 No one remembers. That is strange. P2-P1 is equal to HROG. Yes. Thank God you are remembering. We cannot afford to forget that. This is the variation of pressure with height. Now what if you have point number 1 here and point number 2 is at the same... Sir, we are not able to see your screen. Yes, sir. Sorry. Yes, sir. Okay. Suppose this is point 1, that is point 2. They are separated by a distance of x. So, now tell me what is the variation? Is P1 equal to P2? Yes, sir. They are the same. Yes, sir. Yes, sir. Very cool. The assumption is no acceleration horizontally. No acceleration in horizontal direction. But if the entire beaker accelerates with acceleration A, then it will be like P1 will have higher pressure. So, P1 will be equal to P2 plus rho AX. Are you getting it? This also you should know pressure can change horizontally if acceleration is there in the horizontal direction. But most of the time there is no acceleration in the horizontal direction. So, that is why we don't consider that. Now, we are going to take a quick, very, very simple numerical so that you can refresh from your memories. Suppose this is the beaker and on the beaker you are placing a test tube. And how you are placing the test tube is such a way that you are first putting this test tube completely inside the beaker so that the test tube is filled with the liquid. And then you are pulling the side up which is closed. So, there is no air over here. So, it is close to vacuum. So, pressure is how much over here? So, 0. Okay. Right. So, suppose this rises by a height of H and density of the liquid is rho. Then, here since it is open to atmosphere it will have an atmospheric pressure P0. Okay. And by the way, atmospheric pressure it is good idea to remember 1.013 10 raised to the power 5 Pascal's. Okay. Now, can I find out atmospheric pressure P0 based on the value of H and rho? 0. Atmospheric pressure I am asking. So, tell me if I draw a horizontal line like this. Pressure along all the points along this horizontal line will be same or not? Right. Pressure along all these all the points on this line will be same. Right. So, the same pressure will be over here also. So, over this point at this point pressure should be P0 only because P0 was the pressure over here which has a same level. And I know pressure over there is 0. So, 0 plus rho g H should be equal to P0. So, P0 is rho g H. Alright. So, if you use mercury by the way you should know the density of mercury also because it's a very common fluid. It is 13.6 times the density of the water. So, if you use mercury over here and you substitute the value of g and P0 you are going to get H as how much? 76 centimeter. Right. So, atmospheric pressure corresponds to 76 centimeter of the mercury. So, at times pressure is given in terms of height of mercury. So, pressure will be simply rho g into height of mercury if you want to find out in proper units. Now, can you tell me? By the way, this instrument is called barometer. Barometer is used to determine the atmospheric pressure. Okay. Now, there is one more instrument but you don't need to know the working. You can directly apply the basics and tell me what I am trying to ask you there. Draw this with me. All those who joined late today we have decided to revisit the fluid mechanics. So, we will do as much as possible today and then I will circulate the videos so that you can complete the revision at your end. I will not rush very fast but also I will not go very slow. Okay. So, suppose there is this kind of instrument connected to a chamber whose pressure is P. Okay. There is this chamber whose pressure is P and this YouTube is filled with mercury. It is filled with mercury. It so happens that the difference in the levels of the mercury is H and outside pressure is P0. Now, P0, how do I know what is P0? You can use barometer to determine P0. Once you determine P0, you can also determine the value of P by using this instrument. This instrument is called manometer by the way. Okay. So, can you tell me what is P in terms of P0? P0 plus HROG. P0 plus HROG. So, again you draw a line like this horizontal line along this horizontal line pressure should be same. Okay. I am assuming that the gas is such that for gas pressure does not vary in the vertical direction because the density of gas is very less. Otherwise for gas also you have to use P2 is equal to P1 plus ROGH. Okay. But approximately I can say that pressure over here is P only. So, pressure over here is also P. Alright. So, P should be equal to P0 plus ROGH. Right. So, if you know what is H, if you know what is P0, you will get to know what is P. You can calculate H by just putting a scale over here. You know, it is visible. You can get it easily. Right. So, this is the small portion of the chapter which talks about the variation of pressure. Alright. Now, because the pressure is changing as you go down, the outcome of that is the buoyant force. Please write down. So, buoyant force is due to... Am I going very fast or it is alright? You can text me privately also. You can message me or WhatsApp, whatever. Is it alright? The pace is fine. Yes, sir. So, you are saying yes on behalf of everyone? No, sir. I am... Okay. Alright. Great. So, let's continue. So, due to variation of pressure in vertical direction, that is why buoyant force exists. Alright. Now, tell me if some object is freely falling, will there be a pressure variation? In vertical direction. Okay. See... So, what is it? So, if the bucket of water is falling vertically with... I mean, with the action due to gravity only, as the bucket is falling freely, okay? Like this. Accessory is free. Then, there will not be any variation of pressure vertically. Okay. The pressure variation happens only when the bucket is stationary. Okay. So, when you say that this is 0.1, this is 0.2, P2 is equal to P1 plus rho G H, only when the fluid is stationary. Okay. And suppose there is an acceleration, let's say, acceleration vertically down is A1, then it will be written as P1 plus rho G minus A1 times H. Okay. And if A1 becomes equal to G, the pressure variation doesn't happen vertically. So, if pressure variation doesn't happen, then there will not be any buoyant force also, because buoyant force is because of pressure variation in vertical direction. Okay. Now, let's discuss a little bit about the buoyant force. We'll talk about its properties. Okay. Buoyant force is basically applied by the fluid on any object, which is partially... Please write down with me. Partially or completely submerged. Okay. The direction of buoyant force in... In which direction? Anyone? Direction of buoyant force? Where? Upward. Upward. Buoyant force still upwards. Upward as in opposite to the equation due to gravity. Okay. Vertically upward. There is no exception to it. Magnitude is what? Magnitude. Anyone remember? Magnitude is equal to the weight of liquid displaced by the object. Okay. It is always true. Weight is what? Mass of the liquid displaced into G. And mass of the liquid displaced where it was? Density of the liquid, okay, into volume of the object which is inside the liquid into G. Because if V volume is inside the liquid, then V volume is displaced. Isn't it? The mass displaced will be density of liquid into the volume which is inside the liquid. If the object is sinking, if object is completely inside, then it has essentially displaced the entire volume of