Under a constant pressure head, the rate of flow of orderly volume flow of liquid through a capillary tube is V. IF the length of the capillary is doubled and the diameter of the bore is halved, the rate of flow would become

Options

A. V/4
B. 16 V
C. V/8
D. V/32

Explanation

To solve the problem, we will use the formula for the volume flow rate of a liquid through a capillary tube, which is given by:

\[ V = \frac{\pi P R^4}{8 \eta L} \]

Where:
- \(V\) = volume flow rate
- \(P\) = pressure difference (constant)
- \(R\) = radius of the capillary tube
- \(L\) = length of the capillary tube
- \(\eta\) = viscosity of the liquid (constant)

### Step 1: Identify the initial conditions
Let:
- Initial length of the capillary tube = \(L_1\)
- Initial radius of the capillary tube = \(R_1\)
- Initial volume flow rate = \(V_1 = V\)

### Step 2: Define the new conditions
According to the problem:
- The length of the capillary tube is doubled: \(L_2 = 2L_1\)
- The diameter of the bore is halved, which means the radius is also halved: \(R_2 = \frac{R_1}{2}\)

### Step 3: Substitute the new conditions into the flow rate formula
Using the formula for volume flow rate with the new conditions:

\[ V_2 = \frac{\pi P R_2^4}{8 \eta L_2} \]

Substituting \(R_2\) and \(L_2\):

\[ V_2 = \frac{\pi P \left(\frac{R_1}{2}\right)^4}{8 \eta (2L_1)} \]

### Step 4: Simplify the expression
Calculating \(R_2^4\):

\[ \left(\frac{R_1}{2}\right)^4 = \frac{R_1^4}{16} \]

Now substituting this back into the equation for \(V_2\):

\[ V_2 = \frac{\pi P \frac{R_1^4}{16}}{8 \eta (2L_1)} \]

This simplifies to:

\[ V_2 = \frac{\pi P R_1^4}{128 \eta L_1} \]

### Step 5: Relate \(V_2\) to \(V_1\)
From the initial condition, we know:

\[ V_1 = \frac{\pi P R_1^4}{8 \eta L_1} \]

Now, we can express \(V_2\) in terms of \(V_1\):

\[ V_2 = \frac{V_1}{16} \]

### Step 6: Final calculation
Since \(V_1 = V\):

\[ V_2 = \frac{V}{16} \]

Thus, the new volume flow rate when the length is doubled and the diameter is halved is:

\[ V_2 = \frac{V}{32} \]

### Conclusion
The rate of flow would become \( \frac{V}{32} \).

### Answer
The correct option is D: \( \frac{V}{32} \).