Question

The volume of a liquid flowing out per second of a pipe of length l and radius r is written by a student as

$v=\frac{\pi }{8}\frac{P{r}^4}{\eta l}$

where P is the pressure difference between the two ends of the pipe and $\eta$ is coefficient of viscosity of the liquid having dimensional formula $M{L}^{–1}{T}^{–1}.$ Check whether the equation is dimensionally correct.

Answer 1

Given:

Volume of a liquid flowing out per second of a pipe is given by,

$v=\frac{\pi }{8}\frac{P{r}^4}{\eta l}$

Dimensions of the given quantities: Dimension of v= Volume per second$=\frac{V}{T}=\left[L^3{T}^{-1}\right]$

Dimension of $P=\frac{F}{A}$

$=\frac{\left[ML{T}^{-2}\right]}{\left[L^2\right]}$

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

Dimension of $\text{r}=\left[L\right]$

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

Dimension of $\text{l}=\left[L\right]$

$\therefore$ Dimension of R.H.S. $=\frac{\left[M{L}^{-1}{T}^{-2}\right]\left[L^4\right]}{\left[M{L}^{-1}{T}^{-1}\right]\left[L\right]}$

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

Dimension of L.H.S. $v=\left[M^0{L}^3{T}^{-1}\right]$

Dimensionally, $L.H.S=R.H.S$

As dimensions of both sides are equal.

Therefore, the equation is dimensionally correct

Final Answer:$L.H.S=R.H.S$