Physics, asked by manishapattnaik8301, 1 year ago

Water and kerosene are filled in two identical cylindrical vessels both vessel have small hole in two identical cyliendrical rod

Answers

Answered by Anonymous
5
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

Answered by rtarunraj29
0

Answer:

The correct option is A

v1=v2

Given,

Speed of water

(

ρ=1000kg/m³

coming out of hole = v1

Speed of kerosene

(

ρ

=

800

kg/m

3

)

coming out of hole

=

v

2

We know that for small heights of vessels, speed of efflux is independent of density of fluid.

Speed of efflux for small holes,

v=√2gh

which is independent of density.

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