 Okay, I think we should start now and I hope that you guys have come prepared with the theory because the main thing as you all understand is problem solving, it's not theory. So we can't just keep on reading theory again and again, all right. So let me immediately go to the first numerical. Okay, so this is a fill in the blanks, you have two questions fill in the blanks, try answering them both. Hello sir through Skype could you share screen? Right, this is Ramcharan, we have only Ramcharan on Skype, no sit down please also. Now you can see, okay try solving these two questions, okay should I do it? Do you know that potential energy on a strained wire per unit volume is half stress into strain? If you don't know, please remember this, this is potential energy per unit volume, okay. And whatever work you do will be stored as potential energy, that is what the work energy theorem is, right. So total potential energy is half stress into strain into volume, volume is what? Area into length, okay. Now Young's modulus is given, okay and what else is given is how much is a delta L, fine. The stress is not given, but good thing is that stress divided by strain is Young's modulus, okay. So I can write down stress in terms of Young's modulus, okay. So potential energy will come out to be half into strain square into Young's modulus into area into length, okay. Now strain square is what? Strain is delta L by L, delta L is X, so X square by L square, okay. Multiplied by Young's modulus, multiplied by area into length, okay. So one L will get cancelled and you will get answer as half X square YA by L, fine. So like this you have to do the first question. Remember that the potential energy stored in a strained wire is half stress into strain per unit volume, okay. Now do the second question. You remember bulk modulus, the definition of it, change in pressure divided by change in volume per unit volume. So you will get minus of V dP by dV, this is the bulk modulus. So can you use the definition of bulk modulus to get delta R by R? Assume fractional change is very less. These are some different types of numerical which you might not be solving regularly, but there are things to learn which you can keep in mind. Anyone? Okay, let me solve this. So I will get dV by V is equal to minus of dP by K, okay. Now dP is what? There is extra mass M is kept on the piston, right? So dP is changing pressure because of that. So this I can write as MG by A, this divided by K, okay. Now I am just taking the mod of it as in I am just bothered about the magnitude. So I will keep this as positive only because when you increase the pressure, volume decreases. So that is why there was a negative sign. But since I am bothered only about the magnitude, so I am keeping it like this, okay. So this is dV by V, okay. And what is V? V of a sphere is 4 by 3 pi R cube. Getting it? So from here, I will get dV is equal to 4 pi R squared dr, okay. So dV by V is what? This divided by 4 by 3 pi R cube, all right? So you will get, you know, 3 dR by R is dV by V, okay. Why I am doing all that? Because nobody is asking me dV by V, here dR by R is asked. So that is why I need to find dV by V in terms of dR by R, okay. So I will get 3 times dR by R to be equal to mg by ak, all right? So from here, dR is, you can say, delta R, delta R by R is mg by 3 times ak, okay. So you know, you can solve it like this. So I can understand these are not the routine ones. So you will face some difficulty in solving these questions. Are you able to understand this? Any doubts? Please type in yes or no? No, sir. Those who are online? See, there is a small lag. There is, I think, four or five seconds of lag between those who are in Skype and those who are on YouTube. So you may feel that at times. So guys, those who are on YouTube, no doubts. Okay, let's take up new set of questions. So you may see that these chapters can get easily mixed with other chapters also. So don't get surprised by that. Okay, so try doing these two questions. So I'll give you a small hint for third question. Do you know how V is related? At any temperature, V is equal to V0, 1 plus gamma delta T. Okay, do you know how density is related? Rho, rho is rho 0, 1 minus gamma into delta T. Density will reduce, volume will increase. Try using these. Metal is inside mercury. So if this is the piece of metal, what is given here is that the metal piece is partially submerged. Okay, so it is floating. Okay, now Saim here is getting answered. Okay, no, Khushali is saying, is it 1 plus gamma 2 delta T? Okay, let us see. When you heat, when you change the temperature, okay, what will happen to the mercury's density? The mercury's density will reduce. Okay, so it was now initial density 1 minus gamma 2 delta T. This is now the mercury's density. Okay, now mercury will create a buoyant force. Okay, so how much buoyant force it should create? It should create buoyant force in order to balance out the mg of the metal. Okay, now it doesn't matter whether the metal, this metal expands or contract right now. It will have same mass. Right, so the amount of mercury that will be displaced is determined only by the mass of the metal. Okay, so rho dash into volume that is now submerged. Okay, into G, this should be equal to mg. Okay, this is your actually second equation. Earlier it was like this. Earlier density of mercury into V into G, this was equal to mg of the metal. Okay, so this is the earlier scenario and this is the later on scenario. Okay, now here rho dash is what? Rho of mercury 1 minus gamma 2 delta T into V dash. Okay, V dash is the volume of mercury that is displaced. All right, now in the first equation you have mercury into V that is displaced to G, this is equal to mg. Fine, so if I divide these two equations, what I will get here is 1 minus gamma 2 delta T to be equal to V divided by V dash. Okay, V divided by V dash is actually less than 1. Fine, so volume displaced by the metal piece later on is more than what is volume displaced earlier. Okay, so in order to find the fractional volume of metal submerged in the mercury, you need to divide the fractional volume of metal submerged in the mercury. So, this is actually this V by V dash. So, the V dash by V is 1 minus 1 divided by 1 minus gamma 2 delta T. Okay, so from here you will get how much extra volume it got submerged. That is gamma 2 delta T divided by 1 minus gamma 2 delta T. Okay, now earlier it had, this is let's say equation number 3. So, earlier it had volume V submerged which was simply equal to M divided by rho of the mercury. Fine, now this is the volume that is submerged earlier. So, you can check exactly how much extra volume is getting submerged later on. So, this is the extra volume of the metal that is getting submerged later on. Fourth one, fourth one is straightforward, try to do this. It's an application of Bernoulli's theorem and continuity theorem. Yes, I am here, I can see that you have got some answer. Others, what you are getting? So, is it 3500? Others, those who are online, can you reply online as in YouTube? Okay, so we have now few answers. Let me explain what we have to do here. See, first of all it is a horizontal pipe. Okay, so you can assume the scenario to be something like this. Okay, now this pipe is not open to atmosphere. Had that been the case, then pressure both sides would have been atmospheric pressure only. Okay, so it's a cutout of a bigger pipe, where the cross section area is let us say over here, the cross section