Water flows in a streamline manner through a capillary tube of radius a. The pressure difference being P and the rate of flow is Q . If the radius is reduced to `a//2` and the pressure difference is increased to 2P, then find the rate of flow.

Options

A. 4Q
B. Q
C. `(Q)/(4)`
D. `(Q)/(8)`

Explanation

To solve the problem, we will use the principles of fluid mechanics, specifically Poiseuille's law, which describes the flow of a viscous fluid through a cylindrical pipe.

### Step-by-Step Solution:

1. **Understand the Initial Conditions**:
- Initial radius of the tube, \( r_1 = a \)
- Initial pressure difference, \( P_1 = P \)
- Initial rate of flow, \( Q_1 = Q \)

2. **Apply Poiseuille's Law**:
Poiseuille's law states that the volume flow rate \( Q \) through a capillary tube is given by:
\[ Q = \frac{\pi P r^4}{8 \eta L} \]
where \( \eta \) is the dynamic viscosity and \( L \) is the length of the tube.

3. **Set Up the Initial Flow Rate**:
For the initial conditions:
\[ Q_1 = \frac{\pi P a^4}{8 \eta L} \]

4. **Understand the New Conditions**:
- New radius of the tube, \( r_2 = \frac{a}{2} \)
- New pressure difference, \( P_2 = 2P \)

5. **Calculate the New Flow Rate**:
Using Poiseuille's law again for the new conditions:
\[ Q_2 = \frac{\pi P_2 r_2^4}{8 \eta L} \]
Substitute \( P_2 \) and \( r_2 \):
\[ Q_2 = \frac{\pi (2P) \left(\frac{a}{2}\right)^4}{8 \eta L} \]

6. **Simplify the Expression**:
Calculate \( \left(\frac{a}{2}\right)^4 \):
\[ \left(\frac{a}{2}\right)^4 = \frac{a^4}{16} \]
Now substitute this back into the equation for \( Q_2 \):
\[ Q_2 = \frac{\pi (2P) \left(\frac{a^4}{16}\right)}{8 \eta L} \]
Simplifying further:
\[ Q_2 = \frac{2\pi P a^4}{16 \cdot 8 \eta L} = \frac{2\pi P a^4}{128 \eta L} = \frac{\pi P a^4}{64 \eta L} \]

7. **Relate \( Q_2 \) to \( Q_1 \)**:
From the initial condition, we know:
\[ Q_1 = \frac{\pi P a^4}{8 \eta L} \]
Therefore, we can express \( Q_2 \) in terms of \( Q_1 \):
\[ Q_2 = \frac{1}{8} Q_1 \]
Since \( Q_1 = Q \):
\[ Q_2 = \frac{1}{8} Q \]

### Final Answer:
The rate of flow when the radius is reduced to \( \frac{a}{2} \) and the pressure difference is increased to \( 2P \) is:
\[ Q_2 = \frac{Q}{8} \]