Pressure

Pressure: The force acting per unit area of surface is called pressure.

i.e. Pressure = Force / Area

P = F / A

Unit of pressure: Its S.I. unit 16 N/m² It is known as Pascal (Pa)

Liquid pressure: The pressure exerted by liquid on walls of its container is called liquid pressure.

Measurement of liquid pressure

Let us consider a liquid of density ‘d’ and depth ‘h’ as shown in the figure.

Now, We have,

P = F / A

P = (m x g) / A [Since, F = mg]

P = (d x v x g) / A [Since, d = m/v]

P = (d x l x b x h x g) / A [Since, v = l x b x h]

P = (d x A x h x g) / A [Since, A = l x b]

P = d x h x g

P = hdg

This is the required formula to calculate liquid pressure.

Q) What are the factors on which liquid pressure depend?

Factors on which liquid pressure depend on are

1. Density of liquid (d)
2. Depth of liquid (h)
3. Acceleration due to gravity (g)

Pascal’s law: This law states that,” The pressure applied in an enclosed liquid kept in closed container is transmitted equally in all directions”

Let us consider a machine with two pistons, A and B of area ‘A₁’ and ‘A₂’ respectively, as shown. The force applied in piston A is ‘F₁’ , so that force exerted in piston B is ‘F₂’ .

According to statement,

Pressure in Piston A = Pressure in Piston B

P₁ = P₂

F₁ / A₁ = F₂ / A₂

Also,  F₁ / A₁ = F₂ / A₂ = F₃ / A₃ …………. F_n / A_n

Q) Name the devices in which Pascal’s law is applied.

Pascal’s law is applied in hydraulic machines such as hydraulic lift, hydraulic press, hydraulic brakes etc

Q) Hydraulic machine is called force multiplier. Why? Explain

Let us consider a hydraulic machine with two pistons A and B of small area A₁ and A₂ respectively as shown in figure.

Now,

Force applied in piston A is F₁ so that force developed in piston B is F₂ .

According to Pascal’s law,

F₁ / A₁ = F₂ / A₂

F₂ =  (F₁ xA₂) / A₁

Since A₂ > A₂ , Therefore  F₂ > F₁

Hence, hydraulic machines multiply the applied force and make it larger, so it is called force multiplier.

Q) What is the role of liquid in hydraulic machine?

The liquid transmits applied pressure equally in all the directions.

Solved numerical

a) Calculate the force exerted in piston B

A₁ = 100 cm² = 100 / (100×100) m² = 0.01 m²

F₁ = 150 N

A₂ = 1 m²

We know that,

F₁ / A₁ = F₂ / A₂

150 / 0.01 = F₂ / 1

F₂ = 15000N

Hence, force exerted is 15000 N.

b) Calculate area in piston B and force on piston C from figure.

A₁ = 10 cm² = 10 / (100×100) m² = 0.001 m²

F₁ = 150 N

F₂ = 300 N

A₂ = ?

A₃ = 30 cm² = 30 / (100×100) m² = 0.003 m²

We know that,

F₁ / A₁ = F₂ / A₂ = F₃ / A₃

150 / 0.001 = 300 / A₂ = F₃ / 0.003

Taking first and second ratio we get,

A₂  = (0.001 x 300) / 150 =  0.002 m²

Again, taking first and third ratio we get,

F₃  = (150 x 0.003) / 0.001 =  450 N

Hence, Area in piston B is 0.002 m² and force in piston C is 450 N.

2) Ratio of area of small piston and large piston of hydraulic machine is 2:5. If 200N force is applied in smaller piston, calculate force exerted in larger piston.

Let area of small piston be 2x m² and large be 5x m²

A₁ = 2x m²

F₁ = 200 N

A₂ = 5x m²

F₂ = ?

We know that,

F₁ / A₁ = F₂ / A₂

200 / 2x = F₂ / 5x

F₂ = 500 N

Hence, 500 N is exerted in larger piston.

Study and answer

a) On which law is the given device based on? State the law.

The device is based on Pascal’s law of liquid pressure which states that “The pressure exerted in an enclosed liquid kept in a closed container transmit pressure equally in all directions”

b) What is the role of liquid in it?

Role of liquid is to transmit pressure equally in all directions.

c) Calculate force applied in it.

A₁ = 1 m²

F₁ = ?

A₂ = 3 m²

F₂ = 750 N

We know that,

F₁ / A₁ = F₂ / A₂

F₁ = 750 / 3 = 250 N

Hence, force applied in it is 250 N.