it. Okay. But if it is only partially submerged, ADV. Who is ADV? ADV, you are on mute, I think. Now you can speak ADV. Who is ADV? Right. Okay. Now I am muting you. Alright. So this is rho L VG vertically upwards. And the, if object is completely inside, okay, if object is completely inside, the buoyant force will be rho L VG. And the gravity force will be density of the solid into V into G. Okay. So if the buoyant force is less than the weight of the object or the liquid density is less than the solids density, then the solid will be sinking inside. Okay. But if density of liquid is more than the density of solid, then the solid can float. Fine. Now I am going to ask you a small question. If you have a beaker like this which is made up of steel, can this object float on the water or it will sink? Can it float? It is made up of steel. Density of steel is of course more than the water. Can it float? No, no, no. What does your intuition say? It will float, right? You just take, yeah, you might have done this, right? You have like in, when you were a kid, you were like, you have started playing with the mobile phone now only, like when there was no smart phone and you were playing with whatever come in your hand. So this, you might have done, you've taken a plate of steel and put it on the water which is stationary, the steel will float. Okay. Then how come steel is floating? Steel is floating because steel is able to displace more volume than its own volume. Okay. The volume of steel is very less. It is a very thin material. Okay. But it has displaced this much water. Okay. It has displaced this much water, which is more than its own volume. Okay. But then if you take a steel plate and if you vertically put it down, if you put it down like this, then it will just sink. It will not be able to displace the same, correct. So it will not be able to displace the weight of the water which is equal to its own weight when you're putting the plate like this vertically. But if you put it horizontally, it will able to displace water which has more volume than its own volume. The shape enables that. So I have a question. If we invert the beaker, the container, it will sink. If you invert the beaker, it may sink if there is no air trapped. If there is air trapped, the liquid won't be able to, you know, if you're putting it like this, this is what you mean, right? So if air is trapped here, then it may not sink. It has to get out. If air is out, then it will sink. Okay. This is the fluid static, okay? And one more thing, the buoyant force is right down. Buoyant force point of application is through the center of mass of the submerged portion, okay? It's not the center of mass of the entire object. The buoyant force, suppose you have to apply the torque equation, then you have to take into account the torque due to buoyant force also, okay? So the point of application of buoyant force you should know. Otherwise, how will you write torque which is forced into perpendicular distance from the force? So point of application of buoyant force is center of mass of the submerged portion. For example, if there is a beaker like this and you have a stick which is hanging on like that, it is hinged here and the stick is inside the water like that. So mg will be acting from the center of mass. This is mg, okay? And buoyant force will act from the center of mass of the submerged portion. This will be liquid, the density of liquid volume which is displaced into g. So basically from here to here, you find out the volume of the solid. That much volume of the liquid is displaced. So this much will be the buoyant force apart from mg and the normalization also you can take. So these are the some basic things. Now let us take few questions on fluid statics. Then we'll get into the fluid dynamics, okay? So I've already downloaded it. So you can see. I'll tell you which question to do, okay? Fifth question. You are able to read the question, right? That is not difficult. Is it A, maximum at D and I mean minimum at D and maximum at B? Okay. Fine. Let's see. Yes, maximum at B. The pressure varies like this downwards because of accession due to gravity. Pressure increases in this direction because of this acceleration. Pressure increases backward also, okay? So D will have the least pressure and you can see pressure of A is more than D but pressure of B is maximum, okay? So B is maximum, D is minimum. All right. Let's proceed. Do the sixth. Question number six. Okay. Many have answered since A. So is it D? D. Okay. Fine. I'll quickly talk about it. You draw a horizontal line like this and then you say pressure along this line has to be equal. So pressure over here will be atmospheric pressure which is over there plus density of glycerin G into 0.1 meters, rho G H. This should be equal to atmospheric pressure plus density of oil G into H plus from here to here you have to take, okay? This length will be 0.1 minus H meters, okay? So density of mercury G into 0.1 minus H, okay? P naught and P naught gone, all right? So you know the density of glycerin, density of mercury and oil density is also given. The densities are given in gram per centimeter cube, okay? So if you multiply it with 1000, you'll get in kg per meter cube, okay? So G, G, G is gone. So 0.1 times rho G is equal to rho naught times H plus rho M times 0.1 minus H. You can substitute density in terms of gram per centimeter cube also because left-hand side and right-hand side both have the density so multiplication factor will get cancelled away. So you substitute the values of density and you'll get the value of H, right? I'm confirming this. 666 D, D for Delhi, okay? Do the, okay? No. Do the 10th question. 10th is slightly tricky but it'll be nice if you solve it yourself. Okay, someone got the answer. Brinda saying B. Should I solve? There is a variation of pressure in x-direction as well as y-direction. Simply use the formulas that we have discussed. Is it B? 2 meter per second. Okay. Alright, let us see how to solve this question. So first step, I'll draw a horizontal line like this. So L-shaped tube is kept inside the bus that is moving with constant expression. The motion level of one liquid is 12, whereas the right arm, it is 8 centimeter. When the orientation of the tube is shown, the azimuth diameter is smaller, you need to find the expression of bus. Let's say the expression of bus is A. So this is 45 degrees, this is 90 degrees, this also should be 45 degrees. The horizontal distance, did you find what is the horizontal distance? This distance is how much? Anyone got it? This distance. Okay, so this is 8 since it is 45 degree, this also will be 8 only and this will be 12. So 12 plus 8. 