Question

Is dimensions of the critical velocity of the liquid flowing through a tube?

Answer 1

Dimension of coefficient of viscosity , η = [MLT⁻²]/[L][LT⁻¹] = [M¹L⁻¹T⁻¹]

[Because viscous force = 6πηrv ]

dimension of density of liquid , ρ = [ML⁻³]

dimension of radius of tube , r = [L]

Now, terminal velocity \bf{\propto}η^x ρ^y r^z

[LT⁻¹] = [M⁻¹L⁻¹T⁻¹]^x [ML⁻³]^y [L]^z

[LT⁻¹] = [M]^(x + y) [L]^(-x-3y+z) [T]^(-x)

compare both sides,

x + y = 0 ⇒x = -y

-x - 3y + z = 1 ⇒z = -1

-x = -1 ⇒x = 1

And y = -1

Hence, x = 1 , y = -1 and z = -1