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"Mechanical Properties Of Fluids: Videos"

00:00:02 [Music] [Applause] [Music] [Applause] [Music] so good morning everyone so we are we have looked at in the last chapter we have looked at the mechanical properties of solids in this chapter we are going to talk about mechanical properties of fluids and I will just tell you that what we mean by fluids there are essentially three states of matter namely the solid liquid and the gases and the solids are characterized by having specific shapes and sizes okay and if you apply pressure to it the change in volume is negligible and


00:01:12 sometimes it's so small that it cannot be recorded while the same is true for liquids but the liquids cannot take shear stress at all so it doesn't have any specific shape or size and it takes the shape of the container that it is put into and when we come to gases they have a different property in fact the change in volume by application of pressure in gases is enormous or can be enormous and like when you fill in automobile tire with air and by using you know a device so as to fill in air


00:02:09 the air doesn't go and settle in the bottom of the tire rather it just fills the space available to it uniformly while the liquid actually if you pour in liquid it will just go to the bottom run to the bottom and start building up from there the gases do not have that property and because the gases and the liquids they do not have any specific shape they can flow and that's why they're called as fluids so when we talk about so we are discussing mechanical properties of fluids and in that so far we have


00:03:12 defined solids liquids and gases it's these two collectively are called as fluids so it's the liquids and the gases both can flow and anything that can flow is called as a fluid now there is another state of matter in addition to the three which happens at very large temperature when the atoms are actually stripped of their electrons and they acquire a charge and these are called as ions so atoms either I mean which are devoid of some electrons essentially which are at the outermost shell and


00:04:14 they form ions and this state of matter is called as plasma and some scientists are of the opinion that the colloids which are the suspension of tiny particles in a liquid such as milk should also be considered as a separate state of matter but nevertheless we shall talk about primarily these three states of matter solids liquids and gases and as I have told that we have mostly dealt with solids in the last chapter so we shall be more concerned about liquids and gases or collectively which are known as fluids


00:05:03 now so we have fluids all around us and the air that we breathe in is a fluid human body consists of water which is also a fluid with mostly consists of water and a lot of processes that go on in human bodies or in other living beings even in plants they are mediated by fluids such as water or some other you know liquid so it's very important that we understand the properties of the fluids and the way actually to characterize them so let's just go into understanding fluids and also solids


00:05:57 from an inter atomic or an either molecular perspective point of view so solids as we know that have definite shape and pi do not have do not have definite shape and size and the reason being that the intermolecular force of attraction in solids is very large so those attractions or the interactions that keep the molecules stay together and thus the solids have a definite shape the intermolecular force of attraction in liquids is small but not negligible however the intermolecular forces or inter atomic forces in gases are


00:07:17 negligible so these are the sort of from the microscopic point of view we can distinguish the solids liquids and gases in this particular manner but what is important for us in this chapter is to understand some of the properties which distinguish them mechanically that is some of the mechanical properties which are of relevance to us such as density and specific gravity if I ask this question whether a block of code is heavier or a block of iron you definitely say that a block of iron is heavier but that's not true a large


00:08:20 log of port is definitely heavier than nail or a small block of iron so what property distinguishes these two put and iron so let's talk about density to begin with so the word density means let's denote it by a symbol Rho which is equal to a mass divided by the volume so M is equal to mass of the of the substance and V is the volume so so it density is a property of the substance so whether big or small when a particle or when a particular substance is made of certain materials no matter what shape or size


00:09:52 it takes it will have the same density so and the SI unit of density is kg per meter cube and of course the it also are sometimes it's the CGS units a unit is use of density is gram or centimeter or simply it's written as gram per cc in general pressure and temperature will affect the value of the density for a given substance so while putting the density it is customary to talk about ought to mention about the temperature and pressure at which it's calculated so I'll give you some examples of the


00:11:07 densities of certain substances and you would know that that what are the ranges of densities of solids liquids and gases so let's just give you some examples so so we have solids liquids and castles and let's write the material and row so this symbol is called as room so this is in kg per meter cube again we have material and rope in kg per meter cube and all right so we have island having identity 7 point 8 into 10 cube we have an aluminium which has a density of 2 point 7 into 10 cube kg per meter cube


