Question

A dimensionally consistent relation for the volume $V$ of a liquid of coefficient of viscosity ${}^{\prime }{\eta }^{\prime }$ flowing per second, through a tube of radius $r$ and length $l$ and having a pressure difference $P$ across its ends, is :

• A. $V=\frac{\pi P{r}^4}{8\eta l}$
• B. $V=\frac{\pi \eta }{8P{r}^4}$
• C. $V=\frac{8P\eta }{\pi r^4}$
• D. $V=\frac{\pi P\eta }{8r^4}$

Answer 1

The correct option is A $V=\frac{\pi P{r}^4}{8\eta l}$

$V=$ rate of flow $=\frac{volume}{second}=\left[M^o{L}^3{T}^{-1}\right]$

$P=$ Pressare $=\left[M^1{L}^{-1}{T}^{-2}\right]$

$r=$ radius $=\left[M^0{L}^1{T}^0\right]$

$\eta =$ coefficient of viscosity $=\left[M^1{L}^{-1}{T}^1\right]$

$L=$ length $=\left[M^0{L}^1{T}^o\right]$

Taking (i) option,

$V=\frac{\pi P_{r}4}{8\eta L}$

$\left[M^0{L}^3{T}^{-1}\right]=\frac{\left[M^1{L}^{-1}{T}^{-2}\right][M^0{L}^1{T}^0{]}^4}{\left[M^1{L}^{-1}{T}^{-1}\right]\left[M^0{L}^1{T}^0\right]}$

$[M^0{L}^3{T}^{-1}=\frac{\left[M^1{L}^{-1}{T}^{-2}\right]\left[M^0{L}^4{T}^0\right]}{\left[M^1{L}^{-1}{T}^{-1}\right]\left[M^0{L}^1{T}^0\right]}$

$\left[M^0{L}^3{T}^{-1}\right]=\frac{\left[M^2{L}^3{T}^{-2}\right]}{\left[M^1{L}^0{T}^{-1}\right]}$

$[M^0{L}^3{T}^{-1}=[M^0{L}^3{T}^{-1}]$

Hence proved