Pressure in Fluids and Atmospheric Pressure

Take a vessel filled with a liquid (say, water). Place it on a horizontal surface. Make several small holes in the wall of the vessel anywhere below the free surface of liquid. It is observed that liquid spurts out through each hole. This shows that the liquid exerts pressure at each point on the wall of the vessel.

Consider a vessel containing a liquid with density ρ. Let the liquid be stationary. In order to calculate pressure at a depth h, consider a horizontal circular surface PQ with area A at depth h below the free surface XY of the liquid as shown below.

The pressure on surface PQ will be due to the weight of the liquid column above the surface PQ, (i.e., the liquid contained in cylinder PQRS of height h with PQ as it's base and top face RS lying on the free surface XY of the liquid).

Thrust exerted on the surface PQ

= Weight of the liquid column PQRS

= Volume of liquid column PQRS x density x g

= (Area of base PQ x height) x density x g

= (A x h) ρ x g = A h ρ g

This thrust is exerted on the surface PQ of area A. Therefore, pressure

P = $\dfrac{\text{Thrust on surface}}{\text{Area of surface}}$ = $\dfrac{\text{A h ρ g}}{\text{A}}$ = h ρ g

Hence, Pressure = depth x density of liquid x acceleration due to gravity = h ρ g

The principle of hydraulic machine is that a small force applied on a smaller piston is transmitted to produce a large force on the bigger piston.

Two devices that work on this principle are —

1. Hydraulic brakes
2. Hydraulic jack

(a) The parts are —

X → Press plunger

Y → Pump plunger

(b) When the lever arm is moved down, the valve B closes and the valve A opens, so water from cylinder P is forced into the cylinder Q.

(c) When the lever arm is moved down, the valve B closes due to an increase in pressure in cylinder P. Now pressure from cylinder P is transmitted to the connecting pipe. As the pressure in connecting pipe becomes greater than the cylinder Q, the valve A opens, so water from cylinder P is forced into the cylinder Q.

(d) When the release valve is opened the ram (or press) plunger Q gets lowered and water of the cylinder Q runs out into the reservoir.

(e) When the lever arm is moved up the valve B opens upwards.

(f) When the lever arm is moved up the valve B opens upward because the pressure in cylinder P decreases.

(g) An hydraulic press is used for pressing cotton bales and goods like quilts, books, etc.

When handle H of lever is pressed down by applying an effort, the valve V opens because of increase in pressure in the cylinder P. The liquid runs out from the cylinder P to the cylinder Q.

As a result, the piston B rises up and it raises the car placed on the platform. When the car reaches the desired height, the handle H of lever is no longer pressed.

The valve V gets closed (since the pressure on either side of the valve becomes same) so that the liquid may not run back from the cylinder Q to the cylinder P.

To apply brakes, the foot pedal is pressed due to which pressure is exerted on the liquid in the master cylinder P, so liquid runs out from the master cylinder P to the wheel cylinder Q.

As a result, the pressure is equally transmitted and undiminished through the liquid to the pistons B₁ and B₂ of the wheel cylinder Q. Therefore, the pistons B₁ and B₂ gets pushed outwards and brake shoes gets pressed against the rim of the wheel due to which the motion of the wheel retards.

Since, the area of cross section of piston A in the master cylinder P is less than that in the wheel cylinder Q, a small force applied at the foot pedal produces a large force on the pistons B₁ and B₂ of the wheel cylinder Q.

This is the force responsible for retarding the vehicle. It should be noted that due to transmission of pressure through liquid, equal pressure is exerted on all wheels of the vehicle connected to the pipe line R.

On releasing the pressure on the pedal, the spring pulls the brake shoes to it's original position and forces the piston's B₁ and B₂ to return back into the wheel cylinder Q. As a result the liquid runs back from the wheel cylinder Q to the master cylinder P and thus the brakes gets released.

equal to the weight of the body

Reason — The thrust exerted by a body placed on a surface is equal to its weight. Thrust is a vector quantity.

both (b) and (c)

Reason — The pressure exerted on a surface depends on two factors :

1. the area on which the thrust is applied and
2. magnitude of thrust.

scalar and vector quantities respectively

Reason — Pressure is a scalar quantity whereas thrust is a vector quantity, because pressure is defined in terms of magnitude only, and thrust is defined in terms of both magnitude and direction.

dyne cm^-2

Reason — C.G.S. unit of pressure is dyne cm^-2 where 1 dyne cm^-2 = 0.1 N m^-2 or 1 N m^-2 = 10 dyne cm^-2

10⁵ Nm^-2

Reason — 1 bar = 10⁵ Nm^-2.

760 torr

Reason — torr is a unit of atmospheric pressure after the name of the scientist Torricelli where, 1 torr = 1 mm of Hg and 1 atm = 760 mm of Hg = 760 torr.

increasing the area of surface

Reason — Larger the area on which a given thrust acts, lesser is the pressure exerted by it, as pressure is inversely proportional to the area of surface.

$\text{Pressure} = \dfrac{\text{Thrust}}{\text{Area}} \\[0.5em] \text{or} \\[0.5em] \text P = \dfrac{\text{F}}{\text{A}}$

small, high

Reason — For a given thrust, the pressure on a surface is increased by reducing the area of surface on which it is acting, hence, cutting tools have either sharp or pointed edges so that a small thrust may cause a high pressure at that edges and cutting can be done with less effort.

hρg

Reason — The three factors on which the pressure at a point in a liquid depends are —

1. depth of the point below the free surface (h)

2. density of liquid (ρ) and

3. acceleration due to gravity (g)

Pressure = hρg

It increases with an increase in the density of the liquid.

Reason — As Pressure = hρg, so pressure is directly proportional to the density of the liquid. Hence, it increases with an increase in the density of the liquid.

P₁ < P₂

Reason — As the density of sea water is more than the density of river water, hence, the pressure at a certain depth in sea water is more than that at the same depth in river water because pressure increases with the increase in density of liquid.

