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5.1

5.1 State CharaState Characteristics of cteristics of solid , solid , liquid aliquid and gasnd gas

SOLID

SOLID LIQUID LIQUID GASGAS

Microscopic Microscopic view view Particle Particle Arrangement Arrangement

Tightly packed, usually Tightly packed, usually

in a regular pattern. in a regular pattern.

Close together with no Close together with no regular arrangement. regular arrangement.

Well separated with no Well separated with no regular arrangement. regular arrangement. Shape

Shape Fixed Fixed shape shape Follow Follow container container Undefined Undefined ShapeShape Volume

Volume Fixed volume Fixed volume Fixed Fixed volume volume Undefined Undefined volumevolume Kinetic Energy

Kinetic Energy Content

Content Low Low Average Average HighHigh

Compressibility

Compressibility Hard Hard to to compress compress CompressibleCompressible Easily to compressEasily to compress

Flow

Flow Ability Ability Does Does not not flow flow Flows Flows easily easily Flows Flows easilyeasily

Movement Movement

Vibrate (jiggle) but Vibrate (jiggle) but generally do not move generally do not move from place to place. from place to place.

Vibrate, move about, Vibrate, move about, and slide past each and slide past each

other. other.

Vibrate and move freely Vibrate and move freely

at high speeds. at high speeds.

5.2 Define the Density and Pressure 5.2 Define the Density and Pressure

DEFINITION DEFINITION

The density of a material is defined as its

The density of a material is defined as its mass per unitmass per unit volume.

volume.

SYMBOL SYMBOL •

•The symbol of density isThe symbol of density is ρ (rho).ρ (rho). FORMULA

FORMULA Where: Where:

•ρ (rho) is the density, •ρ (rho) is the density, •

•m is the mass in kg,m is the mass in kg, •

•V is the volume in mV is the volume in m33..

SI UNIT SI UNIT

DENSITY

DENSITY

m

m

 

  

  

(2)

 Different Different materials materials usually usually have have different different densities.densities.

Example 1: Example 1:

Calculate the density and relative of wooden block which has dimensions

Calculate the density and relative of wooden block which has dimensions           and a and a mass of mass of .. Solution: Solution:

 









3 3 2 2 2 2 2 2 3 3 4000 4000 10 10 5 5 10 10 2 2 10 10 1 1 10 10 40 40 m m kg  kg  vv m m                             DEFINITION DEFINITION Relativ

Relative density is e density is thethe ratio of the densityratio of the density (mass of a unit(mass of a unit volume)

volume) of a substance to the density of a givenof a substance to the density of a given refer

reference ence material.material.

Specific gravity

Specific gravityusually means relative density withusually means relative density with respect to water.

respect to water.

SYMBOL SYMBOL

•The symbol of relative density isThe symbol of relative density is RDRD..

FORMULA: FORMULA: SI UNIT: SI UNIT:

No Unit

No Unit

   R

   R

  e

  e

   l

   l

  a

  a

   t

   t

   i

   i

  v

  v

  e

  e

   D

   D

  e

  e

  n

  n

  s

  s

   i

   i

   t

   t

  y

  y

(3)

 A solid metal cyl

 A solid metal cylinder has radiuinder has radiuss     cm and length cm and length       . Its mass is. Its mass is   . Find the. Find the density of the metal and its relative density.

density of the metal and its relative density.

Solution: Solution:

 



3 3 6 6 2 2 2 2 2 2 10 10 93 93 .. 3 3 10 10 5 5 .. 0 0 m m l  l  r  r  Volume Volume                       3 3 6 6 3 3 4 4 .. 16539 16539 10 10 93 93 .. 3 3 10 10 65 65 m m kg  kg  vv m m                     539 539 16 16 1000 1000 4 4 16539 16539 RD RD Density, Density, Relative Relative .. ..  water   water  density of  density of   material   material  density of  density of        DEFINITION: DEFINITION: The

The pressurepressure,, PP, is defined as the , is defined as the ratio of force to arearatio of force to area

SYMBOL: SYMBOL: •

•The symbol of Pressure isThe symbol of Pressure is PP.. FORMULA:

FORMULA: Where: Where: •

•P is the Pressure,P is the Pressure, •

•F is the Force in newton,F is the Force in newton, •

•A is the Area im mA is the Area im m22..

