All right, guys. So today I'm going to talk about, I'm going to start this chapter, mechanical properties of fluid. Okay. So I'll start this topic. Meanwhile, people are joining in. So I can't make everybody wait for people joining late. Okay. So let us start. All of you please write down mechanical properties of fluid. Okay. So we are going to start this chapter, mechanical properties of fluids. Now, when I talk about the mechanical properties of fluid, you will you will understand that just like in the previous chapter, in which chapter was what mechanical properties of solids, right? Mechanical properties of solids. What are the mechanical properties we have talked about? Quickly tell me. Elasticity. And we have actually concerned ourselves with with the with strength of the solids, isn't it? So strength of the solid is something which is the most important physical property of a solid. And why we are focusing on strength of solids? We are between these structures from the solids. Okay. Fine. So what I was saying was that the mechanical properties is very important when it comes to solid, because we are constructing material or constructing, you know, an object using solids.  So the strength is very important to know. Okay. But when it comes to fluids, the strength of fluid is not something which we are concerned about. Okay. So we don't actually care about the strength. So our focus is on the other mechanical properties of the fluids. Okay. Other mechanical properties when I say it could be as simple as that, you know, if I apply some amount of pressure, then with what velocity the fluid will come out. Okay. Or if I go inside the fluid, will the pressure of the fluid changes or how much force a fluid can apply? Okay. So these are the kind of properties that we are dealing with when we talk about the mechanical properties of fluids. Fine. Now, we are going to understand few of the parameters, parameters to understand fluid properties. Now, do you remember when we were talking about the, this circular motion, what kind of parameters we have used to analyze the circular motion? Do you remember? Okay. I'll type names now. Anurag, tell me. Yes, sir. What are the kind of parameters that we have used to analyze the circular motion? So angular velocity. Correct. We have used angular variables. We have used angle. We have used rate of change of angle as angular velocity. Okay. So like that, we have utilized the different types of parameters to understand the circular motion. Fine. And similarly, depending on the scenario, we use different, different kinds of parameters. For example, in circular motion, I'll not be using parameters like displacement, velocity or speed to analyze the motion. Okay. It is much more convenient to use the parameters which are made for it. For example, angular variables are actually made to analyze the circular motion. It becomes very comfortable. Similarly, to analyze the fluids behavior, I need to use some of the parameters of fluids that is generic and that is only for the which defines a particular fluid property. Okay. So in order to understand which parameter it is, there are few experiments. Okay. So let's take up these experiments. So you know that in this chapter, we are going to study about the fluids. Okay. And this, the fluid chapter can be divided into three parts. Please write down. I will first introduce the structure of the chapter itself. That'll be better. Okay. So this entire chapter can be divided into three segments. First is fluid statics. What does it mean fluid statics? Anyone? Fluid when it's not moving. Fluid when it is not moving, when you analyze its physical property, that is what it is fluid statics. Now tell me, if you take a bucket of water and run with it. Okay. Is that the study of that fall under fluid statics or not? You have a bucket. Yes. So as the particles are not moving with respect to each other, that's why the fluid particles are not moving with respect to each other, the stationary. Okay. See, when you say that fluid is in motion, you don't actually mean to say an example like what we have just taken. Okay. So when I say fluid is flowing, the flow of a fluid and the example that we have just taken are two different things. Fluid flow is when, as you said, one particle of a fluid move relative to the other particle. Okay. So that is what it is. Okay. Ariman is messaging. He can't hear anything. Fine. So we'll talk about the fluid statics where the fluid particles are at rest relative to each other. Okay. So what are the kind of properties that we'll be talking about when fluid is at rest? Give me a few examples of fluid at rest. Water in a glass. What? Water in a glass and properties. Okay. Fine. We'll continue. So the example that you have taken the storage of fluid somewhere. Okay. So for example, there's a dam. Okay. And a lot of water is getting stored. And this, this water that we are talking about is huge. And this water body will exert pressure. Okay. And because of that, there can be a force on the dam itself. So we must know to create a dam, how much. Anyways, guys, let's continue. Okay. So fluid statics, we must know these properties of fluids. For example, pressure is a very, very important property of fluid. I must know with what force a static body of water is exerting force on the dam. Otherwise, I'll not be able to construct the dam. And similarly, if someone goes inside water, we know that as a human being, we cannot go beyond let's say 10 or 20 meters. Okay. Otherwise, the pressure of water becomes so big that it will compress our body from all sides. And because of that, we'll not be able to breathe. Because when we breathe, when we breathe in our lungs expand. And because the pressure from the water is so high that it doesn't let our lungs expand. And hence, we don't, we'll not be able to take oxygen in our body. All right. So that is why there's a limit up to which divers can go up to. And similarly, there is a limit up to which every structure, like submarine, submarine can also cannot go beyond a certain limit. Because the pressure due to a fluid soon we are going to learn that it keep on increasing as we increase the depth. Okay. So that is why it is extremely important to understand how much pressure a particular fluid can exert. Because it is a matter of life and death at times, how you design a submarine and how much diver can dive up to. Okay. So that is why a fluid statics is there, which in which we are going to study fluid at rest. Okay. And then there is a fluid dynamics. As a name suggests, in the fluid dynamics section, we are going to study the flow of the fluids. Okay. Now, what will happen is that in the, when the fluid is flowing, it will have velocity, of course. Right. So I should know that what causes the change in velocity. And I should also know that how to write the kinetic energy of the fluid. If I know exactly how to write kinetic energy and potential energy of a fluid, I may be able to use work energy theorem or conservation of energy. And then I can analyze the motion in a much better way. Fine. So this is fluid dynamics. And then there is third section of the chapter, which talks about other properties. Okay. So those who cannot hear me, you can please use this link as well to join. Okay. This is, okay, this is the max I can do. Fine. When it comes to other properties, there are few other properties that we are going to learn. And these are surface tension, right down surface tension is there. Then we are going to learn about viscosity. All right. Then we are going to learn about the type of flows. There are different types of flows and they because of the type of flow, the property of the fluid could vary. You might have heard about turbulence. Yes or no. So when the flow is turbulent, a different kind of property is exhibited by the flow by the fluid. And if the flow is laminar, then there is another kind of property that is exhibited. Okay. So this is how the entire chapter is organized. Okay. So we are going to first learn about the fluid statics. As in, we are going to understand the properties exhibited by the fluid when the fluid is at rest. Write down fluid statics. Now fluid is at rest. And the most common example of fluid at rest is fluid inside the bucket. Okay. So we are going to draw a bucket now and we are going to analyze the property of the fluid inside the bucket. So suppose this is fluid inside the bucket. There are few experiment that we are going to conduct now. A very, very basic type of experiment I'll conduct. Let's say that you have this kind of arrangement. You have a piston like this inside a cylinder and this piston is connected with a spring like that. Okay. Let's say area of cross section is A1. All right. Now I am going to move. So suppose this piston, this entire piston and spring arrangement is kept inside the fluid. Now fluid can exert force from which side? From the top, from the sides or from the bottom? Which side fluid can exert force? All sides. What? All sides. All sides. No, I am talking about suppose spring is getting compressed. Now tell me. From the top. The spring will get compressed only because there is a force from the top. Yes or no? Yes, sir. Okay. So when this spring suppose get compressed, if it get compressed by X, okay. And the spring constant is K. The force with which the fluid is applying force. The fluid force will be K times X only. Isn't it? Right. So what I have seen is that if I move the, if I move this structure in the horizontal direction like this inside the bucket, if I move from this point to