P₂ - P₁ = hρg

Reason — The pressure at a point inside a liquid at a depth h is P₂

P₂ = pressure at the top of a dam (P₁) + pressure due to liquid column

= P₁ + hρg

So, P₂ = P₁ + hρg

Hence,

P₂ - P₁ = hρg

the pressure exerted by a liquid increases with its depth

Reason — A dam has broader walls at the bottom than at the top as the pressure exerted by the liquid increases with it's depth. Thus, as depth increases, more and more pressure is exerted by water on the walls of the dam.

Pascal's law

Reason — The principle of a hydraulic machine is that a small force applied on a smaller piston is transmitted to produce a large force on the bigger piston. This principle is based on Pascal's law.

force multiplier

Reason — The principle of a hydraulic machine is that a small force applied on a smaller piston is transmitted to produce a large force on the bigger piston. Hence, it acts as a force multiplier.

Maximum pressure is exerted when breadth and height form the base

Reason — Pressure is given by the formula : $\text {Pressure} = \dfrac {\text {Thrust}}{\text {Area}}$

For the same weight (thrust) of the block, the pressure is inversely proportional to the contact area. So, larger base area → smaller pressure, and smaller base area → greater pressure.

As, minimum area → side with minimum dimensions → 30 cm breadth and 15 cm height.

Hence, maximum pressure is exerted when breadth and height form the base because it has smallest contact area.

The pressure exerted by the liquid filled in a vessel is same at all points.

Reason — Liquid pressure increases with depth so, pressure is not the same at all points in the vessel as it depends on how deep the point is from the free surface.

As we know,

Pressure (P) = $\dfrac{\text {Thrust (F)}}{\text {Area (A)}}$

Given,

F = 1.5 N

A = 2 mm²

Converting 2 mm² in metre²

As 1 mm = $\dfrac{1}{1000}$ m

So, 1 mm² = (1) mm x (1) mm

= $(\dfrac{1}{1000}) m \times (\dfrac{1}{1000}) m$

= 1 x 10 ^-6 m²

Hence,

2 mm² = 2 x 10^-6 m²

A_A = 2 x 10 ^-6 m²

A_B = 6 x 10 ^-6 m²

Substituting the values in the formula above, we get,

Pressure on A

$\text P = \dfrac{1.5}{2 \times 10^{-6}} \\[0.5em] \text P = 0.75 \times 10^{6} \\[0.5em] \Rightarrow \text P = 7.5 \times 10^{5} \text { Pa} \\[0.5em]$

Hence, Pressure on A = 7.5 x 10⁵ Pa

Pressure on B

$\text P = \dfrac{1.5}{6 \times 10^{-6}} \\[0.5em] \text P = 0.25 \times 10^{6} \\[0.5em] \Rightarrow \text P = 2.5 \times 10^{5} \text { Pa} \\[0.5em]$

Hence, Pressure on B = 2.5 x 10⁵ Pa

As we know,

(a) Force (F) = mass (m) x acceleration due to gravity (g)

Given,

m = 7.5 kg

1 kgf = 10 N

Substituting the values in the formula above we get,

$\text F = 7.5 \times 10 \\[0.5em] \Rightarrow \text F = 75\ \text N$

Hence, F = 75 N

(b) As we know,

Pressure (P) = $\dfrac{\text {Thrust (F)}}{\text {Area (A)}}$

Given,

Area of the base = 12 x 8 = 96 cm²

Converting cm² into m²

100 cm = 1 m

So, 100 cm x 100 cm = 1 m²

Hence, 96 cm² = $\dfrac{1 \times 96}{10000}$

Therefore, A = 0.0096 m ²

Substituting the values in the formula above, we get,

$\text P = \dfrac{75}{0.0096} \\[0.5em] \Rightarrow \text P = 7812.5.5\ \text Pa \\[0.5em]$

Hence,

Pressure = 7812.5 Pa

(a) As we know,

Pressure due to water column of height h = h ρ g

Given,

h = 1.5 m

ρ = 1000 kg m ^-3

g = 9.8 m s ²

Substituting the values in the formula above, we get,

$\text P = 1.5 \times 1000 \times 9.8 \\[0.5em] \text P = 14.7 \times 10^{3}\ \text {N m}^{-2} \\[0.5em]$

Hence, pressure = 1.47 x 10⁴ N m^-2

(b) As we know,

Pressure (P) = $\dfrac{\text {Thrust (F)}}{\text {Area (A)}}$

Given,

A = 100 cm ²

Converting cm ² into m ²

100 cm = 1 m

So, 100 cm x 100 cm = 1 m ²

Hence, 100 cm ² = $\dfrac{1 \times 100}{10000}$

Therefore, A = 10^-2 m ²

Substituting the values in the formula above, we get,

$1.47 \times 10^{4} = \dfrac{\text {Thrust (F)}}{10^{-2}} \\[0.5em] \Rightarrow \text {Thrust (F)} = 1.47 \times 10^{2} \\[0.5em] \Rightarrow \text {Thrust (F)} = 147\ \text N \\[0.5em]$

Hence, Thrust = 147 N

(a) As we know,

Pressure due to water column of height h = hρg

Given,

ρ = 1000 kg m ^-3

g = 10 m s^-2

h = 6 cm

Converting cm to m

100 cm = 1 m

So, 6 cm = $\dfrac{1}{100}$ x 6 = 0.06 m

Hence, h = 0.06 m

Substituting the values in the formula above we get,

$\text P = 0.06 \times 1000 \times 10 \\[0.5em] \text P = 600 \text { Pa} \\[0.5em]$

Hence, Pressure = 600 Pa

(b) As we know,

Pressure (P) = $\dfrac{\text {Thrust (F)}}{\text {Area (A)}}$

A = 300 cm ²

Converting cm ² into m ²

100 cm = 1 m

So, 100 cm x 100 cm = 1 m ²

Hence, 300 cm ² = $\dfrac{1 \times 300}{10000}$

Therefore, A = 3 x 10^-2 m ²

Substituting the values in the formula we get,

$600 = \dfrac{\text {Thrust (F)}}{3 \times 10^{-2}} \\[0.5em] \Rightarrow \text {Thrust (F)} = 600 \times 3 \times 10^{-2}\\[0.5em] \Rightarrow \text {Thrust (F)} = 18\ \text N \\[0.5em]$

Hence, thrust = 18 N

(a) As we know,

Pressure due to water column of height h = hρg

and

Pascal's law, states that the pressure exerted anywhere in a confined liquid is transmitted equally and undiminished in all directions throughout the liquid.

