1. Given data

• Density of liquid: $\rho = 750\ \text{kg m}^{-3}$
• Area of wide section: $A_1 = 1.2\times 10^{-2}\ \text{m}^2$
• Area of narrow section: $A_2 = \frac{A_1}{2}$
• Pressure difference: $P_1 - P_2 = 4500\ \text{Pa}$
• Pipe is horizontal, so heights are same.

We need the volume flow rate $Q$.

2. Use continuity equation

For incompressible flow, $A_1 v_1 = A_2 v_2 = Q$

Since $A_2 = \frac{A_1}{2},$ we get $v_2 = \frac{A_1}{A_2}v_1 = 2v_1.$

3. Apply Bernoulli’s equation

For horizontal flow, $P_1 + \frac{1}{2}\rho v_1^2 = P_2 + \frac{1}{2}\rho v_2^2$

So, $P_1 - P_2 = \frac{1}{2}\rho\left(v_2^2 - v_1^2\right)$

Substitute $v_2 = 2v_1$: $4500 = \frac{1}{2}(750)\left((2v_1)^2 - v_1^2\right)$ $4500 = 375(4v_1^2 - v_1^2)$ $4500 = 375(3v_1^2)$ $4500 = 1125v_1^2$

Thus, $v_1^2 = \frac{4500}{1125} = 4$ $v_1 = 2\ \text{m s}^{-1}$

Then, $v_2 = 4\ \text{m s}^{-1}$

4. Find flow rate

$Q = A_1 v_1$ $Q = (1.2\times 10^{-2})(2)$ $Q = 2.4\times 10^{-2}\ \text{m}^3\text{s}^{-1}$

Now write in the required form: $Q = 24\times 10^{-3}\ \text{m}^3\text{s}^{-1}$

So the required integer is: $\boxed{24}$

5. Comparison with stored answer

Stored correct answer = $24$

Our derived answer also = $24$. Hence, they agree.