Classical Mechanics - Fluid

An open-ended U-tube of uniform cross-sectional area contains water (density 1.0 gram/cm³) standing initially 20 cm from the bottom in each arm. An immiscible liquid of density 4.0 gr/cm³ is added to one arm until a layer 5 cm high forms. What is the ratio h₂/h₁ of the heights of the liquid in the two arms?

A. 3/1
B. 5/2
C. 2/1
D. 3/2
E. 1/1

(GR9677 #30)

Solution:

Given:
ρ_water = 1 gr/cm³
ρ_liquid = 4 gr/cm³
h_a = 5 cm

P₁ = P₂
ρ_lgh_a = ρ_wgh_b 
h_b = ρ_lh_a / ρ_w = (4)(5) / (1) = 20

h₁ = h + h_a  
h₂ = h + h_b 
-------------- −
h₁ − h₂ = h_a − h_b  = 5 − 20 = − 15 (Eq.1)

Initially, h₁ + h₂ = 20 + 20 = 40
Finally, h₁ + h₂ = 40 + 5 = 45 (Eq.2)

(Eq.1) → h₁ − h₂ = − 15
(Eq.2) → h₁ + h₂ = 45
------------------------------ +
2h₁ = − 15 + 45 = 30
h₁ = 15

h₂ = 45 − 15 = 30

h₂/h₁ = 30/15 = 2

Answer: C