 All right welcome everyone waiting for others to join in so probably we'll wait for a minute in a minute we'll start okay which chapter is going on in your school can you tell in your school what is going on started neutral laws of motion that is in Yashonpur in in HSR fluid property is finished okay that is okay so see what we were doing was system of particles chapter okay which is big right so I don't want by the time I finish the center of mass or the system of particles chapter your school goes ahead in some other dimension okay so that's the reason why I have decided that I will take a pause in system of chapter okay so we will we have already done one portion of the chapter that is definition of definition of the system okay so now I'm planning to go to the book number two and take mechanical properties of solids and fluids and after we are done with these two topics and we have caught up with your school what is happening there then I'll go back to the system of particles okay so that we are not different from what is happening in your school right so this is the new chapter yes so do you guys like this idea or you want to continue with system of particles something else going on in school and something else here semester portions are out what is the semester portion what is the portion and when is the exam you don't know yet okay all right anyways okay so this is a new chapter all together and I think some of you have had ut is going on in your school and it just got over right so I can see that some of you might not have completed the homework or at least you have some excuse that you're not done with the assignment because of ut okay so can you submit today by evening by end of the day whichever time you want to take till the night you have time whatever the assignment that is given to you for this week when you complete and submit everyone I have not seen many submissions this week probably because of the because of your ut and all okay so I'm assuming you'll be submitting today take all right so you submitted after the class forgot to send all I got the Pranav got it so Pranav's and when is a semester exam Pranav 16th of September October okay by then we'll be done don't worry all right guys so let us quickly continue with from where we have left now those who have done the assignment or those who want to who those who will be doing it today today is the last day okay I've extended the deadline so you can take today's time also because of your ut you can take a little bit more time and you will keep on seeing these things for example when I do the assignment you will see that there is a term called radius of gyration have you ever encountered it radius of gyration anyone so if radius of gyration is k for any object if radius of gyration is k all right yeah it is there in the assignment so if if the radius of gyration is k with respect to an axis the moment of inertia will be m into k square very simple definition radius of if radius of gyration is k about one axis moment of inertia will be m into k square okay so all of you can you find out k for a disc of radius r how much it is it's a uniform disc uniform mass distribution if radius is r what is the radius of gyration for the disc about the natural axis how much it is do you remember moment of inertia for the disc how much it is okay puja got it others what is the formula for moment of inertia for the disc mr square by two there's a formula for moment of inertia and if radius of gyration is k this should be equal to m into k square so from here you'll get the value of k to be r by root two so it's a very simple concept fine so sometimes when you solve numerical you'll encounter this term radius of gyration and this is how you deal with it okay so all of you do you want to take one numerical for the moment of inertia then start a new chapter or directly start a new chapter because i think that some of you might have forgotten because uts are going on okay take a problem all right fine let's take a question let's take a question on the moment of inertia once we are done we will start a new chapter and you know whatever is the integration that is required in physics it's over now okay so now you will see integration only in the advanced level questions but till now whatever integration we have done integration or calculus whatever we have done if you are comfortable with that you're done with calculus for physics okay so you need to if you're not comfortable with the calculus in physics just revise it once again and you will become comfortable with it because it calculus is a very repetitive thing it's not very tricky you're finding it tricky because you're doing it for the first time that is the only reason okay so i'm taking a question in which calculus is involved so that even calculus is revised okay so suppose you have a hollow cone mechanical property of solid is a very simple chapter and you will see that once i start doing it but before i start let's do a question on system of particles that is which concept moment of inertia concept you need to find moment of inertia about the natural axis okay about this axis for this cone of mass m cone radius is r and the height of the cone h slant height is also given to you l everything is given slant height is l from here to here it is l radius is r everyone start doing it moment of inertia about the natural axis you have to find out this is r this is l this is h which formula will you use moment of inertia will be equal to integral of moment of inertia of a small element dm okay what is a differential element will you take here everyone how that dm will be what do you think everyone dm mass how will it look like rings so take a ring like this ring has a thickness take a ring of radius small r okay radius is small r take this length slant length as x so this thickness of the ring would be dx okay now for ring what is the moment of inertia if mass is dm moment of inertia for the ring will be how much everyone what will be the moment of inertia r square dm for ring right but the radius you have to take is small r not the capital r radius of the ring square into dm this is your di moment of inertia of the ring this you have to integrate to find out the total moment of inertia okay in order to find that you need to write dm in terms of dr okay so how will you do that mass is distributed in area or volume everyone mass is distributed in the area or volume it is a hollow cone mass is distributed in area so mass per unit area will be what m divided by pi rl there is mass per unit area right then into there is no base it is hollow into area of the ring what should I write area of the ring as everyone what it is area of the ring to pi r dr is this correct is this correct area of the ring to pi r dr everyone is this the area of the ring to pi r dx thickness is not dr thickness is dx thickness is along the slant thickness is not along the radius got it that's a big change here so this is your dm okay so I will cancel out pi from here I will get it as 2m 2m rl into r into dx now the problem is although you have written m in terms of r but then dx is also coming in okay can I write dx in terms of dr can you write dx in terms of dr think over it tell me what is dx in terms of dr or first tell me what is x in terms of r what is the relation between x and r everyone all of you agree that o a b and o c d they are similar triangles this triangle is similar to the bigger triangle all of you agree right so I can have a relation between x and r by ratio of the sides x by r is equal to what l by capital R so from here x is equal to l by r into small r everybody understand this this is x in terms of r now when you differentiate it differentiate it with respect to r you will get dx by dr it is a trick it's a common trick okay so first time when you see you'll be like oh something completely different but then you will get used to it once you keep on doing it dx will be equal to l by r dr okay so just have to substitute here in terms of dx sorry in terms of dr you substitute dx so you will get dm equals to 2m by rl into r into dx now dx I have written in terms of dr good thing is l get cancelled so dm is equal to 2m by r square r dr everybody understood how dm is coming type in quickly any doubts please type in directly everybody understood how this dm is coming there is a small trick using similarity of triangles we have to write dx in terms of dr because dx is not dr it's a separate variable getting it okay now substitute this dm over here and get the answer integral of r square dm what is that shashwat 2m by r square r cube dr limit fall for the small r will be what what should I write limit here small r is radius from where to where it is changing what to what 0 to r all of you it was 0 at the top radius of the ring is 0 then it grew in size and became capital R at the bottom so 0 to capital R so what is the answer moment of inertia now tell me the answer get the answer so moment of inertia for the hollow cone about the natural axis is mr square by 2 okay this so I hope you have seen how integral is used here properly you can understand the calculus here see I'm not saying that you should be able to solve it but whatever I have done you have come you have learned it right whatever I have done here you have understood it that is good enough to start with okay yes Pranav fine so the homework is you need to find out the homework is find out the moment of inertia for the solid cone about the natural axis so that is your homework so do it after the class but now we are going to start a new chapter very simple chapter it is it will get over today itself fine so write down the chapter's name mechanical properties of solids have you seen the index of your book to have you seen it anyone so index of book to you will see that there are chapters on properties mechanical properties of solids then you have mechanical properties of liquids or fluids then you will have thermal properties of matter right so you have half of the book book to is about the properties of material fine now there are other half of the chapters which are different from the properties of the matter but my point is that your half of the book is related to properties okay now when I talk about property what does it mean let's say property of someone what all I mean how you describe or what what will make you describe property of anything let's say property of the pen what all thing you will talk about it everyone it's a property of pen what you can say will you say that it can write if I say property of this pen if I write down if if I show you five pens and if I just take one of the pen and ask you what is the property of this one will you say that it can write all five can write yeah then you go to the specifics of it you'll say it is black what else what else you can say it is it has certain length let's say 10 centimeter length or 20 centimeter length right so what what are you trying to do you're good you're getting into something which defines it okay cylindrical conical arrangement you will get into very very specifics of it you can say that it is made up of a particular type of plastic you can name the compound you can go up to that specific okay you can say that the mass of the pen is just 10 grams okay you can say its nib is not very sharp you can go it into very very specifics of it so more and more specific you get into better you are defining it okay so when I say properties of solid forget about mechanical word okay forget about that let's say properties of solid what all thing come