Hence,

Pressure due to water column = Pressure due to mercury column

Hence,

h_w ρ_w g = h_m ρ_m g

Given,

h_m = 70 cm

ρ_m = 13.6 g cm^-3

ρ_w = 1 g cm^-3

From the above formula, we get,

h_w = $\dfrac{\text h_\text m \text ρ_\text m}{\text ρ_\text w}$

Substituting the values, we get,

$\text h_\text w = (\dfrac{13.6}{1}) \times 70 \\[0.5em] \Rightarrow \text h_\text w = 952\ \text{cm} \\[0.5em] \Rightarrow \text h_\text w = 9.52\ \text{m} \\[0.5em]$

Hence, height of a water column = 9.52 m

(b) No, if the cross section of the water column is made wider, the height of the water column will be unaffected.

As we know,

Pressure due to water column of height h = h ρ g

Given,

density of water = 1000 kg m ^-3

g = 10 m s^-2

P_w on ground floor = 40,000 Pa

P_w on first floor Pwf = 10,000 Pa

In order to know the height of the first floor, let us calculate the difference in pressure

P = P_wg - P_wf = 40,000 – 10,000 = 30,000 Pa

Substituting the values in the formula,

$\text P = \text {h ρ g} \\[0.5em] 30,000 = 1000 \times 10 \times \text h \\[0.5em] \text h = 3\ \text m \\[0.5em]$

Hence, the height of the first floor = 3 m.

Given,

ρ_m = 13.6 x 10³ kg m ^-3

ρ_w = 10³ kg m ^-3

Height to which water is poured in one arm, h_w = 13.6 cm

By pouring 13.6 cm of water, the mercury level in the left arm goes down to point A by x cm, while in the right arm, it rises to point C by x cm. Therefore, BC = h_m = 2x cm

By Pascal's law,

Pressure in the water column = pressure in the mercury column

Therefore, P_A = P_B

⇒ h_w ρ_w g = h_m ρ_m g

⇒ 13.6 x 10³ x g = 2𝑥 x 13.6 x 10³ x g

⇒ 1 = 2𝑥

⇒ 𝑥 = $\dfrac{1}{2}$ = 0.5 cm

Hence, the rise in mercury level = 0.5 cm

As we know,

Pressure (P) = $\dfrac{\text {Thrust (F)}}{\text {Area (A)}}$

Given,

A = 10 cm ²

Converting cm ² into m ²

100 cm = 1 m

So, 100 cm x 100 cm = 1 m ²

Hence, 10 cm ² = $\dfrac{1 \times 10}{10000}$

Therefore, A = 10^-3 m ²

F = 2 N at area of cross section 10 cm ²

Substituting the values in the formula we get,

P = $\dfrac{2}{10^{-3}}$     [Equation 1]

Now when, A = 100 cm ²

Converting cm ² into m ²

100 cm = 1 m

So, 100 cm x 100 cm = 1 m ²

Hence, 100 cm ² = $\dfrac{1 \times 100}{10000}$

Therefore, A = 10^-2 m ²

Substituting the value in the formula above we get,

Therefore, P = $\dfrac{\text F}{10^{-2}}$     [Equation 2]

Equating 1 and 2 we get,

$\dfrac{2}{10^{-3}} = \dfrac{\text F}{10^{-2}} \\[0.5em] \Rightarrow \text F = \dfrac{2 \times 10^{-2} }{10^{-3}} \\[0.5em] \Rightarrow \text F = 20\ \text N$

Hence, force at 100 cm ² = 20 N

As we know,

Pressure (P) = $\dfrac{\text {Thrust (F)}}{\text {Area (A)}}$

Let,

area of cross section of the master cylinder = A₁

area of cross section of the wheel cylinder = A₂

force applied on pedal be F₁ = 0.5 N

force applied on brake shoe F₂ = 15 N

By the principle of hydraulic brakes which works on Pascal's law

Pressure on narrow piston = pressure on broader piston

Hence,

$\dfrac{\text F_{1}}{\text A_{1}} = \dfrac{\text F_{2}}{\text A_{2}} \\[0.5em] \dfrac{0.5}{\text A_{1}} = \dfrac{15}{\text A_{2}} \\[0.5em] \dfrac{\text A_{1}}{\text A_{2}} = \dfrac{0.5}{15} \\[0.5em] \Rightarrow \dfrac{\text A_{1}}{\text A_{2}} = \dfrac{5}{150} \\[0.5em] \Rightarrow \dfrac{\text A_{1}}{\text A_{2}} = \dfrac{1}{30} \\[0.5em]$

Hence, the ratio of area of cross section = 1 : 30

As we know,

Pressure (P) = $\dfrac{\text {Thrust (F)}}{\text {Area (A)}}$

and by the principle of hydraulic machine

Pressure on narrow piston = pressure on broader piston

Hence,
$\dfrac{\text F_{1}}{\text A_{1}} = \dfrac{\text F_{2}}{\text A_{2}} \\[0.5em]$

Given,

Area of narrow piston (A₁) = 5 cm ²

Area of wider piston (A₂) = 625 cm ²

force (F₂) = 1250 N

Substituting the values in the formula above we get,

$\dfrac{\text F_{1}}{5} = \dfrac{1250}{625} \\[0.5em] \text F_{1} = \dfrac{1250 \times 5 }{625} \\[0.5em] \Rightarrow \text F_{1} = 10\ \text N \\[0.5em]$

Hence, force acting on the smaller piston = 10 N

Assumption — There is no friction and no leakage of liquid.

