 Alright, so let us move ahead now. There is something called shear modulus or the shear stress as well. And that is more important when it comes to breaking a solid or being very careful about the structure of your building. I done shear modulus. So shear modulus is about the shear stress. So let us take a simple example of how shear stress looks like. So here it is. This is the box of a solid and the lower surface is sixth. You are pushing the upper surface with the force F. When it comes to longitudinal, the force was perpendicular to the surface. Either it was compressing or stretching. When it comes to shear, the force is parallel to the surface. And the kind of deformation you will see is that the upper surface, since it is free to move, goes like this. So the shear stress is basically force divided by area that is parallel to the force. And how do you think we should quantify deformation? This length is let's say 10. You could look at the angle. Or should I say that how much is the displacement? Delta x. Should that be my deformation? Correct. But if I take a larger length of the solid, don't you think that delta x will be different? Delta x will be more and more. So then it has to be angle. Angle is a better representation. Let's say angle is theta. I will not take alpha. Alpha we have been using for other. Theta remains same. Doesn't matter what is the length. So epsilon is basically theta, which is roughly equal to tan of theta. Because theta is very small. And tan of theta is basically delta L by L. Reading it? Yeah. Fine. So this is a ratio between shear stress and the shear strain is called modulus of rigidity. Write down. And this is very important because it will take three times lesser effort to break a solid like this than stretching and compressing. Okay. So write down like this. This is this will be FL by parallel area into delta x. And for the same material for, for, do you think the G and y will be same or different? Different. Yeah, definitely. Roughly G is roughly young modulus divided by three roughly. Okay. For most material. Okay. And this is how the earthquake breaks the solid. In fact, when you use a scissor. So in scissor, what happens is there are two metallic plates that move on top of each other. So when you do that, basically you are applying force parallel to the surface. As in, can I say, so you see this scissor? Yeah. Okay. So you can see the upper plate. This plate is moving on that plate. And suppose there is an object in between. And this edge will, will arrest the objects lower surface and then this surface will apply a force parallel to the surface. Okay. And solid breaks easily when you're using scissor, you're basically applying the sheer stress. Okay. And similarly knives as well. Knife, what it does is that there is, let's say an object like this, you have knife, which is just cutting it like that. So basically you're applying a force parallel to the surface. Yeah. It is getting cut like this. You have always applied force in this direction. Yeah, okay. You don't apply force like this to break apart, let's say potato. Yeah. That will longitude knife. But we are applying, we are using knives. It is much easier to apply sheer to break a solid. And we have been doing it day to day tasks. All right, let's take a question on the sheer modulus as well. The value of G is given. 8.6 into 10 is upon 9. Yeah, so I got an answer. What is the answer? I'm getting 16.1 into 10 to the power of minus four meters. 16 into 10 to minus four. You always miss one factor of 10. Check. Is it 16 into 10 to the power of five? Yes. Okay. So check. Yes. All right, so this is the sheer modulus. I don't need to solve it. You both have it? No. All right. Now there is a small experiment to determine the young's modulus of a material of a wire. Okay, let's discuss that. So I will draw the experimental setup. I think this chapter will be over today. So we don't need to drag it to the other day. So basically what I'm doing is I'm having two similar wires. I'm having two similar wires and both have this weighing plane. And both of them had same length initially. But this particular weighing pan, I'm adding weight because of which the length increased. Fine. So what I'm doing is I'm also attaching a scale like this over here. And then I'm putting this thing, a pointer from the second wire. Okay. So earlier it might have been pointing over here. But as it got stretched, the pointer moves down. So this much will be the delta L. Okay, yeah. So you want to tell for the length this much. Okay, you need to be very careful. For this length, this