that point, point number one to point number two, if I move the value of X or the compression in the spring remains the same. What does it mean? Force in vertical direction is affected by motion in horizontal direction. It means that the force write down force due to the fluid on the piston is unchanged. If moved horizontally. Understood. This is what it means. Force is K X. And if X remains the same, then the force also is the same. It automatically means that yes or no. Now when you move, when you're moving the piston, okay. When you're moving the piston down, then it is seen that the value of X is increasing. What does it mean? Okay. Skanda, what I'm saying is that I have this entire structure. I am going to put it inside. Suppose I'm putting inside here like this. This is structure is putting inside. Okay. There is a spring. This cylinder is inside now and I'm moving the cylinder from point one to point number two. And I'm observing how much spring get compressed. If compression in the spring is unchanged, then I'll say that the force due to the fluid is same. Okay. So this is what I've seen. If I move horizontally, the compression X is same. Because of that, I can say that the force due to the fluid on the piston is unchanged. If I move the piston horizontally. Fine. Understood, Skanda? Okay. Now, see, let us do this experiment further. Suppose I go from point number three to point number four. I am now moving vertically. Okay. When I'm moving vertically, I'm seeing that if I go from three to four, X increases. What does it mean? Force increases. Force increases. Force on the piston increases. As you move and if you increase the depth, the force due to the fluid increases. Okay. These are the findings from this very simple experiment. Okay. Now, suppose I take another piston, which is slightly wider like this. I have a piston like that and I'm using the same spring. Okay. If I have the same spring, this is cross-sectional area A2. Okay. What is observed is that if I move this spring piston from point number one to point two, the X, the compression in the spring increases. What does it mean? That force on this piston at the same level is more than the force on that piston. All of you agree? Sir, could you repeat that? Yes, sir. See, what I'm saying is if I move this one, piston number two. Okay. There is this arrangement, arrangement number one and this one. Second piston spring arrangement. If I move this spring and piston arrangement from point one to two, this one shows more compression in the spring. What does it mean? This experience has more force. Yes or no? I understand because if the area is more, shouldn't the force be less? I'm not saying why it is. I'm saying this is what you have observed. The compression in the spring is more. Does it mean that the force is more? Force is KX. If you have observed that spring compression is more than in this case, the force in this case, let's say it is F2 and that was let's say F1. F2 is more than F1. So force increases if area is increasing. This is what is observed. And another thing which is observed here, which is very interesting is force per unit area F1 by A1 is equal to F2 by A2. Please write this down. So even though the forces experienced by these two pistons are different experimentally, but force per unit area remains the same. All of you understood this? Yes, sir. Yes, sir. Okay, so please speak up quickly.  So I'll move slower and slower if you don't respond quickly and use your mic. Do not use chat. Okay, speak up. Otherwise I have to open the chat window again and again. Fine. So this experiment has proved that the pressure is more fundamental quantity for the fluid and force can change. If you just change the area, force will change. So, but then still the pressure will remain same at a particular level. Okay. So because of that pressure becomes the more fundamental quantity. All right. And hence we are going to use the pressure, the definition of pressure to understand the property which fluid exhibits. Okay. Any doubt till now? No, sir. Okay. All right. So let me start with the first law that we are going to learn in this fluid statics Pascal's law. Please write down Pascal's law. Okay. Pascal's law we are going to learn. So according to Pascal's law, please write down. According to Pascal's law, at a particular point, at a particular point inside a fluid, pressure is same in all directions. At a particular point inside a fluid pressure is same in all directions. So let's say you, you consider a point inside the bucket that has let's say water in it. Okay. Try to find out pressure at this point, whichever direction you go in all direction, the pressure is same. Fine. It is same in all directions. This is what Pascal is saying. All right. Now there is a proof of Pascal's law as in you, we may not call it proof because in this proof, we are going to just take one of the examples and we are going to demonstrate that Pascal's law is valid. Okay. So all of you please draw a bucket. Suppose this is the bucket. Okay. And it is filled with fluid. That's it is filled with fluid. And I'm going to assume draw a triangle like this. It's a right angle triangle. Now this triangle is imaginary. Okay. This triangle actually doesn't exist. Okay. This triangle is made up of the same fluid as that, which is inside the bucket. Okay. Inside the bucket, let's say there is a fluid of let's say density row. Okay. This is the same fluid. Same fluid is there. And this triangular zone is an imaginary part of the fluid inside it. Okay. This doesn't actually exist. It's a hypothetical imaginary object that you are assuming. Understood. Now let's say this angle is theta. Let's say this angle. All of you draw this. This angle is let's say theta. Okay. And can you tell me you have three surfaces surface one. Let us say this is surface two and that is surface three. Okay. Now tell me which direction the surface one will experience the force. This is an imaginary surface. Okay. Now who will apply force on this surface? The fluid which is right next to it will apply the force on it. Right. So can you draw the force? Which direction the force will be? Quickly tell me that one which direction the force will be? All directions. All directions. Okay. The acting on all surfaces. Okay. That is, you know, understandable. See here when I talk about. Towards the triangle. Okay. We have just written that pressure is same in all directions according to Pascal's law. Okay. That is fine. But what is force? Force is pressure into area. Okay. Now on a particular area, let's say surface one is the area. Okay. The surface one will experience the pressure from the fluid which is outside the surface one and the, the force on that surface will be perpendicular to the surface like this. Why it is like that? Because the way the pressure is defined. Okay. You might not be aware of it. But the pressure is defined as perpendicular component of the force divided by area. Okay. This is how pressure is defined. Now you may wonder that we have just said that pressure is same in all directions. But how come we are assuming that force is only perpendicular? Okay. What I'm saying is that if it does not matter what is the orientation of the surface. If surface is like this, the force will be like that. If surface is like that, the force will be like this. Okay. If it is like that, the force will be again like that. So all the time, whenever you are finding the force because of the pressure, it will be perpendicular to the surface. Okay. And why, how it is possible because pressure is same in all directions. Okay. Pressure always compresses the object. Okay. It always pushes the surface. And here also it pushes it like that. Okay. So let's say this is force F1 by the fluid and from here, which direction the force will be on surface three, which direction perpendicular horizontal. It will push it like this. Right. So let's say this is F3 and there will be another force from here. This will be F2. Okay. Okay. One more thing, remember that we are proving the Pascal's law. You cannot apply Pascal's law while you're proving it. Fine. So, but then I know that the force because of pressure will be perpendicular to the surface. So that is what I'm using here. And based on this definition, I'm going to prove that the pressure in all three surfaces are equal. Okay. But right now I'm just drawing the forces. Now, if this is theta, is this triangular object at rest? It is at rest. Right. So some of the all forces should be zero. So can you write down the equation where net force in x direction is zero and net force along y direction is zero. Okay. Just write down the two equations quickly. This is angle theta. Have you written equation? Net force equal to zero only you have to write. Okay. Dhyan is saying F1 sin theta equal to F3. F3 minus F1 cos theta equal to zero along x. Okay. F3 is equal to F1 cos theta. Okay. So let's see here. The component of F1 horizontally is what? In this direction. How much it is equal to this angle? F1 cos theta. F1 cos theta. No. This angle is 91 is theta. 