20 centimeter. Correct. So this is 0.2 meters. Okay. I will reach here from here. Let's say this is 0.1, this is 0.2. So P2 will be equal to P1 plus rho A into 0.2, rho A into X. Fine. Okay. And P1 is basically atmospheric pressure, P0. Fine. And let's say this is 0.3, 3. Okay. So that distance is 12 minus 8, which is 4 centimeter. Okay. So P2 is also equal to, if I take vertically from P3 plus rho G into H, that is 0.04. Okay. And I also know that P3 is nothing but P0. It is open to atmosphere. Okay. So when I use this, okay, I will equate P0 plus rho G into 0.04 to be equal to P0 plus rho A into 0.2, P0 P0 gone, rho is gone. And if you take G as 10, then 0.4 is equal to 0.2 times A. And from here, A is coming out to be 2 meter per second square. Fine. All right. So this is how you have to do this. Now tell me one thing. If you have a beaker. If you have a beaker like that and let's say the beaker dimension is this length is given as L. If this beaker acts forward, the top portion will be horizontal or it will be inclined. Inclined. It will be inclined, right? And it will be inclined like this. You can probably refer to your own experience. It will be like that. Okay. So basically, I am asking you to find out this angle theta. How much will be angle theta accession due to gravity is G vertically downwards. So it's just A by G, right? Theta is A by G. Oh. Solve it like a numerical. If you know already don't refer to your, you know, what you have remembered. Solve it. Even if you know it. Tan theta is A by G. Just correct. So if I draw a line like this and let's say this is H. So tan theta should be equal to H by L. Okay. Let's say this is 0.1, 0.2 and this is 0.3. So P2 is equal to P1 plus rho AL. And P2 is also equal to P3 plus rho GH. And P1 is this atmosphere. P3 is also atmospheric pressure open to the atmosphere. Okay. So P1 is equal to P3. So if I write it down like this P1 plus rho AL equal to P3 plus rho GH. Then P1 is equal to P3. So they go away, rho and rho go away. And I will get H by L is equal to A by G. So tan of theta which is H by L is also equal to A by G. Okay. 13th question number 13. Should I solve? The pressure is not constant throughout the area. Pressure changes. Just use the tan theta is equal to A by G, right? Yes. It could have been done directly. But if you stick to the basics you feel lot in control. No one got 13th. Okay. Brinda is getting B. Kavya is getting A. Others. Okay. So basically there is a spring mechanism over here. So if you draw the free body diagram of this surface it will experience a spring force of K times delta X like this. And there will be a force because of the pressure. Let's say F is that force. Okay. So F minus K delta X should be 0 because this surface is stationary. It is not accelerating. Net force should be 0. F should be equal to K delta X. Now delta X is what is asked. Delta X is F by K. The problem is I don't know what is F. K is given to me. So I need to find the value of F. Okay. How to find the value of F? I'll just multiply pressure with the area. But the problem is pressure at every point is changing. Pressure over here is something else. Pressure over there is something else. Everywhere pressure is different. Okay. So if I look this surface from this side then I will see it as a square plate. Okay. And fluid will be like, you know, outside of the screen. So if I consider a small strip like this, a small strip of width DH at a distance of H, then since this width is so less, I can assume that pressure is practically constant for this width DH. All right. And this is separated by two halves. Okay. So the length and the width of this plate is also A only. Okay. So the length of the strip is A into DH. Okay. So area, the small area is A into DH. Pressure over there is atmospheric pressure plus rho GH. Fine. So basically the force, the small force on that strip will be P0 plus rho GH, which is the pressure multiplied by the area, ADH. Now you need to be very careful here because listen here, listen guys, I'm talking something important. Okay. You need to be careful here because atmospheric pressure is applying force on this direction also. Okay. So in that process, you may ignore that. All right. So there is a force from this side which is equal to P0 into the total area, which is P0 into A squared. Fine. So this equation, you need to, if you consider total force, then you have to write over here F minus P into A squared is that. So delta X will be equal to F minus pressure, atmospheric pressure into A naught, A squared. Okay. So this is the force, small force, total force will be integral of this, integral of H goes from 0 to A. So this will give you P0 into A squared plus rho G, this will give you rho G. I'll make some space here. Rho G into A into HDH, right? So HDH when you integrate becomes H squared by 2. So you will get A cubed by 2 like this. This is your force. This is my value of F. So F minus P A squared is basically rho G A cubed by 2. Fine. So delta X is nothing but this divided by the spring constant. So rho G A cubed divided by 2 times K. This is my delta X. Fine. So this is how you have to solve this. Is there any doubt? No doubts? This one. Do the third question. Third one. Final answer for this 13th question is A, 1.3 centimeter. Okay. Do it now. Question number three. Have you understood the question? Actually the question is not very clear. Somebody has understood. Yeah. It is not very clear. So leave it. Let's not get into that. Let me read the question and then give it to you. Fourth question. Fourth is good. Try solving the fourth one. Question number four. Leave third one. Third is unclear. Scroll down a bit. Any scroll up a bit, sir? Is it not visible? You scroll down a bit. Are you seeing question number four properly? You're not listening. I am asking you to solve four. Okay. Britta, what's the answer? Others? Okay. I will do it now. Is it D? 30 kg. 30 kgs. Let's see. First I will tell you how to go about it. Then I'll show you the answer. Okay. So you need to assume that suppose it is completely sink. Then what are the forces acting on it? There will be a buoyant force. Okay. If volume of this solid is V, the buoyant force is rho GV. Entire volume is displaced. Okay. And then there will be an MG force. Okay. Mass is directly given. So I don't need to write M in terms of density. This is MG. Okay. M which is given to me. And also there is another mass kept on the top. So because of that, a normal reaction will be applied on this volume which is equal to mass which you are adding up into G downwards. Okay. So when you balance all the forces, you are going to get rho GV minus M plus M times G should be equal to 0. Okay. So G is gone and you're going to get small m as density of the liquid into volume of the solid minus capital M as your answer. Additional mass which you have to keep. Now density of the water is 1000. Right. It is 1000. And the volume of the solid, how will you find density of the solid is given and the mass of the solid is given. So volume you can find out by dividing the mass with the density 1120 divided by 600. Okay. Minus 120. This is how much anyone has done like this 200 minus 120 which is 80 kgs. Okay. Like this you have to do. Okay. Solve the seventh question. Question number seven. No options will not be shown. Did you get the answer? Okay. I will solve it now. Are you able to hear me? Yes. Okay. It's MG mu minus one. MG mu minus one. It is not mu it is eta. It's called eta. Sorry, sorry. Okay. Anyways, let's see. Solid sphere has density which is eta times lighter. So density of the solid is density of liquid time divided by eta. Okay. Because it is this much time lighter than water. Okay. So you can find it in water tank by a string tied to the base. If the mass of the sphere is M then the tension is asked. So this is the sphere. Tension will act downward. Okay. And there will be a buoyant force plus there will be a gravity force. Okay. MG. Okay. So the sphere is at rest. So buoyant force minus MG minus T is equal to zero. All right. So T will be equal to buoyant force minus MG and buoyant force is equal to density of liquid mass of the sphere is given. Okay. So density of solid. I'll write in terms of density of solid. It will be better that way. The density of liquid is eta times density of the solid. So buoyant force will be density of liquid which is eta times density of solid a rho VG minus M into G. So M is what? M is density of the solid into V. This is M density into volume. That into G. So you will get eta minus one times rho SVG. So rho S into V is mass. Eta minus one into M into G. Anybody got this as the answer? Let me check the final answer. T is eta minus one. Yeah. There's nothing wrong. Is there such option? Eta minus one into MG. T is the answer. Seventh question is D. Yeah. We have got it correct. Okay. Try doing all these fill in the blanks. 