is the delta L. Yeah. Are you getting it? So if you want to tell for entire length, you should connect a pointer over here at the end. Yeah. Okay. But anyways, that precaution you can take. This one is the one from the top to the bottom. That's delta L. Delta L is for this length. Yeah, okay. This length got changed by delta L. Okay. If you attach the bottom, delta will be more then. Yeah. But ignoring this length, suppose I'm attaching at the bottom only, I can say that the strain is delta L by L. So I found delta L and L. I exactly know what is the load I'm keeping here. So I know the sigma also, which is F by A. F is mg, whatever load I'm keeping there. Okay, so Young's modulus will be equal to sigma by epsilon. Got it? Yeah. All right, so let us move to the last bulk modulus. What do you think bulk modulus will be? The volume. Compressing it from bulk. Yeah. Yeah, it is basically change in volume. Okay. It has to do with change in volume. And what changes the volume? Basically when you apply forces from all the sides. Okay. So this force usually is applied by a fluid. Fluid is something which can apply a uniform compression from all sides. All right. So the bulk modulus, the cause is pressure only. Pressure itself is forced per unit area. Okay. So the cause is pressure. We say that sigma is pressure only. And what about epsilon? That's the same thing. Delta V by V. Yeah. Change in volume by initial volume. Yeah. Letting it. In this case, since you are compressing, delta V will be less than zero. Okay. So the bulk modulus, by the way, sigma is not P. It is delta P. Atmospheric pressure is anywhere there. Yeah. Extra pressure you are putting, that is delta P. That is my sigma. Okay. So the ratio between these two quantities, delta P divided by delta V by V. Is there something wrong in the expression? Can you tell me quickly what it is? Something I have missed. Is it the sign or something? Exactly. Bulk modulus should be positive number. You won't say that it is my strength of a material is negative. Yeah. You automatically put a minus sign. That is why you see a minus. So this can be written as minus V. If I write in a differential format, it will be minus V. Dp by dV. Yeah. And we know that if you go inside the fluid, pressure will be what? Atmospheric pressure. Plus rho gh. So changing pressure is how much? Rho gh only. Yeah. So your delta P is basically rho gh in a fluid. And when you deal with bulk modulus, you automatically assume that you are dealing with a very small object. So the pressure, you will say that it is more or less same for the entire object. Otherwise you will argue that the pressure here is something and here it is something else because there is a height difference. Fine. Assume the object is very small and height difference is negligible. Okay. Okay, whenever we are solving any problem related to bulk modulus until and unless it is specified that it has a height of 2 meter 5 meter like that. We are going to assume that the object is negligible in this. Yeah. Let's take this question now. Both of you write down. The average depth of Indian ocean. You know how much it is roughly? No. Average depth of Indian ocean is around 3 kilometers. It's an average depth. Somewhere it could be around 5 kilometers also. Yeah. It is extremely deep. So 3 kilometers at average depth, you need to find out the compression of the water itself. At the bottom of the sea. Bottom of the ocean. Oh. The bulk modulus is given as 2.2 into 10 raise to power 9 Newton per meter square. Take G as 10 meter per second square. Find out delta V by V. Yeah. How much? I'm getting minus 1.4 into 10 to the power of minus 5 meter cube. 1.4. Wait, no, not me. Yeah. Yeah, minus 1.4 into 10 to the power of minus 5. What? I'm getting 73.33. Is it? Yeah, for some reason. How can we use the edge? What do you have written? Oh, wait, I didn't convert it into meters. Right. Factor of 10 Dhruv is missing. Yeah. Don't miss that when you deal in money. What mistake did I make then? So you have got. Yeah. So now I'm getting minus 1.4 into 10 to the power of minus 2. Now just correct. Negative you should remove. When they are saying find out compression. Okay, yeah. Find out compression. That is correct then. But if I say that what is the fraction compression, then you should tell me a negative number. No, sorry, then you should know. Positive. I don't know what I'm making a mistake. Delta P is H rho G. H is 3000. Rho is 1000. Yeah. G is 10. Delta P. Yeah. Delta V by V is delta P divided by B. Yes. 1.36. 