in your mind property of any solid let's say I tell you I give you a piece of copper and I ask you what is the property of this block you will say about its mass you will say that about a density you will also say that it's a good conductor of heat you will say it is a good conductor of electricity right you will say that if you apply lot of force on it then also doesn't get deformed right so all these kind of things you will talk about okay but this chapter is about mechanical properties of solids now you have probably in your mind you might be having these kind of property thermal conductivity how well it can conduct density what else everyone strength you can say mass correct volume fine then you can say ductility anything else anything else malleability luminous yeah you can keep on talking about various things but then yeah melting point very good rigidity is covered in ductility malleability okay melting point maybe boiling point so you will see that some properties are physical properties for example if I take a person's name for example whom will I take let's talk about Charon okay so if I talk about properties of Charon what will come in your mind you will be like okay Charon's weight is this much Charon's height is this much Charon sir Charon wears this color of the dress okay these are things they are physical attributes of Charon these are the physical properties looks and feel of it okay then someone will say that Charon is a very emotional person Charon gets happy very easily okay Charon is happy go looking happy go lucky person okay all those things you will try to describe that will be its different kind of property do you agree one kind of property is different from the other kind of property physical properties one and then there are these other set of properties which have to do with more of emotional quotient or the mentality and other things fine so there are two different types of property do you think one property has to do with the other one physical attribute does it have to do with anything with the mental attributes of anyone no right they are two different things fine so the the laws which govern the physical attributes are different from the laws which governs the mental attributes of a person okay you can exercise and become very strong so your physically may be very strong but mentally may be very weak okay so there are different kinds of you may have to do meditation to become stronger mentally so what I am trying to say is that different rules govern different kinds of property that is a reason why you have mechanical properties separate from thermal properties the rules which govern the thermal property of matter is different from the rules that govern the mechanical properties of matter okay and in mechanical properties specifically mechanical properties of solid follow one set of rule mechanical properties of liquids and gas they follow different set of rules so that is a reason why you don't have mechanical properties of solids and liquids together in a single chapter but when it comes to thermal properties you have a single chapter it is not thermal properties of solids then thermal properties of fluids it is a thermal properties of matter everything together because rules are same but here rules are different so that's why mechanical properties of solid is different and liquid it is different now when I say mechanical properties what all things are there out of this strength is there mass is there density volume ductility and malleability fine these kind of attributes we are going to learn in this chapter right these are the mechanical properties of the solids all right now can anyone tell me why we are studying mechanical properties of solids why there's a need to study it at all anyone why why you want to understand the behavior of solids that is what I'm asking so I'm asking why you are learning mechanical properties solid answer is to learn the mechanical properties of solids as in what is the usefulness where you use it I'm not taking names anybody's name tell me so you look around yourself all the structures you talk about homes solids you talk about the motorbikes you talk about the fan you talk about the bad books everything you look around yourself everything is solid okay and understanding the strength is very important understanding how solid will behave when you apply a force on it is very very important you don't want your building to crumble under the pressure so that is why you might have not seen it but nowadays metro construction is going on it will go on forever for next 10 15 years but whenever they are constructing the metro station do you know they do soil testing they check whether the soil is good enough for the metro or bridge to be constructed there because even the property mechanical properties of soil is very important do you know that a lake when it is dried you cannot construct a high-story building on the dried lake because the the the soil which is of the dried lake is not conducive for it your instructor may fall okay so soil testing is very important to understand the mechanical properties and also when you construct the bridges or when you construct the or can when you construct the building you might have seen in your basement there are a lot of pillars like where if you're if you're staying in a flat you'll see that on a basement the entire building is stood on a lot of you know pillars now one can say that I don't need to calculate the strength of anything I will have my many pillars I'll have hundred pillars so I'll be safe no no no problem with it okay but the disadvantage will be that that guy will end up spending a lot of money every pillar you stand probably you may have to shell out a lot of money so if you if you're playing very very safe without even calculating you may have to spend a lot of money so you might have seen the bridges every now and then there is a pillar so if you don't let's say calculate after what distance you should put a pillar then you'll play safe and after every two three meters you'll construct a pillar so imagine the kind of wastage you will have lot of money will go for waste so it makes sense to understand the strength of the material and understand how every material behave when a force or a pressure is applied on it so that you can make an informed decision that okay fine I'll be safe if I have only this much thickness of the pillar and I'll put after every 10 meters it is fine so I'll not unnecessarily waste the material okay so strength of material study is very very important so there's a reason there's a reason why inside the pillar there are iron rods you you're aware of it right iron rods they put inside the pillars okay there's a reason to it the reason the simple reason is that the concrete the cement can take lot of load lot of pressure it can take if you have pressure like this if you compress it but as soon as there is a there's a force or there's a pressure which tries to expand it the the cement will break apart it is so brittle that is the reason why you put the iron rod in between so that it can support not only the compression but extension also so there is a reason for each and everything that is done okay so that's the reason why we are studying this chapter it's a very very important chapter if you are aiming to go for I mean if you if you want to become an architect or a civil engineer anybody here who is aiming to become an architect here okay there is at least one so for you it is very very important do you know that if you if you take these civil engineering courses or architectural courses you will be studying for one or two years only this chapter right now this chapter is very very small just basically introduction of it it is there in class 11 but if you go for engineering same chapter you will have to it will become this much thick book okay you have to study quite a lot and detail how everything behaves okay one thing one solid might behave in a dry condition in one way one one way to behave but same solid probably in a moist situation will be different behavior okay so very very important these factors are okay so that's why we are studying it so all of you I think I have answered why we are studying this chapter okay so just give me a second so first of all one thing you need to understand that every solid gets deformed if we apply force or pressure whatever it is we are going to discuss it if you are applying a force on a solid it will get deformed deformation may be negligible you may not be able to feel it or see it but it is there okay but why don't you know if let's say very very small deformation is there if every day I apply the force after 10 days deformation should be visible to us then why it is not visible what do you think is happening little little deformation should happen every day and then you will be able to see it every time we apply force on something even if you are while opening the laptop also small deformation should happen after let's say 100 days deformation should be visible that oh yes this is the total deformation which is visible to us correct it goes back to its original shape and size okay whatever small deformation it has it immediately as soon as you stop applying the force or effort it goes back to its original shape what is this called this behavior of going back to its original shape elasticity this is called elasticity okay so rubber band is more elastic or this pen is more elastic what do you think okay some of you are saying rubber band some of you are saying pen the answer is pen is more elastic because that because the way you are defining elasticity you are saying rubber band because deformation is higher and it is still able to go back but elasticity is not defined like that definition of elasticity is write down write down tendency tendency of an object to go back to its original original shape and size okay now this tendency is more for the pen than a rubber band how is that I can apply 100 Newton on this pen deformation I'm not able to see but it is there and then it still goes back to its original shape but if I apply 100 Newton on a rubber band it'll break it won't be able to go back it's gone fine so tendency is like how much an object wants to go back how much effort it applies to go back to its original shape which is of course more for the pen right so this is called elasticity okay there is one more term called plasticity write down write down plasticity so plasticity is referred as permanent deformation have you ever seen that you apply a force on something and the solid is permanently deformed example any example clay it is written in in your textbook clay what else spring sometimes sometimes spring if you stretch the spring too much will it be able to go back no paper is it breaks very easily right folding paper wire wire if you stretch okay and have you ever stretched a polythene like this have you ever stretched a polythene so when you stretch it does it go back when you stretch it let's say a little bit it doesn't go back permanent deformation so plasticity is plastic is the best example of plasticity fine the name comes from there all right so these are the two behaviors and we are going to study in this chapter we are going to study the elasticity of solid elasticity only okay if you talk about mechanical properties of solid it is a huge okay you need to discuss that if I have let's say 50 moisture in the air then how it will behave if I you know you know if if let's say at this temperature this is the mechanical property at some other temperatures