a water drop of radius 1.5 mm is falling from height 1 km having drag constant 0.5 density of water drop is 1000 kilogram per metre cube density of Air 12.5 kilogram per metre cube find its terminal velocity
Answers
regards
Terminal velocity of the bubble is 8.85 km/hr.
Given, radius of water drop, R = 1.5 mm = 0.0015 m [As 1m = 1000mm]
As the drop is spherical, so its area is given as:
A = Пr², where r is the radius of the drop.
Putting r = 0.0015m, we get
A = 3.14 × (0.0015)² [П = 3.14 approx.]
= 7.06 × 10⁻⁶ m²
Given, density of water drop = 1000 kg m⁻³ and,
Volume = 4/3 Пr³, where r is the radius of sphere
As Density = Mass/Volume
So, Mass = Density × Volume
= (1000) × 4/3 × 3.14 × (0.0015)³
= 1.413 × 10⁻⁵ kg
Terminal velocity v is given as:
v = [(2×mass×g)/(Drag constant×Density of air×Area of sphere)]^(1/2)
We have,
mass = 1.413 × 10⁻⁵ kg,
g = 9.8 ms⁻²
Drag constant = 0.5 (from question)
Density of air = 12.5 kg m⁻³
Area of sphere = 7.06 × 10⁻⁶ m²
Putting these values, we get
v = 2.46ms⁻¹ = (2.46×3600/1000) km/hr = 8.85 km/hr [1m = 1000km and 1 hr = 3600s]
So, terminal velocity is 8.85 km/hr.