00:13:07 would usually it's taken as the Pinewood it has a density about point 5 into 10 Q and a glass it's 2.5 into now coming to the liquids its water and it's very important to say that at 4 degree centigrade or to 77 Kelvin the value is 1 into 10 Q and seawater is one point zero two five and into 10 to the power 3 sea water is a known to be more dense than ordinary water and then you have mercury which is 13 point 6 into 10 cube and now we have it high Lanka whole is equal to zero point 7 9 into 10 Q so


00:14:44 these are for the liquids as you can see that they are as almost one order of magnitude less than than the solids while it is almost of the same density as the wood has in fact a lower density than that of mercury and even glass has something glass has density which is lesser than mercury now let's go to the gases this is air which as you know is a mixture of nitrogen oxygen and other gases and this is 1.29 remember there is no 10 to the power 3 here it's simply 1.2 9 kg per meter cube helium gas is a 0.179 kg per meter cube


00:15:49 and carbon dioxide has 1.98 kg per meter cube so you see that the gases have a density which are pretty less very less in fact income as compared to the solids and the liquids three orders of magnitude less and this as we have discussed is due to the negligible force of attraction between the atoms or the molecules that constitute these gases and just two of that these all these values that are coated they are coated at a temperature equal to zero degrees centigrade which is 273 Kelvin and a pressure of one


00:16:49 atmosphere in fact it is important to as I as we have talked earlier that these densities are actually functions of temperature and pressure so it's important to mention the temperature and pressure at which there they are calculated or there the values are coated except for water the value is equal to 1 into 10 cube kg per meter cube for 1 gram per cc at 4 degree centigrade so let us now do a problem a simple numerical problem so what is the what is the mass of lead sphere of radius radius 0.5 meter given that the


00:18:00 row of lead is equal to 11300 kg per meter cube so in order to calculate the mass we shall use the formula that mass is equal to density into the volume so in order to find the volume it is a sphere so the volume of a sphere is given by 4/3 PI R cube which is equal to 0.5 cube which is equal to 0.5 to 3 meter cube and a mass is equal to the density of lead multiplied by this volume and that is equal to 11,300 kg per meter cubed multiplied by 0.5 to 3 meter cube and if you simplify this it comes out as 5 9 1 0 H so this is the


00:19:20 the mass of a let's fear which has a radius of 0.5 meter you understand that if we replace lead by an iron sphere or aluminium sphere this mass is going to be different because this quantity even if the radius remains the same because the density of iron or density of aluminium are different than that of slit now we shall continue with more problems later let us now define the other quantity which is called as a specific gravity and specific gravity is defined it is defined as the ratio of a


00:20:23 substance divided by the density a ratio of the density of the substance divided by the density of water as for degree centigrade so it's a dimensionless quantity so this is a density of a substance divided by density of water at 4 degree centigrade now the advantage of defining this let us write specific gravity with the SG this is just the abbreviations used for that for them since the density of water at 4 degree centigrade is equal to 1 kg per meter cube so this becomes this is equal to 1


00:21:22 kg per meter cube so the specific gravity of a substance is just the density of the substance multiplied by 10 to the power minus three in a kg per meter cube however it's numerically equal to the density in the CGS unit okay so this is equal to 10 to the power minus 3 into density of the substance and I should write numerical value because this is dimensionless in and this is equal to simply equal to soap so if this quantity is quoted in CGS units then we will not have this 10 to the power minus 3 so


00:22:26 this is how specific gravity is defined so let's talk about an important concept in the context of fluids that is liquids and gases that is the concept of pressure as we know that pressure is defined as the force exerted per unit area so let's write it with symbols P is equal to F by a where F is the force or the load that is given to a certain object and is the area over which the force acts and the SI unit of pressure is is a Newton for meter squared or it also has a name as Pascal's and one Pascal is