some other mechanical property it is endless but in class 11 only study of elasticity is there okay so we are going to see that how the elastic behavior is shown by the object now before getting into the elastic behavior of an object we need to first understand what deforms the object what is the cause cause of deformation is it the force cause of deformation of solids okay in order to study that just take a simple experiment okay I'm going to draw it here see all of this comes with the experience it is not that you know it should be this and it should be that we have observed that this is true that is why it is true suppose you take two wooden planks like this okay you have two wooden planks and have you seen the pencil heels pencil heels have you seen let's say this is the pencil heel of course of some lady and let's say the mass of that person is 60 kg so I am assuming that someone who is wearing pencil heel is lighter so 60 kg okay this is a wood this is also wooden block both are same here the lady is standing on her one leg okay and over here over here one elephant is standing on his one leg elephant weight is couple of thousand kgs okay now where do you think solid will be getting deformed this case or that case case number one or case number two both more chances more chance chance is what suppose it happens only one in only one which one suppose a wooden is wood is placed on the horizontal floor then the first one it is the first one where do you think force is applied more force on the wood which one is more case number two force is more both both of them are applying mg force here mg and here also mg only but small m is the mass of the lady which is very less capital m is mass of the elephant very large but suppose you have doubt that I don't know which one will get deformed then what lady does she stands on the pencil heel only and take a turn like that then there will not be any doubt there will be a ditch here created like that okay definitely here you get deformed but you can see the force is way lesser than the force over there so it is not force that is deforming the solid if you look at these two cases here what is more is force per unit area okay so this is something which is the cause cause of deformation right now so if a solid breaks if a solid breaks solid if it breaks it's due to how densely the force is applied as in force per unit area okay nothing else everybody understood this thing fine so now we know what is the cause all right but the problem is still the solid can get deformed in multiple different ways solids right down can be deformed in many ways so we need to discuss one by one how many ways it is getting deformed and what are the laws which govern all that okay am I going very slow very very slow okay suppose you have a cube like this or a cube like this this is a piece of solid all right now talking about the deformation of this solid cube in what ways you can deform it you can apply force from here and there you get deformed right this is one way of deforming it you can apply force from two opposite sides and it get deformed these two opposite side can be any two opposite sides can be top and bottom can be sideways this side or that side getting it so this is one of the ways can you think of any other way of deforming this solid everyone same but in opposite direction okay good so you can apply the compression like this and it turns out that if you talk about the elastic solid then both ways it is same only whether you apply compression or the extension okay another proper way of deforming it keeping the base fixed see here the force is perpendicular to the face force is perpendicular to the face okay another way is you can have a force can I have force parallel to the face can can that happen which will create deformation anyone parallel to the face there is a way you can keep the bottom face fixed okay this bottom face if you fix it and you apply a force like that parallel to the face okay like this if you apply the force the face will shift forwards everyone getting this you have a pen like this you're pushing it like that rather than stretching it you're pushing it like that everyone understood okay so correct so this face will be shifting little bit forward okay so this is what will be seen it will shift that way so this deformation is different point number one the force is parallel to the face and second point is the angle between the two vertices is not 90 anymore okay these are two deformation another very common deformation is this one rather than force from two sides or one side you can have the force from all sides okay it is very common inside the fluid so you will learn that fluid has a special property that it can apply pressure same pressure in all direction so if an object is inside the fluid it can get compressed from all the sides okay so when you put a solid under a lot of pressure then how it behaves even that we are going to study it fine now of course many other things will come in your mind that what will happen if i rotate this what will happen if i do something else okay the answer to those questions is that these are the three ideal ways you can deform the solid and when get solid get deformed everything can happen together also same solid has force like this like that and it getting crushed also all three together okay but when we are studying it we can't study all of them together we'll be studying it one by one and then even if all of them are together happening then i can apply all the laws which i have learned together but while learning we'll have to study one by one okay everyone understood right now talking about the heart of the chapter as in the most basic thing about the chapter is about stress and strain tell me if you know what is stress already what is it what is stress not your mental stress physics what it is okay force parent area pressure okay so you're telling me stress is what force you apply divide by the area is that what you're saying what force you apply divide by the area understood fine although that's not entirely correct we'll discuss it right now all of you write down that what we are trying to do here is that we are trying to find out what is the action as in what causes deformation and what is the reaction what and how to quantify deformation you know it is one way of saying that okay fine this person runs faster that person runs even faster and this one runs slower okay so until this i tell that what is the velocity all these discussions are meaningless if i say that lot of deformation will happen it will not give you you know sense of satisfaction or you will not understand what it is till you put a number to it till you like we're fine the deformation is a point zero zero one zero zero one should mean something okay so we have to quantify what is the deformation okay we need to also quantify what is the cause of the deformation we have little bit touched upon it when we discuss the two example elephant and the lady one okay but over here small thing is that the stress which is the cause of the resistance sorry cause of the deformation is not what you apply divide by area what it is it is the force of resistance force of resistance which material applies on you it is a pair force divided by the area will force of resistance be always equal to water force you are applying tell me force of resistance is it the it is the pair force of what you are applying will it be always equal to what force you are applying force of resistance it seems like it seems like what force i am applying whatever force i am applying the material will apply equal and opposite force on me and that is a reaction force force of resistance is a reaction to my action so ideally they should be equal and most of the time they are equal also okay most most of the time force of resistance is equal to the applied force now i'll talk about one example which those who have played with stretching the membrane or stretching the polythene for no reason they'll understand more so when you stretch the polythene just before polythene breaks do you feel that suddenly polythene has stopped applying resistance or the resistance has dropped suddenly for a large amount have you felt that right you felt it just before the polythene breaks turn out that kind of nonsense okay don't tell okay you have to keep it formal i'm not here to listen to all that loser kind of thing those who have you're telling that those who have done this experiment they're losers you mean that don't use those kind of words okay so when you stretch it when you stretch the polythene it is about to break the polythene suddenly stop applying the resistance force okay so it is not that the resistance is always equal to what force you apply getting it if an object is weak for example this chair this chair will be able to apply whatever force i'm applying till whatever strength it has if i apply one million newton on this chair chair will break okay so chair will not feel the force whatever i'm applying chair will feel the force only till the point it is resisting it okay so it is a force of resistance which is the cause if if you're talking about a weak material which cannot even feel the amount of force which you are applying before that itself it breaks then no point you know talking about that force you have to talk about only that force which the material can feel and it will feel only till the point it is not getting broken apart okay yeah the newton law third law of motion is still valid because if you apply large amount of force let's say you have a sponge if you apply one million newton on it it'll break immediately it will break but it'll create acceleration so force equal to mass expression when net force is zero then it'll have equal and opposite distance yes aditya all that we'll discuss hold on so all of you understand stress is force of resistance divided by area okay then this is the cause of the deformation and there is this word strain that quantifies the deformation amount of deformation the way we quantify it this is called the strain fine now initially itself we discussed that there are different different ways solids can get deformed so we need to take one by one these ways and understand what is the stress what kind of stress the object has and how to define the strain for that particular scenario okay so first thing that we are going to learn here is longitudinal stress and strain right down so longitudinal stress and strain is a first kind of stress and strain what is that wherein let's say you have an object this object you have okay a thick wire fine you are applying force on it like that equal and opposite force suppose you are applying fine this wire doesn't break so the resistance force will also be equal to whatever force you are applying force f and force f okay just so that you know the resistance force because you'll be dealing with wires a lot in this chapter resistance force i should write that in the previous slide resistance force in a wire is which force any guesses which force is resistance guess yes it is tension good so if you have any confusion what force should i use to create to find out the stress get the value of tension tension is the resistance force okay all right and most of the time tension will be equal to whatever force it is getting applied on it okay so let's say it is a uniform wire of length l the area is a okay what else should be given that's it okay so when you're applying the force what is observed is that this wire get stretched a bit okay this much amount of stretching