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
Image given below

where P is the pressure difference between the two ends of the pipe and η is coefficent of viscosity of the liquid having dimensional formula ML–1 T–1. Check whether the equation is dimensionally correct.

Answer 1

Volume per sec = L^3 T^-1
Now according to above equation
V= P r^4 / ( neta ) l

=( M L^-1 T^-2 ) ( L^4 )
-------------------------------
( M L^-1 T^-1 ) ( L )

= L^3 T^-1

Hence proved !!!

Answer 2

Given:

Length of the pipe=l

Radius of thee pipe=r

Pressure=P

Coefficient of viscosity =$\eta$

To prove:

Whether the given equation is correct or not.

Solution:

Dimensional formula is an expression which shows the power of the fundamental units which are raised to obtain one unit for the derived quantity.

Given, Dimensional formula $= M L^{-1} T^{-1}$

Volume per sec $=L^{3} T^{-1}$

Now, according to above equation,  

To prove the dimensional formula let us follow the given steps below,

$=\frac{\text { Pressure } \text { radius }}{\text { Coefficient of viscosity } \times \text { length }}$

$V=\frac{\pi P r^{4}}{\eta \times l}$

$=\frac{\left(M L^{-1} T^{-2}\right)\left(L^{4}\right)}{\left(M L^{-1} T^{-1}\right)(L)}$

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

Hence, the given question is proved.