00:23:37 equal to 1 Newton per meter square so let's just consider that a person whose weight is 60 kgs and his birthday is equally distributed by his two legs and each foot has an area of a 600 centimeter square so the pressure that he would be giving on the ground so his weight is equal to 60 into G just for now let's take G to be 10 meter per second squared so this is equal to 600 Newton is the force that is exerting and so this is equal to F and so the pressure is f divided by his two feet which has feared area of 10 say 600


00:24:52 centimeter square so this is equal to 600 Newton divided by 2 into a 600 into 10 to the power minus 4 so each feet has an area of 600 centimeter square so there are two feet each foot has a area of 600 centimeter square 2 feet will have a 1200 centimeter square so this is equal to meter squared so 600 will cancel and this will be 0.5 into 10 to the power minus 4 Newton per meter square so that's the pressure that he exerts due to his own weight so now let us look at an important point about pressure


00:25:56 exerted by fluids so fluids exert pressure on a body from all sides and so let's particularly talk about pressure due to static fluids so we have a container is filled up to water till that level and there is a cube and this fluid exerts a force from all sides and this force acts normally on the surface of the of the material of the cube and what I mean by normally is that the forces act perpendicularly as it's shown here had there been a non perpendicular component that is a component which is not normal to the


00:27:16 surfaces then there will be a component of the force which is parallel to the surface such as say this surface and if there is a component that is parallel to the surface by Newton's third law this cube will in turn exert a force on the fluid equal and opposite to that and because of that the fluid will be set in motion which is contrary to what we have assumed that the fluid is static so there can't be any component of force which is acting at a given angle to the surface it has to be always normal to


00:28:04 the surface of the of the material so this is the basic notion of pressure due to fluids let us now compute that how to calculate the pressure due to a fluid for a given body so let's take again an open container as we have taken earlier let's take a level of water and let's take a sea just to be keeping our discussion simple let's take a cue and let this height be H and so we are this cube is of height H the density of the liquid equals to Rho so the liquid exerts the pressure on the bottom


00:29:25 surface of this Q which is of magnitude so F is equal to mg and this is equal to 8 m is equal to V Rho and a G and we are talking about a column of water of volume B so according to this question V is equal to the height times the area of cross section of these bottom sections so this is of area a and so this is equal to H a Rho G since pressure is defined as force by area so the pressure is equal to F over a which is equal to H Rho so the pressure due to a fluid at a height H below the level of the fluid is


00:30:37 H Rho G so which means that greater the height of the object the pressure will be more soapy simply scale as H given that Rho and G remain constant but there is a small problem there we are taking Rho to be constant which means the density of the liquid remains constant which is a pretty correct by and large correct in the context of liquids except for the case of ocean water where there is an enormous mass of water if you consider it a point at a depth 8 which is significantly below the sea level then


00:31:32 there could be a change in the density of the of the water with height but without getting into that problem we can say that even for gases which are largely compressible there could be significant variation of density with with the height or with the distance from where thing measured so we actually need a more direct calculation of pressure how it varies as a function of depth in a fluid not necessarily in a liquid but in a fluid as soon as we talk about a liquid we can safely assume that the Rho is constant however we need to


00:32:24 know this relation so for doing that let us take this case we draw the same drawing as we have done accepting that so this is the open container is filled up to the water is filled or a liquid is filled up to this level and let us take a small disks like water or the liquid that we are considering and which is measured from the bottom which exists at a at a distance Y from the bottom and the slab is of thickness B Y okay so we are going to calculate the pressure due to static fluid for that we are taking a fluid or


00:33:35 a liquid in an open container and we are measuring the distances from the bottom of the container we have taken a certain mass of water which is at a distance Y from the bottom has a thickness dy and the the liquid has density Rho and we need to calculate the pressure so what are the forces acting on this there is a force that is acting upward or let's call it the pressure and multiplied by the area of cross section of this class which is the force acting which is due to the fluid and in the upward direction


00:34:27 there is also a force acting down which is a say a p plus r DP multiplied by a so we have taken pressure at at a height or this is rather a distance why add P and pressure at a height again which what I mean is distance is equal to P plus DP so this is at height y plus dy again measured from the ground is P plus V P so the force due to the fluid acting upward is P a on this bottom surface of this disc the force that is acting downward is P plus DP into a a denotes the area of the slab and of course we also should consider the


