Water is filled in a container upto height 3m. A small hole of area 'a' is punched in the wall of the container at a height 52.5 cm from the bottom. The cross sectional area of the container is A. If a//A=0.1 then v^2 is (where v is the velocity of water coming out of the hole)
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Given :
The height of water in the container = 3m
The height of hole from bottom = 52.5cm = 0.525m
(a/A) = 0.1
To Find :
The square of the velocity of the flux
Solution :
- we know the relation
Square of the velocity of flux(v²) = 2gh / [1 - (a/A)²]
- By substituting the values in the above relation
v² = 2gh / [1 - (a/A)²]
= 2×9.81×(3-0.525) / [1 - 0.1²]
= 50 m² / s²
The square of the velocity of the flux is 50 m² / s²
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