1 sin theta. So F3 minus F1 sin theta is zero. So F3 is equal to F1 sin theta. This is first equation and vertically what I can say. F1. Sorry. F2 minus F1 cos theta is equal to zero. Right? So F2 is equal to F1 cos theta. Yes or no? Vertical direction. All of you understood this? Yes sir. Yes sir. Okay. Now tell me let's say this is surface one area. I'm calling area of surface one to be A1. Area of surface two as A2 and area of surface three as A3. This thickness is uniform. Thickness is uniform and they are forming rectangles only. So is there a relation between A1 and A3? How much is A3 equal to in terms of A1? A3 is equal to A1? Sine theta. Sine theta. Correct. See this length, this length is hypotenuse times sine theta. And length into width is the area width is same across. It is only about length. So that is why A3 is A1 sine theta and A2 is equal to A1 cos theta. Fine. Now if I take ratio, if I divide these two equations, you'll see that sine theta disappears from here and cos theta disappears from there. So if you combine these two equations, you will realize that F1 by A1 is equal to F2 by A2 is equal to F3 by A3. Okay. Now does this prove that pressure in all direction is same? Or at least it demonstrates that if you take a rectangular object, the pressure in this direction is same as pressure in that direction and pressure in this direction. Yes or no? Yes, sir. Okay. Now, tell me what is the assumption that we have used while deriving this? Anyone? Sir, are we ignoring gravity? Exactly. Good. So this is what we are doing. We are ignoring the gravitational force. No force of gravity. Okay. Now, when I can ignore the gravity force, anyone? When can I do that? When the mass is almost zero? Yes or no? Yes, sir. So I am in fact assuming here that this triangular object is very, very small and that is what the Pascal's law is about. Pascal's law tells me about pressure at a particular point. Okay. So we are assuming that this triangular object is extremely small and hence since we are ignoring gravity, I can say it is like a point and hence it is a demonstration of the fact that the pressure at all directions in a particular point is same. All of you understood? Any doubt? No, sir. No, sir. Okay. No, sir. Fine. So we have also demonstrated one more thing about pressure sometime back is that the pressure does not change if you move horizontally. Yes or no? Right. Now, let us talk about variation of pressure with depth, variation of pressure with depth we are understanding now. Now, again, just like previous cases, we have to draw the bucket. So we are going to draw multiple buckets. It is about fluid static. Okay. Now, let us see how the variation of pressure will happen with the depth. Now, this, if I draw one line like this, it automatically means that inside it is fluid. Okay. I'm not going to draw dot, dot lines throughout. Okay. Unless it creates a clutter. So let's say fluid is inside this bucket. I want to see how the variation of pressure happens with the depth. Okay. So what I'm going to do here is that I'm going to assume an imaginary object here. Let's say I assume a cylinder. Okay. This cylinder has height h. This is height h. Let's say that this is point number one. Okay. This is point number one and this is point number two. Okay. And the area of cross section of this cylinder is uniform and it is equal to a. Fine. And let's say the pressure at one is P1 and let's say pressure at point number two is P2. Okay. So can you draw the free body diagram of this cylinder? This cylinder is imaginary. It is made up of the same fluid. Okay. Let's say mass of the cylinder is M only. Can you draw the free body diagram quickly show the forces and try it. So ignoring gravity. Gravity you can't ignore now because you are not assuming it to be a very small cylinder. You're saying that it has height h. It's not small. Okay, sir. Are you done? Yes, sir. Okay. So this is how it will look like. So this is the cylinder, let's say. Okay. Which direction the pressure force will be on the first surface up or down? It has to be down. Remember, pressure always pushes. Pressure never pulls. Okay. So this is P1A. P1A will be the force. Okay. Show me you can speak. You don't need to chat. Your mic is working right. And from the below it will be upward direction. Yes or no? Let's say the force is P2A. What are the force? Gravity. Which is MG. All of you understood this? Yes. So if I balance the forces here, it will be P1A plus MG is equal to P2A. Right? Next force will be zero. So P1A plus MG minus P2A will be zero because this cylinder is at rest. Okay. Now I can rearrange the term and I can write it down as P2 is equal to P1 plus MG by A. So you can see clearly pressure is changing as you go from point one to point two in the vertical direction. Okay. Now I want this expression H in the picture. Why I want H? Because clearly because of H only there is a pressure variation. If I move horizontally the height H is zero. If I move horizontally. So only when I move vertically down then only the H is coming. So what I'll do is that I'll try to write mass in terms of density. So mass will be equal to rho A into H. Okay. So once I substitute this mass. Let's say this is second equation. On to the first equation. I'll get P2 is equal to P1 plus. You can just see if you substitute rho AH here, you'll get rho GH. So a very nice looking expression is coming here. Pressure changes with depth H with this. Okay. And another good thing about it that density rho is coming in the picture. Mass is not coming in the picture. Okay. And density doesn't depend on how much liquid you take. So that is why this expression is much better than this one. Okay. All of you understood? Yes. Yes, sir. Yes, sir. What we have learned till now we have learned the variation of pressure horizontally and we have learned the variation of pressure vertically. Okay. But one assumption that we have taken. Okay. That assumption is that the bucket or whatever the fluid body is not accelerating acceleration of the fluid body is zero. All right. Now I'm going to now change that assumption and I'll make it more generic. All of you understood till here. No doubts. Don't don't hesitate to ask doubts. You can message me. You can pick up. Okay. I'll move forward. I'll move forward. Okay. So let's say that I'm going to find out the variation of pressure variation of pressure. With acceleration. Okay. So let's take an example here. I'll have a bucket. Let's say this bucket has wheels in it. And this bucket is accelerating with acceleration a. Okay. Now let's say that fluid is there inside. So what do you think the fluid surface will be like? Will it be horizontal or it will be little bit inclined? And how will be the inclined if it is inclined? What do you think? Will it be horizontal? It will be inclined. It will be inclined. It will be inclined. It will be inclined like that. No. No. No. Nothing. Nothing. It will be inclined like this only. Like this only. You can take up. You can take let's say a mug of water. And you can even try running with it. You will see that the water will come out from this side. No. Got it. All right. Now in this case I want to see how the pressure will vary horizontally. Now you'll see that the pressure will vary horizontally as well. Okay. So now let us assume an imaginary cylinder which is horizontal. Okay. Let's say the length of the cylinder is L. This is the length of the cylinder. And let's say the area of cross section is a. Okay. Area of cross section is let's say a. Okay. And this side, let's say pressure is P1 that. Let's say this side is P1. That side is P2. So which direction the force will be. It will be like this or not. Is there a correct direction of force? Yes, sir. So it will be let's say P1A. And this way it will be like that. So let's say this is P2A. Now if let's say density of this liquid is low. Can you derive a relation between P1 and P2? The derivation is extremely similar to the way we have done already. That's the most fun. What? Can we take this mass M? You can take this mass M. Later on you need to replace mass with the density. Anybody done? No one. Should I do it or should I wait? Sir, one minute. You need to just draw the free body diagram. Okay. So is P1 is P2A is equal to P1A plus row A L into A. Okay. Yeah, something similar. Let me derive it here. This is the imaginary cylinder. Okay. There is this force P1A. This is the force P2A and the mass is M. Okay. So net force is equal to mass same acceleration. Right. So P2A minus P1A. This should be equal to M into A. Okay. And mass I can write in terms of density. So P2A minus P1A is equal to density into volume, which is A into L like this. Fine. So I can cancel out area and I'll get P2 is equal to P1 plus row A L. So this is a very similar looking expression just like we had in case of vertical variation with gravity. We have P2 is equal to P1 plus row GH. Instead of G, we have A. Okay. Instead of H, we have L. This is what I'm getting here. Now the pressure is changing horizontally as well. Are you getting it? Now tell me one, one question, one answer for this question is will this acceleration change the vertical variation of the pressure? Yes or no? Yes sir. Yes sir. Yes sir. Why? This is the height of the water. See this acceleration is horizontal. Okay. It cannot change things which are perpendicular to itself. Are you getting it? We have learned about vectors, right? A vector cannot affect anything that is perpendicular to itself. So this acceleration has no effect on the variation vertically. It will still follow the same law. P2 is equal to P1 plus row GH. It is going to follow that law itself. Okay. So let's say that. Sir then what about the increased height on the left-hand side? Huh. That is, that you have to take care of. Okay. But the law remains same. I'm talking about the law. Okay. Okay. Let's say this, this distance is h1. Okay. Let's say this is h1 and this distance is h2. And outside it is atmospheric pressure. P0 let us say. So pressure over here will be what? P2 should be equal to what? Row. Can