Question

Consider a small ball of radius (r) falling through a viscous liquid of viscosity with a velocity (v) derive the exact form of relation between viscous force (f) experienced by the ball. Given that F=kr^x viscosity^y v^z. Where; k is a dimensionless constant.

Answer 1

Based on dimensional analysis, the relation between viscous force (F) experienced by the ball is F = krvη.

What is the relation between viscous force, F experienced by the ball?

• The viscous force, F on the small ball is directly proportional to the radius: F∝r
• The viscous force, F on the small ball is directly proportional to the velocity of the ball: F∝v
• The viscous force, F on the small ball is directly proportional to the coefficient of viscosity of the fluid: F∝η

F∝r^av^bη^c

F = kr^av^bη^c

where K is proportionality constant

Substituting the dimensions of each of the quantities in the above equation:

[M^1 L^1 T^-2] = [M^0 L^1 T^0]^a [M^0 L^1T^-1]^b [M^1 L^-1 T^-1]^c

[M^1L^1T^-2] = [M^c L^a+b-c T^-b-c]

Equating the powers of M, L, T on both sides:

c = 1

a + b - c = 1

-b - c = -2

Solving for a and b, we get:

a = 1, b = 1, c = 1

Substituting the values of a, b and c:

• F = krvη

Therefore, the relation between viscous force (F) experienced by the ball is F = krvη.

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