 Alright friends, so here is a very interesting question on fluid statics, so let us solve this one. So you can see the question is in front of your screen, so there is a YouTube section which is which is aligned like the way it is shown over here and inside it there is liquid alright total length of the liquid is 4L alright and yes the both sides the arm lengths are 3L the density is given as rho alright and it is given that the YouTube is going with acceleration A, so of course looking at the diagram you can find out that acceleration has to be in this direction, this acceleration is G fine okay, so now what happens is that the entire YouTube is abruptly stopped okay or you can say acceleration has become 0 maybe it is still moving with constant velocity but acceleration is 0 fine and they are asking us to find out what is the gauge pressure at the middle section of the bottom one fine, now over here we need to understand one thing is that when it is abruptly stopped the liquid will not remain like this fine, liquid will of course start accelerating along the YouTube right, so let us say it starts accelerating like this, so this side if let us say acceleration is A1 over this side also, acceleration is A1 only because liquid is incompressible it cannot be compressed that is what we are assuming here fine, so immediately after the acceleration it stops YouTube the liquid inside will accelerate fine, so there are two cases right, the case number one is when it is accelerating case number two is suddenly when it stops, so let us take these two points over here point number one and point number two okay, outside pressure is atmospheric, so I can write over here fine, now we know that how pressure varies with the height isn't it, so let us assume that this height is h1, this height is h2, so we can write down over here that P1 is equal to atmospheric pressure plus rho g h1 right and P2 will be equal to the atmospheric pressure plus rho g h2, this is for case when it is horizontally accelerating and the liquid is not moving inside the YouTube, liquid is maintaining this shape fine, so we can say that P2 minus P1 if we subtract we will get rho g h2 minus h1 fine, this is the one equation we can have and the second equation we know that if it is horizontally accelerating then pressure will vary horizontally as well fine and we also know according to Pascal's law that pressure in all directions are equal, so whether I talk about point number one and two vertically or horizontally I will be getting the same value, so let us take a cross section of this horizontal length like this okay, so from here this part will experience a force of let us say P1 into A of cross section A okay, from here it will experience P2 into A right, so simply we will be using the Newton's second law over here right, it is accelerating with acceleration of g, so I can say that P2A minus P1A this is the net force in the direction of acceleration, this is equal to mass which is rho A into length of the horizontal section, so rho A L is the mass, mass times acceleration which is g right, so I can have P2 minus P1 is equal to rho g into L alright, so I will name the bottom one as equation number two. Now, looking at equation number one and two P2 minus P1 left hand side is same, so I can say that H2 minus H1 is equal to L right, so this is the, so I am trying to find out basically what are the heights H1 and H2 because they will be needed for the second case right, based on the height H2 and H1 only this acceleration will be created okay and then total length is given to us total length is 4L, so total length should be H1 plus H2 plus the bottom length L this should be equal to 4L this was given to us, so H1 plus H2 is given to us as 3L right, so solving this one and that equation if you just add it up you will get H2 to be equal to 2L and H1 you will get it as L okay, so we have finally found out that what are the values of H1 and H2 okay, so this completes the case number one when the YouTube was accelerating with acceleration due to gravity now the acceleration has become 0 right, now what will happen, now if acceleration of the YouTube has become 0 the liquid will start moving like that fine, now you have to draw the free body diagram of three sections, two vertical sections and one horizontal section fine, so let us do that quickly, so this is the left hand side vertical section, this is the right hand side vertical section okay, so over here the pressure is atmospheric pressure, this is the atmospheric pressure from here again pressure in all directions are equal, so whether we talk about horizontal or vertical the pressure remains same, so P1A this will be P2A fine, this is PA force due to the atmospheric pressure the force will be down like this okay, now acceleration is like that, so this is let us say A1 and over here acceleration is like this A2 fine, so let us write down the equation of the vertical