Question

If nn identical water droplets falling under gravity with terminal velocity v coalesce to form a single drop which has the terminal velocity 4v. Find the number n.

Answer 1

We solve this by using the equation for Terminal Velocity of an object on a liquid,

v = square root ((2*m*g)/(ρ*A*C))

m - mass of the falling object

g - the acceleration due to gravity.

ρ - the density of the fluid the object is falling through.

A - the projected area of the object.

C - the drag coefficient.

Since g, ρ and C is constant, we can say that,

v α square root(m/A)

If we take the water drop as a Sphere,

A = 4πr^2

Then,

v α square root(m/4πr^2)

v α square root(m)/r

If we take the radius of one of the droplets as r, we wanted to find the radius of the droplet when all n droplets get together.

If we take that as R,

Volume of the radius R droplet = Total Volumes of radius r droplets

(4/3)πR^3 = (4/3)πr^3 + (4/3)πr^3 + ...... + (4/3)πr^3

(4/3)πR^3 = n x (4/3)πr^3

R^3 = n x r^3

R = n^(1/3) x r

Then we can write,

v α square root(m)/r -----------------------------(1)

4v α square root(nm)/n^(1/3) x r--------------(2)

By (2)/(1),

4 = square root(n)/n^(1/3)

4 = n^(1/6)

n = 4^6

n  = 4096