00:35:50 effect of gravity so the gravity this will be so just between before we write the gravity so there is a up P a so this will be P plus DP a minus P a so this is downward and this is upward so this is the fluid pressure also due to the weight of so let's write it as due to gravity let's write that as D F and G which is equal to D M into G so this G subscript stands for gravity and this G stands for acceleration due to gravity so DM is the mass of this disc of the liquid and this is equal to


00:37:20 R Rho G and the DV which is equal to Rho G a B Y so again this is acting downward so the net force is equal to P plus BC a minus P a and a plus or we can write it net force upwards will be P a minus P plus VP a minus Rho G a dy now at equilibrium this net force is going to vanish so then we can write that P a minus P plus DP a minus Rho DP dy equal to zero so if we cancel a from all from both the sides that is divide both sides by eight I get a simple differential equation of the form DP dy equal to minus so this


00:39:21 solution of this differential equation solution of this differential equation will give me the variation of pressure as a function of Y there is a negative sign which tells you that the pressure will be more if the height that is this distance from the bottom is less which means that if you talk about the height from the from the top surface then pressure will actually become more as the height or the depth into the water I mean as it increases the pressure will be more since the distance is measured


00:40:08 from the bottom surface we are getting a minus sign which is which make sense because we need to have the as the water column gets bigger and bigger is going to exert more force at a given point so this is my defining equation which should give me the variation of pressure as a function of the distance either measured from the from the bottom of the container or in other words it can be measured at the top of the container so what we are trying to get at is the following that we should we have considered here the fluid pressure and


00:40:56 pressure due to or rather the this is the force due to the fluid pressure and this is the the force due to gravity but there could be an additional pressure that is that could be acting which is usually the atmospheric pressure let us see that how we actually get this atmospheric pressure so you have this part of the discussion is here so now we go ahead and calculate P as a function of Y by solving this differential equation which means that we are going to integrate this equation in order to get P as a function of Y so that is


00:41:41 obtained so a DP integrated from a p1 to p2 these values of p1 and p2 are really arbitrary which can be a fixed according to the given problem and now I am going to write this as minus Rho G dy and integrated from y1 to y2 so there are two points y1 and why to which are say arbitrary y1 is the distance of this point say point eight from the bottom of the container y2 is the distance of the point B from the again from the bottom of the container where respectively the pressures are p1 here at a and p2 here


00:42:37 as B so we have solved this equation this is fairly easy to solve we simply integrate it and we continue writing with a minus sign and so p2 minus p1 equal to minus rho g y2 minus y1 you must have noticed that here we have taken Rho and G to be constants and that's why they are taken out of the integral however as we told that either for gases or for for liquids in a huge water mass such as an ocean you may not have Rho to be constant Rho could be a function of Y and that functional dependent needs to be known in order to


00:43:37 put it into this equation and integrate suppose Rho is a linear function of Y suppose in some problem Rho is equal to alpha Y in which case we should not put Rho to be a constant and take it out of the integral rather this alpha which is assumed to be a constant here which can be taken out of the integral and this will be an integration of Y dy in which case it will not be y2 minus y1 but it will be Y 2 square minus y 1 square divided by 2 in any case we are not specifying the functional dependence of


00:44:15 Rho and it's taken to be a constant here and we can write this equation so this is how the pressure difference between two points which varies with the distance measured of the two points from the bottom of the containers to be like this now let us assume that that my y2 is this entire height of the water column measured from the bottom in which case my p2 becomes simply equal to P R 0 which is the atmospheric pressure so this is the pressure due to the atmosphere which is known as the atmospheric pressure so


00:45:21 that's equal to p2 and my Y 2 for this particular case my y2 is equal to I call it zero or rather let us call it as H and we can we can get or if you if you call this as the total height of the the water column and now we can measure it from here and absorb the negative sign in which case we can write this instead of H we can write this as 0 that is we are now no longer measuring the distance from the bottom surface but we are measuring it from the top surface in which case my y2 becomes equal to zero


