 Okay, anybody else online other than Amog? All of you have seen the official, this thing, have you seen the official answers of your j-mains? How was it Amog? Was it in lines with whatever you have guessed or some surprises were there? Okay, who else have joined the class? Who else? Amog let's start with you. See today's session I wanted to understand in more detail about your experiences. Okay, and all those Ramcharan is also there. So Ramcharan Amog you can throw light on how you went about preparation for j-mains and what was effective, what was not so effective and how we can change this strategy of preparation for j-mains in the next attempt. Okay, now all this is valid for j-advanced also. Don't think that whatever strategy you will be discussing is not valid for j-advanced because the syllabus is same, right? So whatever mistakes or whatever surprises you have seen in j-mains, same surprises we will see in advance as well. Okay, if we continue to do the same thing. Okay, so first let's keep it open and let can you share your experiences like what was something which you could have done better? You know you could have done something which could have immediately improved your marks or something which is like quite visible that oh my god why I did not do this kind of thing? Something like that happened to any one of you? Okay, great. Now the difficulty level let's talk about difficulty level. Okay, so you've seen the difficulty level of whatever mock test you know you guys have taken, right? So let's say that whatever mock test we have taken was of the difficulty level of let's say 6, okay, from 0 to 10. Okay, now can you tell me what was the difficulty level? What was the difficulty level of the actual j-mains that has happened? Was it less than 6? And if it is less than 6 how much? Was it like 2 or 3? Or it was more than 6? Okay, good. See one more thing I wanted to tell you guys that you know you may feel that somebody else's j-mains have come easier, somebody else's have come difficult. Okay, it doesn't matter it's all relative. So in whatever test you have taken in that test what is your relative performance that is where you will stand, okay? So don't worry if your this thing is difficult or easy. Okay, fine. Now tell me the topics in physics that you think that you have a gap on and we can focus on those topics you know going forward. One by one you can list down all those topics, right? Don't tell me the chapter names because that will be like it will be like you are actually not aware of what is your weakness. So tell me the topics for example Doppler's effect like that it should be focused. So did you find out the topics or you are aware of what are your weaknesses are so that we can focus on those weaknesses, okay? So wave motion related problem, okay? What do you mean Aditya? Wave motion means what? Okay wave equation, alternating current equations related to phase differences. Think on it guys you guys have written many many mock tests and you have taken the actual j-mains also. There will be many topics will be there, okay? Don't hesitate just feel free and tell me all those topics, okay? Don't hesitate. That's it. Here comes the mechanics. Yes semiconductor in fact semiconductor we will study again will not focus on theory too much will focus on the problem solving. So semiconductors is something which I will take up completely and as you rightly pointed out semiconductor circuits. That's it. What about EM waves? It's a small chapter so that is why we can talk we can say that entire chapter. No EM waves, right? You don't need specific heat capacity, okay? EM wave. What else quick? There has to be at least 25 to 30 topics otherwise you should be scoring close to 30 sorry close to 13 to 4 120, okay? Standing waves. Yes standing waves will have nodes and anti nodes. Energy in capacitors, great. Experimental physics which you do in your labs. What all topics you can think of you can look back on your J mains actual paper which you have written. Can you think of you know questions which you have got it right which you have got it wrong sorry a question which you have got it wrong from what topics they are from impulse. Finding magnetic field finding magnetic field is it? Why you did? I mean you have lost too many marks in ciliers is it? What about this chapter? Okay, is that it? So your main weakness lie in non-mechanics topic. So you can see that clearly you can see your weaknesses are in non-mechanics topics, all right? So at time we tend to you know spend a lot of time in mechanics and we because of that the other topics get ignored, all right? So you know what you have to do now, okay? Don't just completely engrossed into the mechanics topics, all right? Because physics is 70% of physics is other than mechanics, all right? So make sure you focus on the other 70% also, okay? So you should fall in love