 Let's solve a question on surface energy and one on surface tension for the first one We have a spherical raindrop of radius r which breaks into 27 smaller droplets of the same size So all of these droplets 27 of them. They're of the same size How much work is done on the large raindrop by the external environment and s is the surface tension of water? Okay, as always pause the video and first attempt this question on your own Okay, hopefully you have given this a shot now. Let's see what the question is asking us It is asking us how much work is done on the large raindrop on this raindrop by the external environment Okay, let's make sense of that So the external environment might be applying all of this force on this on this large raindrop And because of that this drop might have it might have Changed into smaller droplets the ones that you see 27 of them when the droplet was large Like this there was some surface energy associated with it, right? And similarly when the droplets are small even then there will be some surface energy associated with it But if we if we think about the work that is being done on the large raindrop by the external environment That can lead to change in some surface energy In fact, we can use the work energy principle and say that work done is equal to change in energy Now in this case we can assume that there is no friction as such So whatever the work that is being done on the large raindrop that is not leading to any increase in thermal energy or there is no friction So all the work that is being done on the large raindrop is purely Leading to change in the surface energy of the raindrops. So there is some surface energy associated initially There is some surface energy associated finally This is EF and this is EI and the change in the surface energy can be written as EF minus EI Now we can think back to what was this surface energy Well, we defined surface tension There was one way of defining surface tension as force per unit length right F by L But this there's one more way of defining surface tension, which is energy surface energy per unit area E by A so E E over here is nothing but S into A S into A And we can write it in this manner because we see that in options There is there is this variable of S. So it's okay if we include S in our equation. Now, let's think about the area Let's see how the area is changing Well, one thing when this raindrop was broken into 27 smaller raindrops We can be sure that one thing that is not really changing is the volume, right? There is no amount of water being lost anywhere. This big raindrop is just changed to smaller raindrops So the initial volume and we can write that as 4 by 3 By r cube. This is equal to the final volume and final volume is 27 multiplied by the volume of one small raindrop so small r cube Let's work this out So this just gets cancelled off and we get when we remove the cubes we get r equal to 3 Small r. So this is how the radius of the large raindrop is related with the radius of a smaller raindrop Okay, so now if we if we if we plug in if we plug in this this value of energy surface energy over here the final the final surface energy really is Surface tension into the final area Final surface area minus surface tension into the initial initial surface area And we are taking S because even a surface tension is not really changing, right? It's water to begin with water that ends with and surface tension is just a property of a liquid It's just like density so it won't it changes from liquid to liquid But it's constant because we are only dealing with water a spherical raindrop. So water in this case So S remains the same final and initial now over here we can take S common and Final area we have 27 raindrops. So 27 Into 4 pi r square that is the surface area of a sphere 4 pi r square So 4 pi and let's write r by 3 in place of small r We can take 3 to the left hand side. This becomes r by 3 So in place of small r we write r square Divided by 9 and we are subtracting 4 pi 4 pi. There's no space left big r square That is the initial area. This is 4, okay So 9 this goes by 3 and 3 into 4 is 12. So this becomes let's write it. Let's write it Over here work is equal to 3 into 4 12. So s 12 pi r square minus 4 pi r square and this is when you work this out This is 8 pi r square s that is option C. All right, let's look at one more question and this one will be on surface tension All right, here we have a u-shaped frame which has a 12 centimeter long sliding wire on the open side So we can see that on a diagram. This is supposed to be 12. So let me change that It's dipped in a soap solution and placed horizontally and Unstressed spring with a spring constant of 0.4. So let's write that this is 0.40 Newton meters has one end tied to a wall and the other end is attached to the sliding wire It's observed that that wire slide to the left this long sliding wire is going to the left Stretching the spring by 2 centimeters before coming to a stop. What is the surface tension of this soap solution? All right now when we look at the situation when we see that the sliding wire is being it's sliding to the left We can ask ourselves why? Well, this is a soap solution and it will have it will have a surface tension Now what was surface tension to begin with we can say that surface tension is a property of any liquid That wants to minimize wants to minimize its surface area So surface tension is the property of any liquid that wants to minimize its surface area and this wire that is sliding to the left Then makes sense because now the surface area There was this initial surface area and now this final surface area slightly less, right? It's not as much as it was initially the wire has moved to the left And the only reason that the wire will move to the left is that there must be some force, right? There must be some force that is acting on the wire. So let's try to draw the free body diagram for a wire over here Now there must be some force that is acting to the left, but we know that there is a spring attached to the right So there is a force which is acting to the right as well This force is the force due to the spring now in the question in the question It says that the wire moves 2 centimeters Before coming to a stop so when it has moved 2 centimeters it has stopped and the only reason that an object that is in motion If it stops from the Newton's second law, we can say that the forces on it are balanced, right? The forces on it must be balanced So when it has moved 2 centimeters then this force due to due to the soap solution This should be equal to the force due to the spring says from from Newton's laws We already know what the force due to a spring is it's k into x and x over here is 2 centimeters That is the extension of the spring and also the distance moved by this wire But what is the force that the soap solution is exerting on this wire? Well to figure that out we can go back to what surface tension was Surface tension mathematically we defined it as the force per unit length force force per unit length and in this case this force right here We can say we can say this is s into l right if we assume that s is the surface tension of this soap solution and L is the length of this fire so in place of f we can write we can write s into l and we can equate it But the story isn't complete right now Turns out for a soap solution we take two interfaces. There is a surface on the top There's a surface at the back if you try to see how it looks like we can have a look at this image right here There's a there's a top surface maybe the ones that we see and there is a surface behind at the bottom and between you have all the water in Bulk you have you have the water So when it comes to soap solutions, we consider two surfaces So there's a force acting on the wire at the top surface at surface one And there's a force acting on the wire at the bottom surface and at the interface two at the surface two So turns out therefore we multiply this with two. We added two over here Because the force acts at surface one and one more force acts at surface two So this is 2 sl and now when you work this out, we need to figure out we need to figure out the surface tension So let's keep s on one side when we do that. This becomes equal to kx kx divided by divided by 2l Now I think we do know everything. We know what length is 12 centimeters. We know x. This is 2 centimeters We also know k that is 0.4 Newton meters So pause the video work out work out the calculation on your own and then and then hit play All right, this will come out to be equal to this will come out to be equal to 0.03 Newton per meters that is the surface tension of this soap solution Okay, you can try more questions from this exercise in the lesson and if you're watching on YouTube Do check out the exercise link which is added in the description