 Yeah, so we were talking about liquid exerts pressure because of its height. Okay, if H differs height difference or depth you can also say if height differs. Then pressure will also be different. Right. Suppose we have, we know that solid particles are solid objects exerts pressure because of its mass. Gaseous particles exerts pressure because of collision. Right. I've discussed all these things. Now suppose you have a tube. Okay. In which, you know, any liquid that is present generally we take mercury for the, you know, for the measurement of pressure. So suppose we have a column here in which a liquid is present of certain density. So liquid that is present here, it exerts pressure. Okay. So because of the weight of the liquid, we have a mass component which acts in downward direction, mg, and in upward direction, we have pressure because of atmospheric. So that is TATM into A, where A is the cross sectional area of the tube. Right. Pressure into area that is the force. It is the cross sectional area at equilibrium. What we can write at equilibrium. We can write this force equals to this force. So P of atmospheric into the area of cross section is equals to mass of the liquid in into the gravitational acceleration. If this liquid is off density row. Right. Row is the density and volume is B we are assuming for the liquid. So area is equals to what we can write. Tell me. Atmospheric pressure is equals to into area related be here only. Fine. Mass of the liquid is nothing but what is nothing but density into volume of the liquid into the gravitational acceleration. As this volume is nothing but it is row area. Right. Into the height of the liquid. This is H is the cross sectional area. So this area into height is nothing but the volume. So volume is nothing but area into H gravitational acceleration is C. This is P atmospheric into the area of the column. This A and this A will get cancelled and hence the relation that we get here is P atmospheric is equals to H row G or row G H. So you see the pressure for any liquid is directly proportional to its height. Okay. That's why as you go deeper into the sea. Right. As you go deeper into the sea. The pressure increases because the liquid column the height of the liquid column is increasing right above you. Suppose this point is this point is a we have here. This point is B we have here and this point is C we have here. Height is increasing right. This is the height of A. This is the height of B and this is the height of C we have here. Above the point C. So we can say the pressure at C because H is more is greater than the pressure at B is greater than the pressure at A. But and the same height the pressure would be same because H is same over there. Yes guys respond. P is equal to H row G. Yeah. Now you see two things here. Yeah. So we have a tube here. This is mercury present in this. So when you place a tube like this with open and down. So tell me here at this point understand it carefully at this point. What is the pressure at this level? At most pressure. Okay. So here we have P ATM. Here also we have atmospheric pressure here also we have atmospheric pressure. And since you've placed the tube like this. So this column is also filled up with HD right. So whatever height we have here. Right. That is the height of the liquid and there's no gas present here right this empty. Either it is completely filled with the HD or it is empty. There is no gas present over here. Okay. So at this point the pressure because of this height of mercury is what we can say pressure at A is equals to we can write row of HD G into H isn't it. Yes. Correct. Can we say that this relation is right. Okay. And this pressure we know P is what P atmospheric. Yeah. This is equals to row HD G into H. Right. Density of mercury if you know you can find out the pressure here the height here that is row HD into G. And this height is observed to be 76 centimeter. Hence we say 76 centimeter of HD is equals to one atmospheric pressure. Correct. This is the concept of barometer we have with the help of the height of the HD liquid we can find out the pressure. Now I am taking a different condition here. The condition is you know we have a liquid present that is HD only. And similarly one tube is inverted with open and down inverted like this but this tube is filled with a gas over here. And here we have mercury column and this is suppose any gas. So at this point again the pressure is P ATM. But at this point if I ask you what is the pressure at this point B. Pressure at B is equals to the pressure of HD column because we have HD column here plus the pressure of the gas. This is the pressure we have isn't it right. So this is the case of if you find out the pressure with respect to liquid HD over here you won't get the correct pressure here. Because of this gas also we have some more pressure. This is the case of faulty barometer. You won't get the correct measurement of pressure here because of HD right. Hence it is a faulty barometer case we have. However you need to find out you need to find out the formula of P is equals to H ROG only in this question. Okay. So you have to keep this in mind that the pressure exerted by pressure exerted by 76 centimeter of HG is equals to one atmospheric. Okay. So by with 76 centimeter of HG. Suppose for HG I'll write down two set of data here. Just try to