Question

A heavy liquid with a density 11 g/cm3

is
poured into a U-tube as shown in the lefthand figure below. The left-hand arm of the
tube has a cross-sectional area of 11.6 cm2
,
and the right-hand arm has a cross-sectional
area of 5.74 cm2
. A quantity of 109 g of a
light liquid with a density 1.4 g/cm3
is then
poured into the right-hand arm as shown in
the right-hand figure below.
h1
h2
11.6 cm2
5.74 cm2
heavy liquid
11 g/cm3
L
11.6 cm2
5.74 cm2
light liquid
1.4 g/cm3
Determine the height L of the light liquid
in the column in the right arm of the U-tube,
as shown in the second figure above.
Answer in units of cm.
008 (part 2 of 2) 10.0 points
If the density of the heavy liquid is 11 g/cm3
,
by what height h1 does the heavy liquid rise
in the left arm?
Answer in units of cm

Answer 1

h₁ = 0.605 cm

U welcome x

Also I'm only 80% sure I'm right

Step-by-step explanation:

Part One is a geometry problem. First, find the volume of the light liquid from the mass and density.V = 92.5 g / (1.2 g/cm³)V = 77.1 cm³Next, find the height of the liquid from the volume and area.77.1 cm³ = 5.16 cm² × LL = 14.9 cmFor Part Two, we first use geometry to find h₂ in terms of h₁. The volume decrease of the heavy liquid in the right tube equals the volume increase of the heavy liquid in the left tube.V₂ = V₁A₂ h₂ = A₁ h₁5.16 cm² × h₂ = 12.2 cm² × h₁h₂ = 2.36 h₁Next, we use physics. The pressure at the bottom of the right tube equals the pressure at the bottom of the left tube.P₂ = P₁ρ₂ g (h₂ + h₁) = ρ₁ g Lρ₂ (h₂ + h₁) = ρ₁ L(8.8 g/cm³) (2.36 h₁ + h₁) = (1.2 g/cm³) (14.9 cm)h₁ = 0.605 cm