A rectangular tube of uniform cross …

Q : A rectangular tube of uniform cross section has three liquids of densities ρ₁​, ρ₂ and ρ₃​. Each liquid column has length l equal to length of sides of the equilateral triangle. Find the length x of the liquid density ρ₁ in the horizontal limb of the tube, if the triangular tube is kept in the vertical plane.

Solution:
At junction force applied by left to coloumn = force applied by light of coloumn.
(1 − x) cos30^o p₁​ gA + x cos30^o p₂ ​gA = (1 − x) cos30^o p₂ gA + x cos30^o p₃ ​gA
{tex}\Rightarrow{/tex} (1 − x)p₁ ​+ xp₂ = (1 − x)p₂ + xp₃
{tex}\Rightarrow{/tex} p₁1 − p₁x + xp₂ = 1p₂ − xp₂ + p_3​x
{tex}\Rightarrow{/tex} xp₂ − xp₁ ​+ xp₂ − xp₃ = 1p₂ − 1p₁
{tex}\Rightarrow{/tex} x(2p₂ − p₁ − p₃) = 1(p₂ − p₁)
x = {tex}\frac {1(p_2 − p_1)​}{2p_2 − p_1 − p_3}{/tex}