I say P2 is equal to P0 plus? P0. Row. H1. Okay. Don't hesitate. This is what we have learned. Vertical direction. The pressure will vary like this. So I'm going from atmospheric pressure inside the fluid here. So this is what I have to write. P2 is equal to earlier pressure plus row GH1. Okay. And P1 should be equal to what? P0 plus? Row GH2? No. Watch P0? At atmospheric pressure. Pressure outside the fluid. Okay. Okay. So what I'm saying is pressure outside the fluid is same as pressure just inside the fluid. I'm assuming that. Okay. Because when you go just inside the fluid, you can say h tends to zero. So whatever is the air pressure outside the fluid, it is just inside the fluid also the same pressure. And then you can use this variation of pressure with height using this law you can find out inside. Now, if I subtract these two equations, I am going to get P2 minus P1 is equal to row G. H1 minus H2. Any doubts? No, sir. No, sir. From equation one, I also know that P2 minus P1 should be equal to row AL. So if you compare the left hand side and the right hand side, you're going to get row G H1 minus H2 is equal to row AL. Okay. And if you rearrange the terms, you'll get H1 minus H2 divided by L to be equal to A by G. Okay. Now H1 minus H2 divided by L is tan of theta. And what is a theta? Theta is this angle. Why I'm saying that is tan of theta because if you draw a line, if you draw a horizontal line like this. Okay. Then this is also theta. This distance is H1 minus H2. This distance and that distance is L. So H1 minus H2 by L has to be equal to tan of this angle, which is angle of the surface. Are you guys fine with it? Any doubt? No, sir. Okay. This is not in your school textbook, but then regularly it is asked in the school exams. By the way, is this chapter started in your school? Yes, sir. Yes, sir. Okay. So you are finished. No, we finished it. You finished it. So you already might be knowing your problems. Okay. Fine. We'll move forward. So we have done a very, very important law as in how pressure changes between one point to the other point. Okay. In both the scenarios where the bucket is at rest or where the bucket is in motion with some acceleration. Okay. Why this law is important to us? Because we can't keep on finding pressure at different, different points by using integration, differentiation like that. If I know pressure at one particular point, this law tells me a way to find out pressure anywhere inside the fluid. Okay. So it is a very, very important law. Okay. Okay. And based on this law itself, we are going to make few instruments. Okay. So please write down barometer. Barometer is an instrument that measures atmospheric pressure. So this is how the setup is. All of you please draw it with me. This time you draw a wider bucket.  It has, let's say a liquid of density row. Fine. Now you, you take a test tube. Let's say this is a test tube. And I put the test tube inside this tub. Okay. So that it is filled with liquid inside the test tube. It is completely submerged inside this tub. And then I'm lifting that test tube so that the, the close end is at the top. I'm lifting it like this. Fine. And I see that the liquid rises up to this level. Okay. So until that level liquid is there, let's say that rise in the liquid has height capital H. Fine. Can you tell me what is the pressure over here? How much is the pressure? Zero. Yes. It is near vacuum. Why it is near vacuum? Because using this exercise, I have removed entire air. Entire air is replaced by the fluid. Fine. And then I'm seeing there's a rise in the water level like that. Okay. Of course, outside, what is the pressure? Atomospheric pressure is there. Let's say P naught is the pressure. Okay. Now, can I say that pressure at this location should also be atmospheric pressure. At this point also it should be atmospheric pressure or not? Yes. What is the reason? Why it should be atmospheric pressure? Transmitted equal. Pascal's law? Not Pascal's law, but the entire tub is at rest. So horizontally there should not be any pressure variation. Okay. So whatever is the pressure horizontally here should be same as there. Horizontally pressure doesn't change, right? That was the first thing we have learned. Okay. Here also pressure should be atmospheric pressure. Okay. Okay. Now, let us say density is rho. If density is rho, according to that equation P2 is equal to P1 plus rho gh. How can I write pressure over this point? This is the second point and this is the first point. Let us say this is first. So what it should be in terms of rho, h and everything else. P1 is what? P1 is 0. It is near vacuum. So P1 is 0. P2 is atmospheric pressure. And this is rho g into capital H. Fine. So you know that if you know the value of capital H, you can calculate the atomic pressure by using this relation. All you have to do is put a scale over here. Okay. Measure what is the reading of this height, h. And you can just multiply with rho and g. You'll get the atmospheric pressure. Stop typing. Please speak up. Okay. I can't open the chat window again and again. Until it is absolutely necessary, you can chat. Otherwise, please speak up. Now tell me if this is a relation, if I use water for water, do I know what is the value of density rho? Rho is equal to what? For water? 1000 kg per meter cube. 1000 kg per meter cube. In terms of liter, how much is the density? One liter is how many kgs? Water? 1000 meter cube. Nobody knows. One kg per liter. Liter is a unit of volume only. One liter is 10 raised to the power minus 3 meter cube. Remember this. Okay. And the density of water can also be written as 1 gram per centimeter cube. Everything is correct. Okay. Again, 1 centimeter cube is 10 raised to the power minus 6 meter cube. Okay. Now tell me, what is the atmospheric pressure? Do you know what is the value of atmospheric pressure? 1.01. Okay. Please write down. Most of you have idea, but you must write down 1.01 into 10 raised to the power 5. And it is Newton per meter square. That is the SI unit of the pressure. This entire unit of pressure can also be referred as Pascal's. Have you heard of Pascal's before? Yes sir. Yes sir. Okay. This is Pascal's. Now tell me, if you are using water as liquid in this experiment, what should be the value of H? 76 centimeter. That is wrong. That is mercury. Please calculate and tell me. Don't remember and tell me. Okay. 10.1. 10.1. 10.1. 10.1. What 10.1? Meter. Meter. 10.1 meters. All of you getting it? 10.1 meters. Yes sir. So if you substitute the value of P not this and density as 1000, you're going to get height as 10 meters. So imagine you, if you are using barometer to measure that massive pressure and we use water, you need to have a test tube which is 10 meter long. Okay. So that's the reason why I need to have a fluid which is more comfortable to use and will give me a lesser height to deal with. Otherwise I can't build 10 meter long test tube. Fine. And we use a very heavy fluid when we conduct the barometer experiment. Okay. And the best fluid which shows near zero viscosity and is very heavy is mercury. Okay. So mercury is the fluid that we are going to use here. Density of mercury, do you remember how much it is? 13.6. 13.6 times density of water. Okay. It is 13.6 times density of water. Fine. So there is a way to tell the density also. There is a concept of relative density. You might have learned this in chemistry already. Relative density is nothing but density of the object divided by density of water. Okay. Same thing is called specific gravity also. Okay. Now tell me what is the value of H if I use mercury instead of water? Calculate quickly. You will get around 76 centimeter. All you will do is divide 10 and 13.6 or you can do it everything altogether again. You get the same thing. Understood. Yes, sir. Yes, sir. So using mercury is better option here. I have to construct 76 centimeter long test tube which is manageable but 10 meter is definitely not manageable. So barometer is also called the mercury barometer because most of the time we are using the mercury itself. Fine. Now there is another instrument read on manometer. Manometer looks like this. Please draw this like this. Okay. See manometer is used to find the unknown pressure by comparing with unknown pressure. Okay. And what is the known pressure? The best known pressure is atomic pressure. Right. So I am going to compare the pressure over here. This is the unknown pressure which is P. I want to calculate this pressure P. Fine. And what is known to me is atmospheric pressure. So what I will do is that I will open this end of this instrument to the atmosphere so that it experiences atmospheric pressure from outside. So this is atmospheric pressure. Atmosphere pressure let's say P naught. Okay. And it so happens. See inside the manometer there is a fluid. Okay. And it so happens that the fluid adjusts itself like this. Suppose this is fluid from here to here. Why fluid is adjusting like this? Because this pressure P is pushing the fluid down. Are you getting it? Yes. Yes. I hope you have understood the scenario. Now tell me one thing. One quick question. Will the pressure change from that point to this point inside from here to here? No sir. No sir. I know. Why no? It's all right. What? What? This is gas only. Okay. This is gas. Liquid is starting from here. Sir that's also fluid. Gas is also fluid. So that derivation which we have used P2 is equal to P1 plus rho gh. This you can use for gas also. If gas is at rest. Okay. You can use it. But the density of gas is extremely less. So even though the pressure is changing from this point to that point, the variation in pressure is so less because density is extremely small. Density of gas is very, very less. I ignore the variation of pressure with height. I ignore the variation of pressure with height for the gases. And usually, you know, in chemistry also, we assume that if gas is there in a room or any container, pressure is same everywhere. But actually it is not the case. Pressure does change with the height. But still for the sake of simplicity, we assume that pressure is same. So we'll continue having that assumption for the gases, that pressure everywhere where the gas is remains same. Now suppose this level of the liquid and that level of the liquid, let's say the difference in height is H. So how can I relate P and P0? Anyone? Can I relate? P0 is equal to rho g H. P is equal to P0 plus rho g H. P is equal to P0 plus rho g H. How? How are you getting that? P is acting on the left side. Sir, because it's a liquid with constant density throughout, so it will have P is equal to P0 plus rho g H. You are using P2 is equal to P1 plus rho g H, right? Yes, sir. Tell me where is 0.2 and where is 0.1? 