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PART ONE
A heavy liquid with a density 8.8 g/cm^3 ispoured into a U-tube as shown in the lefthand figure below. The left-hand arm of the
tube has a cross-sectional area of 12.2 cm^2
and the right-hand arm has a cross-sectional
area of 5.16 cm^2
. A quantity of 92.5 g of a
light liquid with a density 1.2 g/cm^3
is then
poured into the right-hand arm as shown in
the right-hand figure below.
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.

PART TWO
If the density of the heavy liquid is 8.8 g/cm^3
by what height h1 does the heavy liquid rise
in the left arm?
Answer in units of cm.

PART ONE A heavy liquid with a density 88 gcm3 ispoured into a Utube as shown in the lefthand figure below The lefthand arm of the tube has a crosssectional are class=

Respuesta :

MathPhys

Answer:

14.9 cm

0.605 cm

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 cm³

Next, find the height of the liquid from the volume and area.

77.1 cm³ = 5.16 cm² × L

L = 14.9 cm

For 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 = V₁

A₂ h₂ = A₁ h₁

5.16 cm² × h₂ = 12.2 cm² × h₁

h₂ = 2.36 h₁

Next, we use physics.  The pressure at the bottom of the right tube equals the pressure at the bottom of the left tube.

P₂ = P₁

ρ₂ g (h₂ + h₁) = ρ₁ g L

ρ₂ (h₂ + h₁) = ρ₁ L

(8.8 g/cm³) (2.36 h₁ + h₁) = (1.2 g/cm³) (14.9 cm)

h₁ = 0.605 cm

The height L of the light  should be 14.9 cm

The heavy liquid rise should be by 0.605 cm

Calculation of the height & rise in the heavy liquid:

Here first we determine the volume of the light liquid from the mass and density.

So,

V = 92.5 g / (1.2 g/cm³)

V = 77.1 cm³

Now the height is

77.1 cm³ = 5.16 cm² × L

L = 14.9 cm

Now for the increase in the heavy liquid could be measured by considering the volume reduce of the heavy liquid in the right tube equivalent to the volume increase of the heavy liquid in the left tube.

So,

V₂ = V₁

A₂ h₂ = A₁ h₁

5.16 cm² × h₂ = 12.2 cm² × h₁

h₂ = 2.36 h₁

Now

The pressure at the bottom of the right tube equivalent to the pressure at the bottom of the left tube.

So,

P₂ = P₁

ρ₂ g (h₂ + h₁) = ρ₁ g L

ρ₂ (h₂ + h₁) = ρ₁ L

(8.8 g/cm³) (2.36 h₁ + h₁) = (1.2 g/cm³) (14.9 cm)

h₁ = 0.605 cm

learn more about the height here: https://brainly.com/question/13200371

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