(a) As we know,

Pressure (P) = $\dfrac{\text {Thrust (F)}}{\text {Area (A)}}$

and by the principle of hydraulic machine

Pressure on neck = pressure on bottom of bottle

Hence,

$\dfrac{\text F_{1}}{\text A_{1}} = \dfrac{\text F_{2}}{\text A_{2}} \\[0.5em]$

Given,

Diameter of neck (d₁) = 2 cm

Hence, A₁ = 𝜋 $(\dfrac{2}{2})^{2}$ = 𝜋

Diameter of bottom of bottle (d₂) = 10 cm

Hence, A₂ = 𝜋 $(\dfrac{10}{2})^{2}$ = 25 𝜋

Force applied on the cork in the neck (F₁) = 1.2 kgf

Substituting the values in the formula above we get,

$\dfrac{1.2}{\text π} = \dfrac{\text F_{2}}{25\text π} \\[0.5em] \text F_{2} = 1.2 \times 25 \\[0.5em] \text F_{2} = 30 \text{ kgf} \\[0.5em]$

Hence, force exerted at the bottom of the neck = 30 kgf.

(b) The Pascal's law is applied to solve the part (a) which states that the pressure exerted anywhere in a confined liquid is transmitted equally and undiminished in all directions throughout the liquid.

Hence,

$\dfrac{\text F_{1}}{\text A_{1}} = \dfrac{\text F_{2}}{\text A_{2}} \\[0.5em]$

(a) As we know, by the principle of hydraulic machine

Pressure on the smaller piston = pressure on the larger piston

Hence,

$\dfrac{\text F_{1}}{\text A_{1}} = \dfrac{\text F_{2}}{\text A_{2}} \\[0.5em]$

Given,

Diameter of the smaller piston (d₁) = 5 cm

Hence, A₁ = 𝜋 $(\dfrac{5}{2})^{2}$ = 6.25 𝜋

Diameter of the larger piston (d₂) = 25 cm

Hence, A₂ = 𝜋 $(\dfrac{25}{2})^{2}$ = 156.25 𝜋

Force applied on the smaller piston (F₁) = 50 kgf

Substituting the values in the formula above we get,

$\dfrac{50}{6.25\text π} = \dfrac{\text F_{2}}{156.25\text π} \\[1 em] \text F_{2} = \dfrac{50 \times 156.25}{6.25} = 1250 \\[0.5em] \text F_{2} = 1250 \text { kgf} \\[0.5em]$

Hence, force exerted on the larger piston = 1250 kgf.

(i) As we know,

Pressure (P) = $\dfrac{\text {Thrust (F)}}{\text {Area (A)}}$

Given,

Force on the narrow piston A = 4 kg

A_A = 8 cm ²

A_B = 320 cm ²

Substituting the values in the formula above, we get,

$\text P_\text A = \dfrac{4}{8} \\[0.5em] \Rightarrow \text P_\text A = 0.5 \text { kg cm}^{-2} \\[0.5em]$

Hence, P_A = 0.5 kg cm^-2

(ii) As we know, by the principle of hydraulic machine

Pressure on piston A = pressure on piston B

Hence, P_B = 0.5 kg cm^-2

(iii) Thrust on piston B is acting in the upward direction, which is given by

Pressure (P) = $\dfrac{\text {Thrust (F)}}{\text {Area (A)}}$

Substituting the values, we get,

$0.5 = \dfrac{\text{thrust}}{320} \\[0.5em] \Rightarrow \text{thrust} = 0.5 \times 320 \\[0.5em] \Rightarrow \text{thrust} = 160 \text{kgf} \\[0.5em]$

Hence, thrust on piston B = 160 kgf

As we know, by the principle of hydraulic machine

Pressure on piston A = pressure on piston B

Hence,

$\dfrac{\text F_{1}}{\text A_{1}} = \dfrac{\text F_{2}}{\text A_{2}} \\[0.5em]$

Given,

A₁ = 2 cm ²

Converting cm ² into m ²

100 cm = 1 m

So, 100 cm x 100 cm = 1 m ²

Hence, 2 cm ² = $\dfrac{1 \times 2}{10000}$

Therefore, A₁ = 2 x 10^-4 m ²

A₂ = 12 cm ²

Hence, 12 cm ² = $\dfrac{1 \times 12}{10000}$

Therefore, A₂ = 12 x 10^-4 m ²

F₂ = 150 N

Substituting the values in the formula above we get,

$\dfrac{\text F_1}{2 \times 10^{-4}} = \dfrac{150}{12 \times 10^{-4}} \\[1 em] \Rightarrow \text F_1 = \dfrac{150 \times 2 \times 10^{-4}}{12 \times 10^{-4}} \\[0.5em] \Rightarrow \text F_1 = 25\ \text N \\[0.5em]$

Hence, force applied = 25 N

Pressure is the thrust per unit area of surface.

$\text {Pressure} = \dfrac{\text {Thrust}}{\text {Area}} \\[0.5em] \text{or} \\[0.5em] \text P = \dfrac{\text F}{\text A}$

Pressure is a scalar quantity.

The S.I. unit of pressure is newton per metre² . This unit is named as pascal (symbol Pa).

The pressure exerted by a thrust is inversely proportional to the area of surface on which it acts. Larger the area on which a given thrust acts, lesser is the pressure exerted by it.

Example — A brick of weight 4 kgf having dimensions 20 cm x 10 cm x 5 cm, exerts maximum pressure on ground when it is placed with it's longest side (20 cm) vertical, as shown in fig below, while it exerts minimum pressure when it is placed with it's shortest side (5 cm) vertical, even though the thrust is same in each case.

The tip of an allpin is made sharp so that large pressure is exerted through the pointed end and that they can be driven into, with less effort.

(a) It is easier to cut with a sharp knife than with a blunt knife because in a sharp knife even a small thrust causes a great pressure at the edges and hence, cutting can be done with less effort.

(b) Sleepers are laid below the rails so that the pressure exerted by the iron nails on the ground becomes less.

A fluid contained in a vessel exerts pressure at all points and in all directions due to it's weight and this pressure is called fluid pressure.