Upthrust: The resultant upward force exerted by a liquid on a body immersed in it is called upthrust.

i.e. upthrust = loss of weight

Factors on which upthrust depends

1. Density of liquid (d)
2. Acceleration due to gravity (g)
3. Volume of body (v)

a) Where does a ship carry more load either on sea water or on river water. Why?

A ship carries more load on sea water as density of sea water is more than the density of river water so that upthrust exerted by sea water is more than by river water as U∝d.

b) What is the difference between floating of a boat in sea and in river. Why?

The hull of boat sinks more in river than in sea because the density of sea water is more than of river water so that upthrust exerted by sea water is more than by river water as U∝d.

c) Its easier ta lift the bucket of water in side water than in air in the well. Why?

We know that the density of water is maw than density of air so that upthrust exerted by water is more than by air as U∝d. Therefore, its easier to lift bucket of water inside water than in air in the well.

Archimedes’s principle: This principle states that. “when a body is fully or partially immersed in a liquid it experiences an upthrust which equal to weight of liquid displaced by it.”

i.e upthrust = weight of liquid displaced

Study and answer.

a) Which law is verified?

Archimedes principle is verified.

b) What is the weight of water displaced?

Weight Of water displaced 5N

c) How much upthrust is acting on a body. Why?.

5N upthrust is acting on a body as upthrust is equal to weight of liquid displaced as per Archimedes principle.

d) What is the loss of weight of body?

Loss of weight = 5N

e) What is the weight of body in air?

Weight of body in air = Weight of body in water + Loss of weight

Weight of body in air = 10 + 5 = 15N

f) If 1kg is equal to 10 N, what is the mass of body?

1O N = 1 kg

1 N = 1/10 kg

15 N = 15/10 kg = 1.5 kg

Hence, mass of body is 1.5 kg

Laws of floatation: This law states that, “A floating body displaces liquid equal to its weight.”

i.e. weight of body = weight of liquid displaced

Barometer: It is the device which is used to measure atmospheric pressure.

Mercury barometer: The device in which height of mercury column is used to measure atmospheric pressure is called mercury barometer.

Structure of mercury barometer.

Working of mercury barometer:

A trough is taken and 2/3rd of it is filled with mercury. About 1m long glass tube is taken and its completely filled with mercury. The mouth of glass tube is closed and it is inverted and its filled as shown in figure. The mercury level in glass tube first falls and remains constant at a level. This height of mercury in glass tube gives atmospheric pressure at that place.

a) Atmospheric pressure at sea level is 760mm of Hg. What does it mean?

It mean that the height of mercury in mercury barometer due to atmospheric pressure is 760 mm.

b) What is Torricellian vacuum?

The vacuum just above the level of mercury in glass tube of mercury barometer is called Torricellian vacuum.

c) What happens to the level of mercury in mercury barometer if it is taken to higher altitude? 

If mercury barometer is taken to higher altitude, the level of mercury falls down because atmospheric pressure decrease in higher altitude.

d) Can we use water instead of mercury in barometer. Why?

We can use water instead of mercury in barometer but about 10m long glass tube should be used which is difficult to handle.

Atmospheric pressure: The pressure exerted by air of atmosphere on the surface is called atmospheric pressure. Its value is 760 mm of Hg or 10⁵ pascal in sea level.

Application of atmospheric pressure

A) Syringe

The medical instrument which is used to inject medicines into body or to draw out blood from body is called syringe.

Structure of Syringe

Working of syringe:

When nozzle of syringe is dipped in liquid and piston is pulled outward the vacuum is formed in cylinder. The atmosphere pressure pushes liquid in vacuum so that, liquid is filled in it. When piston is pushed inwards the liquid is pushed outwards. In this way, a syringe works.

B) Air pump:

The device which is used to fill air in tyre of vehicles, balloons, balls etc is called air pump

Structure of air pump

Working of air pump:

When piston is pulled outwards, the air enters the cylinder of air pump. When piston is pushed inwards, the air in cylinder is compressed so that it fills in tyre through nozzle and valve in tyre. This process is repeated continuously till the required amount of air is filled in tyre, balls, etc. In this way, air pump works.

C) Water pump:

The device which is used to draw underground water is called water pump.

Structure of water pump

Working of water pump

The water pump work due to repetition of upstroke and downstroke.

a. Upstroke: When handle of water pump is pushed down piston goes upward. This is called upstroke. During upstroke, piston valve closes, the water above piston comes upward and flow through spout. the foot valve opens and water rises.

b. Downstroke: When handle of pump is pulled up, piston goes downwards. This is called downstroke. During this process, the foot valve closes and piston valve opens so that water rises above the piston.

