 Now, let us take a small numerical on whatever we have done till now. This is mass M, this is R, this is T. Focus here, all of you, you have a mass small m, fine, kept over here, alright. Now, this mass is released along the axis. The mass is released from that point and what will happen? This mass will get attracted towards the ring, okay. You need to find out what will be its velocity when it reaches the center of the ring. The hint is it is a direct application of work energy theorem, okay. So, remember potential is potential energy per unit mass. So, mass into potential is potential energy. See, I am not interested to see your final answer, okay. I am interested in seeing how you have solved it. So, you have to send me that, okay. Now, I will solve this question. Basically, this is point number one and that is point number two. I am going to use work energy theorem between these two points, alright. So, work done is U2 plus K2 minus U1 plus K1, okay. K1 is 0, okay. Work done is 0, no external force, alright. K2, let us assume it to be half MV square, alright. U1 is potential into mass, okay. Potential we just derived minus of G capital M root over R square plus D square, right. This is the potential. So, this into small m. This is the initial potential energy and the final potential energy is minus of G capital M small m divided by R, fine. So, just substitute these values, you will get the answer. All of you clear? You can solve the same question by using Newton's second law. You can get the field at a distance D, okay. That field into mass is force, okay. And using that force, suppose you get a field over here, okay. Suppose you get a, now you stop typing WhatsApp message now, okay. Stop typing now. I will tell you when you can send. So, over 0.1, let us say at a distance of R, you can find the field, alright. Then field multiplied by mass will be the force. So, you get force in this direction, okay. Force divided by mass is acceleration. So, you have the acceleration now, which is also equal to V DV by DR, alright. So, using integration also, you can find the velocity. But this is just one step if you use the concept of potential, alright. Let us move, assume that, assume that you have a cut out from the sphere, okay. This cut out from the sphere has total mass of M and the sphere had radius R. This was the center of the sphere of which this is a part of, okay. So, radius was R. So, basically what I mean is this distance was R, okay. You need to find potential due to this part of the sphere at that point, at the center what is the potential. It's a hollow, it's a hollow. Mass of the cut piece is M. Yes, it will be minus of GM by R only, okay. Because you can see that all the mass on this hollow cut out of the sphere is at a distance of R only from this point. The potential is this only, fine. So, you should have an eye on these kind of things, right. You should not start integrating the moment you see something. Alright, let us move to the next concept that is on the, fine. Let's talk about the disc. You have a disc like this, okay. It is perpendicular to the plane of your, this thing of the screen, okay. You need to find the potential along the axis at a distance of D from the center. Over here, what is the potential? This is distance D, okay. Mass M and radius R. Alright, find out the potential over here. It's a disc. You can refer to the way we have integrated before when it was for the field. The hint is that you can consider the entire disc to be made up of concentric circles. You can consider it as like rings of different radiuses. Should I solve it? You need to assume, you need to assume a ring of radius R having width DR, okay. So, the potential due to that ring will be what minus of G mass of the ring, let's say DM divided by root over root over R square plus D square. Any doubts here? Understood this one? This is the potential due to the ring only, a small ring which you have taken at a distance of R, fine. Now, the problem with this is that you have two variables, DM and R, right? So, you need to write DM in terms of R. Can you do that quickly? How much is DM in terms of R? Correct. The mass is distributed over the entire area, right? So, M is distributed over the total area of pi into R square. So, M by pi R square is mass per unit area. Find this multiplied by the area of this ring will be the mass of the ring. Area of the ring is what? Perimeter of the ring which is 2 pi R into DR, okay. So, this is DM, alright? So, pi and pi is gone. So, DM is this. So, you can substitute this DM over here to get this expression only in terms of R. So, DV is equal to minus of G M by R square 2R DR divided by R square plus D square raised to power 1 by 2, okay? So, this is DV. Now, can you integrate this to get the value of V? V will be G M by R square integral of this, okay? The value of small R will go from 0 to capital R, fine? Now, you take R square plus D square as your T. So, 2R DR will become equal to DT. You can differentiate this with respect to T. So, you get 2R DR equal to DT, okay? So, V will be equal to minus of G M by R square. This will be numerator will be DT divided by T raised to power half, okay? And the limits when R is 0, T is D square and from there it when R is capital R, it is capital R square plus D square, fine? So, potential is minus G M by R square. Now, integral of T raised to power minus half is what? T to the power half divided by half. Limit goes from D square to D square plus R square, fine? So, you will get