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Question: A rectangular tube of uniform cross-section has three liquids of densities \({\rho}_{1}\), \({\rho}_...

A rectangular tube of uniform cross-section has three liquids of densities ρ1{\rho}_{1}, ρ2{\rho}_{2} and ρ3{\rho}_{3}. Each liquid column has length l equal to the length of sides of the equilateral triangle. Find the length x of the liquid density ρ1{\rho}_{1} in the horizontal limb of the tube, if the triangular tube is kept in the vertical plane.

Explanation

Solution

Consider the base of the equilateral triangle. Find the pressure at one end. And then, the pressure at the other end of the base. As it has uniform cross-section and the length of each side is the same, pressure at both the ends of the base is the same. Thus, equate the expression for pressure at both the ends. Then, substitute the value of heights in the obtained expression. Thus, obtain the value of length x.

Complete answer:
Let the base of the triangle i.e. the horizontal surface be AB.
Pressure is given by,
P=ρghP=\rho gh …(1)
Where, ρ\rho is the density of liquid
h is the height
Using equation. (1), pressure at point A will be,
PA=ρ1gh1+ρ2gh2{ P }_{ A }={ \rho }_{ 1 }g{ h }_{ 1 }+{ \rho }_{ 2 }g{ h }_{ 2 }
Similarly, pressure at point B will be,
PB=ρ2gh2′+ρ3gh3{ P }_{ B }={ \rho }_{ 2 }g{ h }_{ 2 }^{ ' }+{ \rho }_{ 3 }g{ h }_{ 3 }
But, the pressure at point is equal to that at point B.
⇒PA=PB\Rightarrow { P }_{ A }={ P }_{ B } …(2)
Substituting the values in above expression we get,
ρ1gh1+ρ2gh2=ρ2gh2′+ρ3gh3{\rho }_{ 1 }g{ h }_{ 1 }+{ \rho }_{ 2 }g{ h }_{ 2 }={ \rho }_{ 2 }g{ h }_{ 2 }^{ ' }+{ \rho }_{ 3 }g{ h }_{ 3 }
ρ1h1+ρ2h2=ρ2h2′+ρ3h3{ \rho }_{ 1 }{ h }_{ 1 }+{ \rho }_{ 2 }{ h }_{ 2 }={ \rho }_{ 2 }{ h }_{ 2 }^{ ' }+{ \rho }_{ 3 }{ h }_{ 3 } …(3)
Substituting values in the equation. (3) we get,
ρ1(l−x)+ρ2x=ρ2(l−x)+ρ3x{ \rho }_{ 1 }(l-x)+{ \rho }_{ 2 }x={ \rho }_{ 2 }(l-x)+{ \rho }_{ 3 }x
⇒ρ1l−ρ1x+ρ2x=ρ2l−ρ2x+ρ3x\Rightarrow { \rho }_{ 1 }l-{ \rho }_{ 1 }x+{ \rho }_{ 2 }x={ \rho }_{ 2 }l-{ \rho }_{ 2 }x+{ \rho }_{ 3 }x
⇒xρ2−xρ1+xρ2−xρ3=lρ2−lρ1\Rightarrow x{ \rho }_{ 2 }-x{ \rho }_{ 1 }+x{ \rho }_{ 2 }-x{ \rho }_{ 3 }=l{ \rho }_{ 2 }-l{ \rho }_{ 1 }
⇒x(ρ2−ρ1+ρ2−ρ3)=l(ρ2−ρ1)\Rightarrow x{ (\rho }_{ 2 }-{ \rho }_{ 1 }+{ \rho }_{ 2 }-{ \rho }_{ 3 })=l({ \rho }_{ 2 }-{ \rho }_{ 1 })
⇒x=l(ρ2−ρ1)(ρ2−ρ1+ρ2−ρ3)\Rightarrow x=\dfrac { l(\rho _{ 2 }-{ \rho }_{ 1 }) }{ { (\rho }_{ 2 }-{ \rho }_{ 1 }+{ \rho }_{ 2 }-{ \rho }_{ 3 }) }
⇒x=l(ρ2−ρ1)(2ρ2−ρ1−ρ3)\Rightarrow x=\dfrac { l(\rho _{ 2 }-{ \rho }_{ 1 }) }{ { (2\rho }_{ 2 }-{ \rho }_{ 1 }-{ \rho }_{ 3 }) }
Therefore, length x is l(ρ2−ρ1)(2ρ2−ρ1−ρ3)\dfrac { l(\rho _{ 2 }-{ \rho }_{ 1 }) }{ { (2\rho }_{ 2 }-{ \rho }_{ 1 }-{ \rho }_{ 3 }) }

Note: From equation. (1), it can be inferred that pressure is directly proportional to density of the liquid. As the density of liquid or height increases, pressure on the liquid also increases and when density decreases, pressure decreases. When the pressure increases, molecules of substance come closer and thus density increases. Density is inversely proportional to temperature. As the temperature increases, liquid expands. Thus, the density decreases.