Solved exercise

a) What are two ideal conditions for hydraulic press to function?

Conditions for hydraulic press to function are Liquid is almost incompressible and Liquid transmits applied pressure equally in all directions

b) A hydraulic brake functions because of two properties of liquid used. What are those two properties.

Those properties are: Liquid is almost incompressible and Liquid transmits applied pressure equally in all directions .

c) Is Archimedes principle applicable to gases and solid too?

Yes, it’s applicable to both.

d) Write any two instruments based on Archimedes principle.

Instrument based on Archimedes principle are ship and hydrometer.

e) What are the conditions of an object to float on a liquid?

Conditions of an object to float on liquid is density of object is equal to that of liquid and weight of body is equal to weight of liquid displaced by it.

f) How much weight of air is there over your head (0.0125 m²)

P = 10⁵ Pa

A = 0.0125 m²

F = ?

We know,

P = F / A

F = P x A = 10⁵ x 0.0125 = 1250N

Hence, 1250 N weight of air is over our head.

g) Write two instruments which function due to atmospheric pressure.

Instruments which function due to atmospheric pressure are: syringe and a pump.

h) When does an abject float on liquid?

An object floats on liquid when the object displaces liquid equal to its own weight.

i) During what operation what does water come out of spout of water pump?

During upstroke, water comes out of spout of water pump

j) How many valves are there in water pump. Where do they lie?

There are 2 valves in water pump, they are:

1. Piston valve lies in the piston.
2. Foot valve: It lies between cylinder and pipes of water pump.

k) What instruments are used to measure and maintain air pressure in tyre of vehicle?

Pressure gauge is used to measure air pressure in tyre and air pump is used to maintain it.

l) What length of glass tube is taken for mercury barometer?

About 1m long glass tube taken for mercury barometer

m) How does a hydraulic press function?

It functions on the basis of Pascal’s law. According to which “The pressure applied in an enclosed liquid in closed container is transmitted equally in all directions.”

n) Why does an egg float on salt solution but sink in pure water?

We know that density of salt solution is more than density of pure water so that the upthrust exerted by salt solution is more than that of pure water as U∝d. Therefore egg floats on salt solution but sinks in pure water.

o) A wooden cork remains above surface of water but rubber cork floats by remaining just below the surface of water. Why?

The wooden cork remains above surface of water because the very small portion of it can displace water equal to its own weight but the whole volume of rubber cork displaces more water equal to its weight as it has mare weight than wooden cork. Therefore, a wooden cork remains above surface of water but rubber cork floats by remaining just below surface of water.

p) Differentiate between Pascal’s law and law of floatation.

q) Why does a Submarine remain floating below surface of water?

The ballast tank in submarine is filled with water in submarine to make it heavy so that it can displace water equal to its weight. So, it floats below surface of water.

r) Iron nail sink but a ship made of iron floats in water. Why?

Iron nail can’t displace water equal to its weight but a ship of iron is designed in such a way that it can displace water equal to its weight. Therefore, iron nail sinks but ship floats on water.

s) Write two differences between air and water pump.

t) Why is there vacuum above level of mercury in mercury barometer?

As the atmospheric pressure can’t lift mercury upto 1m in glass tube of mercury barometer. The mercury level in glass tube falls dawn and vacuum is formed above its level. Therefore there’s vacuum above level of mercury in mercury barometer.

u) Why does the height of column of mercury drop as it moves to higher altitude?

Height of column of mercury drops because the atmospheric pressure decreases with increase in altitude.

v) Standard atmospheric pressure: The atmospheric pressure at sea level is called standard atmospheric pressure.

w) Write difference between:

x) Why can’t a gas be used in hydraulic press?

Gas can’t be used in hydraulic press because gas is compressible so that it doesn’t transmit applied pressure equally in all directions.

y) Why is it easier to swim in salty water than in fresh water?

We know that density of salty water is more than density of freshwater. So that, upthrust exerted by salty water is more than that by fresh water as U∝d. Therefore, it is easier to swim in salty water than in fresh water.

Study and Answer

1) A stone is weighed in air, salt, water solution and in saturated salt solution. Weight is given and answer the questions.

a) Identify A, B, C, D?

A-Saturated solution

B-Air

C-Salt solution

D-Water

b) What upthrust experienced by a stone in stone solution and in strong salt solution?

