Two capillary of length L and 2L and of radius R and 2R are connected in series. The net rate of flow of fluid through them will be (given rate to the flow through single capillary,
`("X"=(pi"PR"^(4))/(8eta"L"))`

Options

A. `(8)/(9)X`
B. `(9)/(8) X`
C. `(5)/(7)X`
D. `(7)/(5)X`

Explanation

To solve the problem, we need to determine the net rate of flow of fluid through two capillaries connected in series. We are given the formula for the flow rate through a single capillary:

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

where:
- \( P \) is the pressure difference,
- \( R \) is the radius of the capillary,
- \( \eta \) is the viscosity of the fluid,
- \( L \) is the length of the capillary.

### Step-by-Step Solution:

1. **Identify the Capillaries**:
- The first capillary has a length \( L \) and radius \( R \).
- The second capillary has a length \( 2L \) and radius \( 2R \).

2. **Calculate the Resistance of Each Capillary**:
- The resistance \( R_1 \) of the first capillary (length \( L \), radius \( R \)) is given by:
\[ R_1 = \frac{8 \eta L}{\pi R^4} \]
- The resistance \( R_2 \) of the second capillary (length \( 2L \), radius \( 2R \)) is given by:
\[ R_2 = \frac{8 \eta (2L)}{\pi (2R)^4} = \frac{8 \eta (2L)}{\pi (16R^4)} = \frac{\eta L}{\pi R^4} \]

3. **Calculate the Total Resistance in Series**:
- When two resistances are in series, the total resistance \( R_{\text{eq}} \) is the sum of the individual resistances:
\[ R_{\text{eq}} = R_1 + R_2 = \frac{8 \eta L}{\pi R^4} + \frac{\eta L}{\pi R^4} = \frac{9 \eta L}{\pi R^4} \]

4. **Determine the Net Flow Rate**:
- The net flow rate \( Q_{\text{net}} \) through the combined capillaries can be expressed using the pressure and the equivalent resistance:
\[ Q_{\text{net}} = \frac{P}{R_{\text{eq}}} \]
- Substituting \( R_{\text{eq}} \):
\[ Q_{\text{net}} = \frac{P \cdot \pi R^4}{9 \eta L} \]

5. **Relate to Given Flow Rate \( X \)**:
- We know from the given formula that:
\[ X = \frac{\pi P R^4}{8 \eta L} \]
- Therefore, we can express \( Q_{\text{net}} \) in terms of \( X \):
\[ Q_{\text{net}} = \frac{8}{9} X \]

### Final Answer:
The net rate of flow of fluid through the two capillaries connected in series is:
\[ Q_{\text{net}} = \frac{8}{9} X \]