A small uniform tube is bent into a circular tube of radius R and kept in the vertical plane- Equal volume of two liquids of densities - and - fill half of the tube as shown- - is the angle which the radius passing through the interface makes with the vertical- ThenB- -tan -1 -C- -tan -1 -D- -tan -1 -

The correct option is A θ nbsp;= nbsp;tan 1ρ nbsp; nbsp;σρ nbsp;+ nbsp;σPressure at A from both side must balance. Figure is self explanatory. σh2 ...

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Standard XII

Chemistry

Introduction

A small unifo...

Question

A small uniform tube is bent into a circular tube of radius $R$ and kept in the vertical plane. Equal volume of two liquids of densities $ρ$ and $σ (ρ > σ)$ fill half of the tube as shown . $θ$ is the angle which the radius passing through the interface makes with the vertical. Then

θ = {tan}^{- 1} (\frac{ρ - σ}{ρ + σ})

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θ = {tan}^{- 1} (\frac{σ - ρ}{ρ + σ})

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θ = {tan}^{- 1} (\frac{ρ}{ρ + σ})

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θ = {tan}^{- 1} (\frac{ρ}{ρ - σ})

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Solution

The correct option is A $θ = {tan}^{- 1} (\frac{ρ - σ}{ρ + σ})$
Pressure at A from both side must balance. Figure is self-explanatory.
$σ h_{2} g = ρ h_{1} g σ R \sqrt{2} sin (45 + θ) = ρ R (cos θ - sin θ] σ (cos θ + sin θ] = ρ (cos θ - sin θ] tan θ = \frac{ρ - σ}{ρ + σ}$

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A small uniform tube is bent into a circular tube of radius R and kept in the vertical plane. Equal volume of two liquids of densities ρ and σ (ρ>σ) fill half of the tube as shown . θ is the angle which the radius passing through the interface makes with the vertical. Then

The correct option is A θ = tan−1(ρ − σρ + σ)Pressure at A from both side must balance. Figure is self-explanatory. σh2g = ρh1gσ R√2sin(45 + θ) = ρR(cosθ − sinθ]σ (cosθ + sinθ] = ρ(cosθ − sinθ]tanθ = ρ − σρ + σ

A small uniform tube is bent into a circular tube of radius $R$ and kept in the vertical plane. Equal volume of two liquids of densities $ρ$ and $σ (ρ > σ)$ fill half of the tube as shown . $θ$ is the angle which the radius passing through the interface makes with the vertical. Then

The correct option is A $θ = {tan}^{- 1} (\frac{ρ - σ}{ρ + σ})$
Pressure at A from both side must balance. Figure is self-explanatory.
$σ h_{2} g = ρ h_{1} g σ R \sqrt{2} sin (45 + θ) = ρ R (cos θ - sin θ] σ (cos θ + sin θ] = ρ (cos θ - sin θ] tan θ = \frac{ρ - σ}{ρ + σ}$