assuming that the critical velocity of a viscous liquid flowing through a capillary tube depends upon radius r of a - Physics - Physical World

Question

assuming that the critical velocity of a viscous liquid flowing
through a capillary tube depends upon radius r of a tube,density of
the tube and co-efficient of viscosity of the liquid use the method
of dimensions to obtain a relation between these quantities

Hansraj Rai · Accepted Answer

Dear Student!!

According to the given problem,

$v_{c} = K η^{a} ρ^{b} r^{c} ....... (A)$

K is a constant.

In dimension form we can write,

$[M^{0} L T^{- 1}] = {[M L^{- 1} T^{- 1}]}^{a} {[M L^{- 3}]}^{b} {[L]}^{c} = [M^{a + b} L^{- a - 3 b + c} T^{- a}] a + b = 0... (i) - a + - 3 b + c = 1..... (i i) - a = - 1.... (i i i)$

Solving equation (i), (ii) and (ii), we get

$a = 1, b = - 1, c = - 1$

Putting this value in eq. A

$v_{c} = K η^{1} ρ^{- 1} r^{- 1} = \frac{K η}{ρ r}$

assuming that the critical velocity of a viscous liquid flowing through a capillary tube depends upon radius r of a tube,density of the tube and co-efficient of viscosity of the liquid. use the method of dimensions to obtain a relation between these quantities.