15 to 19. Sure. Do 15 to 19 fill in the blanks. You have let's say 10 minutes to complete. You don't need to do it one by one. Okay. You don't need to do in sequence. Do the first one which is easier than the next one like that. Have you done the 15th one by the way? Anyone done 15th? How you have solved 15th? What you have done? Have you added the mass of the ball? So is your answer 640 gram? What is the answer? So your wing pan will read less now. Something is inside. So wing pan will the mass of everything will the mass of everything will increase or reduce. Just one second. 600. Assuming water was displaced. No. It's not 600. Let's see. You need to Just one second. See you need to understand the free body diagram properly. Here what is happening is that buoyant force is applied on the mass. Right buoyant force applied on it, which is rho g v. Who is applying buoyant force? The liquid is applying buoyant force on this ball. So the ball will apply equal and opposite force on the liquid. So the rho g v force is applied on the liquid downwards. So apart from gravity force, which is masses 0.6 kg. So 0.6 g plus rho g v. This will be done now the net weight. And if you talk about mass, 0.6 plus rho into v. Fine. Now, rho should be density of liquid. So it will be 0.6 plus density of liquid is water. So this is 1000. And the volume of the ball is 40 gram divided by 40 divided by 0.8. This volume you are going to get in cubic centimetre to convert in the metre cube. So 10 raise to minus 6 you have to multiply. Fine. So you will get 0.6 plus 40. So 10 raise to minus 2. Then 0.4 by 0.8 into 10 raise to minus 1. So 0.5. So 0.6 plus 0.05. So you are going to get 0.65 which is actually equal to 650 gram. Yeah, no tension on the string. I am not sunk into the water and a pin of initial volume is shown in the figure keeping it sunk. So what the pin is doing? Oh, yeah, there will be tension due to the pin as well. I thought it is sinking. So if it is sinking there will be tension in the string. So what is the question? Even though there is a string there will be a buoyant force by the liquid on the mass. So equal and opposite force will be applied on the liquid as well. So 0.65 kg or 650 gram should be the answer. Let me quickly verify how much it is. 650 gram that is correct. All right. Now do the 16th one. We will go one by one. They are not very straight forward. So we will go one by one. Let me know if you have done 16th. Question is visible Preeti. Question number 16 I am asking now. 1-6. You can see the question. No Preeti, please solve 16th question. I can't make everybody wait. I will send you the worksheet. Then you can see it after the class. Do the 16th one right now. Then 13 can be taken later on. V by 2. Kavya is getting V by 2. Others, hollow sphere of relative density 2. Floats completely immersed on the surface of water. Sharok is also getting V by 2. Let's see. A hollow sphere of relative density 2. What does it mean relative density 2? Density of the sphere is 2 times density of the water. Which I am representing at as rho L. This is what it means. It is floating completely immersed on the surface of water. If the volume of sphere is V, then the volume of cavity will be what? So basically if it would have been solid sphere. If it would have been a solid sphere. Then even if it is completely immersed. Then also buoyant force will not be sufficient to balance the weight of the mass. So buoyant force will be density of liquid into G into total volume. And the mg will be density of solid into total volume into G. So if density of solid is more than the density of liquid for the same total volume B. It will sink down because the downward force is more. But here is a trick that if I somehow make a cavity here. If I somehow make a cavity over here where there is no material. The mass of this hemisphere will be simply equal to density of solid into the volume of solid. Which is not equal to the volume of the hemisphere. Even though the shape is hemisphere. But the density of solid you should multiply with the volume. In which the material is there. You should not multiply volume of the cavity with it to get the mass. So now they can balance out because Vs could be lesser than the volume. So it can displace the volume of the hemisphere from the surface. So density of liquid G into V. When this become equal to density of solid into volume of solid into G. It is balanced. Density of solid is 2 times density of liquid. So rho L, rho L gone. Then G is gone. So volume of solid should be equal to volume of hemisphere divided by 2. Volume of cavity will be also equal to V by 2. Solid is V by 2. Cavity should also be V by 2. Because total volume is V by 2. Have you understood this question? This is very important question. Slightly different. All of you understood, right? You can see that there is yes, no button below your chat screen. So whenever I ask you yes or no type of question, you can simply click yes. So it will show against your name. So like that you can click. If you have not understood, you can click no. All right. Do the next one. Shorok, if you have got it, you can move on to the next one. Complete all the questions. 12 by 3. Most of you are getting 12 by 3. Okay. Let's see how. Cube of each side L floats in a liquid of density 3 times the density of solid. The length of the cube will be outside the liquid is how much? So let's say that this is the cube. This is submerged portion. Length is X. And this will be L minus X. So the volume of liquid that is displaced is basically X into the area of the base. Right? This much liquid is displaced. So that is X into L square. Fine. And force will be equal to density of liquid into X into L square volume displaced into G. And the MG force, the weight is density of solid into total volume, which is L cube into G. I'm assuming it is like there is no cavity. So density into total volume is the mass. If there is a cavity, then this is not mass. Like the way the earlier question was. Okay. So now if it is stationary, the forces should be balanced. The MG force, which is acting down should be equal to the buoyant force, which is acting up. Okay. So L square is gone. G is gone. So X is equal to density of solid divided by density of liquid. Now, density of the liquid is three times the density of solid. So X will be equal to rho S by rho L times L. Sorry. This is L by three. So L by three is inside. What is asking here is length, which is outside. So L minus L by three, which is two L by three. Do the 18th one. It is interesting. 