won't be there but just to you know clearly show i have increased the length substantially so when you're applying this stretching force f l has become l plus delta l and delta l is very small so i'm assuming area is not changed but in fact area also changes when you stretch it okay the assumption is delta l is very small so area doesn't change fine now all of you the stress over here is called longitudinal stress sigma l what do you think this is equal to everyone this is a symbol for the stress sigma what it is equal to sigma l is equal to what resistance force is f only force divided by area everybody understood this everyone type it is it clear now over here you are talking about the perpendicular component of the force perpendicular to the surface that divided by area okay this is sigma l this is the cause of the deformation cause now what is the effect deformation is how much suppose i have to quantify deformation what should i write what should be the deformation delta l okay that's an obvious choice everyone is it delta l cross section area you have to take this is area of cross section okay a is the area of the cross section suppose i have to quantify deformation what should i write here is delta l sufficient everyone what if what if you have a wire of double the length double the length and same amount of force is applied stress is same that delta l will be same or different what do you think delta l will be same or different if length changes delta l will change or will remain same area is same will the delta l change or remain same everyone answer it it will change if it is double delta l will become double okay got it so you know you can imagine like take a smaller spring take a bigger spring if you stretch the bigger spring it will stretch more for the same amount of force okay or if you have any confusion assume this double length as if it has two lengths two uh two rods of this is uh what i'm saying a single bigger rod is made up of two smaller rods every part will have delta l deformation so total deformation with delta l like that you can think so delta l cannot be the effect of the cause so deformation cannot be can't be delta l why it cannot be delta l if deformation is defined as delta l then for every every time if i apply this cause of deformation which is sigma l longitudinal stress it should give me the same deformation but if i define deformation as delta l for the same force per unit area i'm getting different different delta l so this is not proper definition then i have to check okay fine what is the length and all everything i have to take care of so the effect this is the cause the effect of the stress is strain and we are talking about the longitudinal strain which is written as epsilon l okay epsilon l is a longitudinal strain this is equal to delta l by l deformation per unit length this will remain same no matter what length of the wire you take so this is a better variable to deal with than just delta l everyone understand this i think it's very simple chapter a is the cross section area a is the cross section area everyone a is the cross section area okay this is the cause longitudinal stress is cause longitudinal strain is the effect now can you tell me if i have this kind of cylinder i have this kind of cylinder the radii are this is r1 and the bigger radius is r2 there is a cavity inside the cylinder a cylindrical cavity inside a solid cylinder then if i apply force like that what would be my stress how to write down the value of stress over here longitudinal stress everyone try it do understand the question what is the longitudinal stress over here okay aditya got it others hey arnav you haven't corrected your thing gun still type okay puja got it others no no no you guys are making one silly error okay pranav got something imanshu and puja check what you have done arnav is gone not arnav is gone then pranav is also gone is it sigma l is what force divided by area which area you have to take now you have to take area of the material r2 square minus r1 square pi r2 square minus pi r1 square everybody understand this okay all right so talking about these kind of you know we have discussed the the cause which is stress and the effect which is the strain it is like you know you remember newton's second law force is equal to mass same acceleration force is the cause acceleration is the effect so you have an equation between force and acceleration between cause and the effect there is a relation similarly here also we are trying to find out a relation between cause which is the stress and the effect which is the strain okay that relation you can't derive it like you know some formula or things you need to have some experiment or observations based on which you can find out what could be the relation okay for example if we take a let's say a typical solid for which i can draw stress strain curve right down so we will try to observe the stress strain curve for an object and let's see is there a relation correlation between stress and strain that is the best way to see okay so all of you draw this is the very important part of the chapter so everyone should understand it properly so let's say this is sigma and this is epsilon okay so there is a this experimental setup in which what they do is that they have a machine you when you go to engine go to some college you'll see this experiment if you take mechanical or civil you'll have this experiment in which there's a clamp which which holds the steel or iron bar from top and above and then that clamp stretches the iron bar from both sides okay when it stretches it the iron bar length changes so every time its length changed a little bit the force of resistance which is equal to whatever force you're applying because acceleration is zero slowly slowly it moves up so you will measure the resistance force what is being applied and you will also see the strain so automatically that machine will record everything and plot a graph fine so when lot of materials are used in this experiment a typical kind of curve is observed okay the shape of the curve remain similar and this is how it looks like what we are going to draw here okay so first of all you will see a straight line like that up to some point you'll see a straight line and then it bends little bit like this goes like that and dips down this is the typical graph which is observed all right so if I name a critical points over here let me name it this is 0 this is point a up to which it remains linear I'll talk about what is this point b is c is a random point okay I'll put this d at the peak d is the top most point and this is where the material breaks this is where the fracture happened this is fracture point all of you draw this fracture point okay now this d2 let's say this is e d2 e is the same thing which the example you've taken right for polythene when you're stretching it it breaks before breaking it reduce the stress is reduced stress is a resistance force so it get reduced from maximum to the point where it breaks fine so the observation of the graph is that o2a is a straight line okay o2a is a straight line and in the straight line can I say sigma is proportional to epsilon can I say this everyone can I say that sigma is proportional to epsilon everyone do I have to ask very especially to some people those who choose not to say anything I'm asking something all of you reply can I say sigma is proportional to epsilon everyone reply right I can say that and this this thing this behavior is a you can say this point a is proportionality limit up to point a it is proportional and yes it'll be different for the slope of this line is different for different material okay slope is this tan of theta you can say this slope is different for different material now can you tell me for the slope is more for let's say iron or rubber which one slope is more theta is more for iron or for the rubber rubber iron it is iron only right for large amount of stress also I'm getting lesser deformation so deformation of iron will be lesser no so that is how it is so if a material is rigid its slope will be very high okay anyways so a is a proportionality limit then this point up to point a if you have a stress and you leave it it'll go back to here when you when you're taking it to point a and then leaving it then what will happen the stress will become zero okay stress becomes zero so it will be anywhere along this x-axis and if it comes here what does it mean if I release it and the point comes here it means what everyone this is a point for which stress is zero where I have left the object I have reached here then I leave it it comes here suppose this point means what zero stress but what about strain is deformation is there deformation still there okay they are called permanent deformation correct correct okay so till point a it is proportional and if you leave it then it again goes back to this origin point only where zero stress zero strain and that behavior extended till b also okay so till a though it is proportional but till b if you release from b also you know between a to b it is not a straight line so between a to b stress is not proportional to strain okay but then if you release at b also then also it will come back zero stress zero strain but if you go beyond b okay if you go beyond b let's say point number c after that point it will have a permanent deformation it can maybe you know if even if you release it after point b if you go and then if you release it then also there will be some permanent deformation okay the plastic behavior will start from there from b all right so b there is a name to it this is called the yield point okay this is the yield point everyone till this point till this point if you release it sigma if it becomes zero epsilon can also become zero beyond that it will not become zero fine and then there is one more critical point d what does that represent can anybody guess what should this be representing this point is represent any name that comes in your mind what do you think it should represent maximum strength or the strength of the material right so this is our ultimate strength whatever you may want to call it okay so technical term is tensile strength because we are talking about longitudinal here right tensile strength how much tension it can take okay this is the thing here got it this is the stress strength curve any doubts here anyone has any doubt quickly type in else i'll move on to the next anyone has any doubt huh waiting himanshu why isn't e the tensile point because it should be it should represent strength strength is represented by how much maximum you can take right how much maximum e is not that point after see it will have permanent deformation d is a maximum strength so why why do you think that is contradictory tensile strength would be d yes d is a tensile strength after d it will not break it will still go till e and then break if the object takes some time to come back after sigma is equal to zero is it still plasticity take some time to come back no it is not plasticity you waited long enough the time factor is not there okay you can wait for long enough time if it comes back then it is elasticity position c does not occur okay fine all right no and i'm not saying that exact same graph you will get every time but this is a typical graph i'm saying distance between a and b could be lesser could be more distance between b and d could be lesser or more but this is the typical graph every time one thing you'll see that a solid material there will be a portion wherein