Question

Please help me solve this question that has two parts which are 7 and 8

Answer 1

ANSWERS

7) L = 9.38 cm

8) h₁ = 0.54 cm

Step-by-step explanation

7) First, we have to determine the volume of the light liquid. We know that the mass of the light liquid is 102 g and its density is 1.6 g/cm³. The relation between these two quantities and the volume of liquid is,

`\rho=(m)/(V)`

Solving for V,

`V=(m)/(\rho)`

Replace the known values and solve,

`V=(102g)/(1.6g/cm^3)=63.75cm^3`

The tube is in the shape of a cylinder, and the part where the light liquid is has a cross-sectional area of 6.8 cm². The volume of a cylinder is the product of the cross-sectional area and the height,

`63.75cm^3=6.8cm^2\cdot L`

Solving for L,

`L=(63.75cm^3)/(6.8cm^2)\approx9.38cm`

Hence, the height of the light liquid is 9.38 cm.

8) For the height the heavy liquid rises in the left arm, we have to consider that the volume decrease of the heavy liquid on the right arm is equal to the volume increase of the heavy liquid on the left arm.

As before, both are cylinders,

`\begin{gathered} V_(right)=V_(left) \\ A_(right)\cdot h_2=A_(left)\cdot h_1 \end{gathered}`

Remember that the cross-sectional areas of the right and left arms are 6.8 cm² and 12.9 cm² respectively,

`6.8cm^2\cdot h_2=12.9cm^2\cdot h_1`

Solving for h₂, we have a relationship between h₁ and h₂,

`\begin{gathered} h_2=(12.9cm^2)/(6.8cm^2)h_1 \\ \\ h_2\approx1.897h_1 \end{gathered}`

We still need another equation to be able to solve for h₁.

The pressure at the bottom of the U-tube - and therefore the pressure at the bottom of each arm of the tube, is the same,

`P_(right)=P_(left)`

Replace with the expression to find the pressure of a liquid at a certain depth,

`\rho_(heavy)\cdot g\cdot(h_1+h_2)=\rho_(light)\cdot g\cdot L`

The acceleration due to gravity cancels out,

`\rho_(heavy)\cdot(h_1+h_2)=\rho_(light)\cdot L`

Plugin the expression we found from the volumes that relate h₁ and h₂,

`\rho_(heavy)\cdot(h_1+1.897h_1)=\rho_(light)\cdot L`

Factor out h₁,

`h_1\cdot\rho_(heavy)\cdot(1+1.897_{})=\rho_(light)\cdot L`

And solve for h₁,

`h_1=\frac{\rho_(light)\cdot L}{\rho_(heavy)\cdot(1+1.897_{})}=(1.6g/cm^3\cdot9.38cm)/(9.6g/cm^3\cdot(1+1.897))\approx0.54cm`

Hence, the height that the heavy liquid rises in the left arm is 0.54 cm.