 So, let us consider a second example. Let us say we have concentric cylinders and the speciality of this system is that maybe one of the cylinders is having a relative motion with respect to the other. So, let us take an example where the outer cylinder is stationary, inner one is stationary and the inner one is rotating with a particular angular speed omega. There is a fluid that is present in the gap between the 2 cylinders. This kind of visual example that we have seen in our previous lecture. This type of example is important in many ways. I will give you 2 examples for application of this type of situation. One is if you consider an industrial application, in industry there are shafts, cylindrical shafts which transmit power from say one point to the other. So, think about the inner cylinder like a shaft. So, that shaft is rotating and it is rotating but it has to be housed in a certain position. At the same time, if you constrain it with a direct metal to metal contact that will give rise to a lot of wear and tear. So, there is a lubricating material which is typically like an oil which separates that from outer housing which sometimes is called as a bearing and the whole idea is to create a lubricating layer in between these 2 which prevents the wear and tear because of the metal to metal contact. So, the shaft bearing type of arrangement in industry is very common and that is where you will find lot of application of this kind. Even if you are looking from a more fundamental consideration of fluid mechanics, there is often a necessity to measure the viscosity of fluids and this kind of arrangement may be utilized to measure viscosity of fluids. That is the case where it is known as a rotating type viscometer. Viscometer for measuring the viscosity and rotating type because of its particular nature of motion that you can easily appreciate. We will try to see that what is the physical situation that is going behind this type of example. When the inner cylinder starts rotating, it will try to move the fluid with it because of no slip boundary condition. The fluid immediately in contact with that will be rotating with the or will be moving with the same linear speed at different locations. As you go radially outwards, let us say the inner cylinder has a radius of r 1 and the outer has a radius of r 2. So, as you go from r equal to r 1 to r equal to r 2, what you will find? You will find that the velocity in the fluid goes down and the velocity is 0 at the outer radius r equal to r 2. That is also by no slip boundary condition. Now, because of the presence of the fluid, the cylinder which is rotating, it is not rotating in an unhindered manner. It is being subjected to some resistance. It has to overcome that resistance and maintain its motion. So, it requires an external power to be imposed or so to say a torque to be there, which is continuously rotating it overcoming the viscous resistance. Let us say that we are interested to find out what is that torque or may be power necessary to make it rotate with a uniform angular speed. That is the objective of analyzing this. Therefore, if we apply that particular torque and if we see that it is rotating with a uniform angular speed that may be measured by something like a tachometer, then it is possible to relate these two in terms of the viscosity of the fluid. So, everything else being measured from that expression, we should be able to evaluate what is the viscosity of the fluid. That is the basic principle by which one may measure or evaluate what is the viscosity of the fluid that is there in between. Typically, this gap is very narrow and we will see what is the consequence of that narrowness. Now, let us say that we are interested about a section of the fluid. Let us say at a radius at some intermediate radius say r, which is a local variable small r. Let us say that the length of the cylinders or both the cylinders is l, which is perpendicular to the plane of the board and mu is the viscosity of the fluid which is occupying the annular space. So, when we consider at a location r, we have an imaginary surface of fluid which is having a surface area of what? 2 pi r into l. That surface of fluid is a surface on which there is relative resistance or there is a relative motion between the fluid layers. One is towards the inner and another is towards the outer. Whatever is towards the inner tends to move faster, whatever is towards the outer tends to move slower. So, that is the location where there is a shear stress that is present, which is related to the rate of deformation. So, if we want to write what is the shear stress at the radial location r or we should maybe use a superscript because it is not really tau with r as a subscript meaning we have preserved for something else. So, if we write this, then what would be its corresponding expression in terms of say Newton's law of viscosity? There is a mu, there is some sort of du dy type of term. So, what is let us call it some du, du say now we are using a coordinate of r and if we had used a coordinate of r and if we had used a coordinate of y, the only difference would have been that y is from the wall from the solid from the 0 velocity wall towards the inside towards the inner one. So, that y is just oppositely directed to r. So, whatever is du dy is just adjusted with a minus du dr. There is no other difference because why direction of the inner direction is preserved for the direction which is from this 0 velocity to the fluid and this is the r direction is just opposite to that. That is why this minus sign is there to adjust it and you may think also in a different way as you are increasing with radius, you are having a reduced velocity. So, this is negative. If you want to make it positive, you want to adjust it with a negative sign. That is just a matter of sign convention but we have to be consistent with the sign convention. Whatever we have followed till now, we will preserve that. So, that is the shear stress. If it is a