Glycerine flows steadily through a horizontal tube of length 1.5m and radius 1.0 cm. If the amount of glycerine collected per second at one end is` 4. 0 xx 10^(-3) kg s^(-1) ` what is the pressure difference between the two ends of the tube? (Density of glycerine `= 1.3 xx 10^(3) kg m^(-3)` and coefficient of viscosity of glycerine `= 0.83 Ns m^(-2) ) `

Explanation

To solve the problem of determining the pressure difference between the two ends of a horizontal tube through which glycerine flows, we can use Poiseuille's law. Here’s a step-by-step breakdown of the solution:

### Step 1: Gather Given Information
- Length of the tube, \( L = 1.5 \, \text{m} \)
- Radius of the tube, \( r = 1.0 \, \text{cm} = 0.01 \, \text{m} \)
- Mass flow rate, \( \dot{m} = 4.0 \times 10^{-3} \, \text{kg/s} \)
- Density of glycerine, \( \rho = 1.3 \times 10^{3} \, \text{kg/m}^3 \)
- Coefficient of viscosity of glycerine, \( \eta = 0.83 \, \text{Ns/m}^2 \)

### Step 2: Calculate the Volume Flow Rate
The volume flow rate \( \dot{V} \) can be calculated using the formula:
\[ \dot{V} = \frac{\dot{m}}{\rho} \]
Substituting the values:
\[ \dot{V} = \frac{4.0 \times 10^{-3} \, \text{kg/s}}{1.3 \times 10^{3} \, \text{kg/m}^3} = 3.08 \times 10^{-6} \, \text{m}^3/\text{s} \]

### Step 3: Use Poiseuille's Law
According to Poiseuille's law, the volume flow rate \( \dot{V} \) is given by:
\[ \dot{V} = \frac{\pi}{8} \cdot \frac{P \cdot r^4}{\eta \cdot L} \]
Where \( P \) is the pressure difference we want to find. Rearranging the formula to solve for \( P \):
\[ P = \frac{8 \cdot \dot{V} \cdot \eta \cdot L}{\pi \cdot r^4} \]

### Step 4: Substitute the Known Values
Now substituting the known values into the rearranged formula:
- \( r^4 = (0.01 \, \text{m})^4 = 1.0 \times 10^{-8} \, \text{m}^4 \)

Putting everything together:
\[ P = \frac{8 \cdot (3.08 \times 10^{-6} \, \text{m}^3/\text{s}) \cdot (0.83 \, \text{Ns/m}^2) \cdot (1.5 \, \text{m})}{\pi \cdot (1.0 \times 10^{-8} \, \text{m}^4)} \]

### Step 5: Calculate the Pressure Difference
Calculating the above expression:
\[ P = \frac{8 \cdot 3.08 \times 10^{-6} \cdot 0.83 \cdot 1.5}{3.14 \cdot 1.0 \times 10^{-8}} \]
Calculating the numerator:
\[ = 8 \cdot 3.08 \cdot 0.83 \cdot 1.5 = 3.06 \times 10^{-5} \]
Calculating the denominator:
\[ = 3.14 \times 1.0 \times 10^{-8} = 3.14 \times 10^{-8} \]
Now, dividing:
\[ P = \frac{3.06 \times 10^{-5}}{3.14 \times 10^{-8}} \approx 974.2 \, \text{Pa} \]

### Final Answer
The pressure difference between the two ends of the tube is approximately:
\[ P \approx 974.2 \, \text{Pa} \approx 9.74 \times 10^{2} \, \text{Pa} \]