SI UNI SI UNITT::

Application of pressure: cutting tools, injection

Application of pressure: cutting tools, injection needleneedle and tip of thumbtack.

and tip of thumbtack.

PRESSURE

PRESSURE

 A

 A

 F 

 F 

 P 

 P 



(4)

Example 3: Example 3:

 A hammer suppl

 A hammer supplies a force ofies a force of   . . The hammer The hammer head has head has an area an area ofof       . . What iWhat is thes the pressure? pressure? Solution: Solution: Pa Pa 10 10 9.86 9.86 or or m m  N  N 10 10 9.86 9.86 10 10 0 0 .. 7 7 700 700 5 5 2 2 5 5 4 4                A  A  F   F   P   P 

5.3 Variation Of Pressure With Depth 5.3 Variation Of Pressure With Depth

Relating Pressure

Relating Pressure  in a liquid to the  in a liquid to the DepthDepth  and Density and Density  of the liquid: of the liquid:

Consider a cylindrical container oh

Consider a cylindrical container oh height,height,  and cross-sectional area, and cross-sectional area,   which is filled with a liquid which is filled with a liquid of density

of density 

Volume

Volume of of liquid liquid in in the the container container ::

 Ah

 Ah

Mass

Mass of of liquid liquid in in the the containecontainer r :: mmV V      Ah Ah    

Force

Force on on point point X X ::

 g   g   Ah  Ah mg  mg   F   F            

 weightweightof of thetheliquidliquid

Pressure

Pressure on on point point X X ::

 



 g   g  h h  A  A  g   g   Ah  Ah  A  A  F   F   P   P             

Therefore Therefore ::

h

h

 

 X

 X

(5)

What will be the: What will be the:

(a) the gauge pressure and (a) the gauge pressure and

(b) the absolute pressure of water

(b) the absolute pressure of water at depthat depth  below the surface? (Given that: below the surface? (Given that:       , and, and    ).). Solution: Solution: a) a) 2 2 2 2 72 72 .. 117 117 117720 117720 12 12 81 81 .. 9 9 1000 1000 m m kkN N  m m  N   N   gh  gh  P 

 P  ga ga ugug ee water water 

   b) b) 2 2 72 72 .. 218 218 101 101 72 72 .. 117 117 Pr  Pr  m m kN  kN   P   P   P   P  essure essure  Absol

 Absoluteute  ga ga ugug ee atmosphereatmosphere

          5.4

5.4 Pascal’s PrinciplePascal’s Principle

Pascal’s principle states that pressure exerted on an enclosed fluid (liquid) is

Pascal’s principle states that pressure exerted on an enclosed fluid (liquid) is transmitted equallytransmitted equally to every part of the f

to every part of the f luid (liquidluid (liquid).).

Hydraulic systems can be used to obtain a large force by the application of a much smaller Hydraulic systems can be used to obtain a large force by the application of a much smaller force.

force.

We can turn this phenomenon to our advantage if we alter the areas exposed to equal We can turn this phenomenon to our advantage if we alter the areas exposed to equal pressures, as in an

pressures, as in an hydraulhydraulic lift:ic lift:

Since the pressure must be t

Since the pressure must be the same everywhere:he same everywhere:         Pressure is; Pressure is;  A  A  F   F   P   P  So; So;

 

11 22



2 2 2 2 2 2 1 1 1 1 2 2 1 1  A  A  A  A  F   F   F   F   A  A  F   F   A  A  F   F   P   P   P   P atat atat

2 2 1 1

(6)

This says that the Force at the

This says that the Force at the outlet (at 2) is augmented by the size of the area of outlet (at 2) is augmented by the size of the area of the outlet.the outlet. So if we make the area 1000 times

So if we make the area 1000 times larger, we can lift 1000 times the force we apply at Flarger, we can lift 1000 times the force we apply at F11..

By applying Pascal’s principle on a simple hydraulic system, By applying Pascal’s principle on a simple hydraulic system,

 Applicatio

 Applications of Pascal’s prinns of Pascal’s principle includciple include the hydraulie the hydraulic jack, hydraulic jack, hydraulic lift and hydrauc lift and hydraulic brakes.lic brakes.

Example of Applications of Pascal’s Principle (Hydraulic Example of Applications of Pascal’s Principle (Hydraulic  Lift). Lift).

