Dimension of the given physical quantity is as follows,
[V] = dimension of volume/dimension of time
![=[L^{3}]/[T]](https://www.learnatnoon.com/s/wp-content/ql-cache/quicklatex.com-f87d23ada6286a6eb19905e211cbd0a9_l3.png "Rendered by QuickLaTeX.com")
![=[M^{-1}T^{-2}]](https://www.learnatnoon.com/s/wp-content/ql-cache/quicklatex.com-ec86cd7275159f477f43511a15dadfdd_l3.png "Rendered by QuickLaTeX.com")
LHS ![=[L^{3}T^{-1}]](https://www.learnatnoon.com/s/wp-content/ql-cache/quicklatex.com-0f256c6c21a39f0d52c1f4e219878d19_l3.png "Rendered by QuickLaTeX.com")
RHS
LHS = RHS
Hence, the equation is correct.
The volume of a liquid flowing out per second of a pipe of length l and radius r is written by a student as 
where P is the pressure difference between the two ends of the pipe and η is coefficient of viscosity of the liquid having dimensional formula 
. Check whether the equation is dimensionally correct.
where P is the pressure difference between the two ends of the pipe and η is coefficient of viscosity of the liquid having dimensional formula . Check whether the equation is dimensionally correct.The volume of a liquid flowing out per second of a pipe of length l and radius r is written by a student as where P is the pressure difference between the two ends of the pipe and η is coefficient of viscosity of the liquid having dimensional formula . Check whether the equation is dimensionally correct.
where P is the pressure difference between the two ends of the pipe and η is coefficient of viscosity of the liquid having dimensional formula . Check whether the equation is dimensionally correct.