0.2 is equal to let's say 0.2 is on the left side, 0.1 is on the right side. But that is not what you have learned, right? You have learned that everything should be along the same vertical line. That is what you have derived. We haven't derived the equation for something on the left-hand side and something on the right-hand side. Sir, but it's a fluid and it shouldn't change, sir. See, you are correct only. I'm not saying you are wrong, but you're not applying the concept properly. What you have to do is you draw a horizontal line like this, okay? Then since it is horizontal, pressure over here should be equal to pressure over there. So pressure at this point should be equal to P only. And then between 0.1 and 0.2, I am applying this equation. So at 0.2, pressure is P only because this is the horizontal level to this. So P is equal to P0 plus rho g H. Although you get the same thing, but if I complicate the problem a little bit, you will get confused. So you must know that what you are doing. So P minus P0 is rho g H. So if I know the value of H, if I put a scale over here, I can calculate what is H. If I know P0, I'll be able to calculate the unknown pressure P. So these are the simple instruments that can be used. The first instrument is used to calculate the atmospheric pressure. And the second instrument is used to calculate the unknown pressure by assuming that atmospheric pressure is known to you. Okay? Yes, sir. All right. So just give me a moment. Vinaya, who is Vinaya in our class? No one, right? That's Aditi. Okay, sorry. Fine, so this completes the theory of fluid statics. Now we can solve numericals and see what all we have understood. Fine. So I am assuming that you guys have already done your textbook examples. Is that assumption correct? Yes, sir. In our class, we haven't finished fully. No, fluid statics is the first part of the chapter. So you have done the solved exercises of your textbook? No, sir. Yes, sir, we have not done that. Okay, okay. So let me open up that first. This is your textbook. Okay. Can you do 10.2 quickly? You are able to see, right? Sir, 2.01 into 10.5. Okay. All of you got 2 times the atmospheric pressure? Yes, sir. Yes, sir. Okay. So P2 is equal to P1 plus rho gh. You can see that P2 minus P1 is roughly equal to 10 to the power 5, which is atmospheric pressure only. This tells us that after every 10 meter, one atmospheric pressure is added up. So if I go to 20 meters, the pressure will be 3 atmospheres. If I go to 30 meters, the pressure will be 4 atmospheres. So after every 10 meters, one atmospheric pressure gets added up. Okay. And one atmosphere is a tremendous amount of pressure, which is 10 to the power 5 Newton per meter square. It says that for 1 meter square surface, you are going to experience how much force? 1 lakh Newton. Okay. 1 lakh Newton just to give you an indication, 1 lakh Newton is equivalent to lifting 10,000 kg. Okay. So it's not a small pressure we are dealing with. It's a tremendous amount of pressure. And if you have done when you were kid or still if you're doing it, if you take a water bottle like this, which is made of plastic that drinking water, right?  Yes or no? Yes, sir. The bottle get crushed. Why it get crushed? Because the atmosphere forces are crushing it from outside. But what happens is from inside as well, there is air inside as well, which is pushing it outwards. Because of that, it doesn't get crushed. But if you take the air from inside, then what will happen? The outside air will crush it. Okay. This is what happens. So that is the reason why you can't go below the sea beyond let's say 20 or 30 meters. It is not safe. And do you know that the average depth of Indian Ocean is around three kilometers? So the seas are very dense. Okay, you can see there is a question. This one, can you do it? A, B and C part. You may want to write down gauge pressure is extra pressure. Okay, which is pressure minus atmospheric pressure. So if you subtract atmospheric pressure from it, it will give you gauge pressure. So the absolute pressure is just due to the water, sir. First attempt yourself. Density of water. Density of water. Yes, sir. No, sir. Is it should be considered the same or? Same. Okay. Roughly it is. Sir, is gauge pressure 103 atmosphere? 103 atmosphere. Yes. I got 106 I am getting. Sir, 101.0133 to 1045. See, we have just learned that after every 10 meters, one atmosphere pressure increases. So it's 101 atmosphere. 101 atmosphere is the absolute pressure. Yes, sir. After every 10 meters, one atmosphere gets added up and outside the sea, pressure is atmospheric pressure. So at the depth of 10, it is P plus P. Depth of 20, P plus 2P. Similarly at the depth of 1000, it is P plus 100P, which is 101P. Okay. So 101 atmosphere pressure will be there. That is absolute pressure. So what is the gauge pressure? Is it 10 to the power of 7 Pascal? Yes, sir. 10 to the power 7. 10 to the power 7 Pascal. 100 and 100 atmosphere. 100 atmospheres. Okay. See, what they have done is they have, they have calculated it actually. And density, they have taken 1.03. We have taken 1000. Okay. So if you use 1.03 as density, you'll get 104 atmospheres. You can see the calculations. Yes, sir. Okay. And for the gauge pressure, they have subtracted the atmospheric pressure. Same thing. Okay.  Yes, sir. Yes, sir. Fine. Okay. Now let's take one more numerical. Please write down hydraulic machines. Tell me one thing. If I have a bucket of water like this, I keep on saying water, but whatever I say about the water, same thing holds good for any fluid. Okay. Don't confuse it with that. Is it true only for water? If outside pressure is P naught, the pressure at a depth H will be what? P naught plus rho g H. Now if suppose I increase the outside pressure to 2 P naught, then what is the pressure over here? 2 P naught plus rho g H. That's rho g H. Are you getting it? This is very important finding. As in, although this is an implication of the same equation, but this is used in hydraulic machine. The principle of hydraulic machine is this, that if pressure at a particular point changes by delta P, this is a change in pressure suppose, at a particular point. Okay. Then all the points inside the fluid will experience same delta P change. Okay. Outside pressure got changed from P naught to 2 P naught. So changes how much? P naught. Here the earlier pressure was P naught plus rho g H. Now it is 2 P naught plus rho g H. So again here as well the change is P naught only. Right. So there was a delta P change of P naught outside. Everywhere inside, the P naught pressure has increased. Yes or no? Yes sir. Okay. So what we are going to utilize to build a machine. So we will take this thing as a part of numerical itself. Okay. So please draw this treated like a numerical. So typically we avoid taking online classes, but today there were a lot of time constraints. So that is why we had to take the online. There was a sports event in the NFL. A few of the Rajesh Nagar kids also participated there. May not be from 11th, but from 12th there were. And then only bio student came to the school today. Right. There was no computer students came to school. Only computer science students came to school. And it was completely this thing. We can't do anything about it. Anyways, so suppose this is the arrangement. And these two pistons can move up and down. So this piston can move up and down. Even that piston can move up and down right now. By the way, there is a fluid inside. Okay. There is a fluid inside this right now. Both the piston are at the same horizontal level. Okay. Same level they are in. So pressure over here. Can I say pressure over here is same as pressure over there? Yes. Yes, sir. Now suppose I apply F1 over here. Area of cross section of this piston is A1. And area of cross section of piston A2 is there. Now you need to tell me. You need to tell me how much is the extra pressure that get generated below this piston. And because of that extra pressure, how much is the extra force? So the extra pressure will be F1 by A1. Extra pressure over here and extra force over here. So force will be F1 by A1 into A2. Pressure is F1 by A1. When I apply force F1, how much pressure I am changing? Pressure by F1 by A1. This is delta P. The same change in pressure will be seen over here also. Because of this. Because whatever we have just written, this one. Same change in pressure will be experienced here as well. And because of that and extra force get generated. F1 A2 by A1. F1 A2 by A1. Okay. Force into A1. This is F1 by A1. Pressure into A2. Now what if A2 by A1 is 1000? How does that help me? F2 by F1 will be 1000. So F2 will be equal to 1000 times F1? Yeah. Have I multiplied my force by 1000 times? Suppose this is what I am applying. F1. If I apply 10 newton over here, it will generate 10,000 newton. Understood? Yes, sir. This is something which is amazing about fluid. You cannot do this with solids. Solid will transmit force only. But fluid is there. Flu transmits the pressure. And you can trick the fluid. And you can generate much more force than what you are having. So you have multiplied by your force by 1000 times. If A2 by A1 is 1000. Okay. And have you seen the shops which cleans the car? Sometime have you seen the cars get lifted up? Yes, sir. That is the same hydraulic machine itself. Okay. So you can have a bike at this side. You can place a bike over here. There will be a bike at this side. The bike can lift the entire car itself. Because of this property of fluids. Okay. But then at the same time, if this goes down by, let's say, 1000 units. 