O-Level Physics Notes

1. PRESSURE!
Introduction:
Hello everyone! How are you guys? Today we are going to study in detail about pressure. Okay so what do you know
about it? To all the girls out there, do you realise why your pencil heels would get stuck in a muddy ground but if you
wear a pair of sneakers (flat soles), they won’t? Any idea? Let me try. We know that the force pushing downwards on
the ground by her shoes is due to her weight. The high heels sink easily because her weight is acting on a smaller
area compared to the flat soles. We can see that a force acting on a smaller area exerts a greater pressure.
Definition:
So what is pressure basically? We have just seen it depends on the area (in the previous case, it was that of flat
soles and heels) and on the force (weight of the body). So pressure can be defined as the force acting per unit area.
p = F / A
where p is pressure, F is force in Newtons (N) while A is area in square metres (m2
). The SI unit of pressure is
newton per metre square (N m-2
) or Pascal (Pa).
Q1: Calculate the pressure under a girl’s feet if her mass is 40 kg and the area of her shoes in contact with the
ground is 200 cm2
.
Pressure in Liquids:
Okay so in the above image, we can see that the water in the container is falling the fastest and furthest from the
outlet 3. And the water from outlet 2 is going further and faster from that in outlet 1. So we have reached to our main
concept: the amount of pressure inside a body of liquid increases with its depth. The deeper it is, greater is the
weight of liquid is above it, hence a greater pressure. From this, we can reach to the formula:
p = ρgh
where p is pressure, ρ is the symbol for density and h is the height of the liquid column.
Task: Search for the derivation of this formula so you further clear out your concept.

2. From this formula, we can see that the pressure depends on the depth and density of liquid, and not on the cross-
sectional area or volume of the liquid.
ts.
Examine the image above, what do you see? Same height of liquid in all four containers, right? We have just
discussed that the pressure in liquids depends on density and height (or depth). In this case, it’s water in all
containers so they have the same density, and the height is the same. Hence they all have the same pressure.
Easy? Let’s take a look at a different case.
Okay so you see two wells in the diagram above, HA is the vertical height of both wells A and B. HB is though the
slant height so it won’t count. Both have same vertical height, same liquids so same density: both have the same
pressure!
Now in the above image, though the left lake is large and right one small, but the pressure of right one is greater.
Why? Yes you’re right: because pressure does not depend on the cross-sectional area, it depends on height.
Another important concept: the pressure of two points anywhere in the liquid having the same height of liquid column
above them will have the same pressure. Pretty much related to the previous concept. Okay here, give it a look:

3. The point A is above B and has a less height of water column above it hence pressure in A is less than that in B. Now
we also know that the pressure at C and D (they are on same level) is greater than that at B. And we have also
learned that two or as many as hundred points taken at the same height will have the same pressure, hence
pressure at C and D is same. Easy.
Atmospheric Pressure:
What is atmospheric pressure? Okay we have all studies that Earth is surrounded by a layer of air called the
atmosphere. This layer of air weighs down on the Earth’s surface, including us! As a result, this layer exerts a
pressure on the Earth’s surface which is called the atmospheric pressure. The pressure exerted by this layer at sea is
1.013 x 105
Pa. This value is referred to as one atmosphere and is equivalent to placing 1 kg weight on an area of 1
cm3
.
The pressure at higher altitudes is lower.
Now, how to measure the atmospheric pressure? For this, we can either use a barometer or a manometer.
Barometer:

4. Okay this is how it looks like. Atmospheric pressure acts on the reservoir of mercury (shown with arrows). A vacuum
ensures that no pressure is exerted over the mercury column. The length of the column is measured and it gives the
pressure in cm of mercury (Hg).
The above diagram shows barometers which are kept at normal atmospheric pressure which we just learned is 1.013
x 105
Pa. Now this pressure is equal to 76 cm of Hg or you can say 760 mm of Hg. If we convert the value of
atmospheric pressure in Pascals to the height which should be attained if mercury is used (density of mercury is 13.6
x 103
kg/m3
), by applying the formula:
p = ρgh
1.013 x 105
Pa = 13.6 x 103
kg/m3
(9.8) (h)
h = 0.760 m = 760 mm
Okay we have proved that the length of mercury column should actually be 760 mm of Hg.
The important points here are that the height h in the above diagram will remains unchanged if:
1. the glass tube is lifted up from the dish
2. the glass tube is lowered further into the dish
3. the diameter of the glass tube increases
4. the glass tube is tilted
5. the quantity of mercury in the dish is increased
Now, moving on, take a look at the above diagram.
1. The pressure at Pa is zero.
2. The pressure at Pb is 26 cm of Hg.
3. The pressure at Pe is 76 cm of Hg.

5. 4. The pressure at Pf is 84 cm of Hg.
5. And finally, the pressure at Pb and Pc is same. Similarly, the pressure at Pe and Pd is same. Why? Because
they are on the same level.
Manometer:
● Fig. 2-1. In its simplest form the manometer is a U-tube about half filled with liquid. With both ends of the
tube open, the liquid is at the same height in each leg.
● Fig. 2-2. When positive pressure is applied to one leg, the pressure pushes the liquid downwards and
automatically, the liquid in the other leg moves up. The difference in height, "h", which is the sum of the
readings above and below zero, indicates the pressure.
● Fig. 2-3. When a vacuum is applied to one leg, the pressure on the other leg is greater than the pressure in
the vacuum side so the level of liquid falls and rises on the vacuum side automatically. The difference in
height, "h", which is the sum of the readings above and below zero, indicates the pressure of vacuum.
That’s not difficult at all, right?
Okay now, to further clear out any of your questions, let’s try going in detail of manometer. In the above diagram, a
tank of certain pressure P is attached to the manometer to determine it’s pressure (i.e. pressure at 1). The pressure
exerted by the gas in the tank is greater than the atmospheric pressure and so it pushes the level of the liquid
downwards. The level in the other leg will automatically rise, we know. Okay we know that the pressure at same
levels is the same, so pressure at 1 and 3 will be the same, right? Pressure at 4 will always be equal to 1 atmosphere
i.e. 76 cm of Hg.
Now, we have just seen that the pressure of gas tank is greater, so we’ll simply form the equation:

6. Ptank i.e. at 1 = (P at 4 i.e. atmospheric pressure) + (ρgh using height “h”)
While using ρgh, we’ll apply the density of the liquid used, g is constant and height h. Notw that it is if you want the
pressure in Pascals. If you simply want it in cm of Hg then add the height h in cm to 76 cm of Hg at 4.
One other thing, the pressure at 2, what would that be? It has the same concept: it has a greatest pressure due to
greatest depth, so it’s be the pressure at 3 (or 1) plus the height x. Note the pressure will be in cm of Hg.
Q1. Calculate the pressure of the trapped gas in cm of Hg given that the atmospheric pressure is 76 cm of Hg.
Answers:
Q1. 80.7 cm of Hg

7. Ptank i.e. at 1 = (P at 4 i.e. atmospheric pressure) + (ρgh using height “h”)
While using ρgh, we’ll apply the density of the liquid used, g is constant and height h. Notw that it is if you want the
pressure in Pascals. If you simply want it in cm of Hg then add the height h in cm to 76 cm of Hg at 4.
One other thing, the pressure at 2, what would that be? It has the same concept: it has a greatest pressure due to
greatest depth, so it’s be the pressure at 3 (or 1) plus the height x. Note the pressure will be in cm of Hg.
Q1. Calculate the pressure of the trapped gas in cm of Hg given that the atmospheric pressure is 76 cm of Hg.
Answers:
Q1. 80.7 cm of Hg