stress is proportional to strain and in our chapter in class 11th we are going to work only between zero to a okay where stress is proportional to strain we are not going beyond a otherwise chapter will become huge fine so we are only learning in this chapter about the proportionality up to the proportionality limit not beyond that okay now this is a typical graph but doesn't mean that every material on earth should have the same graph okay so there are you know there are objects which have different kind of graphs also for example if i do you know this tissue which is there on the on the herd aorta have you heard of have you heard of aorta anybody taken bio in the herd so you might be you might know that it expands and contract so many times in a day so this is probably the one object which gets deformed comes back to original shape again goes back comes back so you know multiple times it does so this is this is an this is amazing right and the stress strain curve for this one yes you will learn when you go to grade 11 mother so this is the curve you will see getting it there is nowhere like what typical solid material or metals that we deal with okay so although this is typical but don't assume that it will be always there like that now now that we have observed how things are let's talk about a law here which will give us some sort of equation so that we solve some numerical right till now it is all theoretical stuff do like theoretical chapters like these are the numerical types chapter which one do you prefer numericals numericals nlm you know Newton laws of motion is those kind of that kind of chapter you spend your entire life still there will be few questions you will you will not be once you're done with the advanced level nlm let me know i'll give you a couple of questions of some other level okay but a lot of time is there till you reach there so keep on practicing Hooke's law Hooke's law is a law which came before this observation you know before this graph people knew Hooke's came with a law and what was that law Hooke's told that stress is proportional to strain okay and this is how it is between zero to a proportionality limit okay and this is this is of course valid for a linear zone between zero to a okay so for a particular solid material stress is equal to some constant y times strain anybody has seen this constant y before what is this y it's a constant proportionality between stress and strain in the case of longitudinal okay good you guys know it probably yeah you have done this chapter right in school why is the Young's modulus it is constant for a material does not depend upon a down doesn't depend upon the geometry whether you take this thin wire thick wire cube whatever shape and size you take Young's modulus is going to be constant and it's a you know great great tool because Young's modulus is nothing but a ratio between stress and strain every time for all the cases you find out the stress for all the cases you find out the strain take the ratio between stress and strain it will be constant okay what is a unit of Young's modulus everyone what do you think is a unit Newton per meter square does Pascal come in your mind Pascal is Pascal the unit of the Young's modulus can I say that you cannot say Pascal Pascal is a unit of pressure okay Pascal is a unit of pressure you cannot say that it is although dimensionally they are equal dimensionally the unit of work done is equal to the unit of torque but work done in torque they are not same right so you should not say joules for the torque okay so if if there is a wire if there is a wire like this of cross section area a and you are applying force like this area of cross section is a the length is l it undergoes deformation of delta l can you quickly tell me what is the expression of why all of you do it first stress is this strain is is true for a wire like this and whatever we have done this thing is valid not only for the stretching but also for the compressing Young's modulus will be same all right Young's modulus is typically of the order of for metals of the order of 10 raise to power 5 okay so this is the theory of the Young's modulus the longitudinal stress and longitudinal strain let us take numericals on it we'll be taking up some school level thing then couple of advanced levels so that you don't take the chapter for granted there are some small small details in the numericals okay so I'll first take the school level things best thing is to take the ncrt book part two right this is hooks looks like Newton these are not their real hairs you know they should show their head size bigger rather than that they're showing their hair size bigger do this all of you what is the answer abcd abc three parts are there for the first part what is the answer no one got the answer part a that is strange how much quickly answer okay different different answers people are getting part a Vibhav is correct only Vibhav got it if you think that you're beyond the simple question let me tell you you'll suffer in the exam okay you will see kind of failures you'll make in the exam give respect to every question if it is simple you have to give more respect solve it properly okay aditya got it correct others simply have to do force per unit area which is the stress force divided by area of cross section okay force is how much 100 kilo Newton force 100 kilo Newton is 100 into 1000 divided by area area is pi r square which is 10 mm 10 mm is 10 is a minus 2 so 10 s power minus 4 it comes okay so this is 10 by pi into 10 is a power 8 and you should remember pi square is roughly 10 okay so 10 by pi is 3.14 3.14 into 10 is power 8 Newton per meter square this is your stress okay b part elongation how much elongation it has yes elongation is delta l how much it is others quickly answer okay done fine so stress by strain is Young's modulus which is 2 into 10 is for 11 so from here strain is stress divided by the Young's modulus so strain is 3.14 into 10 is power 8 divided by 2 into 10 is a power 11 okay so this is 1.57 roughly 10 so minus 3 epsilon is delta l by l l is 1 meter so delta l by 1 is this so delta l is that only this okay c part what is the answer for c strain is how much what is the unit for strain what it is unit for strain is what no units okay no units so delta l you should write in meters and delta l by l when they are asking you strain this is dimensionless you just need to write like this 1.57 into 10 is so minus 3 the best way to write this is in terms of percentage you can say 0.157 percent is the deformation okay fine this is one of those simple direct formula based question let's take up another one from your textbook only this one is better see here let me tell you a few things copper wire let's say length is l1 let me draw the diagram you know then I'll represent it so hold on for some time let me draw it and then you attempt it this is copper this one is let's say steel okay you are applying force here and there is force f for copper length is l1 for copper Young's modulus is y1 area of cross section is same for both so a only this is l2 and y2 for steel and this is a you don't need to substitute the values don't need to calculate anything in terms of this you need to find out what is the value of f and yes total elongation is given delta l don't substitute the values right now find out what is the problem you guys are facing here you're not able to proceed then what is the problem you're facing tell me you guys are stuck where are you stuck what is the problem the main problem over here is that there are different materials isn't it there is no single Young's modulus for entire rod then how will you do this if they're different material then what you do any ideas average division yes you divide it you break the rod into two parts one take copper droids free by diagram take steel droids free by diagram that's how you start okay so how much force let's say I'm drawing for copper this is l1 y1 I know this side is f what is this force how much should be this force this one is how much who is applying this force steel is applying how much it should be how much force steel applies on the copper don't know copper is at rest so net force should be zero so it should be f only it is at rest net force should be zero that's the reason why steel action reaction pair f here f there okay now what should I do everyone this is taken from your book only textbook are you finding it very difficult everyone very difficult to you what should I do next tell me the answer now are you done those who are answering what to do are you done what suppose delta l1 is the extension over here delta l2 is extension over there what is your relation among delta l's what should I write can I say delta l1 plus delta l2 is equal to delta l all of you agree right it will be this delta l1 in this case how much it will be 180 newton what is 180 oh answer you're telling me 180 you know answer is there in the book also I don't care about that at all I am asking here something else you're telling something else so delta l1 plus delta l2 is delta l right delta l1 how will you get what is delta l1 everyone answer delta l1 is what namrata what it is delta l1 they haven't given y1 y2 it'll be given don't worry in the question they in your book y1 y2 will be given at the back side don't worry about it okay so f l by a delta l this was equal to Young's modulus right so delta l1 will be equal to f l1 divided by a y1 plus delta l2 is f l2 by a y2 this is equal to delta l okay you can take f common f by a you can take common you'll have l1 by y1 plus l2 by y2 is equal to delta l so from here you just separate the terms you'll get the answer for f cross multiply y1 y2 will be given to you don't worry okay it'll be given everybody understood this typing quickly then I can move to next these are your school level questions clear okay let us see little bit of calculus okay now not calculus alone we'll take this one this question calculus will take up after the break this question but you can do it without calculus suppose you have a rod like this okay you are applying a force over here no no calculus is utilized here also sorry calculus can't leave you so then do this one little bit of calculus is there don't worry about it the moment I say calculus suppose you are applying force f from the end there is this rod massless rod okay it has no mass you're pulling the rod down by a force of f okay it has length l Young's modulus y and cross section area a getting it now I'll write here negligible mass tell me one thing when I apply the force am I doing work on the rod or not everyone when I'm applying the force am I doing the work yes or no I'm pulling the rod down rod is moving delta is a displacement right so I'm doing work whichever direction whichever point I am pulling down it is moving so there's a displacement there's a force there's a displacement so work done is there what is happening to this work done according to the work energy theorem suppose it gets deformed and then stops whatever work I have done where it goes potential energy are you getting it getting it so you need to find work done by the force in this case and that is also equal to potential energy change in potential energy or potential is stored inside this rod find out everyone let us see if you can do it your own it will be very nice have you done this in your school finding potential is stored in the rod is the potential energy is it a constant force if it is a constant force simply f into delta l you can do but is it a constant force if it is slowly and