A dimensionally consistent relation for the volume `V` of a liquid of coefficiet of viscosity `eta` flowing per second through a tube of radius `r` and length `l` and having a pressure difference `p` across its end, is

Options

A. `V=(piPr^(4))/(8etal)`
B. `V=(pieta)/(8Pr^(4))`
C. `V=(8Peta)/(pir^(4))`
D. `V=(piPeta)/(8r^(4))`

Answer

To derive a dimensionally consistent relation for the volume \( V \) of a liquid flowing through a tube, we will use dimensional analysis. The parameters involved are the coefficient of viscosity \( \eta \), the radius \( r \), the length \( l \), and the pressure difference \( p \).

### Step-by-Step Solution:

1. **Identify the Parameters and Their Dimensions**:
- Volume \( V \): The dimension of volume is \( [V] = L^3 \).
- Coefficient of viscosity \( \eta \): The dimension is \( [\eta] = M L^{-1} T^{-1} \).
- Radius \( r \): The dimension is \( [r] = L \).
- Length \( l \): The dimension is \( [l] = L \).
- Pressure \( p \): The dimension is \( [p] = M L^{-1} T^{-2} \).

2. **Formulate the Relation**:
We assume a relation of the form:
\[ V = K \cdot \eta^a \cdot r^b \cdot l^c \cdot p^d \]
where \( K \) is a dimensionless constant.

3. **Write the Dimensions of Each Side**:
The left-hand side (LHS) has dimensions:
\[ [V] = L^3 T^{-1} \]
The right-hand side (RHS) will be:
\[ [\eta^a] = (M L^{-1} T^{-1})^a = M^a L^{-a} T^{-a} \]
\[ [r^b] = L^b \]
\[ [l^c] = L^c \]
\[ [p^d] = (M L^{-1} T^{-2})^d = M^d L^{-d} T^{-2d} \]

4. **Combine the Dimensions on the RHS**:
The dimensions of the RHS can be expressed as:
\[ [V] = M^{a+d} L^{-a+b+c-d} T^{-a-2d} \]

5. **Set Up the Dimensional Equations**:
For the dimensions to be consistent, we equate the dimensions from LHS and RHS:
- For mass: \( a + d = 0 \) (1)
- For length: \( -a + b + c - d = 3 \) (2)
- For time: \( -a - 2d = -1 \) (3)

6. **Solve the Equations**:
From equation (1), we have:
\[ d = -a \]
Substitute \( d \) into equation (3):
\[ -a - 2(-a) = -1 \implies -a + 2a = -1 \implies a = -1 \]
Then substituting \( a \) back into equation (1):
\[ d = -(-1) = 1 \]

7. **Substitute \( a \) and \( d \) into Equation (2)**:
\[ -(-1) + b + c - 1 = 3 \implies 1 + b + c - 1 = 3 \implies b + c = 3 \]

8. **Determine \( b \) and \( c \)**:
We have \( b + c = 3 \). Since the options involve \( r^4 \) or \( r^{-4} \), we can assume:
- If \( b = 4 \), then \( c = -1 \).

9. **Final Relation**:
Thus, substituting \( a, b, c, d \) into our original relation gives:
\[ V = K \cdot \eta^{-1} \cdot r^4 \cdot l^{-1} \cdot p^1 \]
Simplifying, we get:
\[ V = K \cdot \frac{p \cdot r^4}{\eta \cdot l} \]
The constant \( K \) can be determined to be \( \frac{\pi}{8} \) based on empirical data, leading to:
\[ V = \frac{\pi}{8} \cdot \frac{p \cdot r^4}{\eta \cdot l} \]

### Conclusion:
The dimensionally consistent relation for the volume \( V \) of a liquid flowing through a tube is:
\[ V = \frac{\pi}{8} \cdot \frac{p \cdot r^4}{\eta \cdot l} \]