Q.The volume of a liquid flowing out per second of a pipe of length $l$ and radius $r$ is written by a student as $v = \dfrac{\pi}{8}\dfrac{P r^4}{\eta l}$ 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 $ML^{-1}T^{-1}$. Check whether the equation is dimensionally correct.

Concept: Dimensional homogeneity – every term in a physically correct equation must have identical dimensions.

The given equation is $v = \dfrac{\pi}{8}\dfrac{P r^4}{\eta l}$, where $v$ is volume per second (volumetric flow rate).

Step 1: Identify dimensions of the left side.

Volume per second has dimensions $[v] = \dfrac{L^3}{T} = L^3 T^{-1}$.

Step 2: Find dimensions of the right side.

• Pressure: $[P] = \dfrac{\text{Force}}{\text{Area}} = \dfrac{MLT^{-2}}{L^2} = ML^{-1}T^{-2}$
• Radius: $[r] = L$, so $[r^4] = L^4$
• Viscosity: $[\eta] = ML^{-1}T^{-1}$ (given)
• Length: $[l] = L$

The right side becomes: …

The heart of dimensional analysis is this: a physically meaningful equation must be dimensionally homogeneous. Every term that you add or equate must carry the same dimensions, regardless of the numerical constants involved. The constant $\frac{\pi}{8}$ is dimensionless, so we ignore it and focus on whether $\frac{P r^4}{\eta l}$ has the same dimensions as volume per unit time.

This equation is actually Poiseuille's law for viscous flow through a cylindrical pipe. The student has written the volume flow rate $v$ (volume per second) in terms of the pipe geometry ($r$, $l$), the driving pressure difference $P$, and the fluid's resistance to flow $\eta$. Let's verify the dimensional consistency step by step.

Step-by-step dimensional check

1. Identify the dimension of the left-hand side.

   The quantity $v$ represents volume flowing out per second, so its dimension is:

2. Find the dimension of pressure $P$.

   Pressure is force per unit area. Force has dimension $MLT^{-2}$, and area has dimension $L^2$, so:

3. List the dimensions of the geometric quantities.

   The radius $r$ and length $l$ are both lengths:

Therefore:

4. Use the given dimension of viscosity $\eta$.

   We are told:

5. Compute the dimension of the right-hand side.

   The right-hand side (ignoring the dimensionless constant $\frac{\pi}{8}$) is $\frac{P r^4}{\eta l}$. Substitute the dimensions: …

Method: Dimensional Analysis

Concept: For any physical equation to be correct, the dimensions on the left-hand side must equal the dimensions on the right-hand side. This is called the principle of homogeneity of dimensions.

Steps

Step 1: Write the dimensions of each quantity

• Volume per second ($v$):

  $[v] = \frac{\text{Volume}}{\text{Time}} = \frac{L^3}{T} = L^3 T^{-1}$

• Pressure difference ($P$):

  Pressure = Force/Area

  $[P] = \frac{MLT^{-2}}{L^2} = ML^{-1}T^{-2}$

• Radius ($r$):

  $[r] = L$

• Coefficient of viscosity ($\eta$):

  Given as $[\eta] = ML^{-1}T^{-1}$

• Length ($l$):

  $[l] = L$

• Constant $\frac{\pi}{8}$ is dimensionless.

Step 2: Write the dimensions of the RHS

RHS = $\dfrac{\pi}{8} \cdot \dfrac{P r^4}{\eta l}$

Since $\frac{\pi}{8}$ has no dimensions:

Substitute: …

✗ Mistake 1: Forgetting that $v$ is volume flow rate, not velocity

Many students assume $v$ stands for velocity (with dimension $LT^{-1}$) and check dimensions accordingly.

How to avoid:

Read the problem statement carefully: “The volume of a liquid flowing out per second” means $v$ is volume flow rate — its dimension is:

Not $LT^{-1}$.

✗ Mistake 2: Writing the wrong dimensional formula for pressure $P$

Pressure is force per unit area. A common error is to treat $P$ as having dimensions of force alone ($MLT^{-2}$) or energy.

How to avoid:

Always derive step-by-step:

• Force $F$: $MLT^{-2}$
• Area $A$: $L^2$
• So $P = F/A$ gives:

✗ Mistake 3: Miswriting the dimensions of viscosity $\eta$

The problem gives $\eta$ as $ML^{-1}T^{-1}$, but students sometimes forget or mis-copy it, especially under exam pressure.

How to avoid:

Write it down clearly at the start:

Then use it directly in the dimensional check.

✗ Mistake 4: Not checking dimensions of the constant $\frac{\pi}{8}$

Students sometimes try to assign dimensions to $\pi$ or the fraction $\frac{\pi}{8}$.

How to avoid:

Remember: pure numbers (like $\pi$, $8$, $1/2$, etc.) have no dimensions — they are dimensionless. So ignore them in dimensional analysis.

✓ Correct dimensional check (for reference)

Given equation:

Dimensions:

• $[v] = L^3 T^{-1}$
• $[P] = ML^{-1}T^{-2}$
• $[r^4] = L^4$
• $[\eta] = ML^{-1}T^{-1}$
• $[l] = L$

So RHS dimensions: …