minus of 2 G M by R square and when you substitute the limits, you will get root over R square plus D square minus root over D square which is D, all right? So, this is what you will get the potential in this case, okay? All of you please go through this once and let me know if you have any doubts quickly. Others, 2R DR DT, okay, fine? See, you will, when you will learn about integration in class 12, you will see, you will understand that these all things are, you know, routine things which you have to do it again and again, all right? But anyways, this I am assuming, okay? R square plus D square, I am assuming it to be a third variable T, okay? Randomly, I have assumed it. So, if this is T, then I can differentiate this expression with respect to T, both sides. So, I will get 2R DR by DT, differentiation of D which is a constant is 0, okay? This should be equal to 1. Derivative of T with respect to T, DT by DT 1. So, 2R DR will become equal to DT, okay? And when you are introducing a third variable, so you need to change the limits also because these limits should be with respect to the integral you are integrating it, okay? Earlier it was with respect to R, now it should be with respect to T, okay? So, when R is 0, since this is T, you need to change the limits. When R is 0, T is D square, okay? Similarly, when R is R, T becomes R square plus D square, okay? So, like this, you go about solving this. But, you know, you are finding it difficult probably because of lack of practice only, there is no other reason. So, once you solve couple of more questions related to integration, you will understand that these things are routine things, you have to do it again and again. Understood now? So, just one more situation, then we will switch to the next chapter, okay? The last part. Now, we are finding the potential due to the straight wire. So, assume you have a straight wire like this, okay? There is a point over here, okay? Which is at a distance or which is at a perpendicular distance of D from the wire, okay? This distance is D, all right? What else is given is that these, the ends of the wire, the ends of the wire, they make angle theta 1 and theta 2 at that point. This point, this angle is theta 1, and that angle is theta 2, okay? Total mass of this wire is m and the length is given as l, length of the wire. Just look back on a similar situation when we are finding for the field. So, after this topic, we will start the school level optics. I think we have already started, we are in between. Just look at your field due to a straight wire derivation. You will be able to do this one as well. Should I solve? Okay. All of you listen here. First of all, understand that this is a scalar quantity we are dealing. So, we do not need to worry about any components. Let us assume that you are taking DL length. From this entire length, you are taking DL, okay? And this DL makes an angle of D theta. This angle is D theta. Why I am taking D theta like that? Because theta 1 and theta 2 are given in terms of limits. So, I will try to write everything in terms of theta and integrate that so that I can put the limits of theta. So, this is, let us say D theta, fine? And this angle is theta, fine? Now potential due to DL length, DV, is how much? Anyone? Potential due to just this DL is how much? Whatever is a mass of minus of G into whatever is a mass of the DL length, let us say Dm, divide by this distance from here till here, whatever is a distance, okay? That distance is DC theta, this one. Yes. Let us take this small d as capital D, otherwise it will confuse us when we integrate or do some other things. This is DC theta like that, alright? So, this is the potential and I need to integrate this only. But the problem is that there are two variables M and theta. So, I need to write M in terms of theta as well, okay? And I know that Dm should be equal to mass per unit length which is M by L into DL, okay? But if I substitute this as Dm over here, still you have two variables, L and theta, DL and theta, okay? So, that is the reason why I need to see how can I write DL in terms of D theta, okay? So, let us say that this is L from here to here, this is L, okay? So, you can see that small L is D tan theta, fine? So, DL by D theta is D 6 square theta, fine? So, DL, you will get it as D 6 square theta into D theta, fine? So, this is DL which I can substitute over here. So, DV is minus of G into M by L into DL which is DC square theta D theta divided by DC theta, okay? So, DV will be equal to minus of G M by L, C theta D theta, fine? So, total potential will be integral of that, fine? So, this is the total potential, right? Do any one of you know what is the integral of C theta? What is the formula for that? You have to integrate from minus theta 1 to plus theta 2. What is the integral of C theta? Anyone knows? So, potential integral of C theta is log of C plus tan theta, G M by L natural log of C theta plus tan theta, okay? The limits are from minus theta 1 to plus theta 2, okay? So, limits you have to put theta 1 to plus theta 2, fine? So, when you substitute the limits, you have C theta plus tan theta, this is theta 2 divided by C theta 1 minus tan theta 2, tan theta 1. So, this is the potential due to the finite wire, fine? Gravitational potential due to a finite wire, fine? Okay? All of you understood those who were online? Just relook this once. Is there any doubt? Clear? Okay, great. So, we will end this session as in the session on Gravitational Potential Now, okay?