slowly moving down it's not a constant force you want to try yourself or should I do it should I give you some time okay try it try it I'm giving you one minute one or two minutes the moment you are stuck let me know I'm giving you hint assume the rod is extended by x rod is extended by a x amount find out four symptoms of x and then integrate fdx you'll get the answer now should I do it is stuck are able to proceed 30 seconds no I'm not sure that's not correct okay all of you if if rod is extended by x what is the value of force everyone in terms of Young's modulus and everything no not that no what is the value of force if it is extended by x by f f l by a delta l which is x is Young's modulus so the force is a y by l times x this is a force right so this force I have to integrate this force is not a constant force so work done is equal to integral of fdx x goes from 0 to delta l okay so work done is equal to a y by l is constant comes outside integral of x dx 0 to delta l so work done is a y delta l square by 2l everyone understand this that's it this is the work done and this is the potential is stored in the rod everybody understood this okay now tell me how to write this formula for the work done or the potential energy in terms of stress and strain I don't want Young's modulus in it I want in the formula only stress and strain can you modify this formula and tell me what it is total extension in the rod is delta l can you modify this formula in terms of stress and strain anyone getting it write down in terms of stress and strain sigma is stress which is f by a epsilon is strain that is delta l by l right Young's modulus can be written as sigma delta l by l so sigma is Young's modulus delta l by l so you can play around with this and modify this in terms of stress and strain what you can write okay let me do it the potential is stored which is the work done can be written as Young's modulus delta l by l into delta l by 2l multiplied by a into l okay and you see that we can write like this everyone everyone can you see that we can write like that this sigma into epsilon by 2 a into l is what what do you think a into l is volume so potential energy stored is half of stress into strain into volume okay and density of potential energy density of the potential energy u by v is simply sigma e by 2 stress into strain divided by 2 potential energy density everybody everybody understood type it go through it once there is no hurry go through it and tell me have you understood how we get the potential energy stored in it it acts like a spring you know spring can store potential energy same way the rod is also like acting like a spring only so even that will have potential energy so whatever work is done to stretch it will be stored as a potential energy so we have found out the work done and then just modified played around with the variables to write it in terms of stress and strain all of you type it have you understood i'll wait for a minute go through and let me know if you have any doubts you know this chapter 99% of it is simple and straightforward but there are a few things which are tricky okay it is not like a laws of motion chapter where many questions are tricky so you just need to learn couple of tricky questions and then you are done with the chapter it is true for any object this formula stress into strain by 2 although you have derived it for rod but because you have derived this energy density in terms of stress and strain and that has nothing to do with dimensions of the rod this is true for any place any place on the bigger object if there is a point on which there is a stress and there's a strain the density of potential energy at that point is stress into strain by 2 are you able to follow kaushik namrata reshap type it is it clear so raj clear so raj okay all right guys so we will take a break now and we will meet beyond that so right now it is 408 after 15 minutes we will meet then it will be 4 23 p.m little bit of chapter is remaining the chapter will get over today itself this is your break time okay come back all right am i audible can you hear me okay fine so let's take some more numericals one or two more numerical see what all different kinds of questions can come all right so here is one more numerical all of you you have a rod like this rod of mass m and length l the mass is uniformly distributed over it it is a uniform rod of mass m and length l hanging like this okay so it will stretch due to its own weight are you getting it you need to find out elongation due to its own weight total elongation okay young's modulus is why try this all of you i'll give you hints one hint i'll give you where do you think first of all tension force no no that is a wrong answer whatever whoever have answered just now that is a wrong answer okay so listen to hear the stress is because of because of the this thing because of tension the stresses all right tension force is a restoring force tension force creates stress now when you hang yourself let's say you are hanging from your hands like that where do you think stress is maximum on your hands or on your feet where it is more hand or feet just imagine you're hanging your hands hand it is more palm it is maximum so stress is changing stress is not constant right stress is different at different positions at the bottom most the stress is zero okay stress is zero at the bottom most point over here getting it so the hint is that you need to assume a dx element over here this is dx at a distance of x from below you're considering dx all right on this length on dx what is the elongation that you have to find out elongation of dx let's call it as dl how much is that try doing it if you can understand yourself it'll be way better than i directly writing it here do it okay aditya got something others what is the tension at a distance x away from the bottom did you get that tension how much it is okay mother got it others you can just take the x length don't you think that the this force on the top whatever it feels this is a tension the dx also feels do you agree with that this is a tension dx also feels this tension is applied by dx dx also applies the same tension back so this tension is how much this is length x x how much is the mass of this length x total length l has mass m so x length will have what mass this has m by l into x right m by l into x into g is the force any of the force on this x length can you answer any of the force other than these two can you think of any of the force no other force and it is at rest right so m by l x g is equal to zero so tension is equal to m by l g times x okay i think cross section area should be given so section area is a let's say now if this is a tension stress is how much stress is t divided by area of cross section a so it will be m g by a l times x everyone is clear till now type it quick till now is it everyone is fine till now whatever we have done so we know that stress by strain is Young's modulus so strain is so the strain is sigma divided by Young's modulus so mg by a y l times x this is my epsilon epsilon is what epsilon is the strain on dx okay change in dx is dl so that is dl divided by dx dx is a length delta l by l delta l is dl l is dx mg by a y l x so change in dx which is dl is equal to mg by a y l x dx and now you integrate it dl will if you integrate dl which is change in dx you will get net change total change 0 to delta l this will go x will go from 0 to total length l all right so the answer delta l you'll get it as mg l l square will come in numerator there is l in the denominator also one of the l will get cancelled away mg l by 2 a y this is the change in length that will happen in a rod due to its own weight okay anyone has any doubt quickly type in else i will go to the next question no doubts all right let's take up another one the next question is similar to the previous question the kind of calculus involved is similar okay so just make sure that you understand the scenario properly you'll get the answer okay it's a repeat of previous question you can say there is a rod lying horizontally okay there is a force f1 from this side and force f2 from that side okay f1 is more than f2 area of cross section is a young's modulus is why the length of the rod is l got it you need to find the extension in the rod because of these two forces till now our cases where two equal and opposite forces now they are not equal forces got it so your tension you have to find out mass you consider as m very similar way the way we have done the previous question very similar way you have to do this also these are the advanced level questions so i have to tell you all the types of questions so that is why we are doing this is anyone close to the answer or you're stuck find out the tension at a distance x process is very very similar will the rod be at rest or it'll accelerate rod will have acceleration also now isn't it after this question we'll get into other theory also only this question is difficult difficult as in tricky it's not very difficult also should i do it now nobody is able to get it i'll wait for a minute let's see see whatever you can just do it whatever you can it you may get it right or wrong it doesn't matter okay do it yourself okay let us say acceleration is this way right now are you getting anything fine so i can say f1 minus f2 is equal to mass time with acceleration i can say that so acceleration i'll get it as f1 minus f2 divided by m can you tell me what is the tension over here at this point try it all of you if you draw the free bar diagram of only x this is f1 let's say this is t the mass of this block is m by l into x tell me what is the tension all of you how will you get this will the acceleration of this be same as this one or not this piece will also accelerate same acceleration so you will have f1 minus t is a net force on that piece is equal to mass of that piece m by l into x into acceleration f1 minus f2 by m okay m and m get cancelled so t is equal to f1 minus f1 minus f2 by l times x this is your t tension over there now everything is same okay so t divided by area of cross section is stress divided by strain which is dl by dx is Young's modulus okay so yeah it becomes slightly messy but only these kind of questions will differentiate topper from others okay so if you're aiming to do extremely well you must be familiar with all of this this is this divided by the Young's modulus into dx is equal to dl so if you rearrange it and then you just integrate it open the brackets integrate 0 to delta l this will be from 0 to l should we consider the other side the other side also same tension you'll get you'll not get anything different mass will then be m by l into l minus x not into x net force is not zero net force is not zero net force equal to mass m acceleration so you need to find tension not the force which you are applying that is not equal to the restoring force you get the same thing do it yourself you'll get it so after the class if you're getting something different put it on the group we'll discuss it okay but right now whatever we have done here everyone go through it and let me know if you have any doubts no doubts Shashwati is written F1 is greater than F2 but it does not matter you can assume it to be this way the answer won't change okay this is how it will be in J advanced okay I I think okay we'll not take these kind of questions too many of them fine there is one small thing that is there in our syllabus yeah but then dx mass is negligible mass is negligible so there is one small thing write