00:46:29 and my y1 let's say becomes equal to H and then my p1 becomes equal to P that I want to calculate and y1 becomes equal to H so under these two condition I can put it in which case what I want to do is that I want to absorb this negative sign and all YT to be equal to zero because now no longer I am measuring it from the bottom surface are measuring it from the top surface and now my p2 minus p1 which becomes equal to my p0 minus P and my the right-hand side which is minus Rho G y2 minus y1 now becomes


00:47:22 equal to Rho G H and my pressure becomes equal to p0 plus Rho G H so this is the final result that we wanted to obtain which says that the pressure at any given point which is at a depth H measured from the top surface of the of the liquid the pressure at a height eight at a depth age inside the liquid is equal to the atmospheric pressure plus the Rho times G times H and so that is the expression for pressure that we wanted to calculate so this part is due to the atmospheric pressure and this part is due to the fluid having


00:48:31 looked at the expression for pressure which is given as P equal to of p0 plus Rho gh this being the atmospheric pressure as we have discussed and this is the pressure due to the liquid column of height H we now want to do some problems using this formula that is pressure due to liquids which is given by this so let's do a problem which says that the surface of water water in a storage tank is is a 20 meter above the water tap in the kitchen of a house so this is understandable that there is a storage


00:50:34 overhead storage water storage tank which is there on the terrace and the distance where the kitchen is or the kitchen tap is the storage tank is located 20 meters above the kitchen tap so the question is calculate the pressure at the tap that and of course given that density of water 4 to 1 into 10 cube kg per meter cube this density is denoted by a quantity called as Rho which almost looks like P but please do not distinguish this from P it's called Rho R H oh so this is equal to Rho of water so now the


00:51:57 pressure at the surface of the tank that is at the surface of water inside the tank so there is atmospheric pressure and the same atmospheric pressure also is there at when the water is issuing out from the tap so essentially the pressure difference is simply given by so Delta P is the pressure difference which is simply given by Rho G into H where Rho is of that of water so this is equal to 1 into 10 cube kg per meter cube G is 9 point 8 meter per second Square and H here is 20 meters if you do


00:52:55 this it becomes 1.96 into 10 to the power 5 Newton per meter square which is also called as 1.96 into 10 to the power 5 Pascal so this is the pressure difference that is there at the between the surface of the water level inside the tank to the tap the nozzle of the tap from which water issues out so this is the simple plug-in type of example let's do another one again a plug-in type however it's related to human body game so what's the difference in blood pressure between the top of the head and the


00:54:31 bottom of the feet of a 1.60 metre tall person standing vertically so so the person who's 1.6 0 meters tall and you're needed to find the the pressure the blood pressure the difference in the blood pressure between the top of his head from the bottom of his feet and the person is assumably standing vertically now the the input that needs to be given in this case is a density of blood and just keep in mind that this density of blood that I'm going to give you is actually the average density of blood because the


00:55:58 blood consists of the blood plasma which has a little less density as compared to the other cells that make up for the blood which has a little more density so this is the average density of blood which is 1 0 6 0 kg per meter cube just take a note that this value for water is thousand kg per meter cube so a blood is a little more dense than water so again the pressure difference is given by a delta P it's equal to Rho G H Rho for blood is given as 1 0 6 0 kg per meter cube with a G to be 9 point 8


00:57:05 meter per second square multiplied by 1 point 6 0 feet and this comes out as 1 6 6 2 0 point 8 Newton per meter square okay so this is the blood pressure difference between the top of his head to the bottom of his feet so let's do another problem and the problem concerns which all of you might have felt at times that you have either traveled up in a through a hill and all travel down very quickly descended from a hill very quickly or it might have happened when you have traveled in an airplane inside the


00:58:16 airplane the pressure is pretty much taken care of but still one feels uncomfortable at times because there is a pressure buildup in the years and what happens is that there is a pop in the year which means that some air is released to equalize the pressure between the inner part of the year drum to the outer part of the Year drum and this as I told you that this can also happen if you are climbing a hill or you're getting down from a hill very quickly and this can cause this popping of there as it says so if it does not


