1. Given data

• Cross-sectional area of tube: $A = 10\,\text{cm}^2 = 10 \times 10^{-4}\,\text{m}^2 = 10^{-3}\,\text{m}^2$
• Speed of water: $v = 20\,\text{m/s}$
• Density of water: $\rho = 1000\,\text{kg/m}^3$

2. Find mass of water flowing per second

Mass flow rate is $\dot m = \rho A v$

Substitute the values: $\dot m = 1000 \times 10^{-3} \times 20 = 20\,\text{kg/s}$

3. Use rate of change of momentum

The water is moving horizontally with speed $20\,\text{m/s}$ and is brought to rest by the vertical wall.

So, change in velocity per second is from $20\,\text{m/s}$ to $0$.

Hence force on water: $F = \dot m \cdot v = 20 \times 20 = 400\,\text{N}$

Equivalently, $F = \rho A v^2$ $F = 1000 \times 10^{-3} \times (20)^2 = 400\,\text{N}$

4. Force on the wall

By Newton's third law, the force exerted by the water on the wall has the same magnitude: $F = 400\,\text{N}$

5. Option check

• A: $300\,\text{N}$ ❌
• B: $500\,\text{N}$ ❌
• C: $250\,\text{N}$ ❌
• D: $400\,\text{N}$ ✅

Therefore, the correct answer is D.