Answer to Question #200642 in Mechanics | Relativity for Master j

Question

The rate of flow, Q, of liquid in a cylindrical tube is given by the equation

Q = πr⁴ ΔP/(8ηL), where

Q = rate of flow of liquid (m³/s)

r = radius of the tube, (m)

L = length of the tube (m)

ΔP = pressure difference between the ends of the tube (Pa)

i. Find the dimensions of the dimensions of the quantities r⁴ , Q and ΔP.

ii. Determine the dimensions of the quantity η.

Answer 1

Rate of flow, Q, of liquid in cylindrical tube is given by the equation -

$Q=\dfrac{{\pi}r^{4}{\Delta P}}{8{\eta}L}$

Q = rate of flow of liquid $(m^{3}/s)$

r = radius of tube (m)

L = length of tube in (m)

${\Delta}P=$ pressure between ends of the tube (pa).

dimensions of all the term given in formula -

$r=L$

P $=ML^{-1}T^{-2}$

${\eta}=ML^{-1}T^{-1}$

1)

Now putting these dimensions in above formula , we get -

$Q=\dfrac{[Ml^{-1}T^{-2}][L^{4}]}{[ML^{-1}T^{-1}][L]}$

$Q=L^{3}T^{-1}$

Now finding the dimensions of $r^{4}$ -

$r^{4}=\dfrac{{\eta}LQ}{{\Delta P}}$

Now putting the dimensions in above , we get -

= $r^{4}=\dfrac{[ML^{-1}T^{-1}][L][L^{3}T^{-1}]}{[ML^{-1}T^{-2}]}$

$=r^{4}=L^{4}$

$\Delta{P}=\dfrac{{\eta}LQ}{{r^{4}}}$

now putting the dimensions , in above equation we get -

${\Delta}P=\dfrac{[ML^{-1}T^{-1}][L][L^{3}T^{-1}]}{L^{4}}=$ $ML^{-1}T^{-2}$

2)

Dimensions of quantity ${\eta}=\dfrac{r^{4}{\Delta P}}{LQ}$

Now putting the dimensions the quantity , in the above question ,we get ,

${\eta}=\dfrac{[L^{4}][ML^{-1}T^{-2}]}{[L][L^{3}T^{-1}]}$ $=$ $ML^{-1}T^{-1}$