2 2 2 2 1 1 1 1

 A

 A

 F 

 F 

 A

 A

 F 

 F 

2 2 2 2 1 1 1 1

 A

 A

 x

 x

 x

 x

 A

 A

 Where: Where: F  F 11 = force at 1 = force at 1 F  F 2 2  = force at 2 = force at 2  A

 A11 = cross sectional Area at 1 = cross sectional Area at 1

 A

 A2 2  = cross sectional Area at 2 = cross sectional Area at 2

 x 

 x 11 = distance moved at 1 = distance moved at 1

 x 

(7)

 A hydraulic car

 A hydraulic car lift has lift has a pump a pump piston with radiuspiston with radius      . The resultant piston has a radius. The resultant piston has a radius of

of      . The total weight of the car and plunger is. The total weight of the car and plunger is      . If the bottom ends of the. If the bottom ends of the piston and plunger are at the same height, what input force is required to stabilize the car and piston and plunger are at the same height, what input force is required to stabilize the car and output plunger?

output plunger?

Solution: Solution:

We need to use the area for circular objects,

We need to use the area for circular objects,      for both the piston and plunger. Apply for both the piston and plunger. Apply Pascal's Principle: Pascal's Principle:

  

  

 



 



 N   N  r  r  r  r   F   F   A  A  A  A  F   F   F   F   B  B  A  A  B  B  B  B  A  A  B  B  A  A 20 20 .. 131 131 150 150 .. 0 0 0120 0120 .. 0 0 20500 20500 22 2 2 2 2 2 2





 

 

 

 





 

 

 

 





 

 

 

 





 

 

 

 



 

 

 

 



 

 

 

 

        5.5 Archimedes' Principle 5.5 Archimedes' Principle Archimedes’ principle

Archimedes’ principle  states that an object which is partially or wholly immersed in a  states that an object which is partially or wholly immersed in a fluid (liquid or gas) is acted upon by an upward buoyant force equal to the weight of the fluid (liquid or gas) is acted upon by an upward buoyant force equal to the weight of the fluid it displaces.

fluid it displaces.

 An object wei

 An object weighs less in wghs less in water than it does in the ater than it does in the air.air.

This loss of weight is due to the upthrust of the water acting upon it and is equal to the weight of This loss of weight is due to the upthrust of the water acting upon it and is equal to the weight of the liquid displaced.

the liquid displaced.

If the weight of the water displaced is less than the weight of the object, the object will sink. If the weight of the water displaced is less than the weight of the object, the object will sink.

Otherwise the object will float, with the weight of the water displaced equal to the weight of the Otherwise the object will float, with the weight of the water displaced equal to the weight of the object.

object.

Archimedes Principle: Archimedes Principle:

The buoyant force is equal to the The buoyant force is equal to the weight of the displaced water. weight of the displaced water.

(8)

Floatation: Floatation:

The principle of floatation states that a floating body displaces its own weight of the liquid in which it The principle of floatation states that a floating body displaces its own weight of the liquid in which it floats.

floats.

 According to Archi

 According to Archimedes’ Princimedes’ Principle:ple:

Therefore; Therefore;

Figure below show four situations of object in a liquid: Figure below show four situations of object in a liquid:

   S    S    i    i    t    t  u  u   a   a    t    t    i    i  o  o  n  n    1    1  F   F  W  W oo  

Buoyant

Buoyant

force

force

object

object

of 

of 

Weight

Weight

However; However;

displaced

displaced

liquid

liquid

of 

of 

Weight

Weight

force

force

Buoyant

Buoyant

 l  l  l  l  o o o o l  l  l  l  o o o o l  l  o o l  l  o o V  V  V  V   g   g  V  V   g   g  V  V   g   g  m m  g   g  m m W  W  W  W                                 

For totally submerged object; For totally submerged object;

l  l  o o V V  V  V      S    S    i    i    t    t  u  u   a   a    t    t    i    i  o  o  n  n    2    2  F   F  W  W oo  

Buoyant

Buoyant

force

force

object

object

of 

of 

Weight

Weight

However; However;

displaced

displaced

liquid

liquid

of 

of 

Weight

Weight

force

force

Buoyant

Buoyant

 l  l  l  l  o o o o l  l  l  l  o o o o l  l  o o l  l  o o V  V  V  V   g   g  V  V   g   g  V  V   g   g  m m  g   g  m m W  W  W  W                         

For totally submerged object; For totally submerged object;