Both liquids and solids exert pressure due to it's weight, however, pressure exerted by a solid acts only on the surface on which it is placed i.e. at it's bottom, but pressure exerted by a fluid acts on the bottom as well as the walls of the container due to it's tendency to flow.

1. depth of the point below the free surface (h),

2. density of liquid (ρ), and

3. acceleration due to gravity (g).

The pressure at a point inside a liquid at a depth h

= Atmospheric pressure + pressure due to liquid column

= P₀ + hρg

where,

P₀ = atmospheric pressure acting on the free surface of liquid

h = depth of the point below the free surface

ρ = the density of the fluid

g = acceleration due to gravity.

The pressure at a certain depth in sea water is more than the same depth in river water because the density of sea water is more than the density of river water.

According to the formula,

Pressure = depth x density of liquid x acceleration due to gravity = h ρ g

Therefore, when density of a liquid is more then the pressure exerted is also more as density and pressure are directly proportional.

It is noticed that as the gas bubble formed at the bottom of the lake rises, it grows in size. The reason is that when the bubble is at the bottom of the lake, total pressure exerted on it is the sum of the atmospheric pressure and the pressure due to the water column above it.

As the gas bubble rises, due to decrease in depth, the pressure due to water column decreases, so the total pressure on the bubble decreases.

According to Boyle's law, the volume of a gas is inversely proportional to the pressure on it. Therefore, the volume of bubble increases due to the decrease in pressure, i.e., the bubble grows in size.

When the bubble reaches the surface of liquid, total pressure exerted on it becomes minimum, just equal to the atmospheric pressure and so the size of bubble when touching the surface becomes maximum.

A dam has broader walls at the bottom than at the top as the pressure exerted by the liquid increases with it's depth. Thus, as depth increases, more and more pressure is exerted by water on the walls of the dam.

A thicker wall is required to withstand a greater pressure, therefore, the wall of a dam is made with thickness increasing towards the base.

In the figure given below, the increasing length of arrows in water represents the increasing pressure on the wall of the dam towards the bottom.

Sea divers need special protective suit to wear because in deep sea, the total pressure exerted on the diver's body is much more than his blood pressure. To withstand it, they need to wear a special protective suit, made from glass reinforced plastic or cast aluminium. The pressure inside the suit is maintained at one atmosphere.

The laws of liquid pressure are —

1. Inside the liquid, pressure increases with the increase in depth from it's free surface.

2. In a stationary liquid, pressure is same at all points on a horizontal plane.

3. Pressure is same in all directions about a point inside the liquid.

4. Pressure at same depth is different in different liquids. It increases with the increase in density of liquid.

5. A liquid seeks it's own level.

We observe that the throw of the liquid from hole A is more than the B i.e., the liquid reaches to a greater distance on the horizontal surface from hole A than hole B. This shows that liquid pressure at a point increases with the increase of depth from it's free surface.

A hydraulic press works on the principle of Pascal's law.

Principle — When a force F₁ is applied on the piston A, it exerts a pressure on liquid contained in the cylinder P. According to Pascal's law, this pressure is transmitted through liquid in tube R to the piston B of the other cylinder Q due to which the piston B tends to move upwards.

Since, the area of cross section of cylinder P is less than that of the cylinder Q, therefore by applying a small force on the piston A, we can lift a large weight kept on the piston B.

When no weight is placed on the piston B, it rises up against a fixed roof with a force F₂ (F₂ > F₁). If a bale of cotton is kept on the press plunger B, it gets compressed.

Use of hydraulic press — It is used for pressing cotton bales and goods such as books, quilts etc.

Thrust is the force acting normally on a surface.

The S.I. unit of thrust is newton (N).

Thrust is a vector quantity.

(a) The physical quantity that is measured in bar is Pressure.

(b) The relation between bar and pascal is 1 bar = 10⁵ pascal.

One pascal is the pressure exerted on a surface of area 1 m² by a force of 1 N acting normally on it.

Thrust is a vector quantity. Its direction of application is normal to the surface.

Pressure is a scalar quantity.

A substance which can flow is called a fluid. All liquids and gases are thus fluids.

(a) As we know,

Total pressure in a liquid at a depth h is P₂

= atmospheric pressure acting on the surface (P₁) + pressure due to liquid column (h ρ g)

= P₁ + h ρ g

Hence,

P₂ = P₁ + h ρ g

(b) From the above expression we observe that,

P₂ > P₁

(i) When the diver moves to a greater depth the liquid pressure increases as liquid pressure at a point increases with the increase of depth from it's free surface.

(ii) When the diver moves horizontally, the liquid pressure remains unchanged as inside a liquid, pressure is same at all points on a horizontal plane.

Pascal's law states that the pressure exerted anywhere in a confined liquid is transmitted equally and undiminished in all directions throughout the liquid.

Applications of Pascal's law are —

1. Hydraulic brakes
2. Hydraulic jack

(a) Pressure at a depth h in a liquid of density ρ is h ρ g.

(b) Pressure is same in all directions about a point in a liquid.

(c) Pressure at all points at the same depth is same.

(d) Pressure at a point inside a liquid is directly proportional to its depth.

(e) Pressure of a liquid at a given depth is directly proportional to the density of liquid.

assertion is true but reason is false

Explanation

Assertion (A) is true because this is a well-known phenomenon in which at high altitudes, sometimes nose bleeds when atmospheric pressure decreases significantly.

Reason (R) is false because at high altitudes, atmospheric pressure drops, but blood pressure inside the body remains the same. So now, blood pressure > atmospheric pressure, and the capillaries in the nose may burst, leading to nose bleeding.

assertion is false but reason is true

Explanation

Assertion (A) is false because water is not used as a barometric liquid mainly due to it's low density, so the height of the water column would be very large to balance atmospheric pressure and it also sticks to glass, which is not desirable in a barometer.