18th is important. Sigma by rho minus one into H. Sigma into rho minus. Sigma by rho minus one into H. If you have got it, then move to the next one. Others continue solving. The next one is... I am not asking the answer for 19th. When I ask, you can tell me. Okay. So you can see on your screen other questions. Question number four is there. Should I solve 18th? All of you. 18th is important. Okay. Let me solve 18th question. So there is a wooden ball of density rho immerse in a liquid of density sigma. This is sigma and wooden ball has density rho up to a height capital H. It is at a height of capital H. It is then released the height above the surface of water through which the ball rises. The ball reaches here. It moves up. Let's say maximum height attained is this. You need to find out what is this small edge. Okay. The assumption over here is that I am assuming that the volume is very small. Okay. So it cannot be when it is coming out. It cannot be some portion is inside the liquid. Some portion is outside the liquid. Either it is inside completely or outside completely the volume I am assuming it to be very less. Okay. And let's say that volume is V. Volume of the solid is V. So if you see that it talks about releasing and it moving up. So basically buoyant force is more than the gravity. Otherwise it will not go up. Okay. So buoyant force is rho Vg and mg force is sigma Vg. Right. So buoyant force that is acting upward. The total force that is acting upward is rho minus sigma times V into g. All right. Because of that the acceleration is rho minus sigma times Vg divided by mass. Mass is what? Sigma V. Right. So V and V gone. And hence I am getting acceleration as rho minus sigma by sigma times g. And this is a constant acceleration. So if I release it with initial losses zero I can find out the final velocity over there. Okay. The final velocity how will I get V square is equal to u square plus 2 as u square is zero. V square is equal to 2a which is sigma minus rho by sigma g into h. Fine. So this is velocity square when it is just when it has just started to come out. Okay. Now I need to assume this point which is after it leaves the liquid. Let's say this is point number one. This is point number two. So at point number one this is the velocity. So what I will do is I will apply work energy theorem between point one and point two. So work energy theorem says that w is equal to u2 plus w is equal to u2 plus k2 minus u1 plus k1. Gravity is doing work for that I am considering potential energy. Blue is zero. k2 is zero because finally it stops. u1 is zero because this line I am assuming to be zero potential energy. So u2 is equal to k1. k1 is what? Half mv square. Now half mv square can be written as if you multiply half into m to the v square you get half mv square right? So half into m into v square then two and two goes and you will get rho minus sigma g into h. This is equal to u2. u2 is equal to mg into small h. Fine. So g and g gone. m is also gone. So small h will be equal to rho by sigma minus 1 times h. This is the answer for small h. Okay. Have you understood this? Those who did not, I mean those who are not getting many questions don't worry because sometime people do a lot of practice and they tend to remember what they have done earlier and questions get repeated. So don't get intimidated by the fact that someone is getting I am not getting the answer. Okay. There can be several reasons why somebody is able to answer some question. Okay. Don't lose confidence. I have seen many such cases. So all you should focus on is your learning whether you have understood or not. Is it clear? All right. Okay. So let's do one thing. We will now start talking about fluid dynamics. Otherwise time will not permit us to cover the fluid dynamics. Please write down fluid dynamics. Let's keep it very simple. Fluid dynamics basically talks about the two equations. One is continuity equation which basically says that mass flow rate at steady state should be constant. What does it mean? It means that suppose there is a pipe. Okay. And two kg per second is coming from here. Then at steady state two kg per second should go out from there. It will never happen that two kg is coming from here and only one kg is going out per second. If that is true then every second one kg is getting stored inside the pipe which cannot happen practically. Okay. Initially when pipe has no water it will accumulate but after it has accumulated and reached the steady state then whatever water it is getting from the left hand side it will reject from the right hand side. Fine. And in order to make sure continuity holds good across the two ends of a single pipe I can use this law. Rho 1, A1, V1 this is basically mass flow rate. Okay. Mass flow rate at point one Rho 1, A1, V1 this should be equal to Rho 2, A2, V2. Okay. And if fluid is incompressible okay. If the fluid cannot be compressed which is usually the assumption in our syllabus then Rho 1 is equal to Rho 2 you cannot compress and increase its density. So area into volume sorry area into velocity sorry area into velocity pipe should be same. This is the continuity equation it says that multiplication of area into velocity and on that cross section of the fluid is constant. Now that this is true for if you have a single pipe like this but if you have this kind of scenario tell me what you will say suppose this is what is happening A1, V1 velocity A2, V2 velocity A3 and V3 velocity tell me what does continuity equation look like here A1 plus A2 is equal to A2 I can't hear it's very feeble voice speak a bit loud how are you saying something tell everyone A3 V3 is equal to A1 V1 plus A2 V2 A1 V1 plus A2 V2 is equal to A3 V3 Kawe is saying Andrew A1 V1 A2 V2 is equal to A3 V3 see you need to check from which side fluid is coming this side it is going in and these two side it is going out so the amount of fluid coming in should be equal to amount of fluid that is going out so that is the reason why A1 V1 should be equal to A2 V2 plus A3 V3 okay so the rate at which the fluid is coming in should be equal to rate at which fluid is going out by the way area into volume this is called volume flow rate volumetric flow rate V is velocity meter per second meter per second into area is meter cube per second the volume which is flowing per second so that is sometime in question you will see volumetric flow rate so that is why so this is the continuity equation that you need to take care in the fluid dynamics there is application of work energy theorem there is application of work energy theorem in fluids that is called Bernoulli's Bernoulli's equation you can say Bernoulli's equation is nothing but application of work energy theorem okay now work energy theorem we already know W is equal to U2 plus K2 minus U1 plus KY but the problem is that it is very difficult to apply in this format when it comes to fluid because when it comes to fluid we are more comfortable dealing with pressure heights and density instead of mass I would like to see density when it comes to fluid instead of force I would like to see pressure because it is easy to deal with pressure as the basic laws in the fluid is with respect to pressure and density so I would like to have this work energy theorem for fluid in terms of density and pressure okay and this is how this entire equation transforms please write down between the two points p1 plus half rho v1 square plus rho gh1 is equal to p2 plus half rho v2 square plus rho gh2 