
Pressure at the point A is the same on both sides of the tube,
ρ1gr(cosθ−sinθ)=ρ2gr(sinθ+cosθ)
⇒ρ2ρ1=cosθ−sinθsinθ+cosθ=1−tanθtanθ+1
⇒ρ1−ρ1tanθ=ρ2+ρ2tanθ
⇒(ρ1+ρ2)tanθ=ρ1−ρ2
θ=tan−1(ρ1+ρ2ρ1−ρ2)
JEE Main 2018 — Physics Mechanics
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:
Held on 15 Apr 2018 · Verified 6 Jul 2026.
θ=tan−12π(ρ1−ρ2ρ1+ρ2)
θ=tan−1[(ρ1+ρ2ρ1−ρ2)]
θ=tan−12π(ρ1ρ2)
θ=tan−1π(ρ2ρ1)
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