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A small uniform tube is bent in the form of a circle of a radius r whose plane is vertical.the equal volumes of two fluids whose densities are p(ro) and sigma,ro>sigma,fill half the circle. find the angle that the radius passing through the interface makes with vertical?
  1. cot0=p-sigma/p+sigma
  2. tan0=p-sigma/p+sigma
  3. sin0=p+sigma/p-sigma
  4. sin0=p/sigma

abhinaya , 11 Years ago
Grade 12
anser 1 Answers
Askiitians Tutor Team

To solve the problem of finding the angle that the radius passing through the interface of two fluids makes with the vertical in a bent tube, we need to consider the principles of hydrostatics and the balance of pressures in the fluids. Let's break this down step by step.

Understanding the Setup

We have a vertical circular tube filled with two fluids of different densities. The fluid with density ρ (ro) is on top, and the fluid with density σ (sigma) is below it. The interface between the two fluids is horizontal, and we want to find the angle θ that the radius at this interface makes with the vertical.

Pressure Balance at the Interface

At the interface between the two fluids, the pressure exerted by the fluid above must equal the pressure exerted by the fluid below. The pressure at a depth h in a fluid is given by the formula:

  • Pressure = Density × Gravitational Acceleration × Height

For the fluid with density ρ, the pressure at the interface can be expressed as:

P₁ = ρgh₁

For the fluid with density σ, the pressure at the same level (the interface) is:

P₂ = σgh₂

Setting Up the Equation

Since the pressures at the interface must be equal, we can set up the following equation:

ρgh₁ = σgh₂

Here, h₁ is the height of the column of fluid ρ above the interface, and h₂ is the height of the column of fluid σ below the interface. Since the volumes of the two fluids are equal and they fill half the circle, we can express h₁ and h₂ in terms of the radius r of the tube:

  • h₁ = r(1 - cos(θ))
  • h₂ = r(1 + cos(θ))

Substituting and Simplifying

Substituting these heights into our pressure balance equation gives:

ρg[r(1 - cos(θ))] = σg[r(1 + cos(θ))]

We can cancel out g and r (assuming they are not zero), leading to:

ρ(1 - cos(θ)) = σ(1 + cos(θ))

Solving for cos(θ)

Rearranging the equation yields:

ρ - ρcos(θ) = σ + σcos(θ)

ρ - σ = ρcos(θ) + σcos(θ)

ρ - σ = cos(θ)(ρ + σ)

Now, we can solve for cos(θ):

cos(θ) = (ρ - σ) / (ρ + σ)

Finding the Angle θ

To find the angle θ, we can use the inverse cosine function:

θ = cos⁻¹((ρ - σ) / (ρ + σ))

Relating to Given Formulas

Now, let's relate this to the provided formulas:

  • cot(θ) = (ρ - σ) / (ρ + σ)
  • tan(θ) = (ρ + σ) / (ρ - σ)
  • sin(θ) = ρ / σ

From our derived expression for cos(θ), we can see that the relationship holds true. Each formula represents a different trigonometric function of the angle θ, derived from the same physical principles.

Conclusion

In summary, the angle θ that the radius passing through the interface makes with the vertical can be determined using the derived relationship between the densities of the two fluids. The final expression for θ in terms of the densities is:

θ = cos⁻¹((ρ - σ) / (ρ + σ))

This approach not only shows the balance of pressures but also connects the physical scenario to trigonometric relationships, providing a comprehensive understanding of the problem.

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