When a liquid moves steadily under some pressure through a horizontal tube, it moves in the form of cylindrical layers coaxial to the ends of the tube. The velocity of different layers is different. The velocity of the layer is maximum along the axis of the tube and it decreases as one moves towards the walls of the tube. According to Poiseuille, the rate of flow of liquid through a horizontal capillary tube varies as the relation `V alpha (pr^(4))/(eta l)` where `l` and `r` are the length and radius of the tube and `p` is the pressure different between the ends of the tube and is the constant of proportionality.
`(pi pr^(4))/(8 etal)` is dimensionally equivalent to (R is the resistance)

Options

A. `p//R`
B. `p//R^(2)`
C. `p//R^(3)`
D. p/R

Answer

To solve the problem, we need to analyze the given equation from Poiseuille's law and find its dimensional formula. The equation given is:

\[ V \propto \frac{p r^4}{\eta l} \]

Where:
- \( V \) is the volume flow rate (in \( m^3/s \)),
- \( p \) is the pressure difference (in \( N/m^2 \) or \( kg/(m \cdot s^2) \)),
- \( r \) is the radius of the tube (in \( m \)),
- \( \eta \) is the viscosity of the fluid (in \( N \cdot s/m^2 \) or \( kg/(m \cdot s) \)),
- \( l \) is the length of the tube (in \( m \)).

### Step-by-step Solution:

1. **Identify the Dimensions of Each Variable:**
- Volume flow rate \( V \):
\[ [V] = [L^3 T^{-1}] = [m^3/s] \]
- Pressure difference \( p \):
\[ [p] = [M L^{-1} T^{-2}] = [N/m^2] = [kg/(m \cdot s^2)] \]
- Radius \( r \):
\[ [r] = [L] = [m] \]
- Viscosity \( \eta \):
\[ [\eta] = [M L^{-1} T^{-1}] = [N \cdot s/m^2] = [kg/(m \cdot s)] \]
- Length \( l \):
\[ [l] = [L] = [m] \]

2. **Substituting Dimensions into the Poiseuille Equation:**
The equation can be rearranged to express the volume flow rate \( V \):
\[ V = \frac{\pi p r^4}{8 \eta l} \]

The dimensions of the right-hand side can be calculated as follows:
\[ [V] = \frac{[p] \cdot [r]^4}{[\eta] \cdot [l]} \]

Substituting the dimensions:
\[ [V] = \frac{[M L^{-1} T^{-2}] \cdot [L^4]}{[M L^{-1} T^{-1}] \cdot [L]} = \frac{[M L^{3} T^{-2}]}{[M L^{0} T^{-1}]} = [L^3 T^{-1}] \]

3. **Conclusion:**
The dimensions of the volume flow rate \( V \) match the expected dimensions of \( [L^3 T^{-1}] \), confirming that the equation is dimensionally consistent.