Pressure Measuring Devices

Solids And Fluids of Class 11

Pressure Measuring Devices

A manometer is a tube open at both the ends and bent into the shape of a “U” and partially filled with mercury. When one end of the tube is subjected to an unknown pressure p, the mercury level drops on that side of the tube and rises on the other so that the difference in mercury level is h as shown in the figure.

According to Pacal’s Law, when we move down in a fluid pressure increases with depth and when we move up the pressure decreases with depth. When we move horizontally in a fluid pressure remains constant.


p + ρogho - ρmgh = po

where po the atmospheric pressure, and

ρo is the density of the fluid inside the vessel

Pressure Measuring Devices

Example: 12.10

As shown in the figure(12.15), a column of water 40 cm high supports a 31 cm of an unknown fluid. What is the density of the unknown fluid?


The pressure at point A due to the two fluids must be equal (or the one with the higher pressure would push the lower pressure fluid away). Therefore,

Pressure due to water = pressure due to known fluid

Pressure Measuring Devices

h1ρ1g = h2ρ2 g

from which ρ2 = Pressure Measuring Devices

Example: 12.11

For the arrangement shown in the figure(12.16), determine h if the pressure difference between the vessels A and B is 3 kN/m2.

Pressure Measuring Devices


Let pressure in the horizontal tube is P

So in left vertical tube

P + ρkgh1 + ρwgho = PB

P + ρkgh2 + ρwg(ho – h – 0.2) = PA

here, pB - pA = 3 × 103 N/m2,

ρw = 103 kg/m3

ρk = 800 kg/m3.

Thus,h = 0.5 m = 50 cm

Pressure Measuring Devices

The Mercury Barometer

It is a straight glass tube (closed at one end) completely filled with mercury and inserted into a dish which is also filled with mercury as shown in the figure(12.17). Atmospheric pressure supports the column of mercury in the tube to a height h. The pressure between the closed end of the tube and the column of mercury is zero, p = 0.

Pressure Measuring Devices

Therefore, pressure at points A and B are equal and thus

po = 0 + ρmgh

At the sea level, po can support a column of mercury about 76 cm in height.

Hencepo = (13.6 × 103)(9.81)(0.76) = 1.01 × 105 Nm-2 (or Pa)

Example: 12.12

What must be the length of a barometer tube used to measure atmospheric pressure if we are to use water instead of mercury.


We know that

po = ρmghm = ρwghw

whereρw and hw are the density and height of the water column supporting the atmospheric pressure po.

∴hw = Pressure Measuring Devices

SincePressure Measuring Devices = 13.6 and hm = 0.76 m

∴hw = (13.6)(0.76) = 10.33 m

Example: 12.13

In the figure shown , find

(a)the total force on the bottom of the tank due to the water pressure.

(b)the total weight of water .

Why is there a difference between the two ?

Pressure Measuring Devices


(a)Pressure at the base due to water is

p = ρwg [ 5 + 1] = (103)(10)(5 + 1) = 6 × 104 N/m2

∴Force = pA2 = (6 × 104)(100 × 10-4) = 600 N

(b)Weight of water = ρw g [5A1 + A2] = 104 [5 × 10 × 10-4 + 100 × 10-4] = 150 N

Talk to Our counsellor