area is 10 centimeter square. Okay, and this is let's say a1 and velocity over here is 1 meter per second. Okay, over there, okay, here the pressure is also given. In fact, p1 is 2000 Pascal, fine. Over there, for the other end, the cross section area is a2 is 5 centimeter square. Okay, so we need to find out what is the pressure over here? p2 is what? Okay, now between 1 and 2, I can use continuity equation. Okay, so a1 v1 is equal to a2 v2 because it is incompressible fluid. All right, so since a2 is half of a1, v2 has to be double of v1. So v2 will be equal to 2 meter per second. Okay, now since it is horizontal pipe and if I use Bernoulli's theorem that is p1 plus half rho v1 square plus rho gh1 to be equal to p2 plus half rho v2 square plus rho gh2. Now rho gh1 and rho gh2 will get cancelled off because it is an horizontal pipe, so h1 is equal to h2. Okay, so their level from the ground is same. Now this kind of scenario comes again and again in fluids where you're dealing with horizontal flow. So you should get familiar with such situations where rho gh has no role to play. So p2 will be equal to p1 plus half rho times v1 square minus v2 square. Okay, now p1 is 2000 Pascal. We'll keep it like this plus half rho is what it's water, right? So yeah, it is water. So half 1000 into v1, v1 is 1, so this is 1 square minus 2 square. Okay, so we will get 2000 plus 500 into, now you'll get minus 3. Okay, so you'll get 500 Pascal as pressure at this end. Getting it like this, you have to solve this particular question. In case of any doubts on this question, please type in, I'll move to the next one. See many a times what happens is that when the pressure comes, the pressure of exam comes, your expected performance suddenly goes down, okay? So you need to understand it is not because of your capability issue, it is only and only it's a psychological issue. So you need to have some confidence in yourself, okay? So don't lose hope or just remember that you were doing good when you've taken a lot of Kmont tests. So you need to just repeat your performance and get what you deserve, okay? So no need to get pressurized because J mains is coming very close, all right? So these are the two questions. These two questions are just true and false statement. You need to check whether these two statements are true or false, okay? Those who are on Skype, first one is true or false? Okay, this is the person, okay? Let's say this is the person sitting in a boat, this person is sitting in a boat, okay? Now how much volume of water it will displace? It will displace the volume of water whose weight is equal to the weight of boat and the man, getting it? So if V0 is the volume displaced, then rho V0 G which is the buoyant force is the mass of the person plus mass of the boat into G. Any doubt on this? Now if the person drinks some water, okay? It drinks let us say volume V1, okay? So if it drinks the volume V1 and let us say boat remains wherever it is, okay? So the level of this water has to go down if boat remains wherever it is because the amount of water in the pond has decreased because the man has drank some water, okay? But actually what happens is the mass of the man increases by mass of water it has drank, all right? So if the mass of the man increases by the mass of water it has drank, so how much extra volume of water it will displace from the pond in order to have the force balance? It will displace the extra volume of water displaced will be equal to the volume of water that is drank. Getting it because now the buoyant force has to balance out more weight and the extra weight is coming because of water inside the stomach of the person, okay? So it has to displace now the same volume of water which it has drank, okay? So if the boat remains wherever it is then the water level will go down but what happens is that boat will occupy that extra volume of water that was taken out. So water level will remain wherever it was, okay? So this statement is false, right? So these kind of questions they are, you know, they have been asked many a times where the water level changes depending on different scenarios. So you need to be comfortable with answering such questions because they are the favorite questions of J question paper setter. All of you understood this? Any doubt on the first one? If you don't have any doubts please proceed to the second one. So when the temperature is increased the tube will not expand. Yes, even the second one is false. What happens when you increase the temperature density of mercury will reduce? It will become rho naught 1 minus gamma delta t, okay? So if this is the height of the barometer, height h then atmospheric pressure let's say pA has to be equal to rho gh, okay? Now if let us say initially density is rho naught, height is h naught, then since atmospheric pressure remains the same it has to be equal to the later on density which is rho naught 1 minus gamma delta t into g into nu height h. Now g is unchanged, g doesn't change with temperature, okay? But density has reduced. So in order to make sure that multiplication of this g and h remains pA, h has to be more than h naught, fine? So that's how you have to solve this particular question. Any doubt on this one? See those routine type of questions which we have been dealing with when we were reading a chapter it is like known to everyone. Everyone solves a question where there is a bucket, there is an object, go and force acts on it, engine need to just balance the forces or when water is flowing in a pipe, you just need to find the pressure or volume or volume flow rate, okay? Those are the like bare minimum or something which is known to everyone. So we need to force ourselves to identify questions which are different from others, okay? Then only you yourself will be different from others and you'll be solving some of the unique type of questions, right? So when you practice any chapter's questions, make sure you don't look for how many questions you have solved but what are the varieties of questions you are solving because that is more important. Now, can you do another question? I'll just draw a figure for that. So here is a question, draw with me. This is a pipe that is filled with water and let us say there is a lamp over here. So I have lit an up a lamp which is heating one edge of this, okay? Let's say this is A and this is B. Water was completely filled earlier, okay? Now because of this lamp, water begins to circulate and it will circulate in counterclockwise direction like this. You need to tell me whether it is true or false. This entire thing is vertical, okay? So gravity is acting downwards. This is gravity. Why? What is the correct answer? So it should be clockwise now because if you heat it then the particles will rise up. Why it will rise up? It should have because like hot particles rise up and the cold ones come down. Okay, see what happens is when you heat this part of the fluid, it becomes lighter, okay? This is denser. This is also denser. So when this becomes lighter, of course, because of the gravity, the heavier particles will move down, okay? And this particle will go up, okay? When this moves up, then there is a flow from B towards A like this, okay? This is also called natural convection, okay? So that's how it will be. The flow will be actually clockwise, not anti-clockwise. Now another question, all of you. So you have a cube of ice. You have a cube of ice and you have, let's say, a metal piece inside. A