 three on fluid statics. In the first two sessions we have seen in detail that pressure is a scalar quantity then pressure is important parameter in a for fluids for solids we use the equilibrium condition for equating two forces but in fluids we go for the pressure. Now in the third session let us see what are the outcomes that I expect you will see that at the end of the session learner will be able to show that a liquid may exert thrust greater than its weight and define Pascal's law and enumerate its application. Now let us see first we will discuss about the Pascal's law and then we will see about the thrust exerted by the body is greater than the weight of the fluid that is contained. Now very important thing is pressure exerted. Suppose I take this particular system in which the left limb has small diameter and right limb has a large diameter as already we have studied that if the liquid is same if the liquid is same in this container then the pressure at this point and pressure at this point is same. Now pressure at a and pressure at b is same so what happens if I apply here some force some force so f upon a a is small so because f upon a 1 this area area of this section is a 1 area of this section is say a 2. So f upon a 1 when I apply it then what happens the pressure develop is large because a is small this pressure is applied over here and this pressure multiplied by so pressure at b is equal to pressure at a which is equal to f upon a 1 then force at b is equal to pressure at b into area at b now pressure at b is f upon a 1 into a 2 now I know that a 2 is greater than a 1 therefore f b is greater than f a so what happens by small force I can lift here a big object that is put over here now for this we are use a very important property of a fluid that is pressure exerted or pressure developed is immediately communicated to entire liquid without any loss and that law is known as Pascal's law. Now first of all before going to see various applications of Pascal's law one more thing I want to make it clear that whenever we want to evaluate exactly how the pressure is calculated as we move deep inside the liquid let us see suppose I take this as a container in which say at this suppose there is a atmospheric pressure I denote by p0 then I take some element here whose width is say dy and at a height of y so what is my concept now from the surface where the atmospheric pressure is acting I am coming down by distance of y meters and I take a element whose say thickness is dy and suppose its area is say a so what is the volume of this element volume of this element is a into dy volume of this element is a into dy so if I multiply this volume by rho density of the particular this fluid and g so this becomes my mass or this is the weight this is the weight now what happens the pressure at this point is say p and pressure at this point is say p plus dp and this is the w coming downward now see what is happening we are now going to derive a simple expression for finding out the static pressure from the surface of a liquid now once you understand this that as we move down the liquid pressure increases a natural corollary that as we move up pressure will decrease and that is evident say when we go up in the atmosphere pressure goes on decreasing because atmosphere becomes rarer and rarer now if I take this element under consideration and show that it is p here it is p plus dp and it is w if I if I write its say Newton's law or free body diagram and the equations so it will be p plus dp p plus dp into area so that is a force minus p into area is equal to w that is the weight which is equal to dy which is equal to dy then rho g dy rho g and a because I know the total weight of this now if I do this I will get dp into a is equal to here I will get rho g dy into a so I will get cancel and I will get dp is equal to rho g dy now if I integrate this integration of dp is equal to integration of rho g dy from p naught to p if I want to find out the pressure at any point arbitrary point p for p naught to p and if I integrate this from say 0 to y 0 to y or suppose 0 to h rather if I call it as h then what will happen this will become p minus p naught is equal to rho g h so finally what I get this is equal to p naught plus rho g h so this is my derivation for static pressure now one important thing about this pressure that is the static pressure that we have is evident from another property of fluid which you must know at this particular moment so that I am taking this property here now see what is written here fluids in equilibrium exert pressure at right angles to the surface of contact now actually you have answered this question when a balloon was there it was filled with water when the piston is pressed water oozes out and it comes necessarily in the radial direction it indicates that it is in the normal direction now we will prove this with the help of a mechanics now suppose this is a container in the container there is liquid and assume that the p to a is the force exerted direction is p a exerted now by newton's third law there is a immediate reaction along the same direction so the force exerted will be along this direction if I assume the force exerted by the liquid in this direction because these are equal and opposite forces if this is the case I will have a net resultant force in this direction which will have a component along this direction and along this direction so the x component will move the liquid in the horizontal direction when the x component moves the liquid in the horizontal direction there is no scope for the liquid to move because it is contained here and we are restricted this motion so it is clear that whenever we have whenever we have the fluid in equilibrium it always exerts pressure at right angles and that is evident suppose I have a container like this suppose I have a container like this in which I feel the liquid in which I feel the liquid so the pressure exerted is shown essentially along this direction along this direction it is along this direction so pressure is always normal to the surface it is always normal to the surface so this is a very important property now next important point is a liquid may exert thrust greater than its weight now this is a very important thing in a paradoxical concept and students find it difficult when they want to give an answer now you can see here I have given you the container that container is such that its height is h and the area is restricted here at the bottom at the top then what happens if I see the total weight of the liquid total weight of the liquid is weight of the liquid present in this part plus weight of the liquid present in this part okay but the height is h height is h so whether this part is present or not will not make any difference on the pressure exerted here so the pressure is going to be rho g into h at this point and multiply by this the area of the base will give the force okay now if this is the force you will find that the liquid weight is going to be less than the volume of the cylinder if its height would have been capital H so this particular reduction in weight which is present gives you the result that weight of the liquid is less than the thrust exerted by the fluid on the bottom surface now this is important in designing of certain fluid static equipments and you must keep this particular in mind now with this you have learnt some important fluid properties and if you want to do further reading once again I refer fluid mechanics by white you can go through or you can refer Bunsell for certain simple numerical examples and with this say will session session thank you