Question

pls do it fast.
dimensional formula
physics class 11
nootan isc physics ​

Answer 1

$\textsf{\underline {\Large {VISCOSITY}}}$ :

Given, Length = l

Radius = r

Pressure difference = p

Poiseuille's Formula, $\boxed{\mathsf{Q\:=\:{\dfrac{V} {t}\:=\:{\dfrac{\pi{p{r} ^{4}}} {8{\eta{l}}}}}}}$

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

$\boxed{\mathsf{\eta{\:=\:{\dfrac{\pi{p{r} ^{4}t}}{8lV}}}}}$ ---> ( i )

Now,

Dimensional Formula of p ( Pressure) = $\mathsf{[\:M{L} ^{-1}{T}^{-2}\:]}$

Dimensional Formula of r ( Radius - always Length ) = $\mathsf{[\:L\:]}$

For $\mathsf{{r} ^{4}\:=\: [\:{L}^{4}\:]}$

Dimensional Formula of l ( length) = $\mathsf{[\:L\:]}$

Dimensional Formula of V ( Volume ) = $\mathsf{[\:{L} ^{3}\:]}$

Dimensional Formula of t ( time ) = $\mathsf{[\:T\:]}$

Now, Putting these formulae in equation ( i ),

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

$\mathsf{\eta{\:=\:{\dfrac{[\:M{L} ^{-1\:+\:4}{T}^{-2\:+\:1}\:]}{[\:{L}^{1\:+\:3}\:]}}}}$

$\mathsf{\eta{\:=\:{\dfrac{[\:M{L} ^{3}{T}^{-1}\:]}{[\:{L}^{4}\:]}}}}$

$\mathsf{\eta{\:=\:[\:M{L} ^{3\:-\:4}{T}^{-1}\:]}}$

$\mathsf{\eta{\:=\:[\:M{L} ^{3\:-\:4}{T}^{-1}\:]}}$

$\boxed{\mathsf{\eta{\:=\:[\:M{L} ^{-1}{T}^{-1}\:]}}}$

➡️ Dimension of Coefficient of viscosity = $\boxed{\mathsf{\eta{\:=\:[\:M{L} ^{-1}{T}^{-1}\:]}}}$

$\boxed{\mathsf{NOTE}}$ : ⚫$\mathsf{\pi, \:8}$ are constant here, so they are dimensionless having no dimensional formula.