Reason (R) is true because this is the essential property of a good barometric liquid as it should neither wet nor stick to the glass for accurate readings and clear visibility of the level.

both A and R are true and R is the correct explanation of A

Explanation

Assertion (A) is true because when the brick is placed with its longest side vertical, the contact area with the ground is minimum, and for a given weight (thrust), smaller area means higher pressure as

$\text {Pressure} = \dfrac{\text {Force}}{\text {Area}}$

​Reason (R) is true because this is a direct consequence of the formula above — pressure is inversely proportional to area for a constant force, since the Reason clearly explains why the pressure is maximum when area is minimum (i.e., brick standing vertically), it is the correct explanation.

both A and R are true and R is not the correct explanation of A

Explanation

Assertion (A) is true because the pressure in a liquid increases with depth and is given by

$\text P=\text {hρg}$

So, the bottom of the dam experiences the highest pressure, and therefore the walls must be made thicker to withstand this pressure.

​Reason (R) is true because this is the statement of Pascal's law which states that at a given point in a liquid at rest, pressure is exerted equally and undiminished in all directions.

But R is not the correct explanation for A because the dam walls are thicker at the bottom due to increase in pressure with depth, not because pressure is the same in all directions.

both A and R are true and R is the correct explanation of A

Explanation

Assertion (A) is true because hydraulic jacks are commonly used in oil extraction processes to apply large pressure over a small area to squeeze oil from seeds.

​Reason (R) is true because Pascal's principle states that :

“When pressure is applied to a confined fluid, it is transmitted equally in all directions throughout the fluid.”

This principle allows a small force applied on a small piston to produce a large force on a larger piston, making the jack effective for pressing operations like oil extraction, since the hydraulic jack works because of Pascal’s principle, the Reason correctly explains the Assertion.

A barometer is an instrument which is used to measure the atmospheric pressure.

A simple barometer uses mercury as the barometric liquid.

Construction — It uses a hard glass tube of about 1 m closed at one end. The tube is completely filled with pure mercury such that no air bubble remains inside the tube. The open end of tube is closed with thumb and the tube is then made upside down several times so as to force out any air bubble which might have entered in it.

The completely filled tube with it's open end closed by thumb is then inverted into a trough of mercury in such a way that the open end of tube is well immersed in mercury in a trough and the tube stands vertical as shown in figure. Now the thumb is removed. Care is taken that no air enters in the tube.

It is seen that the level of mercury in the tube falls till it's height above the level of mercury in the trough becomes h (nearly 76 cm) as shown in the diagram below.

The demerits of a simple barometer are —

1. There is no protection for the glass tube.
2. A scale cannot be fixed with the tube to measure the atmospheric pressure.

Fortin barometer removes these demerits in the following ways:

1. The glass tube is protected by enclosing it in a brass case.
2. For accurate measurement, a main scale alongwith a vernier scale are provided.

An altimeter is an aneroid barometer, and it is used in aircraft to measure it's altitude.

It's principle is as follows —

As atmospheric pressure decreases with the increase in height above the sea level, therefore a barometer which measures the atmospheric pressure, can be used to determine the altitude of a place above the sea level.

It's scale is calibrated in terms of height of ascent with height increasing towards left because the atmospheric pressure decreases with increase of height above the sea level.

To measure the atmospheric pressure, first the mercury in the leather cup is raised up or lowered down with the help of the screw S so that the mercury level in the glass vessel just touches the ivory point I. The position of mercury level in the barometer tube is noted with the help of the main scale and the vernier scale. The sum of vernier scale reading and the main scale reading gives the barometric height. The barometric height shows the atmospheric pressure.

Aneroid barometer has no liquid. It is light and portable. It is calibrated to read directly the atmospheric pressure.

Construction —

The figure above shows the main parts of an aneroid barometer. It consists of a metallic box B which is partially evacuated. The top D of box is springy and is corrugated in form of a diaphragm as shown in fig (b).

At the middle of the diaphragm, there is a thin rod, toothed at it's upper end. The teeth of rod fit well into the teeth of a wheel S attached with a pointer P which can move over a circular scale.

The circular scale is graduated and is initially calibrated with a standard barometer so as to read the atmospheric pressure directly in terms of the barometric height.

Working —

When the atmospheric pressure increases, it presses the diaphragm D and the rod L gets depressed. The wheel S rotates clockwise and pointer P moves to the right on the circular scale.

On the other hand, when atmospheric pressure decreases, the diaphragm D bulges out due to which the rod L moves up and the wheel S rotates anticlockwise.

Consequently, the pointer shifts to the left and the pressure is read over the calibrated scale.

atmospheric pressure

Reason — The thrust exerted per unit area on the earth's surface due to column of air, is called atmospheric pressure on the surface of earth.

76 cm of Hg

Reason — As we know that, barometric height at normal temperature and pressure at sea level is 76 cm of Hg and barometric height is the measure of the atmospheric pressure. Hence, normal atmospheric pressure = 76 cm of Hg.

1 torr = 1 mm of Hg

Reason — torr is a unit of atmospheric pressure after the name of the scientist Torricelli where, 1 torr = 1 mm of Hg.

P₁ < P₂

Reason — Atmospheric pressure increases with depth. Hence, P₁ < P₂.

Torricelli

Reason — In 1643, Torricelli first designed a simple barometer using mercury as the barometric liquid.

all of the above

Reason — Measuring atmospheric pressure, weather forecasting, measuring heights, all these are valid uses of a barometer.

only 30% of the atmospheric pressure at sea level

Reason — At the summit of Mount Everest, the atmospheric pressure is only 30% of the atmospheric pressure at sea level.

all of the above

Reason — Uses of mercury as a barometric liquid are —

1. The density of mercury is greater than that of any liquid, so only 0.76 m of height of mercury is needed to balance the normal atmospheric pressure. Use of other liquids will require a much longer tube.
2. The vapour pressure of mercury is negligible, so the vapours in the torricellian vacuum does not affect the barometric height.
3. Mercury neither wets nor sticks to the glass tube, therefore it gives accurate reading.

It neither wets nor sticks to the glass tube

Reason — Water sticks to the glass tube and wets it, so the reading becomes inaccurate.

a possibility of rain.