okay the assumption is streamline flow streamline incompressible and no viscosity if viscosity is there there will be energy loss then this equation will not get transformed to that it will not be so straight forward and you can apply this equation only for a streamline flow we will talk about streamline flow later on if you get time okay now this term p1 is called pressure energy per unit volume this is called kinetic energy per unit volume this is called potential energy per unit volume so basically we are dealing with the density of the energy itself when we talk about the fluid so this is the application of work energy theorem in the fluids any doubts till now any doubts great so let's take up a question and see whether we can solve it draw this with me so there is a beaker which is closed from the top so pressure of the air over here is given as capital P pressure of the atmosphere is p0 okay fine so this is the scenario and let's say this distance this is the level of water okay this is the level of water the water density is given as rho okay the level of water is at the height of h2 from the floor the hole is also there the hole of area a1 of course the hole area will be lesser than the area of the beaker itself so a1 is the hole area and that hole is at a height h1 from the base okay you need to find velocity with which the water will come out from this hole okay so all of you try attempting it you just have to use Bernoulli's theorem and the continuity equation to solve this just attempt it it's okay if you get it wrong also by the way in case of Bernoulli's theorem h in the Bernoulli's theorem when you write rho g h that h is taken from the base okay but when you use p2 is equal to p1 plus rho g h h is taken from the top so there is a difference between the h which you have seen earlier and the h in the Bernoulli's theorem so can you repeat the question no you can see the diagram find the velocity word word don't worry about whether it is a law or not just attempt it okay if you attempt it honestly then you learn many things alright I will solve it now let's draw a streamline like this along which I will apply Bernoulli's theorem okay this is 0.2 and let's say that is point number 1 so between 1 and 2 I am applying the Bernoulli's theorem okay so let's say at point number 1 velocity is v1 point number 2 let's say the velocity of this is v2 even the surface will go down right if water is coming out the surface will go down so pressure at point 2 is p so p plus rho g h have to take from the bottom let's say I am taking this as my h equal to 0 so rho g h2 plus half rho into v2 square this should be equal to when you come out here outside let's say outside the pressure is atmosphere pressure so which is p0 okay plus rho g h1 plus half rho v1 square alright the continuity equation tells me that a1 into v1 should be equal to a2 into v2 right whenever you see a fluid dynamics problem you cannot think of any other equation other than Bernoulli's theorem alright so blindly whenever you see a fluid that is in motion they ask me something just apply Bernoulli's theorem and continuity equation so v2 I will write v2 in terms of v1 so that I get the value of v1 so v2 is equal to this which I will substitute over there so I will get p minus p0 I am getting p0 that side and rho g h1 also plus rho g h2 minus h1 equals to half rho v1 square minus half rho v2 square v2 square is a1 by a2 like this okay I know that h2 minus h1 is this length which I let's say assume as y okay h2 minus h1 let's say it is the depth at which the hole is from the surface of the water so I am getting half rho v1 square 1 minus a1 by a2 hole square that is equal to p minus p0 plus rho gy okay and then I can further simplify this I will get v1 square to be equal to p minus p0 2 times of that rho gy divided by rho then 1 minus a1 by a2 hole square so v1 will be square root of this now if container is open to the atmosphere then basically p will be equal to the atmospheric pressure and if the area of the hole is very small compared to the area of the beaker I can ignore a1 by a2 hole square also fine so this is gone that is also gone so this will get transformed to this root root over 2 gy v1 but this is an approximation if you would have attempted it properly you will get this as the answer fine so beaker is open to the small hole if in the question it is written it is a small hole then velocity will be root of 2g h where y is the depth at which the hole decides any doubts guys anything you want to discuss any doubts let me give you a question all of you please attempt this draw this it is a portion of a pipe it is not a complete pipe it is a portion of the pipe like this ok let say this is point a and that is b water is flowing through the horizontal pipe let say water is flowing like this the pressure of the water changes by 600 newton per meter square between a and b so between a and b please write down between a and b the pressure change is 600 newton per meter square ok the area of cross section the area of a surface area of a cross section area of a 30 centimeter square cross section area of b is 15 centimeter square fine cross section area of b is also given you need to find rate of flow of water rate of flow of water whenever somebody ask what is rate of flow of water they are asking you volume flow rate ok it means they are asking volume flow rate which is nothing but area at any moment into velocity at that moment area into velocity is constant anyways so area into velocity at any cross section if you find that is the answer can you solve this now anyone got it this is 3.6 into 10 is to power minus 3 no 1800 centimeter cube per second yes anybody else got it ok this try it quickly I will attempt this now so between point this and that point a and point b I can use Bernoulli's theorem ok so pressure at a let's say pa plus half rho va square now how should I write rho g h anyone I can take this as my ground 0 height ok so h will be 0 for a as well as for b so I don't need to write rho g h because height is equal for both this is equal to pressure at b plus half rho vb square ok now naturally velocity at b will be higher so pb will be lower than pa so pa minus pb will be half times rho into vb square minus va square ok pa minus pb is given as 600 right so this is equal to 600 and rho is density of the water which is 1000 1000 divided by 2 is 500 so vb square minus va square is 6 by 5 ok this is equation number 1 and second equation is continuity equation which says that area of a into velocity of a should be equal to area of b into velocity of b now if area has the same units centimeter square and centimeter square you don't need to convert in meter square conversion factor will get cancelled away from left hand side and right hand side so basically I will write 30 as sa into va is equal to 15 into vb from here velocity of b is equal to 2 times the velocity at a this is my second equation so I will use it there and when I put vb over there I will get 4 va square minus va square so I will get 3 va square is equal to 6 by 5 alright so I will get va as root over 2 by 5 which is root over 0.4 ok so this much is the velocity and once velocity is known that velocity into cross section area at a will be the this thing are you getting it so va if you sa you will get volumetric flow rate which is rate of flow of water that into 30 is it clear any doubts