Question

physics class 11
dimensional analysis.
nootan isc physics​

Answer 1

$\textsf{\underline {\Large {Consistency of Dimensions}}}$ :

Given, V = Volume of a liquid

$\mathsf{\eta}$ = Coefficient of viscosity

Radius of tube = r

Pressure difference = p

Relation : $\boxed{\mathsf{V\:=\:{\dfrac{\pi{p{{r} ^{4}}}}{8{\eta{l}}}}}}$

Actually, the above relation will be written as

$\boxed{\mathsf{V\:=\:{\dfrac{\pi{p{{r} ^{4}t}}}{8{\eta{l}}}}}}$

Because it is for the rate of flow of liquid, i. e. $\mathsf{\dfrac{V} {t}}$. 't' will always be there in the denominator to show the rate of flow.

Now,

Dimension of Volume, V = $\mathsf{[{L} ^{3}]}$

Dimension of Radius, r = $\mathsf{[L]}$

Dimension of Coefficient of Viscosity Viscosity, $\mathsf{\eta}$ = $\mathsf{[M{L} ^{-1}{T}^{-1}]}$

Dimension of Pressure difference, p = $\mathsf{[M{L} ^{-1}{T}^{-2}]}$

Putting these Dimensional Formulae in above relation :

$\Rightarrow{\mathsf {L.H.S.}}$ = $\mathsf{[{L} ^{3}]}$

R. H. S. = $\mathsf{\dfrac{[M{L} ^{-1}{T}^{-2}]\:[{L}^{4}]\:[T]}{[M{L} ^{-1}{T}^{-1}]\:[L]}}$

R. H. S. = $\mathsf{\dfrac{[M{L} ^{-1\:+\:4}{T}^{-1}]}{[M{L} ^{-1\:+\:1}{T}^{-1}]}}$

R. H. S. = $\mathsf{[{M} ^{1\:-\:1}{L} ^{3}{T}^{-1\:+\:1}] }$

$\Rightarrow{\mathsf {R.H.S.}}$ = $\mathsf{[{L} ^{3}]}$

➡️ Hence, L. H. S. = R. H. S.

So, Given relation is correct or consistent.

$\boxed{\mathsf {NOTE}}$ : $\mathsf{\pi}$ and 8 are neglected as they are dimensionless.