Reason — If the barometric height gradually falls, it indicates that the moisture is increasing i.e., there is a possibility of rain.

decreases, decreases

Reason — Atmospheric pressure decreases with altitude due to (1) decrease in height of air column, (2) decrease in density of air.

altitudes

Reason — An altimeter is an aneroid barometer, and it is used in aircraft to measure it's altitude.

As we know,

Pressure = h ρ g

Given,

ρ = 13.6 10³ kg m^-3

g = 9.8 m s ^-2

h = 1 mm = 0.001 m (As 1 m = 1000 mm = 0.001 mm)

Substituting the values in the formula above we get,

$\text P = (0.001) \times (13.6 \times 10^{3}) \times (9.8) \\[0.5em] \Rightarrow \text P = 133.28 \text{ Pa} \\[0.5em]$

Hence, 1 mm of Hg = 133.28 Pa

We know that,

Pressure = h ρ g

Given,

ρ = 13.6 x 10³ kg m^-3

g = 9.8 m s ^-2

h = 0.70 m

Substituting the values in the formula above we get,

$\text P = (0.70) \times (13.6 \times 10^{3}) \times (9.8) \\[0.5em] \Rightarrow \text P = 93.3 \times 10^{3} \text { Pa} \\[0.5em]$

Hence, P = 93.3 x 10³Pa

Now, if we use a water barometer,

P = 93.3 x 10³ Pa

ρ = 1 x 10³ kg m^-3

g = 9.8 m s ^-2

Substituting the values in the formula above we get,

$93.3 \times 10^{3} = (\text h) \times (1 \times 10^{3}) \times (9.8) \\[0.5em] \text h = \dfrac{93.3 \times 10^{3}}{1 \times 10^{3} \times 9.8} \\[0.5em] \text h = 9.52 \text { m} \\[0.5em]$

Hence, height of water colum = 9.52 m

Given,

Atmospheric pressure = 76 cm of Hg

Change in pressure = 76 cm of Hg - 70 cm of Hg

= 6 cm of Hg

Pressure falls by 10 mm (= 1 cm) of Hg for every 120 m of ascent,

Therefore, for 6 cm of Hg we get,

$120 \times 6 = 720 m \\[0.5em]$

Hence, the height of the hill = 720 m

The assumption made is that the atmospheric pressure falls linearly with ascent.

We know that,

pressure = h ρ g

Given,

P = 1.04 x 10⁵ Pa

g = 10 m s^-2

density of air (ρ) = 1.3 kg m^-3

Substituting the values in the formula above, we get,

$1.04 \times 10^{5} = \text h \times 1.3 \times 10 \\[0.5em] \text h = \dfrac{1.04 \times 10^{5}}{1.3 \times 10} \\[0.5em] \text h = 8000 \text { m} \\[0.5em]$

Hence, height of the atmosphere = 8000 m

As we know,

decrease in pressure = h ρ g

Given,

density of air = 1.295 kg m^-3

h = 107 m

Therefore,

Decrease in pressure = (107) x (1.295) x (g)    [Equation 1]

Let, decrease in mercury height = H

Therefore, decrease in barometric height = (H) x (13.6 x 10³) x(g)    [Equation 2]

Equating 1 and 2 we get,

$107 \times 1.295 \times \text g = \text H \times 13.6 \times 10^{3} \times \text g \\[0.5em] \Rightarrow \text H = \dfrac{107 \times 1.295}{13.6 \times 10^{3}} \\[0.5em] \Rightarrow \text H = 0.0101 \text { m of Hg}$

Converting m into mm, we get

1 m = 1000 mm

Therefore, 0.0101 m = 0.0101 m x 1000 = 10.1 mm

Therefore, fall in barometric height = 10 mm of Hg.

The thrust exerted per unit area on the earth's surface due to column of air, is called atmospheric pressure on the surface of earth.

We do not feel uneasy even under the enormous pressure of atmosphere above as well as around us because the pressure of our blood balances it. The blood pressure is slightly above the atmospheric pressure.

However, at high altitude, the atmospheric pressure becomes less because the height of air column above that altitude is less than at the earth's surface. As a result, at high altitudes the blood pressure becomes much more than the atmospheric pressure and nose bleeding may occur due to excess blood pressure.

We take a thin tin can fitted with an airtight stopper. The stopper is removed and a small quantity of water is boiled in the can.

Gradually, the steam occupies the entire space of can by expelling the air from it, as shown in figure below. The stopper is then tightly replaced and simultaneously the flame beneath the can is removed. Cold water is then poured over the can. It is observed that the can collapses inward, as shown below.

The reason is that, initially the pressure due to steam inside the heated can is same as the air pressure outside the can.

But, on pouring cold water over the can, fitted with a stopper the steam inside the can condenses, producing water and water vapour at a very low pressure. Now the air pressure outside the can exceeds the vapour pressure inside the closed can.

Consequently, the excess atmospheric pressure outside causes it to collapse it inwards. This demonstrates that the atmosphere outside the can exerts a pressure which is known as atmospheric pressure.

A balloon collapses when air is removed from it because as the air is released from the balloon, the pressure inside the balloon is much less than the atmospheric pressure, hence, the balloon collapses.

Water does not run out of a dropper unless it's rubber bulb is pressed because the pressure inside the dropper is same as the atmospheric pressure when the rubber bulb is not pressed. But, when we press the rubber bulb the pressure inside the dropper increases and hence the water flows out.

Two holes are made in a completely filled sealed tin can to take out oil from it because there is no air inside a completely filled and sealed oil can. When the can is tilted, the pressure due to the column of oil (inside the can) at the hole is much less than the atmospheric pressure outside the can, so the oil does not flow out of the hole.

But if one more hole is made at the opposite end on the top cover of the can, air from outside the can will enter in through this hole and will exert atmospheric pressure on the oil from inside along with pressure due to oil column.

This results in the increase in the pressure on oil, and so it easily flows out through the hole of the can.

The syringe is kept with it's opening just inside a liquid and the plunger is lowered till it's base. Now when the plunger is pulled up in the barrel (as shown in figure), there is no air inside and thus the pressure inside the barrel below the plunger is much less than the atmospheric pressure acting on the liquid surface. As a result, the atmospheric pressure forces the liquid to rise up in the syringe.