is it clear alright let us take the next question do this I tell you when you should start drawing the diagram now please start drawing the diagram with me for me to finish draw with me alright so the area of cross section of large tank this area is given 0.5 meter square it has an opening near the bottom the area of this s1 area of the tap is given as 1 centimeter square ok fine 20 kg is applied on the water at the top so the mass this mass is 20 kg 20 kg is kept on the top ok this 20 kg mass is kept on the top of this lead which can move up and down this the surface on which this mass is kept this surface can move up and down and this surface is mass less it is a thin surface ok 20 kg is applied on the water like this you have to find out the velocity with which water comes out from here this height you can take 50 centimeters and take g as 10 density of water you anyway know 1000 so I want you to solve it from scratch ok it doesn't matter how many questions you solve whatever we are solving let's solve it properly ok so it is very similar to derivation of the torsely's law which we have done sometime back so another opportunity for you to apply the equations properly and the velocity of efflux anyone got the answer so this is 3.3 meters per second who is this hello who has told the answer Shriram others no one I will solve it now should I wait or solve ok isn't here let's say this point just below the surface is 0.1 just outside the hole is 0.2 ok so I am going to apply continuity equation and the Bernoulli's equation between the two points fine the Bernoulli's equation says that pressure at 0.1 plus rho g into the h which is 0.5 meter 50 centimeter right plus half rho into v1 square should be equal to p2 plus rho gh is 0 for that plus half rho into v2 square fine and p1 can be written what p1 can be written as atmospheric pressure plus the weight which we have kept mg divided by the area of cross section which is s1 isn't it so if you don't place any weight over there the pressure will be atmospheric pressure only but since you place the weight the pressure is atmospheric pressure plus weight divided by surface area ok and p2 is the atmospheric pressure only so when you substitute the value of p1 and p2 you are going to get rho g into 0.5 half rho v1 square p0 and p0 will get cancelled away so you will end up with 20 g by s1 this is equal to half rho v2 square so basically we need to find v2 ok and I also know that s1 into v1 is equal to s2 into v2 ok s1 is 0.5 meter square that into v1 is equal to 1 centimeter square which is 10 is power minus 4 meter square v2 ok so v2 will be equal to 5 10 is power 4 so it's like 5000 so this is 5000 times v1 so when you substitute v1 is equal to v2 by 5000 over here over here you will get the value of v2 getting it so you may not have got the answer the calculation is not straight forward so others have used the calculator but this is how you have to solve it is it clear to all of you so we will take a small break now we forgot to take a break ok oh it's already close to 7 so no need to take a break now it's already close to 7 fine let's take up questions question number 10 can you solve quickly 10 Krishna Bajaj is saying B others B Saan is also saying B alright so let's check here pressure at points along the axis now tell me pressure over here will be more or over there will be more where the pressure will be higher 0.1 or 0.2 we know according to Bernoulli's theorem p plus half rho v2 pressure plus this expression p1 plus half rho v1 square is p2 plus half rho v2 square so at 0.2 so pressure will be lower ok at 0.1 velocity is lower so pressure will be higher ok so it cannot be B and pressure cannot be constant also because velocity is changing if this is increasing this should decrease that's why A let's take up a question which is coming in my mind all of you draw this a bucket you have to draw here right and bucket is completely filled with water at the bottom there is a small hole ok bottom there is a small hole and area of cross section of this hole is S1 area of cross section of the bucket is S2 S1 is very small compared to S2 fine you can use the final expression of torsually directly at any moment so let's say this height is H ok water keeps coming out from here you need to find how much time it will take for bucket to get empty alright how much time it will take for the bucket to become empty find out have you attempted should I solve how many of you want to take a break you can click the tick button no break great attempt discussion then will continue should I solve it now ok let me solve it no one is telling me the answer so let's say that after some time the height becomes this much ok let's say this is H small H ok the velocity of efflux according to the this thing root 2gh torsually's law ok and I know that S1 into V which is 2gh is equal to S2 into velocity of this surface why I am finding the velocity of the surface because if I get the velocity with which it is coming down I will know velocity with which the height is decreasing then I can talk about time when the entire bucket will be empty ok this velocity is more helpful the velocity with which this is going down ok so velocity with which it is going down times root of 2gh ok and this velocity you can write it as dh by dt fine the dh over here is negative by the way whenever some velocity comes out h is decreasing so dh is negative so minus dh by dt you can put or you can take care of signs later on as well no problem dh by dt is equal to minus of S1 by S2 root of 2gh fine so dh divided by root h will be equal to minus of S1 by S2 dt now you have to integrate it why I have to integrate because velocity keeps on changing it changes with h so I have written velocity in terms of h and then equate it to dh by dt 0 to t and height earlier was capital H now it should be 0 if I am talking about a total heat, total water loss integral of dh by root h is 2 root h from h to 0 this is equal to minus of S1 by S2 times t ok so t will come out to be 2 times root of capital H S2 by S1 something is missing dh by dt is root 2g also there is 2h by g fine alright let's get into the problem solving of this now look at the question number 14 to 15 this one by solving it anybody is able to solve this 14th question 14 these are actually very good no one but 14th ok I think because break is not given you guys are like totally drained out we will take a small break otherwise it will drain out so we will take 5 5 6 minutes break right now it is 7 0 6 we will meet at 7 7 15 ok let's meet again for 15 minutes we will solve couple of questions take a small break ok come back refreshed if you want to attempt the question during the break please do so are you guys back from much needed break are you guys there yes sir ok so now we have of the class left lot of disturbance from somewhere who is disturbing Shiramya who is disturbing Shiramya I think it was from someone now you can hear alright anybody got the answer 14th should I solve by the way these worksheets were given as a homework last year so look like you have not done your homework yes I love it now a syringe as shown in figure contains an ideal fluid of density rho the cross-section area of plunger the small orifice at the end of the cylinder has area the velocity the velocity of outlaw of liquid on the syringe if the plunger moves the constant velocity under the action