Consider a water tank as shown in the figure. It's cross-sectional area is 0.4 m². The tank has an opening B near the bottom whose cross-section area is 1 cm². A load of 24 kg is applied on the water at the top when the height of the water level is 40 cm above the bottom, the velocity of water coming out the opening B is v ms⁻¹.
The value of v, to the nearest integer, is ___________. [Take value of g to be 10 ms⁻²]

Solution

Physics sir · 4 steps

The load just adds extra pressure on top; spread its weight over the whole tank area, then use Bernoulli.

1. Extra pressure. The 24 kg load presses on the full water surface of area 0.4 m². It adds a pressure mg/A = (24 × 10)/0.4 = 600 Pa on top of the atmosphere.
2. Top speed negligible. By continuity, v_A = v × 10⁻⁴/0.4 = 2.5 × 10⁻⁴ v. So the water surface at A is almost at rest, and its kinetic energy term can be dropped.

   P₀ + mg/A + ρ g h = P₀ + 1/2ρ v²

3. Put in numbers. ρ g h = 1000 × 10 × 0.4 = 4000 Pa, so 600 + 4000 = 1/2 × 1000 × v². This gives v² = 4600/500 = 9.2 m² s⁻².
4. So. v = √9.2 ≈ 3.03 m s⁻¹, which rounds to 3.

The common slip: 69. Dividing the 240 N load by the opening area 1 cm² instead of the tank area 0.4 m² gives 2.4 × 10⁶ Pa and a speed of about 69, not 3.

The idea to keep. A weight or piston on a liquid surface acts like extra pressure mg/A, using the area it actually presses on. Add it to ρ g h in Bernoulli and P₀ cancels from both sides.

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