small metal piece is inside the ice, okay? Now this ice is floating in the water. Ice is floating in the water, okay? Okay, so now everything is maintained at zero degrees Celsius, fine? The temperature is not changing during the entire process. It remains zero degrees Celsius only and now ice melts completely. Ice melts completely. You need to tell me what happens to level of water. This level will move up, go down or it will remain at that level only. Assume density of ice to be equal to density of water, okay? Purvik is saying it will remain at the same level. So I think you might have missed one small detail that metal piece is inside the ice. Purvik has retracted his message, but I have read it. Anyone, what is the answer? Khushali is saying it will increase. The level will go up. Okay, Amoghi is saying it will decrease. Nishchal is saying increase but slightly. Okay, let's see how we can do this question. Now consider the situation when metal piece is inside the ice. When metal piece is inside the ice, okay? See one thing I think all of you can appreciate is that what there is a change that is because of this metal piece inside. Had this metal piece not there, then everything will be, I mean remain the same. But if this metal piece makes any difference, then only there will be some change in the level. Okay, now when metal piece is inside the ice and everything is floating, okay? By the way, I said density of ice considered to be equal to be density of water. That is not correct. Otherwise it will not float with the metal piece. So density of ice is actually less than density of water. But anyways, when ice melts, it becomes water with that density. So now coming back to this point about the metal piece. So when the metal piece, when metal piece inside the ice, how much volume of water it is displacing? Let's say volume of water is Vw and volume of metal piece is Vm. What do you think Vw is greater than Vm? Yes or no? Volume of water that is displaced by the metal piece because of the metal piece is greater than or less than volume of metal. So Vw is the volume of water that is displaced because of the metal piece. Then the weight of the water displaced should be equal to the weight of the metal piece, right? And of course, metal piece has higher density, okay? So volume of water displaced will be more than the volume of the metal itself when it is floating because it has to balance the weight of the metal. But when the metal piece sank inside, then the volume of water displaced will be equal to volume of metal, isn't it? Because the metal piece will now sink, it will displace the same volume of water that is its own volume. So earlier it was displacing more volume, okay? Now it is displacing less volume. So the level will go down. Fine, so we'll move to the next question. In case you have any doubts, please type in quickly or you can speak up. Now we'll be taking regular objective type questions. Single option correct right now. Try solving these questions. No rhetoric, okay? So we have seen question number one scenario many a times, all right? So here it is of no surprise that you're able to solve it quickly. So option C is correct over here. In fact, we use as a rule that horizontally pressure should not change. But if system is accelerating, then horizontally also the pressure will change. So if inside the water, if this is 2 and horizontally, let's say this is 0.1 at a distance x, then P2 should be equal to P1 plus rho Ax, okay? So horizontally pressure is changing. Now it means that pressure at 0.2 is more than pressure at 0.1, okay? So if I consider vertically, then pressure at 2 is P atmosphere plus rho G into H, this height. This height should be more than that height. Then only pressure at 2 will be more than 1. That is the reason why the slope has to be like this. So option C is correct over here. Try solving the second one. So when I have taken properties of material, I have taken both fluids as well as solids for today. So at times you'll see properties of solids, questions as well. Okay, two options. A, everybody is saying, let's see how we can solve the second question. See, whatever is the case, whatever is the dimensions, one thing remains constant. That is Young's modulus because the material is not changing. Young's modulus is what? Sigma by epsilon, okay? Now tension is same, okay? Now tension is same, but stress could be different because of the surface area. Sorry, because of the cross section area. So I can write sigma as tension divided by area. Now area I can write in terms of diameter because diameter is what is getting changed. So I'll be writing it as pi D square by 4 divided by epsilon. Let's say epsilon is change in length divided by original length L, fine? So Young's modulus into change in length divided by the original length is 4T divided by pi D square. Okay? So delta L will come out to be equal to 4TL divided by pi into y square, pi into y into D square, all right? Now you can say that delta L is proportional to L by D square, okay? So whichever options L by D square is maximum, delta L will be maximum for that, okay? So for the first option, delta L is proportional to 50 divided by 0.5 square, so which is coming out to be 200, okay? I'm not taking care of units here because when I compare, I'm taking same units for length and diameter everywhere. So this is 200 delta L by D square when you do for B, it'll come out to be 100. It'll be less than 100 for C option and D option, so I'll just pick option A here, okay? Now I'll take this opportunity to introduce a few things. So at times, you will have wire which is hollow like this, okay? So only this shaded portion has material, okay? And when you consider the area to apply for Young's modulus, you need to only consider this area which has material in it. So pi B square minus A square, if B and A are the radius, will be used as area of cross section, okay, to find this stress, all right? So we can move to the next one, right? Try this one. Anyone? Should I do it? This question came in 1983. Here when Simon here got born, okay? Can you tell me one thing? The top surface, do they remain at the same level? Yes or no? They'll remain at the same level because at the top, it is atmospheric pressure, right? The pressure is same. So along horizontal line only, pressure will remain same. If horizontally, the system is not accelerating, okay? So they both have to be in the same level. If they are not, let us say right hand side liquid is slightly more, then pressure over here will be atmospheric pressure plus rho g h of this liquid. But that is not the thing. Pressure both side has to be atmospheric pressure. Now, let us say second liquid is up till here. So from here, I can draw a line like this, okay? Then, since it is horizontal line, pressure over here, let's say is P1, pressure here, let's say is P2, P1 has to be equal to P2, okay? Now P1 is atmospheric pressure plus rho of the second liquid g into h, where this distance is h, okay? And P2 is what? P2 is atmospheric pressure plus density of the first liquid g into h, right? And both of them are equal because they are along the same horizontal line. So since P1 is equal to P2, this implies density of first liquid has to be equal to density of the second liquid. So if specific gravity of first one is 1.1, for the second one also it has to be 1.1. Okay? So like this, you have to do, you know, this question. Any doubts on this? Okay, I'll move to the next question. So you can