1000 meters. This goes up only by 1 meters. So amount of work done will be same. This side or that side. Force into displacement will be same. But the force will be multiplied. Like a pulley. So pulley. You can use string and pulley arrangement to multiply your force. It becomes two times or three times. Okay. But then displacement also changes. Have you understood all of these concepts hydraulic machine? Yes, sir. Yes, sir. Let's see whether you can do this numerical. You can see 10.6. Solve 10.6. And do not use calculators. Okay. You will not, you're not going to get calculators in any exam that you write. Is it 1500? Have you used approximation? Have you used GS10? Yes, sir. What do you got? 4500. Calculation error. 1470. 1480. 1470. Around that. 1470. Okay. And the extra pressure needed will be how much? That force divided by area? 1.910 to power 5 Pascal. Okay. So here it is. 1.910 to power 5 Pascal. Fine. This is almost double the atmospheric pressure. Fine. Now see what I'm planning to do is that I will not rush and complete the chapter. Because there are so many final points. I'm going to stop here as in with respect to theory. And we will solve a lot of numericals. Fine. So I have a set of numericals with me. Let's take one by one. Start solving this. Yes. That is correct. No one. Okay. Is it C? C. Others? C. C. Okay. I'll solve it now. Suppose this is a straw. The water will move till there is a pressure difference. Okay. So he is drinking water from the glass. This is let us say glass. Okay. Let's say the straw is just touching the surface of water. It doesn't get too much inside. And let's say this is height. So he's drinking the water from the glass from this height and water should reach here at least then only will be able to drink the water. Okay. And let's say this is point number one and this is point number two. So P2 should be equal to P1 plus density of water into G into H. Okay. So P2 minus P1 should be equal to density of water G into H. P2 minus P1 is what? The pressure difference and the pressure difference is 750 mm of mercury. 750 mm of mercury is the way of telling the pressure. If I tell you that 750 mm of HG is the pressure. So actually the pressure is rho mercury G into this height of mercury which is given to me. So P2 minus P1 is actually equal to density of mercury into G into height of mercury which is 750 mm. This P2 minus P1. This should be equal to density of water G into H. Fine. So G and G get cancelled away. So capital H will be equal to density of mercury divided by density of water which is 13.6 into 750 into tens of minus 3. All of you understood this? So you will get around 10 centimeter. Are you guys understanding what I am doing here? Yes sir. Sir could you explain I got disconnected for some time. You got disconnected? For a few minutes because next clock went down for a second. Okay see what this person is doing is using a straw. So the liquid should reach at least this point. So P2 and P1 are related like this. P2 is equal to P1 plus rho W G H. And P2 minus P1 is given as 750 mm of HG. 750 mm of HG means that pressure is density of mercury G into 750 mm. It is a way to tell the pressure that height of mercury. So P2 minus P1 is this. This I substitute as rho W G H and I will be able to calculate the value of H. Okay sir got it. This I think you can do it quickly. All of you try this. Sir G should we take 9.8 or 10? I think it's 6.25 Newton C. Okay. Sir is it B? Hmm? 62.5. Yes sir B. 62.5. B. 62.5. Okay. 62.5. All of you are getting that. Good. So F2 should be equal to F1 by A1 into A2. This is what we have derived. Or you can simply use that pressure should be same. F2 by A2 should be equal to F1 by A1. Because pressure has to be same. So F2 should be equal to F1 into A2 by A1. Now the one arm of hydraulic lift is four times the radius of other. The four should be applied on a narrow arm to lift 100 kg. Okay. So F2 is actually 100 correspond to 100 kg. So we need to find F1. So F1 will be equal to F2 times A1 by A2. So F2 is how much? MG 100 into 9.8. That into A1 by A2. Now the radius is four times. So area will be 16 times lesser. So this is how much? Anybody calculated this? 61.5. Okay. This is 61.5 Newton. So I'll pick 62.5 from the options. Probably they have used 10. Yes. They have used 10. We are using 9.8. So when you write any exam, typically the instructions are written at the start itself. So there itself, they might be writing that wherever required, take the value of GS10. So like that, pay attention to the instructions. Okay. Yes. That's correct. Fine. So let us continue. And I have few more questions, but I just realized that I missed a small concept that I should be covering before I take up the other numericals. So I'll cover it quickly. This concept is called buoyant force. Have you heard of buoyant force? Yes, sir. It is, it gives you a feeling that something can be lifted inside a fluid easily compared to when it is outside the fluid. Okay. Try lifting a heavy object inside the water. Okay. You'll be able to lift it easily, but as soon as it comes out of water, you will suddenly feel that it is not heavier. Okay. Because buoyant forces is supporting your lift. Now let's see what is the cause of the buoyant force, how much it is and when it acts. Suppose there is a bucket of water. Okay. Suppose this is a bucket of water and inside the bucket of water, I imagine a volume of water, which is irregular shape. Okay. This is an imaginary volume of water, but volume I know volume is V. Okay. Inside the water, it is made up of water itself. Okay. And this is at rest. Okay. So if I, if I draw the force balance on this, since it is at rest, what is the value of gravity force? Mg. And how much is Mg? Density into volume. Rho Vg is the density. Okay. So Rho Vg is the gravity force. And since it is at rest, there should be a force upward also. This force should be equal to Rho Vg. Yes or no? Equal and opposite? Otherwise it should accelerate? Yes sir. All of you understood? Yes sir. Yes sir. Yes sir. Okay. Now this is the buoyant force. Suppose this is a scenario. Tell me who is applying this buoyant force? Who it is? The fluid. The fluid. Fluid which is outside this. Yes or no? Yes. So outside this white line, whatever is a fluid, they are applying this buoyant force. Okay. Now, since they are applying the buoyant force, suppose I, by doing some magic, if I replace this imaginary object with a real object, if I keep a steel object here, with same volume, will the buoyant force change? Does it matter to the outside fluid, what it is inside? No sir. No. Does not matter. The buoyant force will still be equal to Rho Vg itself. Getting it? Yes sir. All of you right? Yes sir. Yes sir. Okay. So, from where this buoyant force is coming, what is the cause of this? How this is getting generated? If this is a buoyant force that is there, why we have not considered buoyant force, when we have imagined a cylinder like this, we have just used P1 and P2, right? P1A, P2A and MG force. We have not considered buoyant force. So, was this thing wrong? Anyone? Sir, isn't P2A acting from the buoyant force? You are saying this P2A? Yes. But there is a P1A also, right? You are saying buoyant force is P2A, but this is not the buoyant force. Is that what you are saying? Yes sir. Okay. What about here? Let's say this is the volume of the object. There will be pressure force from the top as well? Yes sir. So, if you push it down, so we are not considering that force? So, what it is? What is the cause of the buoyant force? But then, at least you got one thing correct, that the cause of buoyant force, right down, is variation of pressure downwards. Okay. This is the cause only. The total force upward, which is P2 minus P1A, P2 minus P1 times A. This is buoyant force only, nothing else. Okay. MG is applied by the earth. We are not worried about MG force. But buoyant force is applied by the fluid, which is nothing but outcome of the pressure force. So, whether you consider the pressure force or the buoyant force, it is one and the same thing. But here is a trick. Buoyant force gives you total force. Okay. Net force is buoyant force. So, buoyant force is P2A minus P1A. But if you use pressure forces, you need to subtract one pressure force from the other pressure force. Getting it, buoyant force tells you directly what is the net force. And it has to be upwards. All right. Force, magnitude of that is