The volume of a liquid (V) flowing per second through a cylindrical tube depends upon the pressure gradient `(p//l`) radius of the tube (r) coefficient of viscosity `(eta)` of the liquid by dimensional method the correct formula is  

Options

A. `V prop (Pr^4)/(eta l)`
B. `V prop (Pr)/(eta l^4)`
C. `V prop (Pl^4)/(eta r)`
D. None

Answer

To solve the problem of finding the relationship between the volume of a liquid (V) flowing per second through a cylindrical tube and the given parameters (pressure gradient \( \frac{P}{L} \), radius of the tube \( r \), and coefficient of viscosity \( \eta \)), we will use dimensional analysis.

### Step-by-Step Solution:

1. **Identify the Variables**:
We have the following variables:
- \( V \): Volume of liquid flowing per second (dimension: \( [L^3 T^{-1}] \))
- \( P \): Pressure (dimension: \( [M L^{-1} T^{-2}] \))
- \( L \): Length (dimension: \( [L] \))
- \( r \): Radius (dimension: \( [L] \))
- \( \eta \): Coefficient of viscosity (dimension: \( [M L^{-1} T^{-1}] \))

2. **Formulate the Relationship**:
We assume that the volume \( V \) is proportional to the parameters as follows:
\[ V \propto r^a \left(\frac{P}{L}\right)^b \eta^c \]
where \( a \), \( b \), and \( c \) are the powers we need to determine.

3. **Write the Dimensions**:
Substitute the dimensions into the equation:
- Dimensions of \( V \) = \( [L^3 T^{-1}] \)
- Dimensions of \( r^a \) = \( [L]^a = [L^a] \)
- Dimensions of \( \left(\frac{P}{L}\right)^b \) = \( \left[M L^{-1} T^{-2} L^{-1}\right]^b = [M^b L^{-2b} T^{-2b}] \)
- Dimensions of \( \eta^c \) = \( [M L^{-1} T^{-1}]^c = [M^c L^{-c} T^{-c}] \)

Combining these, we have:
\[ [L^3 T^{-1}] = [L^a] [M^b L^{-2b} T^{-2b}] [M^c L^{-c} T^{-c}] \]

4. **Combine Dimensions**:
This gives us:
\[ [L^3 T^{-1}] = [M^{b+c} L^{a - 2b - c} T^{-2b - c}] \]

5. **Equate the Powers**:
From the above equation, we can equate the powers of \( M \), \( L \), and \( T \):
- For \( M \): \( b + c = 0 \) (1)
- For \( L \): \( a - 2b - c = 3 \) (2)
- For \( T \): \( -2b - c = -1 \) (3)

6. **Solve the Equations**:
From equation (1), we can express \( c \) in terms of \( b \):
\[ c = -b \]
Substituting \( c \) into equation (3):
\[ -2b - (-b) = -1 \implies -2b + b = -1 \implies -b = -1 \implies b = 1 \]
Now substituting \( b = 1 \) back into equation (1):
\[ c = -1 \]
Finally, substituting \( b = 1 \) and \( c = -1 \) into equation (2):
\[ a - 2(1) - (-1) = 3 \implies a - 2 + 1 = 3 \implies a - 1 = 3 \implies a = 4 \]

7. **Final Relationship**:
Now substituting the values of \( a \), \( b \), and \( c \) back into the proportionality:
\[ V \propto r^4 \left(\frac{P}{L}\right)^1 \eta^{-1} \]
This can be rewritten as:
\[ V \propto \frac{P r^4}{\eta L} \]

### Conclusion:
The correct formula for the volume of liquid flowing per second through a cylindrical tube is:
\[ V \propto \frac{P r^4}{\eta L} \]