 All of you please type in your names, those who are online Good morning Okay, so again Tuesday is holiday for you When we had last classroom session I think it's about a month now No, last four sessions have been online But there were three sessions in a week itself It's not a month Ever since Nikhil joined Yeah, I think couple of students joined I think Nikhil was there And two more students joined Preeti was one And one more So if number of students touches I mean near about 40 if it touches Then we'll have two batches One will exclusively focus on J advanced And other will focus on the other compter exams And school curriculum Okay But let's see We are hopeful that we could make two batches Okay, looks like the attendance is little less today Just you have some exams going on Any exams going on right now? Good morning Oh, okay, okay, okay Understood, tomorrow is maths Oh, so you must be tensed Anyways, so see The best part about these online sessions Is that it is getting recorded So you can watch it many number of times And you can pause in between And then go and take your lunch or dinner and come back You can't pause me when I'm in a classroom I'll dictate and I'll be little more sarcastic in the class Right, so that makes you little I'm really nice in online classes, isn't it? Okay, so let us start this topic Gravitational potential It is not a school level topic Okay, so we are taking up We have already done the school We have already done the school level thing Okay, gravitation long back We are moving ahead from the school level And we are discussing the some advanced topics Okay, like for example last class Towards the end for about an hour We had a discussion on gravitational field Okay, good morning Fine, so gravitational field we discussed What is the gravitational field? It is force per unit mass Okay, so if you have a gravitational field Given you can find the force By just multiplying mass with that gravitational field Alright, so we have the formula For the force between two point masses Or two spherical masses Okay, but when it comes to different kind of structure We need to actually integrate and find out Alright, so that is why we have defined Something called gravitational field So that we find field with respect To different different kinds of shapes and sizes Which are regular shapes and sizes And once you find the field Then if you put a point mass over there Then the field multiplied by the mass Of that point mass will be the force Okay, so the gravitational field is a useful concept Alright, so now you should be comfortable With respect to that But then we have not discussed The situation where two bigger masses are there For example one ring is there Another disk is there Okay, so that is a very complicated scenario That is actually not in J advance syllabus also But that we may take it up in class 12 later on But not now What we have done is We have found out the force between two point masses And we have found the force between one bigger mass Which could be ring, disk, sphere Or whatever shape and size could be a wire One bigger mass and a point mass We have not considered the situation of one bigger mass And another bigger mass, fine Now in this particular section Gravitation potential We are going to introduce another physical concept Okay, so I will just talk about the need of this Why there is a need of this concept Good to see that attendance is around 17, 18 Even though tomorrow is maths Nice to see that I hope you will be there till the very end of the class Now anyway I know who all are there in the class Okay, so suppose you have two masses Okay, two point masses are there M1 and M2 They are separated by a distance of r Okay, the potential energy between the two masses Or this system The potential energy is what minus of g M1, M2 by r This is the potential energy The issue with this particular formula for potential energy Is that it requires the presence of both the masses Fine, alright So only if M2 is there Okay, only M2 is there Then there is no potential energy Potential energy is 0 Okay And if you put a mass M1 over here Suddenly potential energy appears Okay, so there must be something over here Because of only M1 Because of only M1 Due to that only the potential energy is coming Fine, so we are going to find out What is the effect of M1 at the location of M2 Because of which potential energy is coming in Alright, because only if M2 is there There is no potential energy Suddenly you keep M1 and potential energy appears Okay, so we are in search for a function Which could represent the energy And it is only a function of M1 Okay So what we do, we already have a formula for potential energy We say that potential energy per unit mass At that location This particular thing is a function of only M1 Okay, so I can say that Potential energy Could be because of M1 and M2 But potential energy per unit mass That is only a function of that mass M1 Okay, so it is good to have this expression Okay, because if you multiply this particular expression With whatever mass you keep here You will automatically find the potential energy Okay, just like if you multiply mass with the field You get the force Similarly, if you multiply this particular expression With the mass you get potential energy Okay, so the name of this thing is potential It is not potential energy Fine, so write down potential So again I am telling you We are going ahead of your school syllabus We are not restricting ourselves to just schools Okay, so that is the reason why We are taking up few advanced topics We are towards the end of this chapter So just one more r and this chapter will be over Alright, so potential Write down potential due to Due to a point mass At a distance of r From the point mass Yes, Shushant Due to a distance of r From the point mass Is minus of g m By the way, this potential and field They are not as such You know, they are just a concept So we have just assumed that There are multiple scenarios and we have derived it In fact, you can treat whatever derivation we have done As if we have solved few numericals So when you talk about concept That's it, I could stop here and say That derivation potential is