Newtonian fluid then what is the shear force which acts on this elemental surface? Let us say df is the shear force on elemental surface. We have already identified what is the elemental surface that is the surface of the imaginary fluid with the dotted line as its radial envelope. So, df on that one is minus mu du dr. We may just call it f. There is no necessity to call it df because it is not like an elementary small volume that area that we are talking about. So, minus mu du dr into 2 pi r l. So, this force is a tangential force. So, this force is like typically you will have this type of force which is tangential to these elements. So, this force will have a moment with respect to the axis of the cylinder. So, what is the moment of f with respect to axis that is f into r? We are just writing it in a scalar form not bothering about the vector nature because the moment vector is perpendicular to the plane of the board that we can understand very easily. So, this is something now you have to understand physically what is happening. There is a particular power that is imposed say a motor is driving this. That means there is a torque that is being input to the system and the same torque is transmitted across different fluid layers. Otherwise it will not be able to rotate with a uniform velocity. So, what it means is that if you call this as say m, then m is something which is a sort of an input and it is balanced with the resistance moment that takes place at various sections so that you have a particular number a particular value associated with that and that is dictated by the input power of the motor. So, you have m is equal to minus mu d u d r into 2 pi r l into r. So, now you can separate the variables in the two sides. So, you can write d u equal to minus m by mu l 2 pi into r to the power minus 2 d r. I am very bad in algebra so whenever I am mistaken please correct it. Now, when you integrate this you can get a sort of variation in u. Remember one very important thing this u is the velocity in the fluid. So, it has a variation from the inner to the outer. For the inner cylinder within the cylinder there is no variation in velocity because it is a rigid body. So, of course linear velocity is varying but angular velocity is the same. Now, the outer cylinder also is stationary but in between there is a difference in linear velocity because the fluid is deforming it is not a rigid body. So, at the inner radius that is at r 1 what is the velocity sorry r 1 on the right side we should write. So, r 1 equal to the velocity at this is omega r 1 right and at r 2 this is 0. So, very quickly we can write that minus omega r 1 is equal to minus m by 2 pi mu l this is 0. That minus sign will get absorbed. You can simplify this and write as say m equal to 2 pi l omega r 1 then another if you just simplify this another r 1 r 2 divided by r 2 minus r 1. If there is any mistake please let me know. Now, so you can see that there is a difference between r 1 and r 2 and here you are actually having r 1 square r 2 in this expression but if you neglect the variation of or if you neglect the velocity profile variation from the inner to the outer and assume that the gap is very thin. So, that it is a linear profile then what would be the difference in expression that you get. So, if it is a linear velocity profile that is taken that is the velocity is varying from omega into r 1 from inside to 0 outside in a linear manner. If the gap is thin then that is valid. So, for in a thin gap limit in the thin gap limit you have the tau equal to mu into omega into r 1 divided by r 2 minus r 1 omega r by h d v d y if it is a linear velocity profile it is just the ratio of the change in the 2. So, this is if this is the tau then what is the moment. So, m is equal to mu omega r 1 by r 2 minus r 1 that should be multiplied with r 2 pi. Now, because the gap is thin you can write 2 pi r 1 l 2 pi r 2 l or if you want to be little bit more accurate may be r 1 plus r 2 by 2 l or whatever it will not make a lot of difference. Let us write may be 2 pi r 1 l into into r 1. So, you can clearly see that as the difference between r 1 and r 2 tends to 0 these 2 expressions lead to almost the same thing. So, if the gap is narrow then the second approximation will give you a very quick estimation of what is the situation. And from this or the more involved expression you can clearly see that if you now know what is the power input to the shaft the power input is this m times the omega. So, if that is known that means m is known the dimensions will always be known. So, r 1 r 2 that means r 1 r 2 minus r 1 l that will be known omega can be measured with a tachometer. So, that can give you what is mu from this expression. So, if you are having a careful experiment where you are having the proper estimate of the power input as well as what is the angular velocity at which this inner cylinder is rotating it will give you some good estimation of what is the viscosity of the fluid that is there inside provided it is Newtonian. And that is how you may estimate the viscosity of an unknown fluid that is a fluid for which you do not know the viscosity. Let us consider a third example. We consider that there are 2 plates for example 2 circular plates. So, these are the sectional views if you draw the other view this will be a circle. So, if you draw the other view of say the top plate it will be something of a circular nature. The problem is whenever I want to draw a circle it becomes an ellipse whenever I want to draw an ellipse it becomes a circle. So, assume this a circle although looks more like an ellipse or may be not even a ellipse. So, this is the other view of the plate there is a fluid which is there in between our objective again is to see that what is the torque or power required if we want to rotate this one with a given with a particular angular speed. The bottom one is stationary situation is quite similar to the previous one. So, we should be able to work it out quite quickly. We have