Water at `20^(@)` C is escaping from a cistern by ways of a horizontal capillary tube 10 cm long and 0.4 mm in diameter ,at a distance of 50 cm below the free surface of water in the cistren .Calculate the rate at which the water is escaping .Coefficient of viscossity of water at `20^(@)C` =0.001 decapoise .

Answer

To solve the problem of calculating the rate at which water is escaping from a cistern through a horizontal capillary tube, we will use the Hagen-Poiseuille equation, which describes the flow of a viscous fluid through a pipe.

### Step-by-Step Solution:

1. **Convert Given Measurements to SI Units:**
- Length of the capillary tube, \( L = 10 \, \text{cm} = 0.1 \, \text{m} \)
- Diameter of the capillary tube, \( d = 0.4 \, \text{mm} = 0.0004 \, \text{m} \)
- Radius of the capillary tube, \( r = \frac{d}{2} = \frac{0.0004}{2} = 0.0002 \, \text{m} \)
- Height of water column, \( h = 50 \, \text{cm} = 0.5 \, \text{m} \)

2. **Calculate the Pressure Difference:**
- The pressure difference due to the height of the water column is given by:
\[ \Delta P = \rho g h \]
- Where:
- \( \rho \) (density of water) \( = 1000 \, \text{kg/m}^3 \)
- \( g \) (acceleration due to gravity) \( = 9.81 \, \text{m/s}^2 \)
- Substituting the values:
\[ \Delta P = 1000 \times 9.81 \times 0.5 = 4905 \, \text{Pa} \]

3. **Use the Hagen-Poiseuille Equation:**
- The volume flow rate \( Q \) through the capillary tube is given by:
\[ Q = \frac{\pi r^4 \Delta P}{8 \eta L} \]
- Where:
- \( \eta \) (coefficient of viscosity) \( = 0.001 \, \text{Pa.s} \)
- Substituting the values:
\[ Q = \frac{\pi (0.0002)^4 \times 4905}{8 \times 0.001 \times 0.1} \]

4. **Calculate the Volume Flow Rate:**
- First, calculate \( r^4 \):
\[ r^4 = (0.0002)^4 = 1.6 \times 10^{-16} \, \text{m}^4 \]
- Now substitute into the equation:
\[ Q = \frac{3.14 \times 1.6 \times 10^{-16} \times 4905}{8 \times 0.001 \times 0.1} \]
- Calculate the denominator:
\[ 8 \times 0.001 \times 0.1 = 8 \times 10^{-4} \]
- Now calculate \( Q \):
\[ Q = \frac{3.14 \times 1.6 \times 10^{-16} \times 4905}{8 \times 10^{-4}} \approx 3.08 \times 10^{-8} \, \text{m}^3/\text{s} \]

5. **Final Result:**
- The rate at which water is escaping from the cistern is approximately:
\[ Q \approx 3.08 \times 10^{-8} \, \text{m}^3/\text{s} \]