00:59:00 pop then there is a pressure that builds up or rather there is a force that is developed and that's why the year task speaking so the question is what is the so so when you not what is when you are run up a tall thin or run down the hill quickly the years fall and as I told op means that some air is released from the years and this is due to the build-up of the pressure due to that that the body takes a little while to get accustomed to suppose you're climbing down a hill very quickly or you're running down a hill very 01:


00:18 quickly so the pressure difference that you initially had and after you have gotten down quite a few say thousand feet quickly so then this pressure buildup can happen question is suppose this did not happen did not happen what would be the force force on the year drum here drum of area 0.5 centimeters square if our change in altitude which means height altitude or you can just simply write it as height of thousand meter so if there is a height difference of thousand meters that takes place and if 01:01:57 the years don't pop so what's the pressure developed and because of that pressure developed what is the force that is exerted on the eardrum so again P equal to the pressure is equal to H Rho and G now it has to be given that the row of air has the density of air it's one point two nine kg per meter cube so thousand meter multiplied by 1 point 2 9 kg per meter cube multiplied by 9 point 8 meter per second square and this thing when you calculate it it comes out to be 1 2 6 4 2 Newton per meter square so this is the 01:03:08 pressure that is developed between the inner part and the outer part of the ear drum because of this pressure there is going to be a force which is equal to the pressure multiplied by the area which is 1 2 6 4 2 Newton per meter square and you multiply it by 0.5 into 10 to the power minus 4 meter square so the meter square will cancel and this becomes equal to 6 point 3 2 Newton so this fixed point 3 2 Newton is is the force that is exerted on the year now you can take this as suppose just for the argument's sake or making 01:04:07 models simple let's take G equal to 10 for now and which means that there is a weight of 0.6 kg that is exerted on the on the years and this is most of the times this is not an unbearable situation but however you would find actually children crying inside the airplane and the reason more often than not is going to be because of this pressure that develops and that creates an ache in this and the child cries so we have looked at mainly so far the density of liquids we have defined density and we have also looked at 01:05:02 density of solids liquids and gases and we saw that the density of gases is at least something like three orders of magnitude less than that of the other solids and liquids and we have also learned about specific gravity and also we looked at the pressure that is exerted by a liquid column of height H or the pressure that is felt at a point inside the liquid at a depth H from the surface and used that result to compute some simple problems so far you

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VIDEO LECTURES FOR MECHANICAL PROPERTIES OF FLUIDS (CLASS 11)

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Mechanical Properties Of Fluids (Density, Pressure due to Static Fluids) | Video #1 | Neet Physics

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Mechanical Properties Of Fluids (Atmospheric Pressure and Gauge Pressure, Measurement of Pressure) | Video #2 | Neet Physics

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Overview

NEET Class 11 Physics NCERT Chapter 4 Mechanical properties of Fluids

In this video, we will cover:

1. Fluids

2. Pressure

3. Density and Relative density

4. Density of a mixture

5. Effect of temperature and pressure on density

6. Variation of pressure with depth

7. Hydrostatic Paradox

8. Pascal’s Law

9. Hydraulic Lift

10. Hydraulic Brakes

11. Atmospheric pressure

12.  Various Units of Atmosphere pressure

13. Measurement of Pressure

14. NCERT Class 11 Physics Chapter 4 NEET Practice Questions

15. Mechanical properties of Fluids NEET for 2024

16. NEET Physics Questions

17. Mechanical properties of Fluids NEET

18. Mechanical properties of Fluids NEET Coaching

19. Mechanical properties of Fluids NEET 2024

20. NEET from Chapter 4 Class 11 Physics

21. Neetbharath.in NEET pe Jeet LH Physics

22. Physics for NEET 2024

23. NEET Class 11 Chapter 4 Physics Mechanical properties of Fluids

24. NEET Mechanical properties of Fluids Chapter 4 Class 11 Physics

25. Class 11 Chapter 4 Mechanical properties of Fluids NEET Coaching

26. Mechanical properties of Fluids NEET Class 11 Chapter 4 Physics

27. FREE NEET Mechanical properties of Fluids

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29. Neetbharath.in NEET Physics

