒ σ = 1 g/cm³
⇒ σ = 1 × 1000 kg/m³
⇒ σ = 1000 kg/m³
coefficient of viscosity of air is 1.8×10⁻⁴ poise
⇒ η = 1.8 × 10⁻⁴ poise
⇒ η = 1.8 × 10⁻⁴ × 0.1 kg/m-s
⇒ η = 1.8 × 10⁻⁵ kg/m-s
acceleration due to gravity = 980 cm/s²
⇒ g = 980 cm/s²
⇒ g = 9.8 m/s²
To find,
the radius of the drop
∴ The radius of the spherical drop is 0.006 cm