assumed that this gap is narrow the viscosity of the fluid is mu. Let us say radius of the plate is the top plate is r. Because the gap is narrow obviously it is expected that we may approximate it with a linear velocity profile from the bottom to the top. But a key factor here is that that linear velocity profile is now radially changing. So, if you consider say a particular radial section like this here you have 0 velocity here and what is the velocity that you will have here omega into local r. So, let us say small r is the local r. So, omega into local r will be the velocity therefore the velocity gradient at section r will be omega r by h and this is because of linear velocity profile assumption from the bottom to the top. So, if you take different radial sections this will be different. So, you cannot if this was a constant we could have easily calculated the shear stress by multiplying whatever constant it was with the total surface area of the plate that is being exposed to the fluid. But now this being a variable we must take it as summations of constants over small small elements and that is how or that is why we have to choose small elements and integrate over that elements. So, we take a small element at a radius r thickness dr. So, whenever we are solving any problems these are very common situations. Many times because of systematically practicing problems we are habituated in taking elements in certain cases doing integration and so on. But many times we forget why we are doing it and it is very important to keep in mind that why we should do it. So, here since it is continuously varying we are interested to obtain estimation for the shear stress or the shear force which is our objective and that shear force is locally varying because the shear stress is locally varying we should take a small element at least over which it is a constant. So, within this dr it does not vary significantly therefore it may be treated like a constant over dr. So, if we consider this area we can multiply the local shear stress with that area to get the local shear force. So, what is that local shear force? So, let us say df we call as local shear force on this dA what is that? So, first you write the expression for shear stress mu into omega r by h that is tau times the area dA. So, what is that area? 2 pi r dr. So, with this as the shear elemental shear force this elemental shear force will have a moment with respect to the axis. So, what is that elemental moment? This into r. So, that is df into r. So, that will be 2 pi mu omega by h r cube dr. So, the total resistive moment should be the integral of this 0 to r. So, that will become pi mu omega by 2 h r to the power 4. Now, why it should be 2 pi r? No, no, no it is distributed over the entire surface. So, it is like it is when you consider a radial location it is not a point it is like entirely distributed and that distributed force has a moment with respect to the axis. So, it is like over the entire element. You can think it even more fundamentally do not just consider the full 2 pi r but consider a small angular element with between theta and theta plus d theta and between r and r plus dr. And then if you integrate that from theta equal to 0 to 2 pi that 2 pi term has automatically been taken care of. So, you should not take it take care of it doubly by considering 2 pi here also. So, this is a very simple expression but it again tells that like there can be situations of variable velocity profiles and those may be taken care of in this way. So, what we will do? We will post you some assignment or homework problems on viscosity may be very much related to these types of problems or may be slightly different and you will find that in the course website the homework problems and may be we will give you a deadline in the next class that when to submit those problems. So, we can sum up with our studies on viscosity. We will study more effectively viscosity later on in one of our related chapters that is equations of motion for viscous flows when we will learn viscosity effect more mathematically. But just now we can sum it up to see that here there is some highly viscous gel and this highly viscous gel is being stirred and you can see that when it is being stirred it tends to get broken and separated in parts. So, when it is doing that of course it is a highly viscous gel there is an important additional force that is coming into the picture which is making it to behave in that type of way and that force is nothing but a surface tension force. So, the next fluid property that we are going to learn is surface tension and what surface tension can do and what it cannot do first let us look into some images before we go on to the more mathematical description of the surface tension. So, first see this example. So, this example is like you have fluid droplet which is interacting with another fluid and see that what type of behavior is taking place and many times it looks like a magical behavior and surface tension really can create magical behavior. So, it can create instabilities in jets and droplets and this type of instability is very very common. I will tell you that qualitatively why should you have it seems you are liking this very much but I am not sure that you will be liking the mathematical details which go behind this very much. So, as a teacher my objective is to first addictive with this and then put a heavy dose of mathematics that goes behind. So, that is what we are trying to do. So, we look into second example. Some of the examples later on may appear to be even better than what you saw earlier. So, this is breaking up or kind of making of a liquid column. So, there is a liquid column it is getting broken up and the reason is that see what are the forces which are acting here one is viscosity viscous force another is surface tension force we will come across that and there are competitions between these 2. So, if the viscous resistances overcome may be surface tension driven instabilities