A liquid of density `750kgm^(-3)` flows smoothly through a horizontal pipe that tapers in cross-sectional area from `A_(1)=1.2xx10^(-2)m^(2)` to `A_(2)=(A_(1))/(2)`. The pressure difference between the wide and narrow sections of the pipe is 4500 Pa. The rate of flow of liquid is ________ `xx10^(-3)m^(3)s^(-1)`.

Explanation

To solve the problem step by step, we will use the principles of fluid dynamics, specifically the continuity equation and Bernoulli's equation.

### Step 1: Identify the given data
- Density of the liquid, \( \rho = 750 \, \text{kg/m}^3 \)
- Cross-sectional area at the wide section, \( A_1 = 1.2 \times 10^{-2} \, \text{m}^2 \)
- Cross-sectional area at the narrow section, \( A_2 = \frac{A_1}{2} = \frac{1.2 \times 10^{-2}}{2} = 0.6 \times 10^{-2} \, \text{m}^2 \)
- Pressure difference, \( P_1 - P_2 = 4500 \, \text{Pa} \)

### Step 2: Apply the continuity equation
According to the continuity equation, the volume flow rate must be constant throughout the pipe:

\[ A_1 V_1 = A_2 V_2 \]

From this, we can express \( V_2 \) in terms of \( V_1 \):

\[ V_2 = \frac{A_1}{A_2} V_1 = \frac{1.2 \times 10^{-2}}{0.6 \times 10^{-2}} V_1 = 2 V_1 \]

### Step 3: Apply Bernoulli's equation
Using Bernoulli's equation between the two sections of the pipe, we have:

\[ P_1 + \frac{1}{2} \rho V_1^2 = P_2 + \frac{1}{2} \rho V_2^2 \]

Rearranging gives:

\[ P_1 - P_2 = \frac{1}{2} \rho V_2^2 - \frac{1}{2} \rho V_1^2 \]

Substituting the pressure difference:

\[ 4500 = \frac{1}{2} \rho (V_2^2 - V_1^2) \]

### Step 4: Substitute \( V_2 \) in the equation
Substituting \( V_2 = 2 V_1 \):

\[ 4500 = \frac{1}{2} \rho ((2 V_1)^2 - V_1^2) \]

This simplifies to:

\[ 4500 = \frac{1}{2} \rho (4 V_1^2 - V_1^2) = \frac{1}{2} \rho (3 V_1^2) \]

### Step 5: Solve for \( V_1^2 \)
Now, substituting the value of \( \rho \):

\[ 4500 = \frac{1}{2} \times 750 \times 3 V_1^2 \]

Calculating the right-hand side:

\[ 4500 = \frac{2250}{2} V_1^2 \]

\[ 4500 = 1125 V_1^2 \]

Now, solving for \( V_1^2 \):

\[ V_1^2 = \frac{4500}{1125} = 4 \]

Thus,

\[ V_1 = 2 \, \text{m/s} \]

### Step 6: Calculate the volume flow rate
Using the volume flow rate formula:

\[ Q = A_1 V_1 \]

Substituting the values:

\[ Q = 1.2 \times 10^{-2} \times 2 = 2.4 \times 10^{-2} \, \text{m}^3/\text{s} \]

### Step 7: Convert to the required format
To express the flow rate in \( \text{m}^3/\text{s} \) in the form of \( xx \times 10^{-3} \):

\[ Q = 2.4 \times 10^{-2} = 24 \times 10^{-3} \, \text{m}^3/\text{s} \]

### Final Answer
The rate of flow of liquid is \( 24 \times 10^{-3} \, \text{m}^3/\text{s} \).

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