Buoyant force Buoyant force Weight Weight Rising Rising

<

<

Buoyant force Buoyant force Weight Weight Rising Rising

>

>

Buoyant force = W

Buoyant force = Weight of liquid displacedeight of liquid displaced

V Vgg mg mg F F displaced displaced liquid liquid of  of  Weight Weight = = force force Buoyant Buoyant          

(9)

   S    S    i    i    t    t  u  u   a   a    t    t    i    i  o  o  n  n    3    3 However; However;

displaced

displaced

liquid

liquid

of 

of 

Weight

Weight

force

force

Buoyant

Buoyant

 l  l  l  l  o o o o l  l  l  l  o o o o l  l  o o l  l  o o V  V  V  V   g   g  V  V   g   g  V  V   g   g  m m  g   g  m m W  W  W  W                         

For totally submerged object; For totally submerged object;

l  l  o o V V  V  V      S    S    i    i    t    t  u  u   a   a    t    t    i    i  o  o  n  n    4    4  F   F  W  W oo  

Buoyant

Buoyant

force

force

object

object

of 

of 

Weight

Weight

However; However;

displaced

displaced

liquid

liquid

of 

of 

Weight

Weight

force

force

Buoyant

Buoyant

 l  l  l  l  o o o o l  l  l  l  o o o o l  l  o o l  l  o o V  V  V  V   g   g  V  V   g   g  V  V   g   g  m m  g   g  m m W  W  W  W                         

For totally submerged object; For totally submerged object;

l  l  o o V V  V  V   NOTE :

NOTE : ooobject, l object, l liquid disliquid dis pla placeced d 

Archimede

Archimedes' Principle explains why s' Principle explains why steel ships float.steel ships float.

 Applications o

 Applications of Archimedes’ prinf Archimedes’ principle can be ciple can be founfound in ships, d in ships, submarines, hot-air ballonsubmarines, hot-air ballons and thes and the hydrometer. hydrometer. Buoyant force Buoyant force Weight Weight Floating Floating

=

=

=

=

Buoyant force Buoyant force Weight Weight Floating Floating •

• Displaced water weight < ball weightDisplaced water weight < ball weight Sink

Sink

• Displaced water weight = hull Displaced water weight = hull weightweight Float

(10)

Example 6: Example 6:

The buoyant force acting on t

The buoyant force acting on the object will decrease when the:he object will decrease when the:

 weight weight of of the the object object decreasedecrease

Example 7: Example 7:

 A

 A concrete concrete slab slab weight weight isis , when it is fully submerged under the sea, its apparent weight is, when it is fully submerged under the sea, its apparent weight is 

. Calculate the density of the sea water if the volume of the sea water displaced by the. Calculate the density of the sea water if the volume of the sea water displaced by the concrete slab is concrete slab is   .. Solution: Solution:  N   N  48 48 102 102 150 150 ight ight Apprent we Apprent we weight weight Actual Actual force force Buoyant Buoyant            According to Archi

 According to Archimedes’ Princimedes’ Principle:ple:

  

 



 



 



 



3 3 6 6 6 6 1019 1019 81 81 .. 9 9 10 10 4800 4800 48 48 81 81 .. 9 9 10 10 4800 4800 48 48 displaced displaced water water sea sea of  of  weight weight force force Buoyant Buoyant m m kg  kg  Vg  Vg  mg  mg   F   F  B B                                       Object Object Liquid Liquid Buoyant Force Buoyant Force

(11)

Figure below shows a boat loaded with some goods floating on the sea. The density of the sea is Figure below shows a boat loaded with some goods floating on the sea. The density of the sea is 

..

(a) Calculate the weight of the boat. (a) Calculate the weight of the boat.

(b)

(b) Figure below shows the Figure below shows the situation of maximum loading of situation of maximum loading of the boat.the boat.

Calculate the additional weight of goods that has to be added to the boat to reach this Calculate the additional weight of goods that has to be added to the boat to reach this situation. situation. Solution: Solution: (a) (a)  N   N  Vg  Vg  14994 14994 8 8 .. 9 9 5 5 .. 1 1 1020 1020  water   water  sea sea of  of  Weight Weight  boat  boat the the of  of  Weight Weight                   (b)

(b) For For maximum maximum loading;loading;

 N   N  Vg  Vg  44982 44982 8 8 .. 9 9 5 5 .. 4 4 1020 1020 displaced displaced water water sea sea of  of  Weight Weight weight weight Additional Additional  boat  boat of  of  Weight Weight                     Therefore; Therefore;  N   N 

29988

29988

14994

14994

44982

44982

added

added

 be

 be

to

to

goods

goods

of 

of 

weight

weight

Additional

Additional

     

(12)

Tutorial 5a (Density and

Tutorial 5a (Density and Pressure)Pressure)

1.

