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 $v=\frac{\pi}{8} \frac{P r^{4}}{\eta l} \quad$ where $\mathrm{P}$ is the pressure difference between the two ends of the pipe and $\mathrm{n}$ is coefficient of viscosity of the liquid having dimensional formula $\mathrm{ML}^{-1} \mathrm{~T}^{-1}$. Check whether the equation is dimensionally correct.
Solution:
Dimension of the given physical quantity is
[V] = dimension of volume/dimension of time = [L3]/[T]=[ML-1T-2]
LHS = [L3T-1]
RHS = [L3T-1]
LHS = RHS
Therefore, the equation is correct.