done Alright, but I am also assuming Different kinds of scenario And keep on doing derivations for them So there can be endless such situations Fine, so that's what it is You need to solve a lot of questions Your own related to potential and field To get comfortable with these topics Alright, so this is the potential Can you tell me what is the direction of the potential? Which direction it is? Correct, it's a scalar quantity It's a scalar quantity The best part? No direction So whenever you find the potential Of anything, you don't have to worry about The components You don't have to find out the x component Y component, all that You just treat it like a number And you just add them up Getting it What is the unit of the potential? What could be the unit of the potential? Give me one of the obvious units of the potential Quick Potential energy is joules Potential energy per unit mass is joules per kg Any doubts till now? Now we will be taking up different kinds of scenarios And try to find potential for those Any doubts till now? Quickly type in So we will be taking up one by one different scenario Just like we have taken up with respect to the gravitational field So if I ask you to do some scenarios When I ask you to answer, then only you type in So that others are also comfortable solving it So let's do it for ring For ring of mass m and radius r What is the potential? All of you? That's a valid question You have to ask where Where I have to find the potential Okay Potential at the center How much it is? Okay, if anybody got the answer, please type in Why it is zero? What is the reason for it to be zero? Yeah, it's a scalar So you can't say that one direction Then opposite direction will cancel out So it cannot be zero Had it been field, then yes Because of this point mass, let's say If you take a small mass over here The field will be like this And because of this one, the field will be in opposite direction So they will cancel out each other Because it's a vector But potential is not a vector So there is no question of direction So I think all of you are Not getting it Anyways So here Tell me If you take a small mass over here dm Its distance is R? Yes, that is correct Okay, so the potential Due to this dm mass Is what, minus of g dm by R This is the potential And the best part is that The distance of all the point masses Is R So you can just Put the integral dv To get the value of total potential And minus g by R Will come out of integral It will be simply integral of dm That is m only So minus of g Into capital M Divided by R So this is the potential due to a ring At the center Okay, understood all of you Why are you getting wrong? Any doubts? Now write down something Which is very important Write down If all the mass Of a system Is At a distance Of d Okay All the mass of a system Of mass let's say capital M Is at a distance of d Then Potential or gravitational potential I am talking about Then the potential v Will be simply equal to Minus of gm By d If you are seeing that All the mass Of the entire system Is at the same distance Is at the same distance From the point where you are finding the potential Then you just directly write Minus gm by d As the potential Because when you integrate G by R will come outside G by d will come outside all the time It will be simply integral dm Which is m only Because which is Similar to this Okay So another scenario is This one You have a ring of Mass m And Radius r You need to find out the potential Along the axis At a distance of d This is mass m And radius r This is the plane of your screen That's wrong Bharat Others This is where you need to find the potential Now answer me one thing Is all the mass All the mass Of this ring Is at the same distance from that point Yes or no Right Now it's like a cone If you could visualize in 3d It's like a cone Bharat negative into negative is positive So this Distance Is root over r square plus d square Okay So all the mass Is at a distance of Root over r square plus d square only Okay So the potential In this case Is minus of gm Divided by root over r square plus d square Understood Now let's take a small numerical On whatever we have done till now See I am again and again asking the same question Because I tend to forget that So when is your Final exam Start date When it starts Then I will know exactly how much detail I can go into This chapter 18th feb When you are planning to attend the classes This is mass m r till 18th Sure Okay fine so I think we have How many weeks classes left We have then One more week in Jan And minimum 2 weeks In feb So 3 weeks Classes are left And if I take 2 more Extra classes so it will be 5 classes Fine I think it can be managed So let's proceed the way We are doing it Alright Focus here all of you You have a mass Small m Fine kept over here Now this mass is released Along the axis The mass is released from that point 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 a direct application Of work energy theorem Okay so Remember potential is Potential energy per unit mass So mass into potential is potential energy Yeah you can send over whatsapp Please send your final answer over whatsapp That will be easier Don't type in complicated expressions I just made it up It's not a standard question from A textbook 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 And Bharat improve your handwriting Okay now I Will solve this question Basically this is point number 1 And that is point number 2 I am going to use work energy theorem Between these 2 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 Mv2 Alright So K1 is Potential into mass Okay potential we just Derived minus of G M R2 Plus D2 This is the potential This into small m This is the initial potential energy And the final potential energy Is minus of G M Divided by R Fine we 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 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 VDV By BR 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 You have A cut out from the sphere Okay This cut out from the sphere Has total mass of Had radius R This was the center of the sphere Of which this Is a part of 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 