Question

The rate of flow of a liquid

Answer 1

Answer:

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

Explanation:

Given :-

$\displaystyle{\sf{V=\frac{\pi Pr^4}{8\eta L} }}$

Then,

$\displaystyle{\sf{\eta=\frac{\pi Pr^4}{8VL} }}$

To find :-

The dimensions of the coefficient of viscosity η of the liquid.

Solution :-

• Volume rate(V) = $\displaystyle{\sf{L^3T^{-1}}}$

• Pressure(P) = $\displaystyle{\sf{ML^{-1}T^{-2}}}$

• Viscosity = η

• Length = [L]

Now, the dimensional formula of viscosity is :-

$\sf{\displaystyle{\frac{[ML^{-1}T^{-2}][L^4]}{[L^3T^{-1}][L]} }}$

On cancelling the terms, we get :-

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