with the marks. You should not fall in love with a topic, okay? Doesn't matter from where you're writing the marks, you should just fall in love with the marks, okay? Nucleus. Anything that I am missing? Give me an example Aditya Mishra. What do you mean when you say Carnot engine application? What was it? Can you quickly tell me what was it? You sent me a WhatsApp message. Which question number Aditya? Question number 30. You guys have any doubts on the physics paper that you have written? Please send me that. We can discuss that as well today. That is heat engine coupling, all right? So you don't have any doubts or any questions on the J mains paper that you have written? Okay, so let us make a small planner over here, all right? So this is something which is, which has to be taken up as theory, all right? So this we can say that requires theory, then this one, this topic also, it requires theory, all right? I think when it comes to AM wave, beads, energy in capacitor, all these you need problem practice, right? If I'm not wrong, please interrupt me if I'm not correct. And then you need theory with respect to, you know, this thing, semiconductor circuits, okay? Then theory is required over here, all right? And this one and probably this one as well, okay? So other than these, I guess everything else will come from the problem practice, right? You need different kind of problems in those topics. Yes or no? Okay, Amog has sent me few questions as doubts, okay? So let us take those questions. Others, if you have any doubts, please WhatsApp me, I'll be taking up today only, okay? Send me your doubts. Now this one, what do you think the answer will be? All of you attempt this one? Amog, what is your doubt in this question? Okay, that is Amog's doubt. But of course, that is not correct. That is what Amog is asking us. Did official answer keys are released? Is official answer keys are released for J mains? That is yet to come. Only your answers are released. Fine. So let's see what will happen here. There is a particle of mass m revolving around the earth with velocity v, okay? So another particle, this is mass m, okay? This is earth, okay? This one is earth. This is earth. Another particle of mass m comes from the space really towards the earth. It comes radially like that, okay? And elastically collides with revolving particle, what will be the path of both stuck particles, okay? So it collides elastically, all right? So along the radial direction, along the radial direction, you can conserve the, this thing. You can conserve the momentum, right? So this is the initial momentum. And if it is colliding inelastically, it should be equal to 2m into v. I am talking about radial velocity, okay? So this is vr. So radial velocity should be equal to v by 2 inelastically, yeah, okay? So it will gain some velocity. Now the mass is 2m. 2m is the mass and it gains some velocity along the radial direction, right? So net, net velocity will be what? The net velocity will be like this, okay? The velocity tangentially will not get affected by the collision. But because of collision, there is a radial velocity also, okay? So it will be like this, okay? So I would say that it will go in the elliptical path, okay? Because it goes like this, okay? Then it again tries to go away from the earth gravitational pull, but then its velocity reverses in direction and my hunch is it will move in an ellipse like this. Amog, are you getting it? Why it will not be circular? It will be circular only when its radial velocity is zero. If radial velocity is there, it cannot be circular, okay? It cannot be circular. Is this thing clear, Amog? In a circular motion, the radial velocity should be zero? Yes or no? But radial velocity is there. So it cannot be circular. Now we have those three options to pick. It will not tend to move towards infinity because it is coming towards the earth. After collision, it has gained velocity towards the earth, okay? So it is not going away from the earth after collision, right? So it cannot be two, okay? Now it can be either elliptical or parabola. Now it cannot be parabola because whatever happens here has to happen there as well. It is, it should be taking a symmetrical path because of the nature of the gravitation force itself. It should be symmetrical, alright? So you cannot solve and get the answer elliptical. You will not get the equation of the ellipse easily like this. But then you can arrive at the answer with visualizing this thing. All of you clear about it? Any doubts? Okay, let's go to the next one. This is a nice question. It was not as simple as I have been hearing. Alright, this question, haven't we did this question? Are you guys able to recall this one? Did we do this or not? I have done this question somewhere. I think in your batch itself. Yes, Lakshya, this is not from the paper, okay? Anyways, all of you