down experiment to determine Young's modulus now in that chapter is straightforward only these couple of questions I wanted to showcase so that don't take it very lightly so it's a simple experiment to determine the Young's modulus and it is like this all of you you have situation like this everyone see earlier both the both the wires are of same cross section area same material same length same length but what you're doing is you're putting a mass on the second wire and you know exactly how much mass you're putting because of which the second wire length increases and you have a scale over here all right and a pointer pointer which was here earlier but because the length of the wire has been changed the pointer has moved down okay the pointer is moving down because a length of the wire is changed so earlier the pointer was here okay so using the pointer you can find out what is the change in length right earlier it was here now it is here so you can find out what is the change in length keep the pointer very close to the end of the rod fine but then you have to keep very large amount of mass otherwise you'll not notice any change in length it'll be very very small okay so keep a very thin wire if you keep and the large mass you keep then you'll see substantial change in length okay that is delta l fine length of the rod was l so this will give you strain okay and suppose you're putting mass m over here stress will be how much everyone sigma will be what what should I write sigma s yes mother got it others you don't know mother is in class 10th you know that he's answering what happened to others what is sigma if I keep mass m over here everyone should answer waiting for you else I have to take your names Akash is not even speaking a single word I don't know what he is doing you're there here doesn't matter Akash you don't even speak and a thing so being there doesn't mean anything Abhinav you have to start talking and these are all personal discussion you can send a message only I can see it I don't know why you have to feel shy what is the answer here Pranav what it is Pooja Nikhil Kaushik Karthik I'm waiting for your answer nonsense I have to take your names Akash what is your answer I want answer from you you are not listening that is why you don't know okay just being in a class you're wasting your time my time mg by a whatever mass you're putting there the weight of that right is mg mg by a is a stress okay strain is this stresses that everything you have measured so the Young's modulus is stress by strain right so you can utilize this it is mg by a epsilon is delta l by l so using this simple experiment you can get the value of Young's modulus okay fine so this entire discussion was related to the longitudinal stress and longitudinal strain fine now we are getting into a different kind of stress okay there is another way you can deform a solid and this kind of stress is called shear stress okay so like what I have told you already that if you can apply a force parallel to the surface then there will be a shear displacement so first write down shear modulus suppose you have this kind of thing all of you draw with me this okay your base is fixed that you can't move and you are applying a force parallel to the surface like this okay this is how you are applying force f fine what will happen is the face will get displaced let me use another color there on this one face will be displaced all of you draw with me everyone okay so this is the kind of deformation that will be seen over here okay so now let's say this is theta getting it so this displacement the face let's say get displaced by a distance of x all right that is your it is like delta l right when you apply the Young's modulus and the longitudinal stress is there delta l is there right here x is there instead of delta l and this length l is that one area is let's say a of the top surface okay so the cause cause is shear stress cause of the deformation is shear stress this is sigma s which is parallel component of the force divided by the area this is sigma s got it now this is the stress how do you think strain will be defined here strain is shear strain how do you think it should be defined here in case of longitudinal it was delta l by l here what do you think it will be can I say x is a strain will it make sense keeping x this area I told you right that is the area area which is parallel to the force everyone tell what should I write strain as in case of longitudinal it was delta l by l watch it will be here everyone x by l right why it cannot be x first answer me that why that effect cannot be x tell me that does x depends on l more length will give higher x right that is a reason why it is better to have effect as x by l x by l can be taken as the x by l can be taken as tan theta where theta is this angle this line makes the edge makes from the initial position and angle is very small so this is roughly equal to theta only got it angle is very small to tan theta and sin theta both are almost equal to theta itself got it so there is something called shear modulus and modulus of rigidity both are same modulus of rigidity or shear modulus g is sigma s by epsilon s just like Young's modulus your shear modulus the name is modulus of rigidity also we will discuss why it is that way this is your modulus of rigidity okay and this is roughly modulus of rigidity roughly equal to y by 3 Young's modulus of same material divide by 3 will be the shear modulus can you tell me what does it mean what is the meaning of this if shear modulus of the same material is very less compared to the Young's modulus it implies what what is the physical meaning of this what is the effect of this anyone easier to easier to deform it is easier to deform using shear a solid can be deformed very easily if there is a shear force acting on it okay three times easier can you give me an example where this is seen just one example where it is seen as you might have seen piece of paper if I stretch it like this then it is easy to break or if I do it like this then it is easy it is easy that way right this is shear okay and when you are using scissor when using scissor scissor offers the shear stress or the longitudinal stress what do you think when you see that to cut it is shear or longitudinal it doesn't compresses it does like this okay it is a shear force using scissor so using scissor it is very easy to cut and your knife when you're cutting potato the force you apply is parallel to the cross section which comes after the cutting it right so if you use knife it is easy to chop the potato but if you stretch the potato like this if you apply longitudinal force will it be easier that way no so there are a lot of examples the earthquake that comes earthquake shakes the ground like this this kind of load is shear which is very difficult for the buildings to take so that is why many of them collides sorry collapse fine so that's the reason why understanding shear is the most important thing if you're talking about any failure in the solid because if you take care of shear the longitudinal automatically takes care of it because one third of the longitudinal is shear getting it let's take a question from your book okay even our bones you know this guy look at this guy don't know why he's doing all that look at this crazy guy this guy is holding on his two legs is holding person one person two person three and there are a lot of tables and chair which is equivalent to another person so effectively on his two legs there's a load of four people around 400 kgs so your bones also have a shear modulus so you sorry the the young's modulus so the bones will get compressed and if it get compressed beyond the yield point it'll not be able to regain permanent deformation can happen okay so yeah crazy fellow but then these are they have a lot of practice they do it for the living i'll give the young's modulus here sorry the shear modulus here which is 5.6 into 10 raise to power 9 Newton per meter square do it first thing is draw the diagram draw the diagram okay aditya got something others no aditya check your calculations akash others okay i'll do it now everyone so square lead of side 50 centimeter so the this side is square 50 centimeter 50 centimeter okay the thickness is 10 centimeter this thickness is 10 centimeter okay okay shearing force on the narrow face so where is the shearing force in the diagram all of you where is the shearing force in the diagram this is the lead slap on it where is a shearing force is like this okay the bottom is fixed this is fixed this is a shearing force okay so which area you have to take area of this you have to take isn't it so the area which you should take is 50 into 10 centimeter square this area you should take fine then um huh so the stress is sigma s is the force which you are applying 9 into 10 is power 4 divided by area that is 550 into 10 centimeter square so that you have to multiply by 10 is power minus 4 to convert in meter square okay so 9 by 5 into 10 is power 6 Newton per meter square this is your stress and stress by strain is your modulus of rigidity so strain which is x by l will give you sigma s by modulus of rigidity g g is this so from here x will be equal to this everybody understood this is l from this is your l right clear to everyone you done it like this those who got the wrong answer you have you seen it all right calculation error don't do that it is very very it create a big trouble calculation errors okay all right so the last last type of deformation is the one in which you're compressing the object from all sides okay that has to do with bulk modulus the name is bulk modulus okay now bulk modulus is a situation in which there is pressure all sides that is compressing the object when you compress the object will the volume decrease or not overall volume will decrease everyone agree will density decrease or increase if volume is decreasing what happens to the density density increases very good okay so over here the volumetric stress it is called volumetric stress okay volumetric stress it is nothing but delta p only okay this is a volumetric stress and the strain what do you think strain will be everyone volumetric strain any guess is what it will be change in volume delta v by v now tell me delta v is a positive quantity or negative quantity delta v is a positive or negative quantity everyone delta v is what final volume minus initial volume delta of anything any variable is final minus initial so it's a negative quantity strain negative strain doesn't mean anything negative deformation what does it mean deformation is deformation right so you put a minus sign here to make deformation positive so minus delta v by v you have to write okay so the bulk modulus b is written as delta p divided by minus of delta v by v okay it can be written as minus of v delta p by delta v okay and in a differential form it can be written as minus v dp by dv that way also you can write okay so this is the definition of the bulk modulus all right everybody clear about it can pressure be pulling the object expanding it yeah it can expand also but the common scenario is this only compression so we'll be talking about this only expansion you know no it doesn't happen I mean very very rarely how the it has to be hollow first of all it can't be solid inside the gas is there it expands so hypothetical scenario you have to generate okay all right so all of you listen here we know that things get can things can get compressed right volume can change mass is density into volume okay if you