Calculate the rate of flow of glycerine of density `1.25xx10^(3) kg m^(-3)` through the conical section of a pipe, if the radii of its ends are`0.1` m and `0.04` m and the pressure drop across its length is `10 Nm^(-2)`.

Explanation

To calculate the rate of flow of glycerine through the conical section of a pipe, we can follow these steps:

### Step 1: Identify the given values
- Density of glycerine, \( \rho = 1.25 \times 10^3 \, \text{kg/m}^3 \)
- Radius of the larger end of the pipe, \( r_1 = 0.1 \, \text{m} \)
- Radius of the smaller end of the pipe, \( r_2 = 0.04 \, \text{m} \)
- Pressure drop across the length of the pipe, \( \Delta P = 10 \, \text{N/m}^2 \)

### Step 2: Calculate the cross-sectional areas
The cross-sectional areas \( A_1 \) and \( A_2 \) at the two ends of the pipe can be calculated using the formula for the area of a circle:
\[ A = \pi r^2 \]
- For the larger end:
\[ A_1 = \pi (0.1)^2 = \pi \times 0.01 \, \text{m}^2 \]
- For the smaller end:
\[ A_2 = \pi (0.04)^2 = \pi \times 0.0016 \, \text{m}^2 \]

### Step 3: Use the equation of continuity
According to the equation of continuity, the flow rate must remain constant:
\[ A_1 V_1 = A_2 V_2 \]
From this, we can express \( V_2 \) in terms of \( V_1 \):
\[ V_2 = \frac{A_1}{A_2} V_1 \]
Calculating the ratio \( \frac{A_1}{A_2} \):
\[ \frac{A_1}{A_2} = \frac{\pi \times 0.01}{\pi \times 0.0016} = \frac{0.01}{0.0016} = 6.25 \]
Thus,
\[ V_2 = 6.25 V_1 \]

### Step 4: Apply Bernoulli’s equation
Using Bernoulli's equation between the two ends of the pipe:
\[ P_1 + \frac{1}{2} \rho V_1^2 = P_2 + \frac{1}{2} \rho V_2^2 \]
Rearranging gives:
\[ \frac{1}{2} \rho V_2^2 - \frac{1}{2} \rho V_1^2 = P_1 - P_2 = \Delta P \]
Substituting \( V_2 = 6.25 V_1 \):
\[ \frac{1}{2} \rho (6.25 V_1)^2 - \frac{1}{2} \rho V_1^2 = \Delta P \]
Simplifying:
\[ \frac{1}{2} \rho (39.0625 V_1^2 - V_1^2) = 10 \]
\[ \frac{1}{2} \rho (38.0625 V_1^2) = 10 \]
Substituting \( \rho = 1.25 \times 10^3 \):
\[ \frac{1}{2} (1.25 \times 10^3)(38.0625 V_1^2) = 10 \]
\[ (1.25 \times 10^3)(19.03125 V_1^2) = 10 \]
\[ 19.03125 V_1^2 = \frac{10}{1.25 \times 10^3} \]
\[ V_1^2 = \frac{10}{1.25 \times 10^3 \times 19.03125} \]
Calculating \( V_1 \):
\[ V_1 \approx 0.020 \, \text{m/s} \]

### Step 5: Calculate the rate of flow
The rate of flow \( Q \) can be calculated using:
\[ Q = A_1 V_1 \]
Substituting \( A_1 \) and \( V_1 \):
\[ Q = \pi (0.1)^2 (0.020) = \pi \times 0.01 \times 0.020 \]
\[ Q \approx 6.28 \times 10^{-4} \, \text{m}^3/\text{s} \]

### Final Answer
The rate of flow of glycerine through the conical section of the pipe is approximately \( 6.28 \times 10^{-4} \, \text{m}^3/\text{s} \).

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