1. An An object has object has a a mass mass ofof  and a volume of and a volume of . What is the . What is the density of the object?density of the object? 2.

2. A A substancsubstance e having having a a density density ofof . What is . What is the volume of the substance if the mass isthe volume of the substance if the mass is 

?? 3.

3. A A room room with with a a dimension ofdimension of          is filled with is filled with  of air. What is the mass of of air. What is the mass of the air?

the air? 4.

4. A A measuring cylinder measuring cylinder is is filled up filled up with with aa  liquid having a mass of liquid having a mass of . What is the density. What is the density of the liquid?

of the liquid? 5.

5. A A liquid having liquid having a a density ofdensity of . If 1cm. If 1cm33  of the liquid turn into  of the liquid turn into  vapors, find the vapors, find the density of the vapor.

density of the vapor. 6. Liquid

6. Liquid   and and  having a density of having a density of  and and  respectively. Without any  respectively. Without any changeschanges of volume, a

of volume, a  liquid liquid   is added to is added to  liquid liquid . Calculate the additional density of the liquid.. Calculate the additional density of the liquid. 7. A

7. A    pressure is exerted on the pressure is exerted on the  floor. Calculate the force acting on the floor. floor. Calculate the force acting on the floor. 8.

8. A A wood block wood block with a with a dimension ofdimension of               height having a mass of  height having a mass of 

. Determine the pressure exerted by the wood block.. Determine the pressure exerted by the wood block. 9.

9. A A blade wblade with a ith a dimension ofdimension of      having a force of having a force of  to cut a meat. What is the to cut a meat. What is the pressure exerted by the blade?

pressure exerted by the blade? 10.

10. A A tank tank with with a a dimension dimension ofof                 is filled with paraffin.  is filled with paraffin. ((         ))

Calculate : Calculate :

a) The pressure exerted on the base of the tank a) The pressure exerted on the base of the tank b) The force that acted on the

b) The force that acted on the base of the tankbase of the tank 11. A density of seawater is

11. A density of seawater is . What is the pressure exerted by the seawater at a vertical. What is the pressure exerted by the seawater at a vertical depth of

depth of . ( Given:. ( Given:     ))

Tutorial 5b

Tutorial 5b (Pascal’ Principle)(Pascal’ Principle)

1.

1. Figure shows a simple hydraulic system. PisFigure shows a simple hydraulic system. Piston A & B ton A & B has a cross sectional area ofhas a cross sectional area of   

   and and      respectively. respectively.

a)

a) What is What is the pressure the pressure at pat piston A, iston A, whenwhen  load is placed on it? load is placed on it? b)

b) What is What is the the pressure at pressure at piston B.piston B. c)

c) If load W placed aIf load W placed at piston B, determine t piston B, determine the load W that cathe load W that can be pushed by n be pushed by piston B.piston B.

2.

2. Figure shows a Figure shows a pipe system that pipe system that is filled wis filled with oil. If ith oil. If the piston A the piston A is pushed is pushed withwith  force: force: (a) What is the pressure exerted by the oil

(a) What is the pressure exerted by the oil (b)

(13)

Cross sectional area of

Cross sectional area of  ,,       and and Cross sectional Cross sectional area area ofof       

a)

a) What What is is the the pressure at pressure at pistonpiston , if, if  force acting on a piston force acting on a piston   ?? b)

b) If If loadload  placed at piston placed at piston , determine the load, determine the load  that can be pushed by piston that can be pushed by piston .. c)

c) If the distance moveIf the distance moved by the piston A is d by the piston A is 1.5m, what is the d1.5m, what is the distance moved istance moved by piston B?by piston B? d)

d) If If loadload  is being replaced with 210kg load, what is the force acted on piston is being replaced with 210kg load, what is the force acted on piston    in order to in order to support the load at piston

support the load at piston ??

Tutorial 5c

Tutorial 5c (Archimede(Archimedes’ s’ Principle)Principle)

1. An empty boat having a weight of

1. An empty boat having a weight of  is floating statically . is floating statically . ((

        ))

a)

a) What What is is the the buoyant buoyant force?force? b)

b) What is What is the volume the volume of of the displaced the displaced water?water? 2.