please solve this. As you are going forward, all of you must appreciate that for us, all the topics are now, you know, we can take entire syllabus together. So we can, during the class, let us be open that we can take up any question from any nook and corner of the entire syllabus, okay? So that is how the session will be today. We'll be taking up questions from anywhere in the entire syllabus. Try solving this one. All of you, I think this can be done easily. Solve few, I mean, create some equations. See direction perpendicular to this inclined plane is, let us say this. Fine. Let us say the velocity, this is u, let's say finally after some time, velocity is v, okay? Okay, all right. Now, we know that horizontally, velocity should not be changed, right? So in the horizontal direction, whatever was the earlier velocity has to be the final velocity, right? So in the horizontal direction, velocity should be same, okay? So if this is velocity u, all right, this is alpha, then this angle will be 90-alpha, right? Okay? So horizontal direction, initial velocity is u cos 90-alpha, which is u sin alpha, fine? This should be equal to, this is v, this angle is beta, this one is 91 is beta, it should be equal to v sin beta, all right? So from here, you get v as u sin alpha divided by sin beta, okay? Now, I need to find time, right? So I can find the time using vertical components. So initial velocity along y direction is u sin 90-alpha, which is u cos alpha. So u cos alpha is the initial velocity along y direction, final velocity along y direction, it has to be in the downward direction, okay? It cannot be in upward direction. So there has to be minus sign coming up. So that is equal to v, similarly it will be equal to v cos beta, right? So v cos beta, that will be equal to minus of u sin alpha by sin beta into cos of beta, all right? So now all you have to do is use this, v is equal to u minus gt to get the time. Is it clear to all of you? Let us take up another one. Semiconductors, I will be taking up as a different topic itself, I will be teaching from start. So this one, this is, okay, this is again, I think from Lux only because question number 41 in physics cannot be there in J mains from mock test, all right? So you have a uniformly tapering conical wire made up of a material of Young's modulus y and has a normal unextended length l, radiate the upper and lower end of the conical wire have values r and 3 r, the upper end of the wire is, okay? All of you please try this. I will draw a diagram. So this is how it is, it is a conical wire, radiate of the upper one, this radiate is r and that one is 3 r, okay? So this length is 3 r, okay? Capital M is suspended at the lower end, M, okay? Capital M is suspended like this, all right? Now this material has Young's modulus y, okay? Let me represent everything whatever is given, this is the length l. Now here we are ignoring, we are ignoring the mass of the wire. We are saying mass of wire is 0, fine? Now Amogh is asking how to find the stress, this is what he is not getting. So all you have to do here is to assume, let me take a black. So you assume a dx length of strip, okay? This is a dx length of strip, fine? This is, let us say, this is at a distance of x from the bottom, fine? Now draw the free-body diagram, draw the free-body diagram of this dx strip. This is the dx strip, okay? Draw its free-body diagram. So this force, if you get, all right? This force, if you get that force divided by the area of cross-section of this dx element. So let us say its radius is whatever, let us say small r, okay? So the stress is f divided by pi r square, okay? So strain in the dx is what? Strain in dx is let us say dy, okay? dy is change in dx, okay? dx is the length, dx will become dy, sorry, dx will become dx plus dy because dy is change in dx, okay? So the strain is dy by dx, all right? So stress by strain is Young's modulus, all right? So when you integrate dy, dy is what? Change in dx. So when you integrate dy, you get change in length, okay? Now try this out, all of you please try this quickly. It is a J advanced level question. Okay, anyone got it? Should I do it? No one is trying? Okay, did you get the radius in terms of x? What is the value of r in terms of x? Anyone? You can use similar triangles, right? No one got the value of radius at a distance x from the base. No one got, okay? So let us do it. You draw a line like this. Yeah, this is more of mathematical question than a physics one, okay? Because physically we have already solved it. Concept-wise there is nothing left. It's all mathematics going forward. So let's say this is x, the base radius is 3r, this is r. So this one will be 2r, total length is l, okay? So we have 2r by l. I am using similar triangles, okay? Will be equal to this distance is let us say, this distance is d, okay? So d divided by l minus x, okay? So I am using