30. Mechanical properties of Fluids NEET Questions for 2024

31. Questions from NEET Mechanical properties of Fluids

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33. Mechanical properties of Fluids NEET Basic Concepts

34. Mechanical properties of Fluids NEET PYQs

35. Mechanical properties of Fluids NEET

Chapter 4 Mechanical properties of Fluids Class 11 NEET Physics deals with:

  • Fluids can be defined as any substance which is capable of flowing.
  • They don’t have any shape of their own.
  • For example:-water which does not have its own shape but it takes the shape of the container in which it is poured.

But when we pour water in a tumbler it takes the shape of the tumbler

  • Both liquids and gases can be categorised as fluids as they are capable of flowing.
  • Volume of solids, liquids and gas depends on the stress or pressure acting on it.
  • In this chapter we will study if we apply force on the fluid how does it affects the internal properties of fluids.
  • Fluids offer very little resistance to shear stress.
  • We will also study some characteristic properties of fluids.

Pressure

  • Pressure is defined as force per unit area.
  • Pressure = Force/Area
  • For Example:-
    • Consider a very sharp needle which has a small surface area and consider a pencil whose back is very bluntand has more surface area than the needle.
    • If we poke needle in our palm it will hurt as needle gets pierced inside our skin.Whereas if we poke the blunt side of the pencil into our hand it won’t pain so much.
    • This is because area of contact between the palm and the needle is very small therefore the pressure is large.
    • Whereas the area of contact between the pencil and the palm is more therefore the pressure is less.
  • Conclusion: Two factors which determine the magnitude of the pressure are:-
  • Force – greater the force greater is the pressure and vice-versa.
  • Coverage area –greater the area less is the pressure and vice-versa.

Example:-

  • Consider a stuntman lying on the bed of nails which means there are large numbers of nails on any rectangular slab. All the nails are identical and equal in height.
  • We can see that the man is not feeling any pain and he is lying comfortably on the bed. This is because there isa large number of nails and all the nails are closely spaced with each other.
  • All the small pointed nails make large surface area therefore the weight of the body is compensated by the entire area of all the nails.
  • The surface area increases therefore pressure is reduced.
    • But even if one nail is greater than the others then it will hurt. Because then the surface area will be less as a result pressure will be more.

Stuntman lying on bed of nails.

Problem:- A 50 kg girl wearing high heel shoes balances on a single heel. The heel is circular with a diameter 1.0 cm. What is the pressure exerted by the heel on the horizontal floor?

Answer:-

Mass of the girl, m = 50 kg

Diameter of the heel, d = 1 cm = 0.01 m

Radius of the heel, r = d/2 = 0.005m

Area of the heel = πr2 = π (0.005)2 = 7.85 × 10–5 m2

Force exerted by the heel on the floor:

F = mg = 50 × 9.8 = 490 N

Pressure exerted by the heel on the floor:

P=Force/Area =490/(7.85x10-5)

= 6.24 × 106 N m–2

Therefore, the pressure exerted by the heel on the horizontal floor is 6.24 × 106 Nm–2.

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All About: Mechanical Properties Of Fluids NEET Physics NEET Chapter 9

NEET Physics seems to be a quite difficult subject for a lot of students. But, if you get a very good conceptual understanding of the subject, it can be very interesting for you. We, at Neetbharath Tutorials Delhi, will give our best to make NEET Physics Mechanical Properties Of Fluids NEET Chapter 9 super-duper easy for you. Learn about Mechanical Properties Of Fluids NEET Physics in detail with experts at neetbharath.in is what we promise. Learning is a lifelong process. As a student, your focus should always be on learning new facts, concepts and tricks so that you can apply them in real-life scenarios. If your conceptual understanding is clear, you will anyway perform well in your exams. So, please do not just study for the fear of exams or just to score marks, instead learn for the love of learning. We have seen many students mugging up concepts without understanding, which, in turn, makes the subject more difficult and boring for them in the future. Always remember, if your basics are strong, you will be able to understand even the most complex concepts.