see there are two questions on your screen. Try attempting both of them. It is written that it's a large open tank. So you can consider that the area of process of tank is much more than the area of the holes. So you can use Toshery's law. What is Toshery's law? Quickly introduce Toshery's law. So is it option A? Wait, if this is area of procession A and the hole is very small and at a depth y, then the velocity of the liquid that is coming out will be equal to root 2 g y. Assuming A is very large compared to small a. So this is Toshery's law. Remember this. Or you can derive it also, but it is good to remember. Now quantities of water flowing out is what? Quantity of water that is flowing out per second is nothing but dm by dt. This is mass flow rate, which is rho A into v. So if density is same, I can say that A1v1 is A2v2. This is not continuity equation, but what is given here is that if this is square hole and this is circular hole, A1v1 here will be equal to A2v2 there. That is what it means. If mass flow rate coming out, both sides are same. Now the quantities of water coming out is same. Circular hole has radius r and depth 4y. Circular is at a depth of 4y. So this will be a square, which is at y and this will be circular, which has a depth of 4y. So let's say this is v1 and this is v2. So v1 is root of 2gy, 2g into 4y is v2. So this will come out to be two times under root 2gy. This is v2. Now A1 is what? A1 is the area of the square hole, which has side length l. So l square is the area this into v1 under root 2gy1. This should be equal to A1 of the square hole, which has radius r is pi r square. This is area of this circular hole. This into velocity of the water coming out from there. So when you equate it, root 2gy get cancelled out. And r will come out to be equal to l divided by under root 2pi. So many of you have solved this question before. So others, those who are not able to get it from the first go, because you are seeing it for the first time, don't get unnecessarily intimidated by that fact. There will be many people who have solved this question four or five times also at times. Anyways, sixth one. What is the answer? Okay, Vaishnavi is asking how you arrive dm by dt is rho av. Vaishnavi, velocity is what? Meter per second. When you multiply with area of cross section, you get volume per second. Area into meter per second, volume per second. Now that volume per second, when you multiply with density, so density into volume becomes mass. But then it becomes mass per second, so rho av it comes. So I am telling you in a very crude way, in case you want to get into the details of it, you can refer the notes where we have derived the Bernoulli's theorem. There we have discussed in greater detail. Okay, sixth one, any answer? No one, should I do it? Okay, let me do this now. There is a hemispherical portion of radius r removed from the bottom of the cylinder of radius r. Okay, so this is the cylinder, hemispherical portion is removed. The volume of the remaining cylinder is v and the mass is m. So volume of the cylinder is given. Okay, it is suspended by the string in the liquid of density rho. It stays vertical. The upper surface is at a depth h. The force on the bottom of the cylinder by the liquid is what? Okay, as a first of all, we know that net force due to liquid in this case is what? Bohand force, okay, which is what? v rho g, isn't it? This is the net force due to the liquid. Now, if you draw a free by diagram of this cylinder, due to the liquid, there will be a force from the top and there will be a force from the bottom. Okay, there will be sideways force also, but they will get cancelled away. Okay, but these two forces will not get cancelled. All right, so we can say that pressure over his is P1. So P1 into a is the force from the top. But here the problem is that this hemispherical portion, every point is at different depth. So let's just keep it as f only. Okay, we know that net force is v rho g. So f minus P1 into a is v rho g. Okay, and of course, there will be, you know, this you might be wondering what about the force due to the tension, which will be from here. Let's say this is tension. Okay, and what about the mg force? So first of all, you need to understand, I'm not doing the force balance here. Okay, I am not writing force balance equation. What I'm writing here is the force due to the liquid equation. Okay, so for the liquid, the net force is in upward direction. That is the buoyant force, which is v rho g. Okay, now this v rho g, you can split it into two components. The force from the top and force from the below. Okay, so this is not force balance. So f will be equal to P1a plus v rho g. Okay, now P1 is not the total pressure here. P1 is just the pressure due to the fluid. Okay, so what I'm trying to say here is that, you know, there will be a atmospheric pressure also here, P8 plus rho gh is P1. So this is P1. Okay, similarly, the atmospheric pressure will be acting from below as well. Okay, but we are trying to find out the force due to the liquid, not because of the atmospheric pressure. Okay, so since when you subtract, the pressure due to the atmosphere will anyway get cancelled away. So I'll just consider the rho gh pressure due to the liquid. Okay, that is simply rho gh only, rho gh. So f will become equal to P1 into a. Now a is what, pi r square. So rho gh into pi r square plus v rho g. This is the answer for the force from the bottom, option number D. So if you understand this question properly, many concepts will be clear because here we have, actually utilize the fact that bow and force is the outcome of all the forces because of the fluid. Any doubts on this question? Anything? Fine, let's move to the next one, these two. So for the seventh, see you have seen a similar question before also, isn't it? So when the coin is on top of the wooden block, when the coin is on top of wooden block, then it is displacing more volume than its own volume. Isn't it? So when the coin is at the top of the wooden block, it is displacing more volume than its own volume. Okay? But when the coin is fell, when the coin fell down, then it is just displacing, you know, it is just displacing volume equal to its own volume. Are you getting it? So earlier it was displacing more volume. Okay? And now it is displacing lesser volume. So H will decrease, H will go down. Are you getting it? Because the water is getting more volume to occupy, so H will reduce. Okay? So what about L? See, L is nothing but this lambda only which is shown over here. So even this will reduce because now the wooden block doesn't have to displace more volume. Earlier it was getting compressed below because of coin. Okay? So both H and L will decrease. Why this will increase? Both will decrease, right? What about 8th one? What is the answer for 8th? 7th answer, D is correct. D for Delhi. All of you understood why 7th D is correct? Type in yes or no? Okay, 8th. Ramchandran Kondi, you are there, right? Yes, sir. Okay. Okay, 8th one. So this is showing a graph between delta L and the load that is connected or to the other end. The cross section of the area of the wire is this, calculate the Young's modulus of the material. So Young's modulus is simply given as sigma divided by epsilon, right? Sigma is what? Force per unit area divided by this is delta L by L. Okay? So you will get fL divided by a delta L. Now here we have L given as 1 meter. So it is simply f divided by a times delta L is Young's modulus, okay? And area of cross section is also given. So Young's modulus is simply f by delta L multiplied by 10 to the power 6. So I think this is a straightforward question where f is 20 newtons, okay? And delta L is 10 to the power minus 4, this into 10 to the power 6. So you will get 2 into 10 to the power 11. A is the correct answer. At times they will give you just, rather than weight, they can give you mass. So you need to multiply that with G also, all right? So we'll move to the next one. Any doubt, guys? Please type in. In case you have any doubts, no doubts. We already discussed the definition of bulk modulus, right? We'll write it again. This is the bulk modulus. So is there option A? Option A. Others? Okay, Purvik is also getting A. Yes, option A is correct here. So volume changed by 10%. It means that delta V by V is actually 0.1, okay? And delta P is also given. Delta P is P2 minus P1, okay? So that is 1.65, 64 in fact, 1.64 into 10 to the power 4 pascals, right? 