similar triangles. This, the upper triangle is similar to the bigger triangle, isn't it? So d will be equal to 2r times l minus x divided by l, fine? So the radius will be equal to r plus d, right? So r plus 2r l minus x by l, okay? So that can be written as 1 minus x by l, fine? So small r is 3r minus 2r times x by l. This is small l, okay? So you can check also. Once you get the expression, in physics, you know, good thing is that you can always check the final expression whether that makes the physical sense or not. So this is the expression, you put x equal to 0, you get r equal to 3r, right? That is what it should be, okay? And when you put x equal to l, you should get small r to be equal to r. So that is what you are getting, okay? So this is the radius at a distance of x, fine? So the area is pi r square, all right? So the stress is, now what is the force F? Can anyone tell me what is this value of F? F is what? This one? How much it is? Anyone? What is F? What is this F? F is a force with which the lower part is pulling the dx element down, okay? They should be same as the force with which the dx element is pulling the lower part in upward direction, okay? So just draw the free board diagram of the lower part. No one? Okay, let me solve this. See guys, here I am drawing free board diagram up to a distance of x only, okay? So this is a truncated cone, all right? The lower side is getting pulled by the weight of the block, mg, all right? This has to be equal to, let us say F, all right? These two should be equal and opposite, right? Newton's third law, isn't it? They should be pairs. So F, since this part is at rest, F will be equal to mg, okay? So the stress will be the force divided by pi r square, where r you have already found out in terms of x. Why are we finding everything in terms of x? Because we have to integrate. So pi into 3r minus 2r x by l the whole square. This is sigma, fine? Are you guys following me? Is there any doubts? Tell me. Are you able to understand what is going on? Type in quickly, yes or no? No doubts? Good. So now we'll be using this relation that stress divided by strain is a constant, which is Young's modulus, all right? So strain is, strain is stress divided by Young's modulus, all right? So let us write here strain is stress divided by Young's modulus and strain is what dy divided by dx, where dy is change in the dx length. dx is changing by a certain amount. It is getting stretched, okay? So dy represents how much dx has stretched, okay? So dy will be equal to sigma by y into dx, all right? Sigma you already found out in terms of x. So substitute here and then integrate it. The limit should be from 0 to l, the x goes from 0 to l. This dy will be let's say from 0 to change in length delta l, okay? The final length will be equal to whatever is the initial length plus change in length. Understood how to solve? Should I go to the next question now? Okay? In case you have any doubts, quickly type in this one. All of you try it. Amog, what is your doubt here? You have any targeted doubt, any specific thing you want to ask expanding due to the electric repulsion? Oh, nice question. Okay, so what you have used work energy theorem, this came in your mains paper. Oh, it's a pretty nice question. Now tell me when r is becoming infinite, okay? Just answer this question. When the radius is becoming infinite, what will be the velocity? Tell me, the velocity will tend towards what? It will tend towards, it should keep on increasing or it should become constant. What it should be? Velocity should become constant, because the repulsion has decreased by a great extent. So the force becomes 0 and if force becomes 0, the acceleration should go to 0. And if the acceleration is 0, the velocity can be like velocity will become constant. So it cannot be one and three, isn't it? The velocity should tend towards a constant value. Understood? How you got the expression, Amog, exactly what expression you got? Can you send me your WhatsApp? See, look at this in 4, in 4th option, initially, the velocity is close to 0 for small amount of time. It doesn't increase quickly. Regarding it, in option 4, it says that velocity should not increase quickly. Initially, there is some resistance in the increase of the velocity, fine? Now is that what should happen? Is this what should happen? Should this happen? Initially, the velocity will be close to 0 for some time. That should be correct or when the charges are very close to each other, then there should be velocity should suddenly increase from 0. What should happen? In a way, it is a plot between, although the plot is between velocity and the radius, but with time, radius is increasing. So r can be a representative of time also. So you can say that the similar kind of graph is between velocity and time also. So if you see the option 4, in option 4, when r is 0, when r