Consider a cylindrical tank of radius 1 m is filled with water. The top surface of water is at 15 m from the bottom of the cylinder. There is a hole on the wall of cylinder at a height of 5 m from the bottom. A force of 5 × 10⁵ N is applied an the top surface of water using a piston. The speed of ifflux from the hole will be : (given atmospheric pressure P_A=1.01 × 10⁵ Pa, density of water ρ_W=1000 kg / m³ and gravitational acceleration g=10 m / s² )

Solution

Physics sir · 4 steps

Piston pressure plus 10 m of water, minus the air outside, becomes 1/2ρ v² at the hole.

1. Pressure at the top. The piston pushes on the water with 5 × 10⁵ N over an area π (1)² = π m². So the pressure on the top surface is (5 × 10⁵)/π = 1.59 × 10⁵ Pa. The hole opens to the air, so the pressure there is P_A = 1.01 × 10⁵ Pa.
2. Apply Bernoulli. Join the top surface to the hole with one streamline. The tank is wide, so the water at the top moves very slowly and its speed can be taken as zero. The top is 10 m above the hole.

   F/A + ρ g h = P_A + 1/2ρ v²

3. Put in numbers. 1/2ρ v² = 1.5915 × 10⁵ - 1.01 × 10⁵ + 1000 × 10 × 10 = 1.5815 × 10⁵ Pa. Then v² = (2 × 1.5815 × 10⁵)/1000 = 316.3 m² s⁻².
4. So. v = √316.3 ≈ 17.8 m s⁻¹, which is option (c).

The idea to keep. For efflux, the top pressure minus the outside pressure, plus ρ g h, equals 1/2ρ v². Measure h from the free surface down to the hole, not from the bottom.

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