of force is what so if the plunger is moving with constant velocity it means that the force from this side is equal to force from that side ok so this force which you are applying to the plunger is equal to the force the fluid is applying on the plunger because of that the pressure of the fluid becomes force per unit area f by a ok so over here pressure is pressure because of the force is f by a plus atmospheric pressure p0 so total pressure over here is p0 plus f by a ok just from between this 0.1 and 0.2 ok I will get p1 plus half rho v1 square is equal to p2 plus half rho v2 square fine and one more thing I am saying that there is a small orifice velocity of 0.2 is very high compared to velocity of 0.1 so I can ignore the velocity of 0.1 ok this is the same thing which we have done when we have derived for the tossery's law so p1 is atmospheric pressure plus f by a this is equal to p2 which is atmospheric pressure outside the orifice atmospheric pressure plus half rho v2 p and p gone so velocity will be equal to root of 2f by a rho option a 15th see work done by a force is force into displacement simple ok so force you are applying is capital F so capital F into displacement should be the work done ok so the work done should be equal to force into the displacement and displacement will be what velocity with which this is going for let's say velocity is u u into t, t is a time taken to empty so assuming the velocity with which the plunger is moving with constant which is written over here it moves with the velocity because of that velocity into time should be the displacement so work done is force into u into t but the options are given in terms of density and volume fine so let's try to simplify this further first of all this is the velocity with which the water will come out so the velocity of the plunger so velocity of plunger let's say u and that into a is a1 v1 this should be equal to small a into this velocity which you have found out this will be 2f by capital A rho so velocity u will be equal to a by capital A divided into 2f by a rho so this is the velocity u so you can substitute that over here so f into u a by capital A root over 2f by a rho into t this is the work done now how to simplify this further anyone the volume of fluid is given as v so the volume of fluid is how much u into t into the area of cross section this should be equal to v alright so from here you will get t equals to volume of fluid divided by u into a ok in fact you can directly substitute u into t so u into t should be equal to v by a so work done is a force into volume divided by area of cross section a fine now basically we need to find out what is the value of capital F fine so capital F if t is given you can find out that um I have to erase this ok anyways so the capital F whatever it is that divided by a should be equal to half rho into v square this is what it says ok so you can substitute this velocity in terms of the velocity of plunger and u into t is the v divided by a so you can use all of that I mean it will take a lot of time over here so you just have to basically modify this expression and get it in terms of density rho, volume and time you need to eliminate F from here ok so use Bernoulli's theorem and continue theorem to eliminate F from this equation you will get the answer ok we can try out couple of more questions surface tension so surface tension we could not do so I will send you a set of my videos so please watch it so that you are comfortable with the entire chapter ok so once you watch the surface tension videos also you can say that the chapter is done um fifth question try solving the question number 5 that is the last question for today try solving it ok somebody got some answer pithi saying A see here what is happening is that the water is coming out because of that there will be an action and reaction I will show you first all of you attempt this alright so assuming all of you have attempted it at least so light container so container has no mass because of that it is written light container so whatever is the mass that is the mass of the liquid only it is kept on a horizontal rough surface coefficient of friction is this a very small hole is made at the depth h a very small hole means that I can use a torsely's law and I can say velocity is root over 2g h directly I can say that water of volume v is filled in the container let's see so water of volume v is filled in a container the friction is not sufficient to keep the container at rest the expression of container initially is how much ok so what will happen is let's try to find out what force the cylinder will experience because water is coming out velocity which is root of 2g h ok so dm by dt rate at which the water is coming out kg per second is rho into area of cross section s into v rho a v this is rate at which the water is coming out dm by dt and rate of change of momentum can be written as velocity into dm by dt over here velocity is constant at that moment we can assume that ok so dm by dt into velocity is rate of change of momentum velocity is root 2g h so this can be written as rho s into 2g h root 2g h becomes 2g h ok this is the dp by dt rate at which momentum is changing right this you can say is also the force between the liquid that is moving out and the rest of the container so the rest of the container will feel a force of this much which is rho s into 2g h ok and on top of it it has a frictional force as well since friction is not sufficient it will create a maximum amount of friction which is mu times n so this is n and that is mg so normal force is equal to m into g where m can be written as surface area s into h ok into rho 2g i am assuming that the hole is at the bottom ok so surface area h becomes volume into density becomes mass and this is g so normalization is this much ok so force of friction is mu times normal reaction ok so wait wait so when a friction is given in terms of v only so i need not assume that h is the bottom so i can assume the total volume is v so i can write normalization as v into rho into g v into rho is mass so friction force is mu into normal reaction and mu is s h by total volume so mu into normal reaction which is v rho g v and v gone so s h rho g is the frictional force ok so total force will be f minus frictional force f is 2 times s rho g h and friction is s rho g h so f minus fr becomes rho s g h ok this is your force this force should be equal to mass into acceleration right so mass of the liquid is density of the liquid into volume that into acceleration which is a let us say density and density gone so acceleration can be written as s h g divided by v so answer is d understood all of you is it clear just copy down quickly this is important alright so after the class i will circulate a video on the rest of the chapter which is surface tension and viscosity these two topics are very important but today we could not get time but since you guys are in flow because we have touched upon the other portion of the chapter i recommend that you should complete the entire chapter as i mentioned send you the videos watch it if possible maybe tomorrow after your english exam complete it and solve couple of questions on surface tension i will send this worksheet has those questions on surface tension and finish this chapter completely alright so that's it for today we will soon bye thank you sir thank you sir