understand this one for mercury and for any other liquid any other liquid right whose density is row is H I am assuming and G is the constant here. So for HG we'll have you know a certain pressure low is the density of HG I'm assuming for this also it is suppose H I'll write down row one row two then it would be better or height of HG right. So if the two liquid liquid this one on HG if the pressure exerted by HG and the pressure exerted by the given liquid is equal right then we must have a relation between the height of HG and the height of this liquid. Okay. So what we can write here pressure of HG is equals to row of HG into height of HG is equals to row of this liquid into the height of this. Correct. Right. This relation also holds true for this you can find out the height of with the help of the density of the liquid. You can find out what height of a given liquid gives pressure equals to the atmospheric pressure understood based on this you see this question there are different different kinds of question possible. Suppose we have a tube with open and up right we have open and up and here we have liquid column present that is 15 centimeter of HG we have here suppose and here we have some pressure present this gas will exert some pressure P of gas and here since the end is open so it is P of atmospheric because of this liquid column also the gas the liquid column exerts pressure also in downward direction P of HG in downward direction. So if you balance the pressure here then what we can write pressure of gas is equals to the atmospheric pressure plus the pressure of the HG column can we say that way to find out the pressure of gas did you understand this is it correct total pressure upward direction equals to the total pressure in downward direction that is it. So this is equals to atmospheric pressure we can write because it is given in centimeter so PATM is 76 centimeter plus P HG is 15 centimeter so pressure in terms of height we are calculating the answer is pressure of gas is 90 centimeter 91 centimeter of HG yes did you understand this tell me guys quickly clear now you see this question exactly the same thing but we'll invert the tube and I want you to do this the tube is inverted with open and right and here we have a mercury column trapped this is a mercury column trapped here and this is 10 centimeter of height the mercury column here we have gas find out the pressure of gas equals to what this is the gas present here tell me the answer is it 61 okay so just we need to balance the pressure here so pressure of gas in downward direction right pressure of this mercury column in downward direction and we have atmospheric pressure in upward direction PATM so here the relation would be what pressure of gas plus pressure of HG is equals to the pressure of atmospheric right so pressure of gas equals to atmospheric pressure minus the pressure of HG column is 10 so it is 66 centimeter of HG yes what is wrong Pradyum okay understood all of you guys okay sometimes what happens the third type of question is sometimes what happens here they won't give you mercury over here but they will give you the different liquid with their density so you see this we have a mercury column and here we have some liquid present okay here we have the pressure of gas some liquid present here and from the top we have atmospheric pressure PATM for this liquid it is not mercury and the density of this liquid is given rho is equals to 6.8 gram per ml okay H height is given 30 centimeter right obviously when rho is this it is not density because the density of mercury is 13.6 gram per ml okay could you find out the pressure of gas anybody got the answer yeah that's right Pradyum see the best way to do this kind of question is the given height of this liquid you find out if this liquid is if this liquid is mercury then what should be the height of the mercury column so that the both liquid mercury and this given one exerts the same pressure so basically what point I'm trying to make that this height you convert into the height of mercury column correspondingly right so if I take the pressure if you equate pressure of HG column equals to the pressure of this liquid column then what we can write here formula I've given you already rho of HG into this height we do not know why what height of this mercury column equals to exerts the same pressure which this liquid exerts with 30 centimeter height into G is equals to the pressure of liquid is again rho into G into the height here so what is the height of the mercury column here H of HG is equals to density of you know liquid is 6.8 given H is 30 centimeter divided by 13.6 is the density of mercury this we are getting 15 centimeter 15 centimeter height of the mercury means this 30 centimeter height if you replace by 15 centimeter of mercury you will have the same pressure here correct now you see again we'll balance the pressure so PATM is equals to the pressure because of the mercury column is equals to the pressure of the gas right 76 plus 15 is equals to the pressure of the gas hence the pressure of the gas is 91 centimeter of HG no doubt yes tell me guys no doubt in this try this question this is the pressure of gas atmospheric pressure and we have multiple liquids with different different densities suppose the height is L1 L2 and L3 ok density for the top one is given rho 1 3.4 gram per