Physics, asked by PhysicsHelper, 1 year ago

Suppose the tube in the previous problem is kept vertical with A upward but the other conditions remain the same. The separation between the cross sections at A and B is 15/16 cm. Repeat parts (a), (b) and (c) of the previous problem. Take g = 10 m/s².

Answers

Answered by tiwaavi
4

See the Question given in that link, and then watch this solution.

   https://brainly.in/question/8681778  

Now, (a). We know volume rate of the flow does not depends upon the the height of the pipe, since pipe.

∴ Velocity of the fluid will remains unchanged.

∴ (a). At A,

Area × Velocity = 1 cm³/s.

4/100 cm² × Velocity = 1

∴ Velocity = 25 cm/seconds.

________________________

(b). At B,  

Area × Velocity = 1 cm³/s.

2/100 × Velocity = 1

Velocity = 50 cm/seconds.

________________________

(c). Pressure difference, will be affected, this is because, there will be the gravitational work also which will be included.

Using the Bernoulli's equation,

      Using the Bernoulli's Equation,

P₁ + h₁ρg + 1/2 ρv₁² = P₂ + h₂ρg + 1/2 ρv₂²

∴ P₁ - P₂ = 1/2 ρ(v₂² - v₁²) + ρg(h₂ - h₁)

∴ P₁ - P₂ = 1/2 × 1 × (50² - 25²) + 1 × 1000(-15/16)

[Note ⇒ Minus sign is used because h₁ - h₂ = 15/16 but h₂ - h₁ = -15/16.]

∴ P₁ - P₂ = 937.5 - 937.5

∴ P₁ - P₂ = 0

Hence, the pressure difference is zero.

Hope it helps.

Answered by bhuvna789456
1

The speed at A =25 cm/s

The speed at B =50 cm/s

The Bernoulli theorem  = zero

Explanation:

As the discharge is the same through the tube

The speeds at A and B would be the same, i.e. 1 cc / s.  

(a) The speed at A,  

v =1/0.04  cm/s =100/4  cm/s =25 cm/s    

(b) The speed at B,  

v' =1/0.02  cm/s =100/2  cm/s =50 cm/s  

(c) The Bernoulli theorem

Pₐ+ρgh+1/2 ρv² = Pᵦ+1/2 ρv'²  

Taking point B for height as a reference level.

Given that  

h=15/16  cm  

Pₐ-Pᵦ =1/2 ρ(v'²-v²) -ρgh  

=1/2×1000×(0.50²-0.25²) -1000×10×15/1600  N/m²  

=1/2×1000×0.75×0.25 -  1500/16  N/m²  

=93.75 - 93.75 N/m²  

= zero

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