1.55, all right? So bulk modulus will be delta P divided by delta V by V. So it will be 1.55 into 10 to the power 5, okay? That's our option A is correct. What about the 11th one? So here comes the surface tension concept. This is turned. This is hemisphere. This is sub-hemisphere, fine? So can I say that radius of bubble one is less than radius of bubble two? Yes or no? So I'll just show you something over here. Let's say this is the diameter of the pipe, okay? Now let me draw a circle over here. Now here I am drawing a circle whose diameter is equal to the length of the, this thing, length of the pipe itself, the length or the diameter of the pipe itself. Now let me draw a sub-hemispherical shape here. So this line no longer be a diameter, are you getting it? So this is how the scenario will be, okay? So you can see that, I mean I'm not able to perfectly draw this. It will be like that, okay? So you can see that the white circle, okay? Where you can just consider scenario like this. Now tell me which one has bigger radius? The white or the orange one? This is let us say hemispherical, which is R1 radius. This is sub-hemispherical, which is R2, which has more radius, the white one, right? So that is the reason why sub-hemispherical bubble will have larger radius, okay? Now we know that delta P for a bubble is 4s by R, okay? Delta P is what? Delta P is pressure inside minus pressure outside. Outside pressure is atmospheric pressure, okay? This should be equal to 4s by R1, right? Right now my focus is to find P1. So I'll write it as P1 minus PA is 4s by R1. Are you able to see my screen? Okay, so this is let us say my first equation. And my second equation is P2 minus PA is 4s by R2, okay? Now tell me which is more, P2 or P1? Since R1 is less than R2, okay? So P1 minus P2 or P1 minus PA, sorry, this will be more than P2 minus PA, right? Hence P1 is greater than P2, getting it? So pressure here is more than pressure there. And that is the reason why the air will flow like this. So air will go from 1 to 2, from higher pressure to lower pressure. That's its regular tendency, right? And when it does that, then the pressure of 1 reduces, pressure of 2 increases, okay? So air flows from 1 to 2 and the volume of soap bubble at end 1 decreases. So that is why option B is correct over here. Any doubt, guys, on this question, I'll move to next one, meanwhile. Any doubts? Okay, do these two questions. Mishra is saying, no, Mishra is answering 13th one, okay? Yes, 13th is C. So 12th one is clearly a question which will just eat up a lot of your time, okay? So the smart way is to first answer question number 13 and then go to question number 12. So we'll also be smart and we'll answer question number 13 first. One end of the horizontal thick copper wire of length 2L and radius 2R is welded to an end of another horizontal thin copper wire length L and radius R. When the arrangement is stretched by applying forces at two ends, the ratio of elongation in the thin wire to that of thick wire is what is asked. So the thick copper wire has length 2L and radius 2R. The thin one has length L and radius R. Now, they are pulled by some force. Let us say force is F, okay? So if you draw a free-body diagram of the thicker wire, it will experience same force and the thin wire also experience the same force, okay? So we have the stress in the thick wire equal to be force divided by pi into 2R square, okay? So this is F divided by 4 times pi R square and sigma of the thin wire is F divided by pi R square, all right? And we know that stress by strain is constant. So let's say thick is sigma 1 and thin is sigma 2. So sigma 1 by epsilon 1, it's Young's modulus, right? Should be equal to sigma 2 by epsilon 2. So from here, I will get epsilon 1 divided by epsilon 2 to be equal to sigma 1 by sigma 2, right? Sigma 1 by sigma 2 is 1 by 4, okay? Epsilon 1 is delta L, 1 divided by the length of the thicker wire which is 2L divided by delta L2 divided by L. This is equal to 1 by 4, okay? So from here, I will get delta L1 by delta L2 to be equal to 1 by 2, okay? So delta elongation in the thin wire is 2 times the elongation in the thick wire. So don't be in a hurry and you should not mark B, okay? Option C you should mark because to clearly see which one you are finding the ratio, all right? Now let us try to solve question number 12. Question number 12 is actually one of those questions which distinguish toppers from others, okay? So it's not a straightforward one. There are a lot of fine things which people ignore, okay? So anyways, a thin uniform cylindrical shell closed at both the ends. So you have a cylindrical shell which is closed from both the ends partially filled with water. So this cylinder is filled with water, okay? It is floating vertically in the water half submerged state. So it is half submerged, fine? And if rho C is the relative density of the material of the shell rho C with respect to water, then correct statement is that shell is more than half filled, okay? So basically, you know, we should first before even thinking first we should do the force balance here, okay? And then look at the statements because we need to actually evaluate if what happens then what will happen? So if then else statements are the options are in terms of if then else statements, right? So we need to first get some equations, then only we can assess the statements. So if you do force balance, then there is downward force, okay? So there is this downward force, let us say mg, okay? And then there will be a buoyant force, okay? So simply buoyant force has to be equal to mg. Buoyant force should be equal to m into g. Now mg can be divided into two parts. mg of the shell, this is shell, okay? Plus mg of the water. This should be equal to the buoyant force, okay? Now buoyant force is what buoyant force is the weight of the liquid that is displaced, okay? So if let us say if this this is if the total height is h then about h by 2 height we have you know the cylinder occupy because it is written that in half summer state it is floating, okay? So if area of cross section is a, okay? Then I can say that buoyant force is a into h by 2 into rho of water into g, okay? So this is for the cylinder and also there will be some amount of liquid that is displaced because of the thickness of the cylinder, you know? So if let us say the entire volume of the shell, okay? Is v. If you ignore the thickness of the shell then because of the thickness of the shell you can ignore the volume displaced, okay? But