is 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 Potential is this only 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 So we have A disc This is a disc it's not a ring Mass M and radius R You need to find Let me 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 Access at a distance Of D from the center Over here what is The potential This is distance D Okay mass M and radius R 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 Consent You can Consider it as like Rings of different radiuses Sorry if I am wrong Message retracted Should I solve it You need to assume You need to assume a ring Of radius R having width D R So the potential Due to that ring Will be what minus of G Mass of the ring Let's say D M 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 D M and R Right so you need to write D M in terms of R Can you do that quickly How much is D M in terms of R Correct the mass is distributed Over the entire area 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 Mass of the ring Area of the ring is what Perimeter of the ring which is 2 Pi R Into D R So this is D M Alright So Pi and Pi is gone So D M is this So you can substitute this D M Over here To get this expression only in terms of R So D V is equal to Minus of G M by R square 2 R D R Divided by R square plus D square raise to power 1 by 2 So this is D V Now can you integrate this to get the value of V V will be G M by R square Integral of this The value of small R Will go from 0 to R Fine R square plus D square As your T So 2 R D R Will become equal to D T You can differentiate this with respect to T So you get 2 R D R equal to D T Okay So V Will be equal to minus of G M By R square This will be Numerator will be D T Divided by T raise to power Half Okay And the limits when R is 0 T is D square And from there When R is capital R It is capital R square plus D square Fine So potential is minus G M By R square 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 Alright 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 2 R D R D T Okay fine See you will When you will learn about integration In class 12 You will understand that these all things are You know routine things Okay But anyways 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 2 R D R by D T Differentiation of D which is a constant Is 0 This should be equal to 1 D T by D T 1 So 2 R D R will become equal to D T 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 When R is 0 T is D square Similarly when R is capital R T becomes capital R square plus D square Okay So like this you go about Solving this but 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 The last part Now we are finding the potential Due to the straight wire So assume you have a Straight wire like this There is a point over here Which is at a distance Which is at a perpendicular distance of D From the wire Okay This distance Is D Alright 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 First of all understand that This is a scalar quantity we are dealing So we don't need to worry about any Components Let's assume that You are taking D L length From this Entire length You are taking D L Okay and this D L 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's say D theta Fine and this angle Is theta Fine Now potential due to D L length D V Is how much Anyone? Potential due to just this D L Is how much Whatever is a mass of Minus of G into whatever is a mass Of the D L length Let's say D M From here till here Whatever is a distance Okay that distance is D Seq theta This one Let us take this Small d as capital D Otherwise it will confuse us When we Integrate Do some other things This is DC theta like that So this is the potential We can 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 D M Should be equal to mass per unit length Which is M by L Into D L Okay but if I substitute this as D M over here Still you have two variables L and theta The reason why I need to see how can I write D L in terms of D theta Okay So let's 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 D L by D theta Is D Seq square theta So D L You will get it as D Seq square theta Into D theta Fine So this is D L Which I can substitute over here So D V Is minus of G into M by L Into D L Which is D Seq square theta D theta D divided by DC theta D theta Okay So D V will be equal to Minus of G M by L Seq theta D theta Fine so total potential will be integral of that Fine So This is the total potential Do Any one of you know what is the integral of Seq theta, what is the formula for that You have to integrate from minus theta 1 To plus theta 2 What is the integral of Seq theta Anyone knows So potential Integral of Seq theta is log of Seq plus tan theta G M by L Natural log of Seq theta Plus tan theta Okay the limits are From minus theta 1 to plus theta 2 Okay so limits To put Theta 1 To plus theta 2 Fine so when you substitute the limits You have Seq theta plus tan theta This is theta 2 Divided by Seq theta 1 Minus tan theta 2 Tan theta 1 So this is the potential due to the finite Wire Fine gravitational potential due to the finite wire Fine You can see the recording Ashutosh And anyway we are going to take a small And watch it All of you understood those Whoever was Online Just relook this once Is there any doubt Clear So We will end this session As in the session on Gravitational potential now Okay we will resume After 10 minutes I will Circulate a new link That link will be on the ray optics I want to keep This The video part separate So that is why I will create a new link So that gravitation and optics do not get mixed Up So You will see a new link in 5 minutes I will post it on your group You have to rejoin Not this link I have to end this session And I will circulate a new link after 5 minutes And we will start the session After Right now it is 10 35 So we will start the session at 10 45 And I will circulate the link At 10 40 So we will continue The optics from where we left Off we will have The school level discussion only At least for 3-4 hours Left In the optics then only we will go to the next level J mains and advance Okay first we will finish up the school level Alright So we will meet again After 10 minutes Bye-bye