A liquid flows through two capillary tube under the same pressure head. The lengths of the tube are in the ratio 2:1 and the ratio of their diameters is 2:3. Compare the rates of flow of liquid through the tubes ?

Explanation

To solve the problem of comparing the rates of flow of liquid through two capillary tubes under the same pressure head, we will use the formula for the flow rate (Q) through a capillary tube, which is given by:

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

where:
- \( P \) is the pressure difference,
- \( r \) is the radius of the tube,
- \( \eta \) is the viscosity of the liquid,
- \( L \) is the length of the tube.

### Step-by-Step Solution:

1. **Identify Given Ratios**:
- The lengths of the tubes are in the ratio \( L_2 : L_1 = 2 : 1 \).
- The diameters of the tubes are in the ratio \( d_2 : d_1 = 2 : 3 \). Hence, the radii will be in the ratio \( r_2 : r_1 = 2/2 : 3/2 = 2/3 \).

2. **Express Flow Rates**:
- For tube 1:
\[ Q_1 = \frac{\pi P r_1^4}{8 \eta L_1} \]
- For tube 2:
\[ Q_2 = \frac{\pi P r_2^4}{8 \eta L_2} \]

3. **Set up the Ratio of Flow Rates**:
- To compare the flow rates, we take the ratio \( \frac{Q_1}{Q_2} \):
\[ \frac{Q_1}{Q_2} = \frac{\frac{\pi P r_1^4}{8 \eta L_1}}{\frac{\pi P r_2^4}{8 \eta L_2}} = \frac{r_1^4}{r_2^4} \cdot \frac{L_2}{L_1} \]

4. **Substitute the Ratios**:
- From the given ratios:
- \( L_2 : L_1 = 2 : 1 \) implies \( \frac{L_2}{L_1} = 2 \).
- \( r_2 : r_1 = 2 : 3 \) implies \( \frac{r_2}{r_1} = \frac{2}{3} \) or \( \frac{r_1}{r_2} = \frac{3}{2} \).
- Therefore, \( \left( \frac{r_1}{r_2} \right)^4 = \left( \frac{3}{2} \right)^4 = \frac{81}{16} \).

5. **Combine the Ratios**:
- Now substituting these values into the flow rate ratio:
\[ \frac{Q_1}{Q_2} = \frac{81}{16} \cdot 2 = \frac{81 \cdot 2}{16} = \frac{162}{16} = \frac{81}{8} \]

6. **Final Result**:
- Thus, the ratio of the rates of flow of liquid through the two tubes is:
\[ Q_1 : Q_2 = 81 : 8 \]