differentiate mass with respect to x you know the chain rule right dm by dx x is any variable it will be rho you know this or not everyone chain rule you are aware of it everyone chain rule okay so you can cancel out dx what is dm dm is what any cases what is the value of dm change in mass in compression dm what it is will mass change when you compress it will mass change dm will be zero it will be zero okay so rho dv is equal to minus of v d rho so from here you get a very important relation dv by v minus of dv by v is equal to d rho by rho okay so you can write your volumetric strain which is d delta v by v delta v by v is roughly equal to delta rho by rho also so in terms of density also you can write okay chain rule we have done in the bridge program right now let's look back look back at it again in the bulk modulus did we make an yes approximation we have made the approximation is change in volume is very less and that is fine that is a reasonable assumption you can't compress a piece of steel from two meter cube to one meter cube right very small change only will happen okay whatever you're doubted directly type in don't say I have a doubt then I'll ask you what is your doubt v is equal to minus v delta p by delta v okay now here this expression in the numerator is delta p the total pressure or the extra pressure what do you think total pressure or change in pressure everyone what do you think delta p is total or extra pressure it is the extra pressure okay it is a change in pressure fine okay and what is the common scenario most common scenario wherein you will get a change in pressure here you have done the fluids right when you go inside the fluid of density rho let's say you go by a distance of h what will be the change in pressure from here to here at a depth of h density is rho delta p will be delta p will be equal to rho g h and what will be the total pressure over there total pressure is what p naught plus rho g h atmospheric pressure plus rho g h okay but then in the formula you have to write delta p only okay water can exert tremendous amount of pressure okay it is you can't believe kind of pressure water can exert in fact there the professional divers they can't go beyond couple of meters inside the sea otherwise water will compress their chest from all sides they won't be able to breathe they have oxygen cylinders and everything but they can't expand the lungs and take oxygen in so they will be dying right in fact the kind of machines that are made submarine for example i i don't whether you guys know it there was an accident in russian russian submarine because of some loss of fuel or something it got submerged a couple of kilometers down below the sea and at that level the compression delta p is so big that the entire submarine got crushed okay because you can't maintain high pressure inside submarine so inside submarine it is lesser pressure because humans are there outside pressure is very large so it compresses it and when submarine will get compressed screws bolts and everything will just fire like bullets and it will get totally crushed okay so it and these guys were maintaining the communication from outside world they could talk to them and everything but nobody can come and save them because pressure is so high none no one can go inside and save them okay so that is it is a very very important parameter the change in pressure inside the water okay in fact there is a question in our textbook let's attempt that isn't vdp is equal to pdv what what are you asking pranam this is the thing okay so i ignore your doubts then all right so let's take that question which i was talking about do this look at the average depth of indian ocean and that is the true value it's not a false value three kilometers below find out density of the water you can take thousand what is the answer everyone okay aditya got something puja swaraj pranam can't be minus 70 that's very very large okay this is what happens pranam you take things for granted you take you assume the simple questions i will not pay that much attention you will get those questions wrong which lakhs of students will get it right okay imagine the kind of damage all right so here everyone delta p you have to find out first which is rho gh rho is 10 to the power 3 g you can take it as 10 h is 3 into 10 this power 3 so delta p is sorted 3 into 10 this power 7 newton per meter square you know what is the atmospheric pressure roughly what it is atmospheric pressure is roughly 10 s per 5 newton per meter square so we are talking about 300 times more pressure than the atmospheric pressure below the sea one atmospheric pressure itself is very large we're talking 300 times more than that anyway so this is delta p the bulk modulus is given bulk modulus is delta p divided by delta v by v okay so minus sign is there but i'm ignoring minus sign because i know it is compression so delta v by v is delta p which is 3 into 10 is power 7 divided by the bulk modulus which is 2.2 into 10 this power 9 so this is roughly you can say 1.39 or 1.4 whatever into 10 is power minus 2 okay in terms of percentage 1.39 percent change or 1.39 percentage of the compression is there fine so volume will decrease by this much percentage density will increase by the same percentage delta v by y v is equal to delta rho by rho this is also equal to delta rho by rho we have proved it everyone clear everyone clear this was the last piece of the kind of deformation which we are studying here okay anyone has any doubt in this okay clear so just a small thing leftover in the chapter it'll take just two minutes then this chapter will be over okay so we are talking about application of the oxygen cylinder also blast if we go very deep no shashwat simple experiment you have a let's say a plastic bottle okay it will not get crushed because of the atomic pressure because inside the bottle air is there but if you suck the air out from the plastic bottle will it get crushed or not if you suck air out of a plastic bottle will it get crushed same way oxygen cylinder won't get crushed because inside oxygen is already there at high pressure okay but then yes if oxygen pressure inside the oxygen cylinder is lesser and outside pressure is high then oxygen cylinder will get crushed okay and there's a high chance that will happen because 300 atmosphere I don't think that much pressure will be there in oxygen cylinder so if you go very deep down then even oxygen cylinder can get crushed okay we are talking about great depths here write down application of and in inside submarine you cannot afford to have higher pressure the moment you increase it to like five or six atmosphere inside the submarine then you will not be able to breathe okay lungs won't be able to expand because from outside pressure is very high you have to maintain low pressure inside the submarine similarly in the aircraft you have to maintain higher pressure inside because outside pressure is very less okay so in the movies you might have seen whenever some let's say a small hole is also there or door is open in the aircraft which is in the air outside pressure is very less inside it is very high so it is like vacuum outside which will try to suck whatever is there inside the aircraft so everybody will fall you know throw outwards that you might have seen in movies if you see movies behavior of material in a very briefly they have been discussing about all of that we have discussed that already in fact uh suppose talk about a bridge you see bridge has this kind of design you've seen this kind of design for the bridges top most portion will have higher area or not top most portion where the load is there the area is big so that force per unit area is smaller there if you increase area force per unit area will be lesser and why why not keep it like this what is the problem in keeping like this straight any guesses why not like that to save the money yeah save the cost right you using this material you can create another this thing all right so you have already decreased the stress you don't need to make so you know strong like that so that's the reason simple reason wastage of material and have you seen eye beams have you seen eye beams eye beams are very popular have you seen it this kind it can break in the middle but it can with a single if you take like this it will break much faster than an object like this this will withstand tremendous amount of load this can't even though amount of material is similar you just have to increase little bit area here huh there's the eye beam where have you seen posters where else have you seen anywhere else have you seen railway tracks railway tracks their eye beams only okay they can also withstand large amount of pressure without crumbling down okay because this this area is larger okay so different applications are there and there is one little formula that is given in your curriculum is this one suppose you take a beam like this okay and you hang a weight in the middle we hang a weight in the middle what will happen to the beam beam will bend beam will bend let's say bends by an amount of delta let's say this is delta you want this delta to be less or more less you don't want deformation in your solids right so the expression for the delta is w w is a weight into l cube divided by 4b d cube into Young's modulus this is an empirical relation proof of this is not in our curriculum l is the length l is a length this is the d this is d okay d is the width and d is the depth okay so you can see here if you want your deformation delta to be lesser what you can do is you can you can keep the length same you can increase the depth you increase the depth or you can increase the d increasing d will will help lot more because this is d cube in the denominator right so this using this relation there might be one or two numericals here and there in your school okay so that is all about the chapter guys and i hope you have you are now very clear about it this chapter is a simple chapter very simple okay what is depth depth is inside right there will be it is not a two-dimensional thing aditya it is a three-dimensional thing there will be one dimension going into your screen right rod has width so width and depth both so yes so what i was saying was whatever varieties of numericals that exist more or less everything we have done today itself okay it is not like Newton laws of motion wherein different different varieties thousand thousands and thousands of different varieties of questions are there here varieties of questions are very few whatever difficult type of varieties are there i have already done it with you guys okay so all you have to do is get to know these difficult kinds and be comfortable with them and this chapter is over at all levels school level j means level j advanced level need level whatever level you're aiming at do not leave this chapter at all because this chapter ensures that if a question comes and you have done little bit of practice you will get full marks so do not think of leaving it and i'm going to give you assignment make sure you do it and finish it off in this week itself so that you never have to look back at this chapter for next two years okay and this chapter will be solid if you do the assignments immediately after class maybe after one or two days okay so that's it guys from my side and we will meet next week we will start mechanical properties of fluid that is a big chapter it'll take three to four classes to complete that bye for now rigid body dynamics will take later first i have to catch up your school