Weight in air = 21N

Weight in salt solution = 19 N

Weight in strong salt solution = 17 N

Upthrust in salt solution = Weight in air – Weight in salt solution = 21N-19N = 2N

Again,

Upthrust in strong solution =  Weight in air – Weight in strong solution = 21N-17N = 4N

c) Calculate volume of stone.

Upthrust in strong = Weight in air – Weight in water = 21N-20N = 1N

We know that,

U = d.g.v

1 = 1000 x 10 x v

v = 0.0001 m³

Hence, volume of stone is 0.0001 m³

d) Calculate mass of stone.

Weight of stone = 21N

10N = 1 / 10 kg

21N = 21 / 10 kg = 2.1 kg

Hence, mass of stone is 2.1 kg

2) Density of 3 solids are given. Study and answer.

a) If equal mass of each A, B, C is taken and which of them has minimum volume?Why?

B has minimum volume because its density is also maximum as d=m/v

b) If equal volume of each is taken, which has maximum mass. Why?

B has maximum volume because its density is also maximum as d=m/v

c) If A, B and C are separately placed in water, which of them sinks and floats? Give reason.

If A is kept in water it floats on it by keeping its same part above water and some part below because its density is less than density of water.

If B is kept in water, it completely sinks because its density is more than density of water.

If C is Kept in water, it just float on water as its density is equal to density of water.

d) Calculate relate relative density of each.

Density of A  = 800 kg/m³

Density of B  = 1200 kg/m³

Density of C =  1000 kg/m³

Density of water = 1000 kg/m³

We know,

Relative Density of A =  Density of A / Density of water = 800/1000 = 0.8

Relative Density of B =  Density of B / Density of water = 1200/1000 = 1.2

Relative Density of C =  Density of C / Density of water = 1000/1000 = 1

Give reasons

a) A bucket can be filled faster on ground floor than on upper floor in a building.

We know that liquid pressure increases with increase in depth as p∝h as depth of water is more on ground than upper floor, so water flows with higher pressure on ground floor. So, a bucket can be filled faster on ground floor than upper floor in a building.

b) Atmospheric pressure decreases with altitude of a place.

When altitude of a place increases, the thickness of air above it decreases. Therefore, atmospheric pressure decrease with altitude of a place.

c) Pascals law is not applicable to gas. It is applicable only to liquid?

Pascals law is not applicable to gas because gas is compressible so that it cannot transmit applied pressure but its applicable in liquid because liquid incompressible so that it transmit applied pressure equally in all direction.

d) A balloon (inflated) pressed in water tries to come up itself?

A balloon pressed in water tries to come up itself due to upthrust of water.

e) An air-filled balloon burst when it reaches a significant height.

We know that atmospheric pressure decreases with increase in altitude when air-filled balloon reaches a height, or pressure inside it becomes greater than atmospheric pressure. Therefore, balloon bursts when it reaches a significant height.

f) A hand operated water pump cannot lift water higher than 10m.

A hand pump works with the help of atmospheric pressure but atmospheric pressure can lift water only upto 10m height. Therefore, a hand operated water pump cant lift water higher than 10m.

Solved numerical problems.

1) Calculate pressure exerted by object of 100 kg on its bottom if base of abject = 250 cm²?

F = 100kg = 100×9.8 = 980N

A = 250cm²= 250 / (100×100) m² = 0.025 m²

P = ?

We know that

P = F / A = 980 / 0.025 = 39200 Pa

Hence, pressure exerted by object is 39,200 Pascal.

2) Calculate pressure exerted by water on bottom of deep dam of 12m from its surface.

h= 12m

d = 1000 kg/m³

g= 9.8 m/s²

P = ?

We know that ,

P = hdg = 12 x 1000 x 9.8 = 117600 Pa

Hence, pressure exerted by water is 117600 Pascal.

3) Calculate upthrust experienced by A, B, C in given diagrams.

For first figure,

d = 1000 kg/m³

g= 9.8 m/s²

A = 12 m²

We have,

A = 12 m²

l² = 12

l = 2 root 3 m³

Now, v = l³  = (2 root 3)³ m³

Again, U = dgv = 1000×9.8×24 root 3 = 407378.34 N

Upthrust experience is 407378.34 N

For second figure,

d = 1000 kg/m³

g= 9.8 m/s²

A = 1.5 m²

h = (5-2)m = 3m

U = ?