if you say here if you look from the top if this cylinder is like this it has some thickness, okay? If it has some thickness then area of cross section is this inner area, okay? Why am I distinguishing the distinguishing two areas? Because the relative densities are different, okay? So this plus the area sorry the volume which is displaced let us say v by 2 where v is the volume of the shell, okay? This volume of the shell divided by 2 into density of water into g, okay? This is the total buoyant force, this is because of the cylindrical hole and this is because of the thickness of the cylinder. This is the weight of the liquid that is displaced. Now this will be equal to the weight of the shell which is volume the shell divided by now you have to take total volume the shell. Volume the shell into density of the shell into g, okay? Plus you have mg of the water inside. So let us say up to x level the water is inside, okay? So I can say that area into x is the volume the water inside into density of water into g, okay? So let me tell you that this question may not have been solved by the toppers also. They might have also left it, okay? So this is meant to learn the concept but not to actually solve. So I mean this is a good opportunity for you to understand how the concepts are applied, okay? Okay, Purvik is asking me to explain again. See when the cylinder is submerged inside the water, okay? It will displace the volume of the water, okay? Now what I am doing is I am treating the cylinder, okay? I am treating the cylinder as if it is hollow but it has some finite thickness, okay? So area of cross-section is the area of the whole of the cylinder, okay? So I am treating that the hollow cylindrical part separately here because I have to use the density of the material over here, okay? So the volume of the water that is displaced because the cylinder is going down is equal to the volume of the hollow portion of the cylinder plus the volume of the material which is inside the water. So that is the reason why we have let us say volume of the shell to be equal to Vs which is the actual volume of the material. Then V shell divided by 2 into rho w into g. This is the volume or weight of the water that is displaced which corresponds to the volume of the material only, okay? But then not only that much volume is displaced but also the volume corresponding to that hollow portion is also displaced. I hope I am making sense here. Let me, are you able to understand this or should I explain again? See this is the scenario. What I am saying is that this cylindrical thing has some thickness, okay? So let us say this is the thickness of the cylinder. So if I just consider this thickness itself, it will have some volume. Yes or no? Just this thickness or the volume of the material that is used to build the cylinder. If I just consider that it has some volume. So that volume the material used in the cylinder is Vs, fine? But that volume will just give you the volume of the material used. But when this cylinder is inside the water, it will not only displace the volume corresponding to the material, but it will displace the entire volume from here till there, getting it? So the volume, the water displaced, how I am finding? I am first finding the volume which is corresponding to this and then I am finding volume which is corresponding to the material which is there. I am separating them apart because the density of the material is given separately and the conditions or the options are given with respect to that. So that is the reason why we are doing it. And it is okay for it that you are not able to do it right in the first go. So it is expected, but then you should understand how it is done right now. That is more important. So you will get an equation corresponding to X now. So once you get an equation corresponding to X, all you have to do now is to evaluate those options. So now here onwards, try to now analyze the situation further and get back to me in case you have any further doubts because again it will take more than 15-20 minutes of discussion just on this question itself. So try this out at your end and let me know if you get stuck. So I will move to the next one. Okay, those who want to take a break can take a break. These two questions, those who are taking a break need not solve these two questions. They can solve question number three onwards. Okay, so these first two questions I am putting for those who are not taking a break, right? It's okay. You can take a break, eat something and come back. Others can solve these questions and anyway this is getting recorded. So you will not miss anything. Okay, first one, if you are giving the wrong answer, draw the free boy diagram yourself. You will see. Okay, so if this is the object, all right, this is the object, it is falling under the gravity. So its acceleration will be G, all right. Then it will of course have MG force because of the earth and let's say upthrust is given as B. So MG minus buoyant force should be equal to mass time acceleration, which is MG only. A is G. So B comes out to be zero. Buoyant force will not be applicable when it is falling under the gravity. Okay, so vertically if things are at rest, then only buoyant force is M into G. Otherwise, you have to take G effective just like, you know, the weight of the object changes when it is inside the lift. So just draw the free boy diagram. You'll be able to see what could be the effective G. So buoyant force is rho V G effective. When it is freely falling, G effective is zero. Anyways, question number two. The spring balance A reads 2 kg when the block M suspended from it. A balance B reads 5 kg when the beaker with liquid is put on the... Ramcharan, you're there, Ramcharan and Kondi. Why you're on Skype? You're not talking anything, just keeping silent. You have to say, okay, every two minutes say something, interact more. The spring balance A reads 2 kg with the block M suspended from it. A balance B reads 5 kg when a beaker with a liquid is put on the balance. The two balances are now so arranged that the hanging mass is inside the liquid. In this situation, okay. Second one, you're saying C. The balance A will read less than 2 kg. Of course, it will read less than 2 kg because now the buoyant force is acting. And B will read more than 5 kg. That is correct because if liquid apply buoyant force on mass M upwards, then mass M will apply equal and opposite force on the water downwards. And hence, according to Newton's third law, option C should be correct. Okay, let me, so this is just one of the options. There can be more than one option correct. So can you see whether other options are also just evaluate? Let me quickly check the answer for you. C is correct, one more option is correct. Of course, if C is correct, B is also correct because B is subset of C. So don't forget to mark options which are subset of each other, okay. Anyways, we'll move to next question. See, in case you have already seen these questions, no point attending the session. You can as well tell me, I can send you a few new questions. Otherwise, if you have solved this question a couple of times, you're wasting time. Okay, now do this. Sir, that is ridiculous. Yes, Amogh, that is ridiculous. I can understand your emotions, but J is the exam where you try to reject students. Third one C, Amogh is getting, okay, almost, not almost everyone. Four people have got C as the option. A vessel contains oil over mercury. A homogeneous