ml rho 2 is given 6.8 and rho 3 is 27.2 ok height is also given 80 centimeter 20 centimeter and this one is 50 centimeter find out the pressure of the gas once again out of yeah that's right 206 just few more minutes we'll finishing it off ok we're done with this chapter one two more time I'll show you then finish we need to for all these liquid we need to find out the corresponding height of mercury column correct so for the first set of data the height of mercury column H of HG equals to density into its height that is 80 divided by the density of mercury which is 13.6 20 centimeter for the second set of data H of HG should be 6.8 into 20 divided by 13.6 it is 10 centimeter of mercury column H of HG for the third set of data is 27.2 into 50 divided by 13.6 that is 100 centimeter right now we'll apply the we'll do the pressure balance here so we can say the pressure of gas equals to the atmospheric pressure plus the pressure of the mercury column total pressure atmospheric pressure is 76 100 plus 20 plus 10 right so it is 206 centimeter of it this one you try we have a container and open tube present like this over the container two gases we have here one gas present here and another one is here two gases we have right G1 and G2 we need to find out the pressure of the gas one and the pressure of the gas two this height of the mercury column it is 10 centimeter HG it is so you see here what we need to do here the pressure is the atmospheric pressure PATM and at the same level the pressure of this gas in downward direction it's same as the pressure over here atmospheric pressure okay so what we can write here the pressure of gas two because it here it is here the same pressure of this gas we have that is the pressure of atmosphere it is P atmospheric and that is 76 centimeter of it okay so pressure of gas is 76 centimeter further we can write PG2 is there in the upward direction because the collision is in this direction so PG2 is equals to PG1 the gas pressure in this direction plus the height of the mercury column which is 10 PG1 is equals to 76 minus 10 is equals to 66 centimeter of HG sometimes they also gives this tube inclined tube they may also give you one last question in this type suppose the tube is given like this it is inclined at an angle and the gas trapped here liquid trapped here this height is H it is given this height is H and H is equals to we have 10 centimeter we have given the pressure of gas is this P of gas atmospheric pressure is this and this is mercury this is HG find out the pressure of the gas inclined column see what we write here the pressure of atmospheric plus the pressure of HG column is equals to the pressure of the gas right so this height if you see the height that it comes down which angle I haven't given I'm sorry this angle is given here it is 60 degree okay so the height that comes down here this height is H sin 60 okay so atmospheric pressure HG pressure of gas atmospheric pressure is 76 centimeter this is 10 sin 60 is equals to the pressure of the gas and hence the answer is 76 plus 76 plus 5 root 3 centimeter of HG this is the answer clear understood this without five more minutes I'll take here one last concept here that is the concept of payload okay max to max five minutes okay very small thing what is payload write down payload is the extra weight food is the extra weight that we can put that we can on balloon on balloon so that so that it can effectively fly they last this question is calculate the payload directly okay so if you see the balloon if you have right so here we have because of the air we have force of Beyoncé in a poor direction weight of balloon in downward direction WB is the weight of balloon right plus the balloon is filled so we have weight of gas also here whatever gas is present plus the payload that we put the weight of that payload WB is the weight of balloon weight of gas with which the balloon is filled this is the payload and buoyancy force in upward direction right so because of this the weight of air displays okay air displays equals to the weight of balloon plus the weight of gas plus the weight of the payload that we have equal amount of weight of air will get displaced right so weight of balloon will be given in the question right always WB will be given we need to find out this payload right this is the objective we need to find out this weight of balloon will be given in the question directly or indirectly it will be given right what is the weight of the gas for weight of the gas we can apply PV is equals to NRT for the gas whatever gas we have so from here we can find out the weight of the gas that would be PV into M by RT molar mass of the gas with this formula we can find out the weight of the gas okay what is the weight of air displays displays equals to the density of air into the volume of the volume of air right so with this formula we can find out the weight of air displays you have weight of air you have weight of balloon weight of gas also you can find out with this and then we can find out the payload one last thing here that the weight of the volume of air equals to the volume of balloon so this is it for the chapter okay we have discussed everything given in the chapter so we are going to start the next chapter okay so tell me what all chapter they have done in the 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