We have two (narrow) capillary tubes `T_1` and `T_2` . Their lengths are `l_1` and `l_2` and radii of cross-section are `r_1` and `r_2` respectively. The rate of flow of water under a pressure difference P through tube T is `8 cm 3// sec`. If `l_1 = 2l_2` and `r_1 = r_2` what will be the rate of flow when the two tubes are connected in series and pressure difference across the combinatin is same as before `(= P)`

Options

A. `4 cm^3s^(-1)`
B. `16/3cm^3s^(-1)`
C. `8/17cm^3s^(-1)`
D. `8 cm^3s^(-1)`

Answer

To solve the problem, we need to analyze the flow of water through two capillary tubes connected in series. We will use the principles of fluid dynamics, specifically Poiseuille's law, which describes the flow rate of a viscous fluid through a cylindrical pipe.

### Step-by-Step Solution:

1. **Understanding the Given Information:**
- We have two capillary tubes, \( T_1 \) and \( T_2 \).
- The lengths are \( l_1 \) and \( l_2 \), and the radii are \( r_1 \) and \( r_2 \).
- It is given that \( l_1 = 2l_2 \) and \( r_1 = r_2 \).
- The rate of flow through tube \( T_1 \) is \( Q_1 = 8 \, \text{cm}^3/\text{sec} \).

2. **Using Poiseuille's Law:**
- The volume flow rate \( Q \) through a capillary tube is given by:
\[ Q = \frac{P \pi r^4}{8 \eta L} \]
where \( P \) is the pressure difference, \( \eta \) is the viscosity of the fluid, \( r \) is the radius, and \( L \) is the length of the tube.

3. **Finding the Resistance for Each Tube:**
- The resistance \( R \) for a tube can be expressed as:
\[ R = \frac{8 \eta L}{\pi r^4} \]
- For tube \( T_1 \):
\[ R_1 = \frac{8 \eta l_1}{\pi r_1^4} \]
- For tube \( T_2 \):
\[ R_2 = \frac{8 \eta l_2}{\pi r_2^4} \]

4. **Substituting the Lengths:**
- Since \( l_1 = 2l_2 \) and \( r_1 = r_2 \), we can express \( R_2 \) in terms of \( R_1 \):
\[ R_2 = \frac{8 \eta l_2}{\pi r^4} = \frac{8 \eta \frac{l_1}{2}}{\pi r^4} = \frac{R_1}{2} \]

5. **Calculating Total Resistance in Series:**
- When the tubes are connected in series, the total resistance \( R_{\text{total}} \) is:
\[ R_{\text{total}} = R_1 + R_2 = R_1 + \frac{R_1}{2} = \frac{3R_1}{2} \]

6. **Finding the New Flow Rate:**
- The new flow rate \( Q_{\text{total}} \) through the combination of the two tubes can be calculated using the total resistance:
\[ Q_{\text{total}} = \frac{P}{R_{\text{total}}} = \frac{P}{\frac{3R_1}{2}} = \frac{2P}{3R_1} \]
- Since \( Q_1 = \frac{P}{R_1} = 8 \, \text{cm}^3/\text{sec} \), we can express \( P \) in terms of \( R_1 \):
\[ P = 8R_1 \]
- Substituting this into the equation for \( Q_{\text{total}} \):
\[ Q_{\text{total}} = \frac{2(8R_1)}{3R_1} = \frac{16}{3} \, \text{cm}^3/\text{sec} \]

### Conclusion:
The rate of flow when the two tubes are connected in series is \( \frac{16}{3} \, \text{cm}^3/\text{sec} \).