sphere floats with half of its volume, immersed in mercury and half in oil. The density of material of the sphere is what? So just you need to find out the weight of the fluid that is displaced and equate that to the weight of the sphere, okay. So here, let's say this is the vessel. So we have the lower portion. Of course, the upper one will be oil, right, and the lower one have to be mercury. So if it is half submerged, then V by 2 of the mercury is displaced. So rho mercury V by 2g is the buoyant force because of the weight of the mercury and rho oil into V by 2 into g is the buoyant force because of the displacement of oil. This has to be equal to the mg force, okay. mg is what? The density of the material into V into g, right. The density of the material is rho mercury plus rho oil divided by 2, okay. Which is what? 13.6 plus 0.8, which is 14.4 divided by 2. That is 7.2, okay. So option C is correct over here. What about 4th? Okay, 4th also you're saying C, let's see. Two rods of different material having coefficient of thermal expansion alpha 1, alpha 2, Young's modulus y1, y2, they are fixed between two rigid massive walls. The rods are heated such that they undergo same increase in temperature. There is no bending and alpha 1 divided by alpha 2 is equal to 2 by 3. Thermal stresses, two rods are equal, provided y1 and y2 is what? Now, see this thermal stress, sorry, thermal strain is given as alpha delta T, isn't it? Because delta L is what? Delta L is L alpha delta T. So delta L by L, which is strain is alpha delta T, okay. So epsilon 1 is alpha 1 delta T and epsilon 2 is alpha 2 delta T, okay. Then stress by strain, sigma 1 by epsilon 1 is given as y1. So basically, sigma 1 is y1 into epsilon 1, which is y1 alpha 1 delta T, okay. This should be equal to y2 alpha 2 delta T, right, because thermal stresses are equal. Now alpha 1 by alpha 2 is 2 by 3. So if I remove delta T, so from here I will get y1 by y2 to be equal to alpha 2 by alpha 1, which is 3 by 2, okay. So option C is correct over here. So we will move to next question. Now quickly solve this one. I think this you can do straight away, okay. Again C, the first of all the velocity will increase because of gravity, right. So in order to keep a1 v1 equal to a2 v2, the area will decrease, isn't it? So it cannot be a. So we have a2 to be equal to a1 times v1 by v2, right. V2 is v1 plus, sorry, v2 square is v1 square plus 2 into g, g you can take as 10 to g into s, which is 0.15, okay. So this is v1 is 1 plus 3. So this is 4. So from here, I will get v2 to be equal to 2 meter per second, fine. So a2 will be half of the earlier area. So half of the earlier area is C, okay. So that is how you have to solve this question. I will move to the next one. These two questions, they are from J advance 2015. Now Saimir will explain how he has got that answer. A and B is the correct answer, both A and B. See the curve will end, the curve plotting will end where the material breaks, okay. So Q can take this much stress, okay. Whereas P can take maximum stress of slightly more than Q, all right. So this is the maximum stress P can take, and this is the maximum stress Q can take. And hence, P has more tensile strength than Q. So in order to understand this, you need to have clear understanding of what is tensile stress and what does ductility and all means, okay. And also the strain of P is more than Q, okay. For the same amount of stress, if you have same amount of stress, the strain of P is more than Q, fine. So that is the reason why P is more ductile than Q, okay. So this is how you have to solve this question. Now question number eight, this is again a very different kind of question. See if you know how to approach it. Okay, spherical body of radius R consists of a fluid of constant density as and it is in equilibrium under its own gravity, okay. If P R is a pressure at a distance R, which is less than the radius of the a sphere, then which of the options are correct? So suppose you have, let us say a DR width, you are imagining a strip, okay. You are imagining a shell at a distance R. So let's say the thickness is DR, okay. So there will be pressure, right. So let's say pressure from outside is P, okay. And pressure from inside is let's say P plus DP, fine. So if you consider that thin strip only, consider that thin strip. So from inside P plus DP is applied and from outside P is applied. So pressure will increase as you go down, go towards the center, okay. At the same time, there is a gravitational attraction that is pulling this shell, okay. So let's say that gravitational attraction force is FG, all right. So if I write the force balance equation over here, just taking the shell into consideration, I can write, you know, since the shell is stationary, so net force should be zero. So P plus DP minus P, okay. This into 4 pi R square. This is the net force due to the pressure, okay. Minus the gravitational force should be equal to zero, getting it. So DP into 4 pi R square minus FG is equal to zero. Now you need to find out force of gravity on the shell, fine. So force of gravity on the shell because of the outer material will be zero. So force of gravity will be only because of the inner material, all right. So you just find out force of gravity due to the inner material. Let us say mass of the inner material is small m, okay. So the force due to the gravity on that entire shell will be gm into rho into 4 pi R square dr divided by R square, okay. Where m can be, you know, m can be written as, small m will be written as rho into 4 pi, sorry, 4 by 3 pi R cube, okay. So like that when you substitute, you will get DP and dr in the equation and then when you integrate pressure, okay, when you integrate the pressure from zero to p, the R will vary from capital R to small r. So like that you will get pressure as a function of R and then you can, you know, simply evaluate which of these options are correct. No aditya that is not correct, pressure at R equal to zero is very high. It tends to infinity actually. No, wait, wait. Yeah, pressure at R equal to zero is very high, okay. So option, so I will just tell you the options which are correct. You can try out from here. I think it is just differential equation and I think you should try out after the class and let me know if you do not get it. So B and C are the correct options. So you will get a differential equation, just solve it and after this it will become straightforward. This came in J advance 2015. All right. So we will move to the next question. Okay. So this is the viscous fluid. We actually ignore viscosity most of the time, but then since this is fear in a viscous fluid and if they have velocity, then there will be a viscous force also apart from the buoyant force. Buoyant force will of course act, but apart from that, there will be a viscous force also acting on it, which is given by this formula 6 pi eta R to V terminal. Few of you might have remembered the formula if it is just one sphere alone. So it is 2 by 3, 2 by 9 pi R rho minus sigma times G divided by mu. Okay. This if you remember, it will be fine. Otherwise, you can always derive it. Okay. So if let us say I am talking about P. So I have to take the density of the P and the density of the fluid, let us say sigma 2 and we have coefficient of viscosity here. So two spheres of equal radii have density rho 1 and rho 2. The sphere are connected. L1, L2 are densities, sigma 1 and sigma 2. So this is sigma 1 and this is sigma 2. Coefficient