Physics · Mechanical Properties Of Fluids · NEET
Yes. Torricelli's law gives v = sqrt(2gh), which is exactly the speed a ball reaches after falling from rest through height h (from v squared = 2gh). The liquid at the hole behaves as if it dropped freely from the top surface. This is why the law is so easy to remember: efflux speed = free-fall speed for the same height h.
h is the vertical distance from the top liquid surface DOWN to the hole, not the total tank height and not the height of the hole above the ground. If a tank is filled to 2 m and the hole is 0.5 m above the base, then h = 2 - 0.5 = 1.5 m. Many students wrongly use the hole's height above the ground; always measure h downward from the free surface.
Because the tank cross-section is much larger than the hole, the top surface barely moves (v at top is taken as 0). Bernoulli's equation then links only the depth h to the speed, and the areas cancel out. Area matters for the RATE of flow Q = a*v (volume per second), not for the speed of each water particle. Speed depends only on h and g.
Then you use the full formula v = sqrt(2(P - Pa)/rho + 2gh), where P is the gas pressure above the liquid and Pa is atmospheric pressure at the hole. Extra pressure pushes the liquid out faster. Only when the tank is OPEN (P = Pa) does this reduce to the simple v = sqrt(2gh).
No. For an open tank v = sqrt(2gh) has no density term, so water and a denser liquid leave with the same speed for the same depth h. Density cancels because both the driving weight and the inertia scale with rho. Density only matters when there is an extra pressure P above the surface, through the 2(P - Pa)/rho term.
A small hole of cross-sectional area 2 mm^2 is present near the bottom of a fully filled open tank of height 2 m. Taking g = 10 m/s^2, the rate of flow of water through the open hole would be nearly:
Try the real previous-year questions from this chapter — each with the answer and a full solution.
The speed of liquid flowing out of a small hole in an open tank is v = sqrt(2gh), where h is the depth of the hole below the free surface.
v = sqrt(2(P - Pa)/rho + 2gh), where P is the gas pressure above the liquid, Pa is atmospheric pressure, rho is the liquid density, and h is the depth of the hole below the surface.
Multiply the efflux speed by the hole area: Q = a*v = a*sqrt(2gh). Only the RATE depends on area; the speed of each particle does not.
Yes. As the liquid level drops, h decreases, so v = sqrt(2gh) becomes smaller and the water shoots out more slowly over time.
Yes. Applying Bernoulli's equation between the top surface (speed approx 0) and the hole, and taking both open to the atmosphere, gives v = sqrt(2gh).
