Derive the equation of velocity of efflux ?
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Suppose that a liquid is taken in a container having a large cross-sectional area A1. There is also a small hole of cross-sectional area A2 on the wall of the container.
Suppose that the velocity of efflux of the liquid out of the container is v2 , The velocity of a liquid particle at the surface, at that instant, is v1 . The height of the liquid column at any instant from the position of the hole to the upper end of the tank is h, and the atmospheric pressure is Po
From the principle of continuity, we get
v1A1 = v2A2 ….(1)
Applying Bernoulli’s principle at the free end outside the hole and also at the surface, we get Po + 1/2 d v1 ^2 + dgh = Po + 1/2 d v2 ^2. ….(2)
Therefore, from equations (1) and (2), we get,
V2^2 = 2gh/ [1-(A1/A2)^2] …(3)
If A2<<< A1 , then,
V2 = under root 2gh ……(4)
Equations (3) and (4) give the velocity of efflux of a liquid flowing out of a small hole in a tank.
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Answer:
Page no. 260 of ncert gives the derivation
Explanation:
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