of viscousities are also eta 1 and eta 2. So if sphere P alone in L2, if P is alone in L2, then VP is the velocity. So if P is in liquid 2, then you have to use liquid 2's density and liquid 2's coefficient of viscosity. Okay. Similarly, VQ is Q alone in L1, which will be 2 by 9 pi R square rho 2 minus sigma 1 divided by eta 1. Okay. So this is how you get VP and VQ and now you can compare. First of all, sigma 2 has to be more than sigma 1. Then only you know this liquid will be at the bottom and the upper liquid will be at the top. Okay. So this is the first thing you should know. And the second thing which there is density of the second this thing. Density of Q should be more than the density of the first liquid. Okay. Then only the Q is submerged. Okay. Q is not in the first liquid. So Q is down. And also density of the second liquid has to be less than, oh wait, if string is taught. Okay. And everything is sinking. If everything is sinking, then density of the second ball has to be more than the density of the second liquid as well. Okay. Then only it will keep on going down with the velocity. Otherwise, it will just float. It will not go down. Fine. So therefore, you can say that rho 1 is less than sigma 1. Okay. And you have sigma 2 is less than rho 2. And we anyway have sigma 2 to be greater than sigma 1. Fine. So rho 1 is less than sigma 1 that is required. Otherwise, the string will not be taught. Okay. So Q is essentially pulling everything down. Q is making effort so that P comes down. So if Q doesn't make any effort, P will not come down. So basically, there has to be a tension in the string. Okay. So if there is a tension in the string and if you draw a free bar diagram of P, let's say this is tension. Okay. Then you have mg force and you have the buoyant force. Okay. So mg is what? Density of the material into volume of the material into G. Buoyant force is density of the liquid into volume of liquid into G. Okay. So if suppose velocity is almost zero, so we can ignore the viscous force, then tension is required over and above the mg force. So it automatically means buoyant force is more than the mg force. Okay. Then only it will be pulled down and then it will gain some velocity and then viscous force will also start acting. But prime FAC, the density of liquid one has to be more than the weight. Sorry, density of liquid one has to be more than the density of P. Okay. So when you analyze all this scenario, you will get option A and D to be correct. Okay. So we will move to the next question. We have around 15 minutes. We'll take up. Okay. Let's take a question on surface tension. See, these questions, few are beyond mains level, but our focus is learning the concept. So that is why I'm making you a little uncomfortable with slightly difficult questions, but you'll learn more. Try solving this, this one. Okay. Simon and Purvik, I have left that question after analyzing the scenario. So I want you to get option A and D. I have told you the final answer. Okay. So after that, it is nothing. You just spend some time. You'll get it. I wanted to discuss this question on surface tension. So that's why I'm not putting this one. So let me know. You can WhatsApp me if you're not getting and send me your attempt. Anyone got the fourth one? I think this fourth question is straightforward. Kushal, how are you getting C? Explain quickly. All of you will know that surface tension T is force per unit length. Okay. And this is also surface energy per unit area. So let us say this is the drop, which is formed on a dropper. Fine. Then let us say that this is the center. Okay. I'm connecting the peripheries like this. Okay. Why I'm doing this because the surface tension will act on this circle. Okay, which is from the dropper. Okay. And along which direction it'll act, it'll act like this. Okay. So if this is the surface tension, okay, which is trying to reduce the surface area, then let's say this is theta. Okay. So if this is surface tension T, so T cos theta. Okay. T cos theta into 2 pi r because this surface tension is acting on the entire of this circle, which is the end of the dropper. Okay. So T cos theta into 2 pi r will balance the weight. Okay. This is the vertical force. So we need to just find the vertical force only. Okay. And also we know that cos of theta is small r by capital r, where this is capital R. And if you draw a vertical line like this, this radius of the dropper is small r. Okay. So cos theta, where this is theta, then this angle, this yellow angle is 91 theta, then this angle will become angle theta. So cos of theta is small r by capital R. So this will give you vertical force to be equal to 2 pi r square T by small r. Okay. So even though I have said that this is very easy one, but it was not. It was a tricky one, which I'm sure many students who have seen this for the first time in the exam would not have got it. Okay. This is again a question which differentiates droppers from others. So don't worry if you're not able to get it. So just try to learn the concept here. What about fifth and sixth? Once you get the fourth one, see these passage questions are, you know, they are linked from each other. If you get one of the question right, you will get all of them right because one answer of the one will be used in the next question. So it will detach from the dropper when the vertical force become equal to mg, isn't it? And when it detaches, it will be almost spherical drop. D, Purvak is getting D. Others? So B. B for Bombay? Yes. Okay. No. See the vertical force is what? 2 pi r square T by capital R. Okay. So this should be equal to the weight of the drop which is 4 by 3 pi r cube into rho into g. Right? So from here, you'll get the radius of drop and it will be approximately option A. Fine. What about the sixth one? Sixth, what is the answer? Sixth one is direct. Surface tension is surface energy per unit surface area as well. So you just multiply surface tension with surface area. You'll get the answer for surface energy. Okay. So you have got radius of the drop from here. All right. So r is equal to 1.4 into 10 raise to power minus 3 meters. So surface energy will be equal to surface tension which is given as 0.11. So 0.11 into 4 pi r square, 1.4 into 10 raise to power minus 3 square. This many joules is the surface energy. Okay. So you'll get option B to be correct over here. Fine. So I think today I have made you a little uneasy with this chapter by showing you different kind of questions that can be there other than your routine ones which I have sort of intentionally not shown in this particular session because I wanted you to be aware that there are few other varieties that exist other than what is there in Etsy Verma. Okay. So that's how you have to take care of the variety rather than the quantity when you practice questions yourself. All right. That's it for today. So I hope you have learned many things today. What I'll do, I'll forward you this particular PDF and few questions I have just given you hints so that you can solve it yourself. So immediately get the answer for those and WhatsApp me in case you don't get it. And yes, so we will come back to you again in couple of days with another session on revision and this time I'm planning to take exclusively entire optics. So I think we have already one session on optics. Now the session will be slightly faster. We'll be solving other kinds of numericals on optics in the next class. Okay. Thank you for today. We'll see you.