A horizontal tube has different cross sections at points `A` and `B`. The areas of cross section are `a_(1)` and `a_(2)` respectively, and pressures at these points are `p_(1)=rhogh_(1)` and `p_(2)=rhogh_(2)` where `rho` is the density of liquid flowing in the tube and `h_(1)` and `h_(2)` are heights of liquid columns in vertical tubes connected at `A` and `B`. If `h_(1)-h_(2)=h`, then the flow rate of the liquid in the horizontal tube is

Options

A. `a_(1)a_(2)sqrt((2gh)/(a_(1)^(2)-a_(2)^(2)))`
B. `a_(1)a_(2)sqrt((2g)/(h(a_(1)^(2)-a_(2)^(2))))`
C. `a_(1)a_(2)sqrt(((a_(1)^(2)+a_(2)^(2))h)/(2g(a_(1)^(2)-a_(2)^(2))))`
D. `(2a_(1)a_(2)gh)/sqrt(a_(1)^(2)-a_(2)^(2))`

Explanation

To find the flow rate of the liquid in the horizontal tube with different cross sections at points A and B, we can follow these steps:

### Step 1: Understand the Given Information
We are given:
- Cross-sectional areas at points A and B: \( A_1 \) and \( A_2 \)
- Pressures at points A and B: \( p_1 = \rho g h_1 \) and \( p_2 = \rho g h_2 \)
- The difference in height of the liquid columns: \( h = h_1 - h_2 \)

### Step 2: Apply Bernoulli's Equation
According to Bernoulli's principle, for incompressible flow, we have:
\[ p_1 + \frac{1}{2} \rho v_1^2 + \rho g h_1 = p_2 + \frac{1}{2} \rho v_2^2 + \rho g h_2 \]
Substituting the pressures:
\[ \rho g h_1 + \frac{1}{2} \rho v_1^2 = \rho g h_2 + \frac{1}{2} \rho v_2^2 \]

### Step 3: Rearranging the Equation
Rearranging the equation gives:
\[ \rho g h_1 - \rho g h_2 = \frac{1}{2} \rho v_2^2 - \frac{1}{2} \rho v_1^2 \]
Factoring out \( \rho \):
\[ \rho g (h_1 - h_2) = \frac{1}{2} \rho (v_2^2 - v_1^2) \]
Since \( h_1 - h_2 = h \):
\[ \rho g h = \frac{1}{2} \rho (v_2^2 - v_1^2) \]

### Step 4: Canceling Out Density
Dividing through by \( \rho \):
\[ g h = \frac{1}{2} (v_2^2 - v_1^2) \]
Multiplying through by 2:
\[ 2gh = v_2^2 - v_1^2 \]

### Step 5: Expressing Velocity
Rearranging gives:
\[ v_2^2 = v_1^2 + 2gh \]

### Step 6: Using Continuity Equation
From the continuity equation, we know:
\[ A_1 v_1 = A_2 v_2 \]
Thus, we can express \( v_1 \) in terms of \( v_2 \):
\[ v_1 = \frac{A_2}{A_1} v_2 \]

### Step 7: Substitute \( v_1 \) into the Velocity Equation
Substituting \( v_1 \) into the equation for \( v_2^2 \):
\[ v_2^2 = \left(\frac{A_2}{A_1} v_2\right)^2 + 2gh \]
This simplifies to:
\[ v_2^2 = \frac{A_2^2}{A_1^2} v_2^2 + 2gh \]

### Step 8: Rearranging to Solve for \( v_2 \)
Rearranging gives:
\[ v_2^2 \left(1 - \frac{A_2^2}{A_1^2}\right) = 2gh \]
Factoring out \( v_2^2 \):
\[ v_2^2 \left(\frac{A_1^2 - A_2^2}{A_1^2}\right) = 2gh \]
Thus:
\[ v_2^2 = \frac{2gh A_1^2}{A_1^2 - A_2^2} \]

### Step 9: Finding the Volume Flow Rate
The volume flow rate \( Q \) is given by:
\[ Q = A_2 v_2 \]
Substituting for \( v_2 \):
\[ Q = A_2 \sqrt{\frac{2gh A_1^2}{A_1^2 - A_2^2}} \]

### Final Expression for Flow Rate
Thus, the flow rate of the liquid in the horizontal tube is:
\[ Q = A_2 \sqrt{\frac{2gh A_1^2}{A_1^2 - A_2^2}} \]