O lvl Phy: Pressure

RI 1998 Sec 3 EOY P2 Q5

(a) Draw a labelled diagram of a simple mercury barometer.

State and indicate clearly on your diagram how such a barometer may be used to obtain a value for atmospheric pressure.

(b) A marine expedition makes a survey of an ocean floor which is 0.50 km below sea-level. Assuming that the density of sea-water is 1.0 * 10³ kg/m³, estimate in Pa, the total pressure on the ocean floor.

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Answer:

(a)

How to use barometer:
Place the barometer where the atmospheric pressure is to be measured. Then measure the height of the mercury column from the highest level of the mercury in the tube to the mercury level at the top of the tub, as shown in the figure.

By taking hρg, where h is the height of the mercury column, g is the gravitational value, and ρ is the density of mercury, we can find the atmospheric pressure.

(b) 0.50 km = 500 m
Total pressure = a.t.m. + hρg
= 1.0 * 10⁵ Pa + (500 * 1.0 * 10⁵ * 10) Pa
= 1.0 * 10⁵ Pa + 50 * 10⁶ Pa
= 5.1 * 10⁶ Pa

O lvl Phy: Pressure

(a) The figure below shows a mercury barometer. At atmospheric pressure, the length of the mercury is 76.0 cm. (Density of mercury is 13.6 g/cm³)

(i) Describe how the barometer can be used to measure atmospheric pressure.
(ii) Hence, calculate the value of atmospheric pressure in pascal.
(iii) Explain why water is not a suitable liquid to be used in the barometer.


(b) The figure below shows a design of a hydraulic brake system. The pedal is pushing at the small piston.

(i) Explain how the force at the pedal can be transmitted to the large piston.
(ii) Hence calculate the force on the large piston when the force on the small piston is 110N.
(iii) Suggest why water is not suitable to be used as a brake fluid.

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Answer:


(a)
(i) The pressure due to the length of the mercury column above the mercury level in the bowl is balanced by the atmospheric pressure. Hence, by measuring the pressure due to the mercury column, we can find the value of the atmospheric pressure.

The equation for the atmospheric pressure is given by P = hρg


(ii) Density of mercury = 13.6 g/cm³ = 13600 kg/m³

Hence, P = hρg = 0.76 * 13600 * 10 = 1.03 * 10⁵ Pa.


(iii) Since the density of water is lower than mercury, the height of the water column will be larger (about 10m) given that atmospheric pressure is constant. The size of the barometer would be too large.



(b)
(i) Since the brake fluid is incompressible, pressure at the small piston is transferred to the large piston. Thus, force is transmitted from the pedal (at the small piston) to the large piston in this manner.

(ii) Since P₁ (at the small piston) = P₂ (at the large piston)
F₁/A₁ = F₂/A₂
110/0.00070 = F2/0.0037
F₂ = 581 N

(iii) Brake fluid is better because it can reduce corrosion in the hydraulic system. It can also act as a lubricant for the moving parts in the system.

O lvl Phy: Pressure

The figure below shows 3 liquids, liquid X, liquid Y and mercury, in a manometer. Given the information below, determine the density of liquid X and density of liquid Y.

Atmospheric pressure = 76 cm Hg
Density of mercury = 13600 k/m³
Density of liquid Y is twice of density of liquid X

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Answer:

We take the pressure of mercury by considering the red line as shown:

Let density of X be ρ_x, density of Y be ρ_y.

LHS: Pressure = atm + 0.05 ρ_x g + 0.02 ρ_mercury g
RHS: Pressure = atm + (0.16 - [0.08 - 0.02]) ρ_y g

Both sides pressure must be the same for mercury

atm + 0.05 ρ_x g + 0.02 ρ_mercury g = atm + 0.09 ρ_y g
0.05 ρ_x g + 0.02 ρ_mercury g = 0.10 ρ_y g

Because ρ_y= 2 ρ_x,
0.05 ρ_x g + 0.02 ρ_mercury g = 0.20 ρ_x g
0.05 ρ_x + 0.02 ρ_mercury = 0.20 ρ_x
0.02 ρ_mercury = 0.15 ρ_x
ρ_x = 0.02 / 0.15 * ρ_mercury
ρ_x = 1813 kg/m³

Thus, ρ_y = 2 ρ_x = 3627 kg/m³

O lvl Phy: Pressure

The apparatus shown in the diagram below is used to measure the density of oil The density of water is 1000 kg m^-3 and the atmospheric pressure is 100 000 N m^-3.

(a) Calculate the pressure at the liquid surface A inside the tube in N m^-2.
(b) Hence, find the pressure at the liquid surface C inside the tube.
(c) Determine the density of oil
[Take g = 10 N/kg where necessary]

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Answer:

(a) Pressure at A = 100 000 + 0.08 Χ 1000 Χ 10 (atm + ρgh)
= 100 800 N m^-2

(b) *Star Question*

Pressure at C = Pressure at A (air pressure evens itself throughout in an enclosed space)

= 100 800N m^-2

(c)

Atm pressure + pressure due to oil column = 100 800

Pressure due to oil column = 800

0.12 Χ density of oil Χ 10 (ρgh) = 800

Density of oil = 667 kg / m^-3

O lvl Phy: Pressure

A hydraulic press has frictionless pistons of diameters 0.01 m and 0.05 m respectively.

(a) Are the points X, Y and Z in the hydraulic press at the same pressure? Give a reason for your answer.

(b) A small piston is fitted to a lever system pivoted at A and a force F is applied at the end of the lever as shown in the figure above. You may neglect the weights of the pistons and the lever.
(i) Calculate the force applied at X, given that the load is 500N.
(ii) Hence calculate F.

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Answer:

(a) Yes, they are at the same pressure. This is because x, y and z are in an enclosed space.


(b) (i)

F_y / 0.05² = F_x / 0.01²

500 / 0.05² = F_x / 0.01²

F_x = 20 N


b) (ii)

Sum of anti-clockwise moments = sum of clockwise moments
20 * 0.02 = F * 0.20
F = 2 N

O lvl Phy: Pressure

A uniform capillary tube, closed at one end, contained air trapped by a thread of mercury 85mm long. when the tube was held horizontally, the length of the air column was 50mm. when it was held vertically with the closed end downwards, the length was 45mm. find the atm in N/m²
(take g=10, density of mercury = 14 000kg/m³)

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Answer:

Let atm = P1
Volume of air trapped is proportional to length of air column

When it is horizontal, only atm is involved
when it is vertical, pressure of air in the capillary tube = atm + pressure due to mercury column
= P1 + 14000 * 10 * 0.085 (pgh)
= P1 + 11900

Hence, using P1V1 = P2V2 (Boyle's Law)
P1 (50) = (P1 + 11900) (45)
Solving the equation,

P1 = 107000 Pa (3 sf)

O lvl Phy: Pressure

On a particular day, the atmospheric pressure at sea level was 102 000 Nm². at the same time, a mercury barometer placed at the top of the mountain indicated 0.6m Hg. Assume that the acceleration due to gravity remained constant. (density of mercury = 13 600 kg/m³)

If the density of the air between the top of the mountain and the sea level is 1.2 kg/m³, determine the height of the mountain top above sea level

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Answer:

Pressure at top of mountain = 0.6 * 10 * 13600 Pa = 81600 Pa
Pressure at sea level = 102000 Pa
Difference in pressure = 102000 - 81600 = 20400Pa

This difference in pressure = height * density of air * g
so, 20400 = height * 1.2 * 10
height of mountain = 1700 m