 Good morning myself Sachin Deshmukh working as an assistant professor in the Department of Civil Engineering of Wild Change Stop Technology, Soolapur. Today we are going to learn about Pascal's law and we are going to see the derivation of Pascal's law. At the end of this topic, you are able to calculate the intensity of pressure at any point in a liquid when it is at rest. This law was first initiated by a French scientist, bless Pascal. This law states that a fluid at rest in a closed container, a pressure change in one part is transmitted without loss to every portion of the liquid and to the wall of the container. Pascal also discovered that the pressure at a point in a fluid at rest is the same in all directions, the pressure would be the same in all planes passing through that specific point. Now we will see the derivation. For this particular derivation, we have taken a very small wave shape molecule so that we can find three axes on this x, y and z axis. See here, let us consider a very small wave shape element, B, A, C of a liquid as shown in the figure. This is a B, A and C, this is a wave shape so that we can find this x axis, we can go for this y axis, this is the y axis and we can go for this z axis. This is angle substituted here and particularly for this particular wave shape, we can resolve the forces horizontally as well as vertically. Let small px is the intensity of horizontal pressure on the element of the liquid, p y, small p y is the intensity of vertical pressure on the element of the liquid and p z, small p z is the intensity of pressure on the diagonal side, diagonal of the right angle triangle element and phi is the angle of the element of the liquid. Similarly, capital Px is the total pressure on the vertical side B of the liquid, capital P y is the total pressure on the vertical side A C of the liquid and P z, capital P z is the total pressure on the vertical side of B C of the liquid. Now we can calculate capital Px is nothing but small px into B A, the side similarly P y is the small p y into A C, P z is small p z into B C. As the element of the liquid is at rest therefore, sum of horizontal and vertical components of liquid must be equal to 0 that is we have to resolve the forces horizontally as well as vertically. First of all we will resolve the forces horizontally, P z sin theta is equal to P x, see here P z sin theta this is a P z, P z sin theta this one P x. Now P z is P z B C sin theta is equal to P x into B A P z into B A as P z is equal to P z into B A put this value B C sin theta is equal to B A equate this terms are going to cancel you will get P z is equal to P x, I repeat once again resolving forces horizontally P z sin theta is equal to P x, P z B C sin theta because P z is P z into B C sin theta as it is and P x is P x into B A put this value, put the sin theta is equal to opposite to hypotenuse put this values canceling you will get P z is equal to P x similarly, similarly resolving the forces vertically similarly resolving this forces vertically this P z cos theta is equal to P y minus W where W is the weight of the liquid element. Since the element is very small neglecting its weight we get P z cos theta is equal to P y or P z B C cos theta is equal to P y into A C but B C cos theta is equal to A C that is from the geometry of the figure canceling the sides we will get P z is equal to P y. So, from equation number 4 this is the equation number 4 that is P z is equal to P x and P z is equal to P y we can find P x is equal to P y is equal to P z. Hence at any point and field at rest the intensity of pressure is exerted equally in all directions which is called as Pascal's law. These are some of the applications of Pascal's law we can apply very small force over here this force travelled over here and this particular car is lifted up we often see this scenario in a automobile station this is another application that is hydraulic pump hydraulic jack this is also one of the often seen application of Pascal's law when we apply force with this handle and this small jack will lift the vehicle similarly for hydraulic brake also the same application the same theory is used of Pascal's law now pause the video and answer this questions the Pascal's law states that the liquid at rest applies pressure at a point is dash in all directions and second one the Pascal law is applicable to the liquid which is there are options given choose the correct option now see the answer the Pascal's law states that the liquid at rest applies pressure at a point is same in all direction remaining three options are not correct and in the second one the Pascal law is applicable to the liquid which is incompressible the other options are not correct that is compressible solid in phase and super compressible compressive so these three are not correct incompressible is the correct option now depending on this Pascal law we will solve one example this is the example of the hydraulic press see the diameter of the ramp and a plunger of an hydraulic press are 150 millimeter and 20 millimeter respectively find the weight lifted by the hydraulic press when the force applied at the plunger is 350 Newton's this is a hydraulic press this is the ramp over this ramp a weight is placed this is a plunger through this plunger a force is applied F1 is the force applied this force carried through this particular portion and then this lifted up this w here that this is the weight of this particular we can say box it is lifted up this F2 that we are going to or we can say w that we are going to calculate whenever the problem is given to you write down the data given first diameter of the plunger d small d its area calculate the area diameter of the ramp then calculate the area of the ramp as well as area of the plunger it is 0.0176 meter square that is the area of the ramp area of the plunger find out its areas then next step that is to calculate intensity of pressure due to plunger now pressure pressure we can calculate force upon area here force applied is of 350 Newton's here force applied is 350 Newton's see here now here 350 Newton's divided by area area of the plunger area of the plunger on this plunger the force is applied so we will get 111 4649.6 Newton per meter square always write the units units are very important always write the units since this is the intensity of pressure this is going to transfer this is going to transfer through this particular section now see here since the intensity of pressure will be equally transmitted as per the Pascal law the intensity of pressure at the ramp is also same but we can calculate the intensity of pressure as weight upon area weight is unknown put both the values and calculate weight weight is 19.61 kilo Newton similar problems you can solve from these reference books