density of the liquid multiplied by the volume, which is inside the liquid into G. And it adds vertically upward. Sir, is there a reaction force to buoyant force? Yes. Equal and opposite reaction. On the fluid, but not on the object. Okay. Okay. Any doubt? Assume buoyant force like a normal reaction. An object kept on the surface will experience a normal reaction. Right? Similarly, an object inside the fluid will also experience something similar to normal reaction, which is buoyant force. Okay. So, keep it simple like that. And the magnitude of the buoyant force is fixed unlike normal reaction. Magnitude of buoyant force is density of the liquid multiplied by volume, which is inside the liquid into G. Okay. Now, this is something which is very interesting because if it is a regular shape, let's say if it is a cylinder, then you can write P1A and P2A. But if it is suppose an irregular shape object, pressure force on this will be like that. Here it will be like this. Here pressure will be like that. So like this, you can see that pressure forces keep on changing their directions. And it becomes extremely difficult to write down the force using the pressure force. Okay. And then you can use the buoyant force to, you can use the buoyant force to write down the net force directly, which is rho L V into G. Have you understood why buoyant force is useful for irregular shape object? You can't use P2A and P1A because the direction of forces will vary a lot. You can directly use rho VG. Okay. Yes, sir. Yes, sir. Okay. All right. So now you are fully equipped with whatever numericals I have to solve. Tell me the answer for this. D, I think. D. D, sir. D for donkey. Yes, sir. D. Okay. D. D. P2, we have derived P2 is equal to P1 plus rho GH. Here we have assumed that the fluid is at rest. Suppose this fluid is accelerating down with acceleration A. Okay. Now, can you derive a relation between P1 and P2? Let's say this is an imaginary cylinder you have. Let's say this is P1 pressure here and P2 pressure there. Can you derive a relation like this quickly? It's a straight forward relation. Do it. The factation is downward. Sir, is P2 equal to P1 plus rho into G minus A into H? Derive it, I said. I don't want to find out the answer right now. I'm interested in how you derive it, not on the final answer. So, if you draw the three-parade diagram like this, there will be pressure force from above P1A from below it will be P2A and there will be an MG force downwards and everything is accelerating with acceleration A. So, I can see that P1A plus MG minus P2A. That is a net force downward. This should be equal to M into A. Okay. So, I can say that P2 is equal to P1 plus M into G minus A divided by A. So, if I rearrange terms, I'll get P2 is equal to P1 plus rho G minus A times H. Now, when the object is freely falling, when the object is freely falling, what is the value of A? How much it is? 10 or 9. A is equal to G only. So, if A is equal to G, P2 will be equal to P1. So, there is no variation of pressure when it is freely falling. Okay. And if there is no variation of pressure, vertically even buoyant force will not exist because buoyant force existence is because the pressure varies vertically. Fine. So, it will be 0. Buoyant force will be 0. Okay. Any doubt? Okay. Let's go to the next question now. This one. Sir, is it 3 by 2 A? Who is this? Sir, Prathap. Sir, D. Okay. I'll solve it now. Option D, 3 by 2. Okay. So, there is a mixed response. Fine. I'll solve it. So, reading from the spring balance is what? Is the check. Now, reading from the spring balance means that you are checking how much is the spring force. Okay. When it is suspended in the air with the spring like this, the spring balance says that it is 60 Newton. So, spring balance will counterbalance MG only. Right? So, MG should be equal to 60 Newton. Okay. Now, let's say volume of this object is V because here we are talking about density. So, let us introduce density itself. So, density is let's say sigma. So, sigma into V into G, which is MG, should be equal to 60 Newton. Okay. So, this is first and reading changed to 40 Newton when the block is submerged in water. So, if this entire block is submerged in water like this, it is inside the water. So, how much is the buoyant force now? On this? 20 Newton. 20 Newton. 20 Newton. Correct. But, I am looking for the expression which is density of water, volume of object. Volume. Volume, entire volume I am taking, reason? It is fully submerged. It is fully submerged. Suppose it is submerged only half, then I will write down rho W V by 2G because it is volume of submerged object that tells us how much is the buoyant force, not the entire volume. Here it will be sigma VG which is MG. So, let's say spring force is FS. So, again net expression is 0, so FS plus rho VG mu over 1. Okay, I have done now. This minus sigma VG will be equal to 0. Now, FS will be equal to sigma minus rho W VG which is equal to 20 Newton. Okay. So, I can use this and that relation. I can in fact divide it. I will divide it. What? Can we just divide sigma VG by rho VG? That is what I am trying to do. Okay. So, if I divide this equation, with that equation. So, I will get sigma divided by sigma minus rho W to be equal to 3. Okay. But this is not the specific gravity. Specific gravity is sigma by rho W. All right. So, that will be equal to if I simplify this as 3 sigma minus 3 rho W. Okay. So, 3 rho W will be equal to 2 sigma. So, sigma by rho W will be equal to 3 by 2. So, specific gravity is 3 by 2. Have you understood how I got answer as A? Yes, sir. Yes, sir. Look at it once and copy down quickly. Let me know if you have any doubts. So, how do you get MG as sigma VG? See, I am assuming sigma is a density of the object. Sigma is density of the mass. Okay. The density into volume is mass only. Sigma V is MG. Okay, sir. And here the spring force balances the MG and spring force right now if it is suspended in air is given as 60 Newton. Okay. Any other doubt? No, sir. All right. Let us move on to the next question. There are two questions. First, you solve the first one. Say it again. Sir, it says in cube floats in water with a height of 4 centimeter cube. Height of 4 centimeter you take. Ignore cube. So, you can't have height as 4 centimeter above the surface, sir. Because the side length itself is... Yeah, side length is only 2 point something. Yes, sir, it's cube root of 10. Okay, please ignore the first one. You do the second one. Don't do the first one. Sir, do we take 4 centimeter cube is above the surface of water? Volume. Let's not do this now. It may turn out to be wrong. Do the second one. All of you do the second one. Sir, it's the second one, D. D for donkey. Yes, sir. Others? The one minute, sir. The second one is D. I'm giving couple of minutes. All of you, please try this. Should I solve now? Sir, it's D. Sir, it's D. Okay, so all of you have tried. Let me solve it now. Suppose this is an iceberg which is floating in the water. Okay. Total volume is let's say V. Total volume is V and let's say the volume that is outside. Because they're asking the fraction of total volume above the sea level. So volume that is outside. Let's say that is X times the total volume. So what it will be volume inside will be how much one minus X times V. Yes or no. Yes. Okay, so it's total force will be zero because it is stationary. The expression is zero. So buoyant force will get balanced by the gravity force. Gravity forces how much density of the ice into total volume into G. And buoyant force is what buoyant force is density of the liquid, which is density of water into volume that is submerged. That is one minus X times V into G. Fine. So these two should be equal. Then only it will be at rest. Okay. So row ice V into G should be equal to road water. One minus X times V into G. Okay. So VG get cancelled and you are getting one minus X as row I divided by row W. Right. So from here you get the value of X to be equal to row W minus row ice divided by row water. Okay. So I can substitute the value of density of water. See, since it is fraction, I don't need to worry about the units as long as the units of row W and row I are same. I don't need to convert in SI units because X is dimensionless and both numerator and denominator has the same dimension. So X will be equal to 1.03 minus 0.92 density of ice divided by 1.03. This will be the fraction. Have you done like this only? Yes, sir. Yes, sir. So you get this as 0.106. Fine. And if you want to make it in two decimals, you'll round it off to 0.11, which is fraction. So percentage will be 11% into 100. Fine. So ice, if you see the iceberg, only 10% of the iceberg is visible above the sea level. More than 90% is inside the sea. Okay. All right. Let us move to next question. I hope this is this you can understand. Flotation is done. This one. 30. 30. 30. 