1. Given data

• Pressure difference between sections $A$ and $B$: $P_A - P_B = 700\ \text{N m}^{-2}$
• Density of water: $\rho = 1000\ \text{kg m}^{-3}$
• Area at $A$: $A_A = 40\ \text{cm}^2 = 40 \times 10^{-4} = 4 \times 10^{-3}\ \text{m}^2$
• Area at $B$: $A_B = 20\ \text{cm}^2 = 20 \times 10^{-4} = 2 \times 10^{-3}\ \text{m}^2$

Since the tube is horizontal, gravitational potential terms are same.

2. Use continuity equation

Let flow speed at $A$ be $v_A$ and at $B$ be $v_B$.

For incompressible flow, $A_A v_A = A_B v_B$

So, $4 \times 10^{-3} v_A = 2 \times 10^{-3} v_B$ $v_B = 2v_A$

3. Apply Bernoulli's equation

For horizontal flow, $P_A + \frac{1}{2}\rho v_A^2 = P_B + \frac{1}{2}\rho v_B^2$

Therefore, $P_A - P_B = \frac{1}{2}\rho\left(v_B^2 - v_A^2\right)$

Substitute values: $700 = \frac{1}{2}(1000)\left((2v_A)^2 - v_A^2\right)$ $700 = 500(4v_A^2 - v_A^2)$ $700 = 500(3v_A^2)$ $700 = 1500 v_A^2$ $v_A^2 = \frac{700}{1500} = \frac{7}{15}$ $v_A \approx 0.683\ \text{m/s}$

Then, $v_B = 2v_A \approx 1.366\ \text{m/s}$

4. Find volume flow rate

Volume flow rate, $Q = A_A v_A$ $Q = 4 \times 10^{-3} \times 0.683$ $Q = 2.732 \times 10^{-3}\ \text{m}^3/\text{s}$

Convert to $\text{cm}^3/\text{s}$: $1\ \text{m}^3 = 10^6\ \text{cm}^3$ $Q = 2.732 \times 10^{-3} \times 10^6$ $Q = 2732\ \text{cm}^3/\text{s}$

This is closest to $\boxed{2720\ \text{cm}^3/\text{s}}$

5. Evaluate options

• A: $1810\ \text{cm}^3/\text{s}$ — incorrect
• B: $2420\ \text{cm}^3/\text{s}$ — incorrect
• C: $3020\ \text{cm}^3/\text{s}$ — incorrect
• D: $2720\ \text{cm}^3/\text{s}$ — correct

So the correct option is D.