If l is length of the tube and r is the radius of the tube, then the rate of volume flow of a liquid is maximum for the following measurements. Under the same pressure difference.

Options

A. `l,r`
B. `(l)/(2),2r`
C. `l,(r)/(2)`
D. `2l,2r`

Answer

To determine the maximum rate of volume flow of a liquid through a tube under the same pressure difference, we can use Poiseuille's law, which states that the volume flow rate \( Q \) is given by the equation:

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

where:
- \( Q \) = volume flow rate
- \( \Delta P \) = pressure difference
- \( R \) = radius of the tube
- \( \eta \) = viscosity of the liquid
- \( L \) = length of the tube

### Step-by-Step Solution:

1. **Understand the Variables**:
- We have the length of the tube \( L \) and the radius \( R \).
- We need to analyze how changes in \( L \) and \( R \) affect the flow rate \( Q \).

2. **Identify the Equation**:
- The equation for volume flow rate is:
\[ Q = \frac{\pi \Delta P R^4}{8 \eta L} \]
- Here, \( \Delta P \) and \( \eta \) are constants for this problem.

3. **Rearranging the Equation**:
- We can express \( Q \) in terms of a constant \( K \):
\[ Q = K \frac{R^4}{L} \]
where \( K = \frac{\pi \Delta P}{8 \eta} \).

4. **Evaluating Different Cases**:
- We will evaluate the flow rates for different combinations of \( R \) and \( L \) based on the options given.

- **Option 1**: \( R = R \), \( L = L \)
\[ Q_1 = K \frac{R^4}{L} \]

- **Option 2**: \( R = 2R \), \( L = \frac{L}{2} \)
\[ Q_2 = K \frac{(2R)^4}{\frac{L}{2}} = K \frac{16R^4}{\frac{L}{2}} = 32K \frac{R^4}{L} \]

- **Option 3**: \( R = \frac{R}{2} \), \( L = L \)
\[ Q_3 = K \frac{(\frac{R}{2})^4}{L} = K \frac{\frac{R^4}{16}}{L} = \frac{K R^4}{16L} \]

- **Option 4**: \( R = 2R \), \( L = 2L \)
\[ Q_4 = K \frac{(2R)^4}{2L} = K \frac{16R^4}{2L} = 8K \frac{R^4}{L} \]

5. **Comparing the Flow Rates**:
- Now we compare the flow rates obtained:
- \( Q_1 = K \frac{R^4}{L} \)
- \( Q_2 = 32K \frac{R^4}{L} \)
- \( Q_3 = \frac{K R^4}{16L} \)
- \( Q_4 = 8K \frac{R^4}{L} \)

6. **Conclusion**:
- The maximum flow rate occurs in **Option 2**, where \( R = 2R \) and \( L = \frac{L}{2} \), giving us:
\[ Q_{\text{max}} = 32K \frac{R^4}{L} \]

### Final Answer:
The rate of volume flow of the liquid is maximum for **Option 2**.