can break it up or tear it up into nice pieces. And third example this is like a train of droplets and you see the droplets are like these are not like rigid spheres. So, they are continuously deforming and you see the way in which these droplets are moving. So, initially one big droplet and then you see that there are trains of droplets of different sizes and they are continuously interacting with each other. So, at least you can appreciate that this is beautiful but this is a very complex physical phenomenon it is not such a simple phenomenon it is not like that there is a rigid ball falling from the top towards the bottom and fluid mechanics therefore is something which is fascinating but it is not as trivial as sometimes mechanics of simple particles. So, we will see maybe one more example you see the type of pattern that is being created by a die on a surface. So, we will just play it once more to wrap this visual display up and go to see that what is the fundamental that goes behind. So, let us therefore with this motivation try to understand that what is this surface tension all about and what is its implication. So, we go to the description of surface tension again this is a very involved topic we will try to develop very elementary qualitative feel of what it is about. Let us say that you have a container in the container there is an interface between so called vapor and liquid very common if you may have water and water vapor and liquid water and water vapor and there may be an interface. Now, what we are trying to do we are trying to focus our attention on what happens to the molecules which are there at the interface. So, let us identify some molecule which is there at the interface let us say this molecule is sitting on the interface and what are the forces to which this molecule is subjected let us try to investigate that clearly you can see that when there is a molecule at the interface there are surrounding molecules and this surrounding molecules on one side there is vapor on another side there is liquid. Vapor is obviously much less dense than liquid and what you expect you expect that which side will be pulling this molecule more liquid side or vapor side the liquid side will be pulling it more. So, it is intuitively expected that this molecule will have a net resultant pool that is acting on it when there is a net resultant pool that is acting on it then there would have been a great chance that this molecule will get will get dissolved in the liquid but it does not happen like that because an interface is always formed. So, there is something that make sure that this interface is always formed what is that something that we will now understand. So, one important concept that we can appreciate is that at the interface there is a resultant force on the molecules which are there at the interface despite that resultant force the molecule still holds its presence at the interface that means it has some additional energy or effective energy by which it can by virtue of which it can sustain its position at the interface by overcoming this net interaction and that energy is known as surface energy and whenever you have a surface energy this surface energy is also manifested in form of a force because it appears that the molecules here are in a sort of a state of tension. So, when they are in a sort of state of tension because of this net pulling and pushing that particular force which is responsible for keeping it in tension is also known as surface tension. So, surface tension and surface energy are quite related and typically whenever we express surface tension we express surface tension as force per unit length what is that length the length is the perimeter length on which this force is acting. So, surface tension therefore is a force per unit length. So, in SIS unit it will be Newton per meter. So, this is not the surface tension force but this is surface tension coefficient. So, when we say surface tension we loosely say surface tension but actually it is surface tension coefficient the surface tension force of course is these times the length on which it is acting that is understandable. Typically, we use two symbols to denote this either sigma or gamma these are the common symbols which are used to designate surface tension. Now, if we want to see that what how the surface tension keeps a system in equilibrium. Let us take an example of a droplet or a part of a droplet sitting on a solid surface. So, this is liquid and the right side so this is vapour. So, this is liquid vapour and this is solid. So, you can see that at the interface between these there is a triple junction that is created. You have a place where you have sort of contact between liquid vapour and solid. Now, if you want to see that what is the equilibrium that sort of keeps this in perspective or keeps this in equilibrium then we can see that you have a force in this direction. This force is because of the surface tension between or maybe we let us just show it in the opposite sense. Sense will automatically come out if we write the equilibrium but let us just show it in a opposite sense. Let us say that we show it like this because just to appreciate that it is an element in tension. So, this is because of the interaction between which two phases yes solid and vapour. So, let us give it a name. Let us give it a name sigma S v S for solid v for vapour. So, whenever we are talking about a surface tension coefficient it basically talks about two different phases which are forming an interface and that is where the surface tension comes into the picture. If we are thinking of this one this is between liquid and vapour. So, surface tension is tangential to the interface and we can give it a name sigma l v. You could also write sigma v l. I mean it is just the two phases which are important not the order in which you write is important. And regarding the liquid and solid you have an interface here maybe sigma ls. There is an angle between these two say