The rate of steady volume flow of water through a capillary tube of length ' l ' and radius ' r ' under a pressure difference of P is V . This tube is connected with another tube of the same length but half the radius in series. Then the rate of steady volume flow through them is (The pressure difference across the combination is P )

Options

A. `(V)/(16)`
B. `(V)/(17)`
C. `(16V)/(17)`
D. `(17V)/(16)`

Answer

To solve the problem, we need to determine the rate of steady volume flow through two capillary tubes connected in series, where the first tube has a radius \( r \) and the second tube has a radius \( \frac{r}{2} \). The pressure difference across the combination is \( P \).

### Step-by-step Solution:

1. **Understand the Flow Rate in a Capillary Tube**:
The volume flow rate \( V \) through a capillary tube can be expressed using Poiseuille's law:
\[ V = \frac{P}{R} \]
where \( R \) is the resistance to flow.

2. **Calculate Resistance for the First Tube**:
The resistance \( R_1 \) for a tube of length \( l \) and radius \( r \) is given by:
\[ R_1 = \frac{8 \eta l}{\pi r^4} \]
where \( \eta \) is the viscosity of the fluid.

3. **Calculate Resistance for the Second Tube**:
The second tube has the same length \( l \) but half the radius \( \frac{r}{2} \). The resistance \( R_2 \) for this tube is:
\[ R_2 = \frac{8 \eta l}{\pi \left(\frac{r}{2}\right)^4} = \frac{8 \eta l}{\pi \frac{r^4}{16}} = \frac{128 \eta l}{\pi r^4} \]

4. **Total Resistance in Series**:
Since the tubes are connected in series, the total resistance \( R_{total} \) is the sum of the individual resistances:
\[ R_{total} = R_1 + R_2 = \frac{8 \eta l}{\pi r^4} + \frac{128 \eta l}{\pi r^4} = \frac{136 \eta l}{\pi r^4} \]

5. **Calculate the New Flow Rate**:
The new flow rate \( V_{new} \) through the combination of the two tubes is given by:
\[ V_{new} = \frac{P}{R_{total}} = \frac{P}{\frac{136 \eta l}{\pi r^4}} = \frac{P \pi r^4}{136 \eta l} \]

6. **Relate the New Flow Rate to the Original Flow Rate**:
From the original flow rate \( V \):
\[ V = \frac{P}{R_1} = \frac{P \pi r^4}{8 \eta l} \]
We can express \( V_{new} \) in terms of \( V \):
\[ V_{new} = \frac{V \cdot 8}{136} = \frac{V}{17} \]

### Final Answer:
Thus, the rate of steady volume flow through the two tubes connected in series is:
\[ V_{new} = \frac{V}{17} \]