Question

1 T
. An incompressible liquid is flowing through a
horizontal tube in streamline manner as shown in
figure. The value of his in meters)
> 5 m/s
3 m/s
(1) 1
(3) 3.8
(2) 0.8
(4) 4.2​

Answer 1

SOLUTION.

A liquid flowing through an horizontal tube follows the rule.....

1/2(ρ)v² + P + (ρ)gh = constant

where P is the pressure at that point

1/2(ρ)v² is the Kinetic energy at that point

(ρ)gh is the potential energy at that point

For Potential energy you can choose any reference point.

For your question I choose just below both the tube as the reference point. so that potential energy of second point become zero.

For first point.

1/2(ρ)(3)² + P(1) + (ρ)gh = Constant

For Second point.

1/2(ρ)(5)² + P(2) = Constant

From those eq we get

1/2(ρ)(3)² + P(1) + (ρ)gh = 1/2(ρ)(5)² + P(2)

(1/2)ρ[(3)²-(5)²] = [P(2) -P(1)] - ρgh

Now this is my fav part to explain...

As we know the pressure difference between two point can be written as ρg(∆h) where ∆h is the difference of height.

P(1) > P(2)

-(1/2)×ρ×16 = - ρgh

8ρ = gh

Height = 0.8

#answerwithquality

#BAL