A thin uniform tube is bent into a circle of radius r in the vertical plane. Equal volumes of two immiscible liquids, whose densities are \rho_1 and \rho_2(\rho_1 > \rho_2) fill half the circle. The angle θ between the radius vector passing through the common interface and the vertical is: [Main 15 April 2018 S1]

Correct Answer is : θ=tan−1[π2(ρ1−ρ2ρ1+ρ2)]

\theta=\text{\tan}^{-1}\pi(\rho_1/\rho_2)

\theta=\text{\tan}^{-1}\pi/2(\rho_1/\rho_2)

\theta=\text{\tan}^{-1}[\pi/2(\frac{\rho_1-\rho_2}{\rho_1+\rho_2})]

\theta=\text{\tan}^{-1}[\pi/2(\frac{\rho_1+\rho_2}{\rho_1-\rho_2})]

JEE Main 15 Apr 2018 Shift 1 Solved Paper

Section: Physics

Question : 9 of 90

Marks: +1, -0

A thin uniform tube is bent into a circle of radius r in the vertical plane. Equal volumes of two immiscible liquids, whose densities are ρ1 and ρ2(ρ1>ρ2) fill half the circle. The angle θ between the radius vector passing through the common interface and the vertical is:

[Main 15 April 2018 S1]

θ=tan−1π(
ρ1
ρ2
)
θ=tan−1
π
2
(
ρ1
ρ2
)
θ=tan−1[
π
2
(
ρ1−ρ2
ρ1+ρ2
)]
θ=tan−1[
π
2
(
ρ1+ρ2
ρ1−ρ2
)]

Validate

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