A pressure-pump has a horizontal tube of cross sectional area 10 cm² for the outflow of water at a speed of 20 m / s. The force exerted on the vertical wall just in front of the tube which stops water horizontally flowing out of the tube, is :
[given: density of water =1000 kg / m³]

Solution

Physics sir · 3 steps

The wall stops 20 kg of water every second, and each kilogram loses 20 m/s of momentum.

1. Mass per second. Convert the area first: 10 cm² = 10 × 10⁻⁴ m² = 10⁻³ m². The mass of water hitting the wall each second is ρ A v = 1000 × 10⁻³ × 20 = 20 kg s⁻¹.
2. Momentum lost. The wall brings the water to rest horizontally, so each kilogram loses momentum v = 20 m s⁻¹. Force equals the rate of change of momentum.

   F = dm/dt v = ρ A v² = 1000 × 10⁻³ × 20² N

3. So. F = 1 × 400 = 400 N, which is option (d).

The idea to keep. For a jet hitting a wall and stopping, force is mass flow rate times speed: F = ρ A v². If the water bounced straight back, the momentum change would double.

Answer from the NTA final answer key. Solution by PrepWiser.