In an exeriment with poiseuille,s apparatus , the following observation were noted :
Volume of liquid collected per minure `=15cm ^(3)`
Heat of liquid =30cm, Lrngth of tube =25 cm
Diamter of tube =0.2 cm, Density of liquir =`2.3 gcm^(-3)`
Find the coefficient of viscosity of the liquird.

Answer

To find the coefficient of viscosity of the liquid using Poiseuille's law, we can follow these steps:

### Step 1: Identify the given values
- Volume of liquid collected per minute (V) = 15 cm³
- Height of liquid column (h) = 30 cm
- Length of tube (L) = 25 cm
- Diameter of tube (d) = 0.2 cm
- Density of liquid (ρ) = 2.3 g/cm³

### Step 2: Convert the volume flow rate to cm³/s
The volume flow rate (Q) is given by:
\[ Q = \frac{V}{t} \]
where \( t \) is the time in seconds. Since we have the volume collected per minute, we convert it to seconds:
\[ Q = \frac{15 \, \text{cm}^3}{60 \, \text{s}} = 0.25 \, \text{cm}^3/\text{s} \]

### Step 3: Calculate the pressure difference (ΔP)
The pressure difference due to the height of the liquid column can be calculated using the formula:
\[ \Delta P = h \cdot \rho \cdot g \]
where:
- \( h = 30 \, \text{cm} = 0.3 \, \text{m} \) (convert to meters)
- \( \rho = 2.3 \, \text{g/cm}^3 = 2300 \, \text{kg/m}^3 \) (convert to kg/m³)
- \( g = 9.81 \, \text{m/s}^2 \)

Calculating ΔP:
\[ \Delta P = 0.3 \, \text{m} \cdot 2300 \, \text{kg/m}^3 \cdot 9.81 \, \text{m/s}^2 = 6762.3 \, \text{Pa} \]

### Step 4: Calculate the radius of the tube (r)
The radius \( r \) is half of the diameter:
\[ r = \frac{d}{2} = \frac{0.2 \, \text{cm}}{2} = 0.1 \, \text{cm} = 0.001 \, \text{m} \]

### Step 5: Use Poiseuille’s equation to find the coefficient of viscosity (η)
According to Poiseuille's law:
\[ Q = \frac{\pi r^4 \Delta P}{8 \eta L} \]
Rearranging for η gives:
\[ \eta = \frac{\pi r^4 \Delta P}{8 Q L} \]

### Step 6: Substitute the values into the equation
Substituting the values we calculated:
- \( r = 0.001 \, \text{m} \)
- \( \Delta P = 6762.3 \, \text{Pa} \)
- \( Q = 0.25 \, \text{cm}^3/\text{s} = 0.25 \times 10^{-6} \, \text{m}^3/\text{s} \)
- \( L = 0.25 \, \text{m} \)

Now substituting into the equation:
\[ \eta = \frac{\pi (0.001)^4 (6762.3)}{8 (0.25 \times 10^{-6}) (0.25)} \]

### Step 7: Calculate η
Calculating the numerator:
\[ \pi (0.001)^4 (6762.3) = 3.14159 \times 1 \times 10^{-12} \times 6762.3 \approx 2.12 \times 10^{-8} \]
Calculating the denominator:
\[ 8 (0.25 \times 10^{-6}) (0.25) = 8 \times 0.25 \times 0.25 \times 10^{-6} = 0.5 \times 10^{-6} = 5 \times 10^{-7} \]
Now substituting back:
\[ \eta = \frac{2.12 \times 10^{-8}}{5 \times 10^{-7}} \approx 0.0424 \, \text{Pa.s} \, \text{or} \, 0.0424 \, \text{N.s/m}^2 \]

### Final Answer
The coefficient of viscosity of the liquid is approximately **0.0424 Pa.s**.

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