sections, so we can basically you know what I am doing here, I am trying to find out what is the acceleration of the liquid inside right, so that is why I am doing all of this okay, so for the left hand side we will have P1A minus atmospheric pressure force, this will be equal to the mass which is rho AL times A1 okay, on the right hand side we will have atmospheric pressure okay, wait sorry about it, so we have to account for the gravitational force also, so yeah it is quite common to miss some details, so over here there will be an mg force, m is what rho AL into g right over here there will be an atmospheric sorry the gravitational force as rho A, now length is 2L so 2L into g fine, so now let us write down the equations, so we have we have atmospheric sorry the P1A minus PA into A minus rho ALg this is equal to mass times acceleration rho AL into A right, so I will just simplify it further capital A will get cancelled away, so we will get P1 minus P0 which is PA actually, P1 minus PA is equal to rho LG plus A okay, this is our first equation or first and second are taken, so let us say this is our third equation, now for the right hand side vertical section you will have what, you will have atmospheric pressure force plus gravitational force which is rho A into 2L into g okay minus P2 into A this is equal to the mass times acceleration, now mass is rho A into 2L into A right, so again you can cancel out area of cross section over here, so that is gone now, so we have atmospheric pressure minus P2 this will be equal to rho into 2L into A minus g okay, so this you can say is my equation number 4 alright, now let us say what about the horizontal this thing, now when we draw the horizontal one our point of interest is this middle point what is the pressure over here right, so let us try to draw the free diagram for the middle section and see what we can do okay, so I have taken you can say this one is point number 3 middle 1, I have taken from 3 to 2 alright, so from 2 of course there will be a force due to the pressure P2A right, so this will be equal to P2 into A right over here pressure is let us say P into A okay, so I can write down over here equation now the thing is that it is accelerating also like that, so this is A1 right this one was A1 alright, so I can say that P2A minus PA should be equal to mass acceleration, now mass is rho A it is half of the length, so this is rho A L by 2 into A1 okay, so we will get over here P2 to be equal to P plus rho A L by 2 into A1 okay, now from equation number 4 P2 is also equal to the atmospheric pressure minus rho into 2L into A1 minus G okay, so using this equation the last one which we have found out using this equation you can get the value of the pressure P okay, now when we try to find out the pressure P it will have an expression of A1 also over here right, so we need to determine that as well alright, so in order to find out the value of A1 what we can do is that we can draw the entire free buoy diagram from 1 to 2 okay, so let us do that I will just create some space over here fine alright, so here is the entire 1 to 2 cross section right, so this is point number 1 this is point number 2, so you can have over here P1A force and over there you have P2A force fine and we have assumed that it was accelerating with A1 fine, so we can write down Newton's second law equation over here right, it will be P2A minus P1A okay, this is equal to mass time acceleration that is rho L into A A1 fine, so you will have P2 minus P1 is equal to rho A1 into L fine, this you can say is equation number 5 and from equation number 3 and 4 if you just add them up you will get P1 minus P2 to be equal to rho L G plus A1 plus rho into 2L A1 minus G okay, so now you have to solve equation number 5 and 6 you will get the value of A1 and if you solve it properly you will get acceleration A1 to be equal to G by 4 alright and once you know the acceleration of the fluid inside the manometer is G by 4, you have to pluck it over here alright, if you pluck it over there you will get the pressure P to be equal to the atmospheric pressure plus 1.375 rho GL fine and since in the question the gauge pressure is asked, so the answer is whatever total pressure you get you have to subtract P A from it, so answer is option number D alright, so this is the way you have to solve this question, now when you look back you feel that the question is very twisted and very lengthy as such, you know it is perfectly fine, yes the question is little lengthy but if you carefully analyze the way we have solved it, we have used nothing other than Newton's second law only to solve the entire question, so you need to understand that knowing a basics and applying it is enough to solve the most complicated types of questions also, so always focus on the basics try to solve the question from the you know basics or the first principle and yes make sure you practice well, bye for now.