A capillary tube of 1mm diameter and 20 cm long is fitted horizontally to a vessel kept full of alcohol of density 0.8 gm//c.c. The depth of centre of capillary tube below the surface of alcohol is 20 cm. If the visosity of oil is `0.12 dyn e cm^(-2)` s, find the amount of liquid that will flow in 5 minuts.

Explanation

To solve the problem step by step, we will use the principles of fluid dynamics, specifically the Hagen-Poiseuille equation which describes the flow of a viscous fluid through a cylindrical pipe (capillary tube).

### Step 1: Convert all units to SI units
- Diameter of the capillary tube = 1 mm = \(1 \times 10^{-3}\) m
- Radius \(r\) = \(\frac{1}{2} \times 10^{-3}\) m = \(0.5 \times 10^{-3}\) m
- Length of the capillary tube \(L\) = 20 cm = \(20 \times 10^{-2}\) m = \(0.2\) m
- Density of alcohol \(\rho\) = 0.8 g/cm³ = \(0.8 \times 10^{3}\) kg/m³ = \(800\) kg/m³
- Depth \(h\) = 20 cm = \(20 \times 10^{-2}\) m = \(0.2\) m
- Viscosity of oil \(\eta\) = 0.12 dyn·s/cm² = \(0.12 \times 10^{-1}\) Pa·s = \(0.012\) Pa·s (since \(1 \text{ dyn·s/cm}^2 = 0.1 \text{ Pa·s}\))

### Step 2: Calculate the pressure difference
The pressure difference \(\Delta P\) due to the height of the liquid column is given by:
\[ \Delta P = h \cdot \rho \cdot g \]
Where \(g\) (acceleration due to gravity) = \(9.81\) m/s².
\[ \Delta P = 0.2 \, \text{m} \cdot 800 \, \text{kg/m}^3 \cdot 9.81 \, \text{m/s}^2 \]
\[ \Delta P = 1569.6 \, \text{Pa} \]

### Step 3: Use the Hagen-Poiseuille equation
The volume flow rate \(Q\) through the capillary tube is given by:
\[ Q = \frac{\pi r^4 \Delta P}{8 \eta L} \]
Substituting the values:
\[ Q = \frac{\pi (0.5 \times 10^{-3})^4 \cdot 1569.6}{8 \cdot 0.012 \cdot 0.2} \]
Calculating \(r^4\):
\[ (0.5 \times 10^{-3})^4 = 6.25 \times 10^{-13} \, \text{m}^4 \]
Now substituting:
\[ Q = \frac{3.14159 \cdot 6.25 \times 10^{-13} \cdot 1569.6}{8 \cdot 0.012 \cdot 0.2} \]
Calculating the denominator:
\[ 8 \cdot 0.012 \cdot 0.2 = 0.0192 \]
Now calculating \(Q\):
\[ Q \approx \frac{3.14159 \cdot 6.25 \times 10^{-13} \cdot 1569.6}{0.0192} \approx 5.1 \times 10^{-10} \, \text{m}^3/s \]

### Step 4: Calculate the total volume in 5 minutes
Convert 5 minutes to seconds:
\[ 5 \, \text{minutes} = 5 \times 60 = 300 \, \text{seconds} \]
Total volume \(V\) that flows in 5 minutes:
\[ V = Q \cdot t = 5.1 \times 10^{-10} \, \text{m}^3/s \cdot 300 \, \text{s} \approx 1.53 \times 10^{-7} \, \text{m}^3 \]

### Step 5: Convert volume to cm³
\[ 1 \, \text{m}^3 = 10^6 \, \text{cm}^3 \]
So,
\[ V \approx 1.53 \times 10^{-7} \, \text{m}^3 \times 10^6 \approx 0.153 \, \text{cm}^3 \]

### Final Answer
The amount of liquid that will flow in 5 minutes is approximately **0.153 cm³**.