Now, U = dgv = dgAh = 1000×9.8×1.5×3 = 44100 N

Upthrust experience is 44100 N

For third figure,

d = 1000 kg/m³

g= 9.8 m/s²

A = 1.5 m²

h = (6-2)m = 4m

U = ?

Now, U = dgv = dgAh = 1000×9.8×1.5×4 = 58800 N

Upthrust experience is 58800 N

4) Study and answer.

a) Calculate upthrust experienced by stone.

Weight of store in air = 21N

Weight of store in water = 19N

Loss of weight = 21-19 = 2N

U = ?

We have, U = Loss of weight = 2N

b) Calculate mass of water displaced?

According to Archimedes principle,

U = Weight of water displaced

Weight of water = 2N

Now, mass of water displaced  = (2/9.8)kg = 0.2 kg

c) State law in which figure is based.

Figure is based an Archimedes principle which states that,”when a body is fully or partially immersed in a liquid, it experiences upthrust equal to weight of liquid displaced by it.”

5) Study and answer

Volume of object = 20 cm³ = 20/(100x100x100) m³  = 0.00002m³

Now,

Volume of object = Volume of water displaced

Volume of water displaced  = 20.00002m³

Now,

density of water = mass of water displaced / volume of water displaced

1000 = mass of water displaced / 0.00002

mass of water displaced = 1000 x 0.00002 = 0.02 kg

Since object is floating,

mass of water displaced = mass of object

mass of object =  0.02 kg

Now, density = mass of object / volume = 0.02 / 0.00002 = 1000 kg/m³

a) Calculate relative density of object.

RD = density of object / density of water = 1000 / 1000 = 1

b) Calculate upthrust experienced by object.

We have, U = dgv = 1000 x 9.8 x 0.00002 = 0.196 N

Hence, upthrust experienced by object is 0.196 N.

c) Calculate weight of object in air and water.

Weight of object in air = 0.02 x 9.8 = 0.196 N

Weight of object in water = Weight of object in air – Upthrust = 0.196 – 0.196 = 0N

d) Calculate mass of object.

Mass of object is 0.02 kg

e) State law in which figure is based.

Figure is based on law of floatation which states that, “A floating body displaces liquid equal to its weight.”

6) The ratio of cross sectional area of bigger cylinder to smaller of hydraulic press is 14:3. If the load to be lifted is of 100kg, calculate force to be applied to lift the load?

Let area of bigger cylinder be 14x m² and that of smaller cylinder be 3x m²

A₁ = 3x m²

A₂ = 14x m²

F₂ = 100×9.8 = 980 N

F₁ = ?

We know that,

F₁ / A₁ = F₂ / A₂

F₁ / 3x = 980 / 14x

F₁ = 210 N

Hence, required force is 210 N.

7) In the given diagram, if area of piston A is 0.02 m² and force applied to it is 300N, calculate

A₁ = 0.02 m²

A₂ = 0.1  m²

F₁ = 300N

F₃ = 800N

A₄ = 2×0.02  m² = 0.1 m²

Using,  F₁ / A₁ = F₂ / A₂ = F₃ / A₃  ………….. = F_n / A_n

a) Balanced load at B if area is 0.1 m²

F₂ = ?

Using first and second ratio,

300 / 0.02 =  F₂ / 0.1

F₂ = 1500 N

b) Area of piston if 800N force is balanced on it.

A₃ = ?

Using first and third ratio,

300 / 0.02 = 800 /  A₃

A₃ = 0.053 m²

c) Load balanced at D

F₄ = ?

Using first and fourth ratio,

300 / 0.02 =  F₄ /  0.1

F₄ =  1500 N

8) Volume of piece of ice is 30,000 cm³ . When that ice is kept on water, what portion of ice remains above water surface? Calculate the density of ice is 0.9 gm/cm³

Volume of ice = 30000 cm³

Density of ice =  0.9 gm/cm³

We have,

density of ice = mass of ice / volume office

0.9 = mass of ice / 30000

mass of ice = 30000 x 0.9 = 27000 gm

According to law of floatation,

mass of ice = mass of water displaced

mass of water displaced = 27000 gm

Now,

density of water = mass of water displaced / volume of water displaced

1 = 27000 / Volume of water displaced

Volume of water displaced = 27,000 cm³

We know that,

Volume of water displaced = Immersed volume of ice

Immersed volume of ice = 27000 cm³

Immersed portion of ice = Immersed volume of ice / Total volume of ice

Immersed portion of ice = 9/10

Now, portion above water = 1-(9/10) = 1/10 part.