30. Theoretically solving it or you have actually written down equations. Theoretically. Theoretically. Theoretically. Okay. Like always, I don't, I'm not interested in final answer. You have to prove it. Suppose I tell you these the answer, now prove it. Mathematically. Write down the equations. You can use those theoretical logic and symmetry and everything else in the exam. Right now we are learning a concept. So our focus should be how can I apply a concept. So try to write down everything in equation format. Can I say row two is more than row one? Yes sir. Clearly heavier liquid will always settle down in the bottom, at the bottom. If row one is more than row two, row one has to be below it. So row two should be more than row one. That you can theoretically say. Okay. So since row two is more than row one, B cannot be correct. Row two is more than row one even. Sorry. Okay. Between AC and D we need to pick up now. Okay. I can, you know, I can block the chat also. I know exactly how to do it. Focus here. What we are doing here. Don't worry about the previous questions. Tell me, how can I prove that row three is more than row one but less than row two? Anyone? If sir, you can use the same logic you used to prove that liquid two is below liquid one. That's a smarter way to do it. Okay. If row three is more than row two, then even if this object is completely submerged inside row two, okay, then also the buoyant force will be how much buoyant force will be row two VG. And I know that row three is more than row two. So row three VG, which is my mg force will be greater than row two VG. So it will keep on sinking down and down. Fine. So object will sink because the objects density is more than the density of liquid. So it will not stay float. So clearly row three cannot be more than row two. All right. And hence row three is less than row two. And at the same time, the buoyant force experienced by this is a sum of the buoyant force by row one. Let's say half of it is submerged in row one. V by two is in row one and V by two is in row two. So what is that buoyant force? Anyone? V by two G that is from row one plus row two V by two G. Okay. So you can see that this buoyant force should be equal to mg. mg is what? Row three VG. So I can see here row three is the sum, the average of row one and row two. That is row two by two. Okay. I've assumed half of it is up and half of it is down. A simpler case. So if that happens, then row three is in between row one and row two clearly. Okay. And if suppose it is V by three and two V by three, then also some weighted average, but still it remains in between somewhere between row one and row two. Okay. And that is the reason why option D is correct. Okay. Yes sir. Let us go to the next question. This one. G is the accession due to gravity. G is not gram. Should I do it or should I wait? Wait a second. Sir, one minute. Sir, one moment. Sir, is it B? Okay. I'll solve it now. So a piece of wax weighs 18.03 G in air. So this is the, let's say density is sigma for the object. So sigma VG, this is MG. Any doubt here? MG is 18.03 G. Okay. Yes sir. Okay. Now this is inside water. Inside water, it will be MG, which is 18.03 G, which is MG minus density of water into volume into G. Boo and force will be acting now. This will be equal to 17.03 G. Fine. So row water V into G will be equal to 18.03 minus 17.03, which is G. Okay. This is the second equation. Sir, it says a piece of wax was 18.03 G and a piece of metal in water is 17.03 G. Okay. Yeah. I'm sorry. Piece of wax, a specific gravity of wax is asked. So piece of wax weighs 13.03 G in air. A piece of metal is this. Okay. So piece of wax, let's say is 18.03 G. This is mass of wax into G. Okay. And then piece of metal is found to weigh 17.03 gram in water. So mass of metal into G minus density of water VG should be equal to 17.03 G. This is a second equation. And then it is tied to a wax and both way 15.23 G in water. Okay. So both of them are tied together. So there will be a volume of wax itself. So let me rearrange everything. This is equal to density of the wax into volume of wax into G. First equation, then mass of the metal. I can say it is density of metal, volume of metal into G. So you can just cancel out G, sir. So let me write it first. So this is equal to row water V into G. This is equal to 17.03 G. Okay. So like what somebody said, I can cancel out G. So I can write it as 18.03 is equal to row W VW. And I can also write down row MVM minus row W V. This is whose volume is this? This is the volume of metal only. Inside the water, it is weighing that much. This into G, sorry, G is gone now. It is equal to 17.03. Okay. These are two equations. And then when you tie them together, what will happen? The MG will act down for both of them. So row metal, volume metal into G plus row wax, volume of wax into G. This minus, bow and force will be how much? Anyone? Density of water into VW. This. Total volume that is displaced by the, both of them will be volume of metal plus volume of water. Volume of wax. This should be equal to 15.23 G. So if I cancel out G, I'll get row MVM plus row W VW minus row WVM minus row WVW equals to 15.23. Okay. So row WVW is gone. So I'm getting row MVM minus row WVM is equal to 15.23. Wait. I'm getting the wax. Sorry, I have written wax and water both as W. Yeah. Okay. Okay. This is wax and this is water. So you can't cancel it. So this is first equation, second equation, third equation. This is water or wax? This is water. So yes, the things are becoming messy right now. Okay. Okay. So basically we can, this is VM, right? VM metal. VM VM. I can take VM common from here. So what is asked in the numerical is what specific gravity of the wax. So they are asking you to find out what is row wax divided by row water. This is what. So what I'll do is that I'll take this term to the right-hand side, take that term also to the right-hand side of this and divide it. When I divide it, these two will disappear. These two will be gone. And then I have after division, there will be one equation and one equation will be this. So I'll try to use these two equations to get row wax by row W. So it will take a lot of time right now. So I'll send the solution directly on the group. I want to solve one more numerical. Yes, tell me. Can you go to the question? So basically it says in the metal in water is 17.03g and when you tie the metal and the wax together it is 15.23g. So how many add two masses? The resultant is like lesser. Then adding two masses you're calculating the net force downwards. So when you attach two masses together and it is inside the water. The bow and force will be much more. But it increases. Okay sir. Because of this it decreases. You're not calculating mass. You're not adding mass. You're adding and subtracting forces. Okay sir. Fine. Alright, please note down one homework question. We have very little time. I will just draw a figure here. A homework question which you need to do it. What is today? Immediately after the class you do this homework. Then only you can say that class is over. Here it is. There is a container in which two flutes are there. One is the, let's say this density is four times rho. That density is rho. Okay. And inside this container there is a solid cylindrical object. This is, if total height is h, this is h by three. So h by three is inside the second liquid. And two h by three is inside the first liquid. Fine. And let's say density of this cylindrical object is sigma. It's a solid cylinder. You need to find a relation between sigma and rho. Okay. If you can, you have two minutes to complete. Sir, do the two liquids have the same volume? What? What? It doesn't matter. It doesn't matter. Fine. We can spend two minutes solving it. I think you can get, some of you will at least get, tell him others density of the cylinder. So do we take area of cross section as A? You can take anything, but it should not feature in the answer. What should we find again? Relation between sigma and rho is sigma equal to two rho. Sir, I'm getting sigma equal to three rho. All you have to do is equate the forces, buoyant force and gravity force. You have to equate mg force and buoyant force. There should be equal and opposite. Yes or no? Yes. So buoyant force should be mg. mg is what? Sigma into area of cross section into h into g. And buoyant force is what? Weight of liquid displaced. So this liquid has a volume of how much? Volume is A into two h by three. This is the buoyant force due to the upper fluid plus buoyant force due to the lower fluid is four rho area of cross section into h by three g. Any doubt? Sir, is it sigma equals two rho? I'm asking you any doubt here? No, sir. No, sir. Okay. So I can cancel out A and I can cancel out g as well. In fact, I can cancel out h as well. Right. So sigma is equal to two rho by three plus four rho by three. Okay. So this is two times rho. So sigma is two rho. Yes, sir. Fine. So today we have done fluid statics. When we meet next, we'll be doing fluid dynamics. Okay. All right. See you. Thank you, sir. Thank you, sir. Thank you, sir. Thank you, sir. Thank you, sir.


Welcome to One Minute Tennis Lessons with The Tennis Mentor. I'm going to show you three reasons why hitting cross court is usually the best option. When hitting cross court from corner to corner, you have a longer distance than you would if you were hitting straight down the line. In singles, you have 4.5 feet longer distance than you would if you were hitting down the line. And when playing doubles and the tram lines are involved you have an even bigger distance. Reason number two is when hitting cross court you're hitting over the lowest part of the net. If you were to go down the line instead, you're going over a much higher part of the net, again giving you less room for error. The final reason why going cross court is usually the best option is you don't have to move as far after you've hit the shot. If you're going cross court into this corner, then you can recover slightly off centre on your half of the court. When aiming down the line, you need to recover to the opposite half of your side of the court. Going down the line can be a lot riskier, so do it at the right time when you've got a shorter ball or a slower ball or when your opponent's out of court.