 Let's solve a couple of questions on excess pressure inside liquid drops. So for the first one, we have the pressure inside a water bubble is found to be the same as a point four centimeters underwater. What is the radius of the bubble? We can use these values and we need to choose one answer out of these four options. All right, pause the video and first attempt this one on your own. Okay, hopefully you gave this a shot. Now first let's try to draw what the question is telling us. So we have, what do we have? We have, let's draw a beaker and you have some water inside of it. Okay, now the pressure inside a water bubble. So there is a water bubble outside. Let's say this is a water bubble and because it's a bubble, it has two interfaces and let's also draw water right just like this. All right, so this is your bubble. Now the pressure inside the bubble. So the pressure inside this pressure, it's the same as a point four centimeters underwater. So a point four centimeters underwater, let's say this point, this point right here, this is four centimeters. The pressure inside this bubble is the same as the pressure at the four centimeter mark. So let's, let's write that pressure inside the bubble. Let's write pressure bubble. This is equal to, equal to the pressure at the four centimeter mark. Okay, now we can ask ourselves, what is the pressure at this point, at the four centimeter mark point? We know there is some atmospheric pressure P naught. So P naught plus P naught plus rho G H, that was the pressure in a fluid, right? At a depth H, the pressure at a depth H was given by rho G H. So this is rho G into H and P bubble, we know that there will be an excess pressure inside a liquid bubble and that's really, that's really P naught plus four S by R. It's four S by R, four S by R is your excess pressure. So the excess pressure here, this is P naught plus four S by R. All right, now let's try to work out, let's try to work out this. We know we can see that P naught gets canceled off and we need to figure out the radius of the bubble. So what do we need to know is we need to figure out this. So let's do that. This is, let's keep R on one side, let's take R to the right-hand side and everything else on the left-hand side. So when we do that, this becomes four S divided by rho G H and that is equal to, that is equal to R, this is equal to R. Now let's put in the values, we know what S is, S it's given in the question, we know what rho is and as well as H, H we can take as, we can change it to, change it to meters to keep everything consistent. So when we put in the values, this becomes equal to four into 0.072 divided by 1,000 into 10 rho G into H, which is 0.04. Okay I encourage you to pause the video and work out this calculation. Okay, when you work this out, this will come out to be equal to 0.0007 meters and when you change it to millimeters by dividing it with thousand, you will get the answer as 0.7 millimeters. And that is option B. Okay, let's look at one more question. Here it is. So air leaks out of a soap bubble, reducing its radius by one-fourth. What is the change in the pressure inside the bubble? And we can, we can note that S is a film air surface tension and R is the initial radius. So, so to begin with, we know that an excess pressure inside a soap bubble, excess pressure inside a soap bubble is given by this, this relation right here. The excess pressure is 4S by R, that is your excess pressure. Now this is true for a soap bubble with radius R, but when you reduce the radius by one-fourth, when you make the radius, when the new radius is really, when you change the radius by fourth, then the excess pressure, then that is given by, that becomes Pi, let's write N for new, this is P0 plus 4S divided by R by 4. It's R by 4. This 4 goes to the top and this becomes, this becomes P0 plus 16S, 16S by R. Now we can clearly see that it really increases and by how much we can use Pi new minus Pi old. So this will be, this will be P0 plus 16S by R minus P0 minus 4S by R. P0 gets cancelled and this becomes 12, 12S by R. So the pressure inside increases by 12S by R. Alright, 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.