10 is power minus 2. Okay. Or in terms of percentage 1.36 percent. Yeah. Okay. Now since the volume is. Okay. I'll just talk about. That since the volume is decreasing, of course density will. Increase. Right. So density is what mass by volume. This is my density. And. I can write it as M divided by. The volume will be V naught plus delta V. Okay. So this can be written as M divided by V naught. 1 plus delta V by V naught. To the power minus 1. Okay. And this can be written as M by V naught. This is minimal approximation delta V by V naught. Rho. So Rho is equal to initial density M by V naught. Rho naught 1 minus. Whatever delta V by V naught. Okay. So like this you can deal with change in density also. And since delta V is negative. Rho will be more than Rho naught. Yeah. Any doubt still now? No doubt. Okay. So this chapter is practically over. Just that at the end there are few applications of elastic behavior of the material. And. Will not. I'll. Tell you the. Few expressions. For example, if you have this rod like this. And if you hang a mass at the center. This thing will bend. This is. This you can say is delta. If you put a mass over here. Let's say. It has. Which creates a weight W. So experimentally it is found out. Let's say this is. Lent. L. Experimentally the dip of the center point is found to be. WLQ. Divided by 4B. DQ. Y. What is. D. See the length is the length dimension. Then there is. Width and the height also. Yeah. Okay. That is what it is. B is the. Bread. And B is the. That. Okay. So that is the reason why you have beams having eye shape. Okay. When you use eye shape beams. Not able to draw the symmetrical one. But anyways. If I use eye shape beam. And put it horizontally. Then this becomes my the value of D. Okay. So if I increase the value of D. My dip. Decreases. Oh. That is the reason behind the eye shape of the beam. Okay. Hmm. Yeah. So. That is all. It is. In your syllabus. There are much more of wider application of elastic behavior. The way you make the arch. It is usually curved like this. Why this kind of shape? Why not anything else? So there is a reason behind all of this. Which you're going to learn in the engineering. Fine. So we have 15 minutes. We'll solve a couple of questions. Is there any doubts? No. Here is the question. There are two masses connected. And there are two wires also. Young's modulus of this is Y1. Length is L1. And area of projection let's say same. I'll not complicate it. So this is Y2. L2 area of projection is A. This mass is M1. And that mass is M2. Okay. Yeah. You need to find the elongation on both the wires. The wires mass is negligible. You'd find the water both wires. You need to find the elongation on both. Are you stuck or doing it? I'm doing it. Yeah. So I got an answer. How much? So I got G into M1 plus M2 into L1 over A into Y1 First tell me for this. What do you want? For that, that's what I got. G into M1 plus M2 into L1 on AY1. Stress for this wire is how much? Stress. Wait. I didn't do Without finding stress. How can you find strength? I like directly That is How much stress? Answer me that. There's a force per area. So M1 plus M2 into G by A. See you guys are not paying attention to the basics. Can't you draw a free body diagram and see what is the force acting on it? So let's say here the force is T1. This will also be T1. M1 G This will be pulled up by T1 and this is going to be pulled down T2 T2 as well. This is M1 and then this get pulled up by T2 just pulled down by T2 and then you have M2 This is M2G and this is T2. Everything is at rest. So T2 is what? M2G T2 is M2G T2 minus M2G is 0 so T2 is equal to M2G and in order to find T1 for this M1 I will draw the free body diagram and write the equation T1 minus T2 minus M1G is equal to 0 So this is M1G. So it is M1 plus M2G This T1 that is T2 So Sigma1 is T1 by A So that is M1 plus M2G by A Sigma2 is T2 by A So that is M2G by A Okay Now it is simple Delta L1 by L1 is equal to Young's modulus times of that should be equal to the stress M1 plus M2 by A into G So from here you can get Delta L1 and for the Delta L2 you write like this Okay so if you don't follow these steps you are not going to get marks in any school exam you are writing at least Got it? Yes Fine So I will send you the worksheet for this chapter Okay finish it off once for all it is a small chapter you have to complete it and then you don't have to revisit this chapter again and again Thank you Let's start a new chapter Planetary of gases Any doubts, anything on this chapter No No doubts Okay So let me send the worksheet After the call only I will send you on the group Okay for today we will have soon another class Thank you