2. A A cube of cube of metal having metal having a a volume ofvolume of  is completely submerged in ….is completely submerged in …. a) a) water water ((

   )) b) b) oil oil ((

   )) c) c) Oxygen Oxygen ((

   ))  According to the

 According to the following, wfollowing, what is the buoyant hat is the buoyant force?force? 3.

3. A A metal block metal block having having a a weight ofweight of  is completely being submerged in to the water. The weight is completely being submerged in to the water. The weight of the block when it completely submerged is

of the block when it completely submerged is  a)

a) What iWhat is the s the volume volume of thof the bloe blockck b)

b) What is What is the dethe density onsity of the f the block (block (

   ).).

4.

4. Figure (a) shows an object is weighed in air and found to have a weight of 2.0N. While Figure (b)Figure (a) shows an object is weighed in air and found to have a weight of 2.0N. While Figure (b) shows the object is

shows the object is completelcompletely submerged into the water.y submerged into the water.    a)

a) What iWhat is the s the mass mass of the of the object?object? b)

b) What What is is the the buoyant buoyant force?force? c)

c) What is What is the mass the mass of the of the displaced displaced water?water? d)

(14)

Answer 5a: Answer 5a: 1) 1) 1500 1500 kg/mkg/m33 2) 2) 0.0003 0.0003 mm33 3) 3) 97.5 97.5 kg kg 4) 4) 800 800 kg/mkg/m33 5) 5) 2 2 kg/mkg/m33 6) 840kg/m6) 840kg/m33 7) 7) 20 20 N N 8) 8) 159.6159.6 PaPa 9) 9) 8.3 8.3 x x 101066 Pa Pa 10a) 15696 10a) 15696 Pa Pa 10b) 10b) 188352 188352 N N 11) 11) 303129 303129 PaPa Answer 5b: Answer 5b: 1a)

1a) 200kPa 200kPa b) b) 200kPa 200kPa c)100 c)100 kgkg 2a)

2a) 66.67 66.67 kPa kPa b) b) 13.33N13.33N 3a)

3a) 500 500 Pa Pa b) b) 600N 600N c) c) 0.1 0.1 m m d) d) 140N140N

Answer 5c: Answer 5c: 1a)

1a) 2000 2000 N N 1b) 1b) 0.2 0.2 mm33 2a) 2a) 7.848 7.848 N N 2b) 2b) 6.28 6.28 N N 2c) 2c) 0.12 0.12 NN 3a)

3a) 2 2 x x 1010-5-5 m m33 3b) 3b) 5000 5000 kg/mkg/m33 4a)

4a) 0.2 0.2 kg kg 4b) 4b) 0.2 0.2 N N 4c) 4c) 0.2 0.2 kg kg 4d) 4d) 2 2 x x 1010-4-4 m m33

Minimum requirement assessment task for this topic: Minimum requirement assessment task for this topic: 1 Theory Test & 1 End-of-Chapter

1 Theory Test & 1 End-of-Chapter

Specification of Theory Test: CLO1- C1 & CLO3-C2, A1 Specification of Theory Test: CLO1- C1 & CLO3-C2, A1 Specification of Labwork: CLO2- C2, P1

Specification of Labwork: CLO2- C2, P1 ****************

**************************************************************************************************************************************************************************************** COURSE LEARNING OUTCOME (CLO)

COURSE LEARNING OUTCOME (CLO)

Upon completion of this topic, students should be

Upon completion of this topic, students should be able to:able to: 1.

1. Identify Identify the bathe basic sic concept concept of of solid solid and and fluid (fluid (C1)C1) 2.

2. Apply coApply concept oncept of solid f solid and fluid and fluid to prto prove reove related physilated physics prics principles. nciples. (C2,P1)(C2,P1) 3.

3. Apply the Apply the concept oconcept of solid f solid and fluid and fluid in real in real basic enbasic engineering pgineering problems. roblems. (C2, A1(C2, A1)) ****************

**************************************************************************************************************************************************************************************** Compliance to PLO

Compliance to PLO

 PLO 1, LD1 (Knowledge)-Test 2PLO 1, LD1 (Knowledge)-Test 2

 PLO 2, LD2 PLO 2, LD2 (Practical Skills)- Experiment 3(Practical Skills)- Experiment 3

References

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