Question

4 The critical velocity of the flow of a liquid through a Κη pipe of radius r is given by v=- where p is the rp

density and n is the coefficient of viscosity of the liquid. Check if this relation is dimensionally (Ans. Correct)

see photo answer question 4

Solve itplss explain
​

Answer 1

Answer:

This relation is dimensionally correct.

Explanation:

Refer to the attachment for full explanation.

How we have find dimensions of critical velocity, density and coefficient of viscosity.

Coefficient of viscosity :

We know formula :

→ Coefficient of viscosity = Force/area × velocity gradient.

Dimension of force = $\sf [M^{1}L^{1}T^{-2}]$

Dimension of area = [L²]

Dimension of velocity gradient = $\sf [T^{-1}]$

(Velocity gradient = Velocity/Distance)

Now, Put all dimensions in coefficient of viscosity formula:

$\longrightarrow \sf \dfrac{[MLT^{-2}]}{[L^{2}] [T^{-1}]}$

$\longrightarrow \sf [MLT^{-2}] [L^{-2}] [T]$

$\longrightarrow \sf [ML^{-1}T^{-1}]$

Dimension of Coefficient of viscosity is $\underline{\bold{ \sf [ML^{-1}T^{-1}] }}$

Density :

Formula :

• d = Mass/Volume

$\longrightarrow \sf \dfrac{[M]}{[L^{3}]}$

$\longrightarrow \sf [M][L^{-3}]$

$\longrightarrow \sf [ML^{-3} T^{0}]$

Dimension of density is $\underline{\bold{[ML^{-3} T^{0}]}}$

Critical velocity :

We know, formula :

→ Vc = Reynolds number × coefficient of viscosity/Density × radius

Reynolds number is dimensionless.

Dimension of Coefficient of viscosity is $\sf [ML^{-1}T^{-1}]$

Dimension of density is $\sf [ML^{-3} T^{0}]$

Dimension of Radius is [L]

$\longrightarrow \sf \dfrac{[ML^{-1}T^{-1}]}{[ML^{-3}T^{0}] [L]}$

$\longrightarrow \sf [ML^{-1}T^{-1}] [ML^{3}] [L^{-1}] [$

$\longrightarrow \sf [L^{-2} T^{-1}] [L^{3}]$

$\longrightarrow \sf [L^{1}T^{-1}]$

Or, $\sf [M^{0} L T^{-1}]$

Dimension of critical velocity is $\underline{\bold{\sf [M^{0} L T^{-1}]}}$