The area of cross-section of a large tank is 0.5 m². It has a narrow opening near the bottom having area of cross-section 1 cm². A load of 25 kg is applied on the water at the top in the tank. Neglecting the speed of water in the tank, the velocity of the water, coming out of the opening at the time when the height of water level in the tank is 40 cm above the bottom, will be ___________ cms⁻¹. [Take g = 10 ms⁻²]

Solution

Physics sir · 4 steps

The load only adds extra pressure on the water surface, so put it into Bernoulli with the water column.

1. Pressure from load. The 25 kg load presses on the whole top surface of area 0.5 m². The extra pressure is Δ P = (25 × 10)/0.5 = 500 Pa.
2. Apply Bernoulli. Compare the top surface (pressure P₀ + Δ P, height h, speed taken as zero) with the opening (pressure P₀, height 0, speed v). The atmospheric pressure P₀ cancels from both sides.

   Δ P + ρ g h = 1/2ρ v² ⇒ v = √2((Δ P)/ρ + g h)

3. Put in numbers. With ρ = 1000 kg m⁻³ and h = 0.4 m: (Δ P)/ρ = 0.5 and gh = 4. So v = √(2 × 4.5) = √9 = 3 m s⁻¹.
4. So. v = 3 m s⁻¹ = 300 cm s⁻¹.

The common slip: 283. Leaving out the load and using only v = √2gh gives √8 m s⁻¹, about 283, not 300.

The idea to keep. Anything pushing on the free surface, like a load or a gas above the water, just adds to the pressure there. Convert it to pressure and add it to ρ g h in Bernoulli.

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