 OK, 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. OK, so this is a fill-in-the-blanks. You have two questions fill-in-the-blanks. Try answering them both. OK, 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, OK? 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, OK? Now, Young's modulus is given, OK? 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, OK? So I can write down stress in terms of Young's modulus, OK? So potential energy will come out to be half into strain square into Young's modulus into area into length, OK? Now strain square is what? Strain is delta L by L. Delta L is x. So x square by L square, OK? Multiply it by Young's modulus. Multiply it by area into length, OK? So 1L 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, OK? 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 numericals which you might not be solving regularly. But there are things to learn which you can keep in mind. Anyone? OK, let me solve this. So I will get dv by v is equal to minus of dp by k, OK? 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, OK? Now I am just taking the mod of it as in I'm just bothered about the magnitude. So I'll 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'm bothered only about the magnitude, so I'm keeping it like this, OK? So this is dv by v, OK? And what is v? v of a sphere is 4 by 3 pi R cubed. Getting it? So from here, I will get dv is equal to 4 pi R squared dr, OK? So dv by v is what? This divided by 4 by 3 pi R cubed, all right? So you will get 3 dr by R is dv by v, OK? 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, OK? 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, OK? So you can solve it like this. So I can understand these are not the routine ones. So you'll 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, OK? Let's take up new set of questions. So you may see that these chapter can get easily mixed with other chapters also. So don't get surprised by that, OK? 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, OK? 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. No one. Metal is inside mercury. So if this is the piece of metal, what is given here is that the metal piece is partially submerged, OK? So it is floating, OK? Now, Simehir is getting answer, OK? No, Khushali is saying, is it 1 plus gamma 2 delta t? OK, let us see. When you heat, when you change the temperature, OK? What will happen to the mercury's density? The mercury's density will reduce, OK? So it was now initial density, 1 minus gamma 2 delta t. This is now the mercury's density, OK? Now, mercury will create a buoyant force, OK? So how much buoyant force it should create? It should create buoyant force in order to balance out the mg of the metal, OK? Now, it doesn't matter whether the metal, this metal, expands or contract right now. It will have, say, mass, right? So the amount of mercury that will be displaced is determined only by the mass of the metal, OK? So rho dash into volume that is now submerged, OK? Into G, this should be equal to mg, OK? 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, OK? So this is the earlier scenario and this is the later on scenario, OK? Now, here rho dash is what? Rho of mercury 1 minus gamma 2 delta t into V dash, OK? 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'll get here is 1 minus gamma 2 delta t to be equal to V divided by V dash, OK? 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, OK? 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, OK? 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, OK? 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, OK? Fourth one. Fourth one is straightforward, try to do this. It's an application of Bernoulli's theorem and continuity theorem. Yes, I mean, I can see that you have got some answer. Others, what are you getting? So is it 3500? OK, others, those who are online, can you reply? Online as in YouTube. OK, so we have now a few answers. Let me explain what we have to do here. First of all, it is a horizontal pipe, OK? So you can assume the scenario to be something like this, OK? Now, this pipe is not open to atmosphere. Had that been the case, then pressure both sides would have been atmospheric pressure only, OK? So it's a cutout of a bigger pipe, OK? Where the cross section area is, let us say, over here, the cross section area is 10 centimeters square, OK? And this is, let's say, a1. And velocity over here is 1 meter per second, OK? Over there, OK, here the pressure is also given. In fact, p1 is 2000 Pascal, fine? Over there, over the other end, the cross section area is a2 is 5 centimeters square, OK? So we need to find out what is the pressure over here. p2 is what, OK? Now, between 1 and 2, I can use continuity equation, OK? 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, OK? Now, since it is horizontal pipe, and if I use Bernoulli's theorem, that is p1 plus half rho v1 square plus rho g h1 to be equal to p2 plus half rho v2 square plus rho g h2. Now, rho g h1 and rho g h2 will get cancelled off because it is an horizontal pipe. So h1 is equal to h2, OK? 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 g h has no role to play. So p2 will be equal to p1 plus half rho times v1 square minus v2 square, OK? Now, p1 is 2,000 Pascal. We'll keep it like this. Plus half rho is what? It's water, right? So yeah, it is water. So half 1,000 into v1. v1 is 1. So this is 1 square minus 2 square, OK? So we will get 2,000 plus 500 into, now you'll get minus 3, OK? 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.