Using the method of dimensions, derive an expression for rate of flow (v) of a liquied through a pipe of radius (r ) under a pressure gradient `(P//I)` Given that V also depends on coefficient of viscosity `(eta)` of the liquied.

Answer

To derive an expression for the rate of flow \( V \) of a liquid through a pipe of radius \( r \) under a pressure gradient \( \frac{P}{L} \), considering that \( V \) also depends on the coefficient of viscosity \( \eta \) of the liquid, we will use the method of dimensions.

### Step-by-Step Solution:

1. **Identify the Variables**:
- Rate of flow \( V \) (volume per unit time)
- Radius of the pipe \( r \)
- Pressure gradient \( \frac{P}{L} \)
- Coefficient of viscosity \( \eta \)

2. **Express the Rate of Flow**:
The rate of flow \( V \) can be expressed as:
\[ V \propto r^A \left(\frac{P}{L}\right)^B \eta^C \]
where \( A, B, C \) are the powers to be determined.

3. **Write the Dimensions**:
- The dimension of \( V \) (rate of flow) is:
\[ [V] = \frac{[L^3]}{[T]} = L^3 T^{-1} \]
- The dimension of pressure \( P \) is:
\[ [P] = [M L^{-1} T^{-2}] \]
- The dimension of length \( L \) is:
\[ [L] = L \]
- The dimension of viscosity \( \eta \) is:
\[ [\eta] = [M L^{-1} T^{-1}] \]

4. **Write the Dimensions of the Right-Hand Side**:
Substituting the dimensions into the equation:
\[ [V] = [r^A] \left[\left(\frac{P}{L}\right)^B\right] [\eta^C] \]
This gives:
\[ L^3 T^{-1} = (L^1)^A \left(\frac{[M L^{-1} T^{-2}]}{[L]}\right)^B (M L^{-1} T^{-1})^C \]
Simplifying the pressure gradient:
\[ \frac{P}{L} \Rightarrow [P] \cdot [L^{-1}] = [M L^{-1} T^{-2}] \cdot [L^{-1}] = [M L^{-2} T^{-2}] \]
Thus:
\[ [V] = (L^1)^A (M L^{-2} T^{-2})^B (M L^{-1} T^{-1})^C \]

5. **Combine the Dimensions**:
\[ L^3 T^{-1} = M^{B+C} L^{A - 2B - C} T^{-2B - C} \]

6. **Set Up the Equations**:
By comparing the dimensions on both sides, we get three equations:
- For mass: \( B + C = 0 \) (1)
- For length: \( A - 2B - C = 3 \) (2)
- For time: \( -2B - C = -1 \) (3)

7. **Solve the Equations**:
From equation (1), we have \( C = -B \).
Substitute \( C \) into equations (2) and (3):
- From (2):
\[ A - 2B + B = 3 \Rightarrow A - B = 3 \Rightarrow A = B + 3 \]
- From (3):
\[ -2B + B = -1 \Rightarrow -B = -1 \Rightarrow B = 1 \Rightarrow C = -1 \]
Substitute \( B \) back into \( A = B + 3 \):
\[ A = 1 + 3 = 4 \]

8. **Final Expression**:
Substitute \( A, B, C \) back into the original proportionality:
\[ V \propto r^4 \left(\frac{P}{L}\right)^1 \eta^{-1} \]
Thus, we can write:
\[ V = k \frac{r^4 P}{\eta L} \]
where \( k \) is a constant of proportionality.

### Final Answer:
\[ V = k \frac{r^4 P}{\eta L} \]