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  • Mechanical Properties Of Fluids NEET NEET solutions: All questions from NEET textbook are solved in detail, along with video solutions for each question. Download Mechanical Properties Of Fluids NEET NCERT solutions PDF for free. Students often get stuck while solving the questions in the exercises of NEET textbook. You can watch the NEET solutions in the form of videos with our experts, and can also refer to pdf format on our website as well as app. If you are a student of CBSE or preparing for exams for NEET, JEE, Civil services, NDA etc. NEET textbooks are very important for you. Please note that we have uploaded videos as per the Revised NEET textbooks (2023-24). Therefore, you can completely rely on the content on our website/app.
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Mechanical Properties Of Fluids FAQs

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Class 11 Physics Mechanical Properties Of Fluids NCERT Chapter 9


Revision of Class 11 Physics Mechanical Properties Of Fluids is a crucial aspect of effective learning. Revision plays a vital role in the learning process and is especially important before exams. Here are some key points you can consider emphasizing in your content:

  1. - Retention and Memory: Regularly reviewing and revisiting the material of Class 11 Physics Mechanical Properties Of Fluids helps reinforce the concepts in student's minds. It strengthens memory pathways, making it easier to recall information during exams and beyond.
  2. - Consolidation of Knowledge: When you revise notes, you are essentially consolidating your knowledge. This means connecting new information with what you already know, making the overall understanding more robust.
  3. - Identifying Knowledge Gaps: Revision allows students to identify any gaps in their understanding or areas where they need further clarification. This gives you a chance to seek help or delve deeper into those topics. For detailed understanding, you can always refer to the videos of Mechanical Properties Of Fluids Class 11 Physics NCERT Chapter 9 on neetbharath.in
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  6. - Spacing Effect: Spacing out revision sessions over time, rather than cramming all at once, has been shown to improve long-term retention. Create a revision schedule leading up to the exams to allow for spaced practice.
  7. - Regular Revision over Cramming: Regular and consistent revision throughout the academic year is very important. Waiting until the last moment to cram everything can be overwhelming and less effective than spaced-out revision.
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By highlighting the benefits and strategies of effective revision, you can approach your studies more mindfully and achieve better results in your exams. Best of luck bachhon!

Class 11 Physics seems to be a quite difficult subject for a lot of students. But, if you get a very good conceptual understanding of the subject, it can be very interesting for you.

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Always remember, it is very important to study with full concentration during Revision. Here are a few tips for you on how to revise with full focus:

  1. - Create a Distraction-Free Environment: Find a quiet and comfortable place to study where you can minimize distractions. Turn off or silence your phone, log out of social media accounts, and inform others around you that you need uninterrupted study time. A dedicated study environment will help you focus better.
  2. - Set Specific Goals: Before starting your study session, set clear and achievable goals. Break down your study material into smaller tasks, and plan what you want to accomplish during each session. Having specific goals will give you a sense of direction and purpose, making it easier to concentrate.
  3. - Use the Pomodoro Technique: The Pomodoro Technique is a time management method that involves studying in short, focused intervals, typically 25 minutes, followed by a short break of 5 minutes. After completing four sessions, take a longer break of around 15-30 minutes. This technique can improve focus and productivity by providing regular breaks to recharge.
  4. - Stay Organized: Keep your study materials, notes, and resources well-organized. Having everything you need at hand will save time and reduce distractions caused by searching for materials. Use color-coded folders or digital tools to maintain a structured study system.
  5. - Practice Mindfulness and Meditation: Before you begin studying, take a few minutes to practice mindfulness or meditation. Deep breathing exercises and clearing your mind of distractions can help you approach your study session with a calm and focused mindset.

- Remember, studying with full concentration is a skill that takes time and practice to develop. If you find your mind wandering during study sessions, gently bring your focus back to the task at hand and be patient with yourself. With consistent effort, you can improve your ability to concentrate and make the most of your study time.

Last but not the least, To get the best hold on Class 11 Physics Mechanical Properties Of Fluids Book Chapter 9. Do not forget to check out:

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