A wide vessel with a small hole in the bottom is filled with water and kerosene
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Consider a wide vessel with a hole in the bottom as shown in the diagram. As the density of the water is greater than the density of the kerosene, it will collect at the bottom.
Let the height of water and kerosene be h1 and h2respectively and their corresponding densities be ρ1 and ρ2
The pressure due to water and kerosene level together would be h1 ρ1 g + h2 ρ2 g
As the water flows from the hole at the bottom of the vessel, there is a kinetic energy involved which is ½ ρ1 v2
Hence, h1 ρ1 g + h2 ρ2 g = ½ ρ1 v2
Divide throughout by ρ1 , we get h1 g + h2 (ρ2 / ρ1) g = ½ v2
Hence, v = [ 2 (h1 + h2 (ρ2 / ρ1) ) g ]1/2
Substituting the values of h1 = 30 * 10-2 m, h2 = 20 * 10-2 m, (ρ2 / ρ1)= 0.80
V = (9.016)1/2 = approximately 3 m/s
Let the height of water and kerosene be h1 and h2respectively and their corresponding densities be ρ1 and ρ2
The pressure due to water and kerosene level together would be h1 ρ1 g + h2 ρ2 g
As the water flows from the hole at the bottom of the vessel, there is a kinetic energy involved which is ½ ρ1 v2
Hence, h1 ρ1 g + h2 ρ2 g = ½ ρ1 v2
Divide throughout by ρ1 , we get h1 g + h2 (ρ2 / ρ1) g = ½ v2
Hence, v = [ 2 (h1 + h2 (ρ2 / ρ1) ) g ]1/2
Substituting the values of h1 = 30 * 10-2 m, h2 = 20 * 10-2 m, (ρ2 / ρ1)= 0.80
V = (9.016)1/2 = approximately 3 m/s
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Since, the density of water is greater than that of kerosene oil, it will collect at the bottom. Now, pressure due to water level equals h1ρ1g and pressure due to kerosene oil level equals h2ρ2g. So, net pressure becomes h1ρ1g + h2ρ2g.Read more on Sarthaks.com - https://www.sarthaks.com/216861/wide-vessel-with-small-hole-the-bottom-filled-with-water-and-kerosene-neglecting-viscosity
From Bernoulli’s theorem, this pressure energy will be converted into kinetic energy while flowing through the whole A.
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