Question

A liquid drop of mass m and radius r is falling from great height. It's velocity is proportional to

Answer 1

Hii dear,

# Answer- Mρ/ηr

# Explaination-
Consider,
ρ = density of air
η = viscosity of air
M = mass of liquid drop
r = radius of drop
g = gravitational accln

So, When a drop fall in air the terminal velocity of the drop is given by
V = [Mg-(4/3.πr^3.ρg)]/6πηr

From this we can say that V is
- directly proportional to M, ρ
- inversely proprtional to η, r

Hope you got your answer...

Answer 2

Answer:

m/r

Explanation:

by using the terminal velocity formula....Preety simple