theta which is known as the contact angle. It is if you write the equilibrium along the horizontal direction then you can clearly see that you can write sigma ls plus sigma l v cos theta is equal to sigma S v. So, when you are considering an elemental area that or an elemental length that elemental length is cancelled from all sides. So, only the sigmas remain. It is basically a force balance but it is looks like a surface tension coefficient balance because the corresponding length that is on which this acts like it may be a unit length like that and that gets cancelled. So, you can see that you get cos theta as sigma S v minus sigma ls by sigma l. So, if you know what are the surface tension coefficients between two phases taken at a time then from that you can estimate the contact angle. This contact angle this is known as a static contact angle of course if it is dynamically evolving then the contact angle may change and this is known as Young's equation. Just a simple equation that relates the contact angle with the surface tension coefficients at equilibrium. Now the question is is it like is it the only condition for equilibrium or is it something different. To understand that we will try to consider a more involved situation when you have say a sort of element like this and say you are stretching this surface. This may be a small element of the surface in of a droplet and say we are trying to stretch it. So, let us see that what are the kinematics or what are the even the kinetics of the stretching. So, kinematic aspect we will forget for the time being and we will just concentrate on the forces which are responsible for the stretching and the geometrical change which are responsible for that. So, to do that we will just draw some sketch. So, this kind of thing we are trying to draw to see that as if there is a stretching which takes say the points a b to new locations a prime and say b prime. Similarly, if you draw the same type of radial lines the other points which are forming the boundary of this they will also go to different locations. So, it it is possible that you get a new location for the other points that is c and d because of the stretching. So, you can get a c prime and d prime let us tentatively try to locate some c prime and may be some d prime. So, we have a deformed element what may be similar in terms of the geometrical characteristics because we have used a sort of stretching and it comes to a new configuration. You can clearly see that this element is made up of actually two different types of lines. One sort of parallel lines say a b and c d another sort of parallel line like b c and a d. Therefore, we say that it has two different curvatures in two different planes. So, this like a b and c d let us say they have a particular curvature and a d and b c they have a different type of curvature. So, let us say that we are calling this radius of curvature as r 1 and let us say that with respect to this r 1. Now there is a displacement. So, when we say r 1 we basically mean up to the center. So, you have to imagine that this is like a surface which is which is which is a curved one. So, as if this goes to the center of this one and then from this. So, maybe you can just stretch it like this and then from here to here let us say that this is delta z is the displacement. Let us say that originally the dimensions of these curved curved surfaces were like x and y. Now x becomes x plus delta x and y becomes y plus delta y. So, we can write from the similarity of the entire geometry that r 1 divided by r 1 plus delta z delta z is equal to y by or it should be x or y you tell it should be x by x plus delta x because x is that dimension that is in the same sense of the radius of curvature r 1. The other one will have a different radius of curvature r 2. So, we just concentrate on r 1 and we can therefore, write r 1 by delta z equal to x by delta x. Now we are interested to see what are the dimensions of the curved surfaces. We are interested to see what are the forces which are acting on it say what stretches it. Let us say there is a pressure differential between the outer and the inner of this. So, this is a membrane. So, the membrane has a difference in pressure from the outer and the inner. Let us say that difference in pressure is delta p. When you have a difference in pressure of delta p that gives rise to a force. What is the force that acts on an area x y? This is almost like a rectangle. These are small elements because if you take big elements the local radius of curvature change has to be taken into account. So, this is just magnified for clarification but these are small. So, delta p into x y is the resultant force because of the pressure differential between the inner and the outer surface of the membrane and that times delta z is the work done because of the pressure difference. So, this is work done due to delta p and that contributes to the surface energy. So, this should be the corresponding work because of the surface tension. So, what is the corresponding work because of the surface tension? That is the surface tension coefficient times the change in area. So, the work done because of surface tension let us say sigma is the surface tension coefficient between the 2 phases interacting here that times the delta a. So, what is the delta a? x plus delta x into y plus delta y is the new area minus x y is the old area. This is basically work associated with stretching of a surface and this pressure differential is creating this displacement it has undergone a stretching the surface has some energy now to sustain that and retain its form. So, that is by virtue of the surface tension. So, this will be x delta y plus y delta x delta x delta y that product being small we will neglect in comparison to the other terms. Now, if we equate these 2 that is what we can write. So, on simplification what we will follow delta p is equal to sigma you can write delta x by x right plus delta y by y this entire thing there should