In a water pump when the plunger is pulled up in the piston, there is no air inside and thus the pressure inside the siphon is much less than the atmospheric pressure acting on the liquid surface. As a result, the atmospheric pressure forces the water to rise up in the pump.

In figure, the atmospheric pressure acts at all points outside the tube (such as C) on the surface of mercury in the trough. The pressure at point A is due to the weight of the mercury column AB above it.

The mercury level in the tube becomes stationary when the pressure inside the tube at the point A which is at the level of the point C, becomes equal to that at the point C.

Thus, the vertical height of mercury column in it (i.e., AB = h) is a measure of atmospheric pressure and is called the barometric height.

The space left empty above the mercury column in the tube is called the torricellian vacuum. Ideally there should be no air in this space.

Vertical height of the mercury column from the mercury surface in the trough to the level in tube, is a measure of the atmospheric pressure and the barometric height at normal temperature and pressure at sea level is 0.76 m ( or 76 cm or 760 mm) of mercury.

If the atmospheric pressure increases, the pressure at point C increases and mercury from trough flows into the tube thereby increasing the vertical height of mercury column in the tube so as to equalise pressures at points A and C.

On the other hand, when the atmospheric pressure decreases, the vertical height of the mercury column decreases to balance it.

Thus, the vertical height of the mercury column from the mercury surface in the trough to the level in the tube, is a measure of the atmospheric pressure.

In an accurate barometer, the empty space above the mercury in the tube is perfect vacuum. If somehow air enters the empty space or a drop of water (or liquid) gets into the tube, it will immediately change into vapour in the vacuum space and the air (or the vapours of liquid) will exert pressure on the mercury column due to which the barometric height will decrease.

Such a barometer is a faulty barometer and it's barometric height will be less than the actual atmospheric pressure.

The name given to this space is torricellian vacuum.

Uses of mercury as a barometric liquid are —

1. The vapour pressure of mercury is negligible, so the vapours in the torricellian vacuum does not affect the barometric height.
2. Mercury neither wets nor sticks to the glass tube, therefore it gives accurate reading.

Water is not a suitable barometric liquid because —

1. The vapour pressure of water is high, so it's vapours in the vacuum space will make the reading inaccurate.
2. Water is transparent, so it's surface is not easily seen while taking the observation.

Advantages of an aneroid barometer over a simple barometer are —

(a) Aneroid barometer is calibrated directly to read the atmospheric pressure.

(b) It has no liquid and is portable.

The atmosphere consists of a number of parallel air layers. Each layer experiences a pressure on it due to the thrust or weight of the air column above it. Therefore, as we go up, the height of air column above us decreases and so thrust exerted by air column also decreases, which results in the decrease of atmospheric pressure with increases in altitude.

Since the lower air layers gets compressed due to the weight (or thrust) of the upper layers, therefore, the density of air layers is more near the earth's surface and it decreases as we go higher and higher.

The decrease in density with altitude is not linear. It is rapid at low altitude (near the sea level) and is slow at higher altitude. Due to decrease in density of air with altitude, the atmospheric pressure also decreases with altitude in a non linear way.

The figure below shows the variation of atmospheric pressure with height above the sea level.

The atmospheric pressure decreases as we go up because —

1. Decrease in height of air column which causes a linear decrease in the atmospheric pressure.
2. Decrease in density of air which causes a non-linear decrease in atmospheric pressure.

At high altitudes, a fountain pen leaks because normally a fountain pen filled with ink contains some air also which is at a pressure equal to the atmospheric pressure on the earths' surface. When the pen is taken at an altitude, the atmospheric pressure at this altitude is low so the excess pressure due to air inside the rubber tube forces the ink to leak out.

Our blood pressure is slightly more than atmospheric pressure. However, at high altitude, the atmospheric pressure becomes less because the height of air column above that altitude is less than at the earth's surface. As a result, at high altitudes the blood pressure becomes much more than the atmospheric pressure and nose bleeding may occur due to excess blood pressure.

The numerical value of the atmospheric pressure on the surface of the earth is 1.013 x 10⁵ pascal.

The physical quantity measured in torr is atmospheric pressure.

1 torr = 133.28 Pa

Pressure is expressed in the unit 'atm'.

Value of 1 atm in pascal is 1.013 x 10⁵ Pascal.

(a) On creating vacuum inside the bell jar, the pressure inside it decreases due to the absence of air.

(b) The pressure inside the balloon remains the same as the amount of air inside the balloon does not change. The reason balloon gets more inflated on creating vacuum inside the bell jar is that due to vacuum the air surrounding the balloon has been removed so the pressure surrounding the balloon has decreased. But the air inside the ballon keeps pushing at the same pressure causing the ballon to inflate further.

The barometer is used to measure atmospheric pressure.

The statement 'the atmospheric pressure at a place is 76 cm of Hg' means that the atmospheric pressure at that place is equal to the pressure due to mercury column of height 76 cm .

The value of 76 cm of Hg = 1.013 x 10⁵ Pa.

If the tube is pushed down into the trough of mercury, the barometric height remains unaffected.

If the tube is slightly tilted from vertical, the barometric height remains unaffected.

If a drop of liquid is inserted inside the tube, the barometric height decreases. Reason being, the drop of liquid will vapourize due to vacuum and exert pressure on the mercury column due to which the barometric height will decrease.

The uses of a barometer are—

1. It can be used to measure the atmospheric pressure at a place.
2. It can be used as an altimeter to measure height.

(a) When a barometer is taken to a mine, it's reading increases because the atmospheric pressure increases with depth.

(b) When a barometer is taken to a hill, it's reading decreases because atmospheric pressure decreases with increase in altitude.

(a) If the barometric height gradually falls, it indicates that the moisture is increasing i.e., there is a possibility of rain.

(b) If the barometric height at a height suddenly falls, it means that the pressure at that place has suddenly decreased which indicates the coming of a storm or cyclone.

(c) A gradual increase in the barometric height means that the moisture in air is decreasing. This indicates the coming of dry weather.