also be a delta z at the denominator right. So, now if you use this relationship delta x by x is delta z by r 1. So, this relationship can be utilized similarly if you write in terms of y it will be r 2 that is the only difference where r 2 is the other the radius of curvature for the other elements. So, this can be simplified by taking help of this and similar expression as sigma into 1 by r 1 plus 1 by r 2. So, if you have an element of an interface in equilibrium then this is how you can relate the pressure differential across that with the surface tension coefficient and the radius of curvature or the radii of curvature of the elements that are constituting the surface. We can take examples as special cases which are convenient examples to take we consider first spherical droplets or may be a spherical bubble. So, if you consider a spherical droplet then what is the situation if you have a spherical droplet as an example. So, when you have the interface as a sphere it has same radius of curvature at all points. So, for a spherical droplet of radius r you have r 1 equal to r 2 equal to r and therefore you have delta p equal to 2 sigma by r in place of this droplet if you have a sort of a bubble say what will be the difference between a droplet and the bubble see droplet is like you have this full thing liquid and outside ambient may be vapour. If you have a bubble so you have a thin layer of liquid here you have something outside and something inside. So, you have here say a vapour then you have here also a vapour. So, you have basically 2 vapour liquid interfaces to consider one is as you jump from the inner vapour to this liquid line and then you jump from the outer liquid to the outside the vapour outside location that is the vapour. So, you have 2 interfaces which are formed because of this one. So, may be this should be multiplied twice to get the net pressure difference. So, you can see that depending on the physical situation this needs to be adjusted but this basically talks about that if there is one interface across that interface if there is a pressure difference then how can that pressure difference be related to the surface tension coefficient. Qualitatively if we try to understand that if we have say such a interface let us say that you have vapour on one side you have liquid on another side. Now can you physically tell that on which side the pressure should be more if the interface looks like this vapour or liquid. So, if you think about the surface tension it is acting something like this. So, when you have a pressure from this side you have a pressure from this side say p l you have a pressure from this side p v. See in the surface tension is already acting is having its component downwards. So, the upward component should be more strong to overcome that. That means p l should be greater than p v in this example. So, when we are talking about delta p we are talking about the magnitude the difference between p l and p v but out of these which one is more that should come from your physical understanding of the problem that is quite clear. The other point that we will mention here is that in a very nice way I introduce this equilibrium to you but you must have seen or you if you have noticed it carefully we have not really shown the equilibrium in the y direction. It should come to your mind that yes very nicely we have seen the equilibrium along x but if you think about equilibrium along y there is only a poor chap which has a vertical component there is nothing else to balance it. Then the droplet should go off take off from the surface you have never seen it just like taking it off like that. We are talking about a surface an interface. So, it is an equilibrium of the interface when we are talking about it is not just the it is not the weight but interaction some interaction between the surface and the fluids fluid molecules which are which are manifested in forms of different intermolecular forces. All intermolecular forces are not together brought in the category of surface tension. So, on an effect the net effect is there is a normal reaction just like what you have as a normal reaction on a block on a plane in a very similar way what is normal reaction it is a manifestation of some molecular scale interactions on a larger scale. So, that type of normal reaction is also there we have not drawn it here explicitly but that type of normal reaction is something that takes care of this type of interaction. So, we should keep in mind that this is one of the situations where you are having a statement of equilibrium the normal component we are not always keeping in view but that also has its role to play. So, the two conditions of equilibrium one is this one and of course the other is the expression for cos theta that we have derived but we have to remember that these are necessary conditions for equilibrium but not sufficient. That means that these conditions may be more fundamentally derived by minimizing the surface energy of the system. So, a system any system in equilibrium minimizes its energy. So, that is the stable sort of configuration. So, if we express the surface energy and set out its partial derivative with respect to say r and theta to 0 then we will get the corresponding expression for equilibrium but that these expressions do not automatically ensure that the second derivative is positive that means it ensures 0 gradient but it does not ensure that it is a minimum. Therefore, these are necessary conditions for equilibrium but not sufficient. So, even if these conditions are satisfied still you may get interesting instabilities in the droplets and bubbles and so on and some of those pictures that we have already seen. So, maybe we wind up here for the day and we will just see one very small movie to wind up the study for the day. So, we just play it again and see that what effects surface tension is creating here and maybe that is enough for the day and we will continue with that in the next class. Thank you.