tending to 0, I am saying that the slope of velocity with r is 0 or in a way, I am saying acceleration is 0, which is absurd. It cannot be. So when the charges are close to each other, just give me a second, spherical region having uniform surface charge. Now, actually what will happen over here is they will ripple each other. See this outermost. Yeah, it is like supernova explosion sort of scenario. Suddenly, it will expand. So acceleration initially 0 does not make any sense. So 4 is not correct. I would be marking 2. Fine. Amok, which one you have marked? See, whenever there is a graphical question, you should not try to get some expression and then you look at the graphs. When they give you a graphical question like this, they are in a way, expecting that you will be visualizing the scenario and answering it. Answer is 2. Yes, it should be 2. You should be visualizing it and answering it. So the physics is all about visualization. Mathematically, you may not arrive at the answer here and even if you arrive at the answer, it will take lots and lots of time. So by elimination or by applying some visualization, you will be able to arrive at the right answer. All of you clear about it? Any doubts? Okay, let us go to the next one. These are done. Fine. Let me download a few more questions. Okay, it is taking a little time to load. Okay, Amok has sent Naman. Naman, is this from your J paper? Ramcharan, quickly take a fresh image and send me. I cannot rotate the image. Send me the images with proper orientation. All right, image with proper orientation. Okay, this one, who sent me that question? This is Aditya. Aditya, you have to send me the paper with, you have to send me the question with proper orientation. Like this one, I will not be able to rotate it right now. Okay, here is Ramu's question. Kondi, where is Ramu? Where is Kondi? All of you, please attempt this one, this question. Amok and Naman. Aditya, please attempt this one. Kondi has fever and headache. The hint is that both the parts of the circuit, let us say 0.12, this is 3 and 4. 1 and 2 and 3 and 4, both will experience same potential difference, V. Okay, because they are in parallel, these two circuit elements. You have to find phase difference between i1 and i2. Okay, so another hint is you find out the phase difference between i1 and V, then i2 and V. And then you can find the difference in phase difference between i1 and i2. Ramu, it is not purely inductance and purely resistance, right? Sorry, purely inductance and purely capacitance, that is not there. What is the resistance over here? Let us assume that is R, although it is not given, it should be given. If R from the above and bottom are not given, then the phase difference between i1 and i2 is 180 degree. But resistance is there. This one, I am not sure whether it will be R only, it is, we are just assuming it. Okay, should I solve or you guys are in between? Please message me immediately. Okay, you are not getting, fine. All right, let us try to solve this question. All right, so the current over here is because of the input voltage V. Now i1, you can find out the value of i1 as V divided by Z1. Z1 is the impedance of this lower which is equal to root over XL square plus R square. Okay, fine. And the phase difference between the current and voltage can be given by this expression. Tan of phi is XL divided by R. Okay, so XL is what? R by root 2. This is R by root 2. That divided by R. So this is 1 by root 2. Okay, so when I say XL by R, you are automatically assuming that the voltage is ahead of current. Otherwise, you would have written minus of XL by R. If you assume that the current is ahead of voltage. Okay, so you can write the formula for tan of phi, you can say that is XL minus XC by R. When you write XL minus XC by R, you are assuming that the voltage is ahead of the current. If you write XC minus XL by R, you are assuming current is ahead of voltage. So you should be knowing what you're assuming. Then only you'll be getting it correct. Omega will also be there. Where is Omega? No, Omega, see, directly the value of XL is given. Ramu, if L is given, suppose L was given, then you need to find the value of XL using Omega. But XL is directly given. This tan phi, the phi is equal to pi by 4. Okay, fine. Now look at the upper circuit between 1 and 2, tan of it is theta. Yes, Ramcharan, that kind of mistake is, if you know how to solve it and you're confused that okay, Omega is not given and you're not able to solve it, it's not good if you're not be doing that. Then there is no difference between you who knows how to solve and the one who doesn't know how to solve. There is no difference, absolutely no difference. In fact, you'll be at loss because you'll be spending time thinking that you'll be able to get it and you'll not get it. The other person will just skip it. He'll save time. So XC by R is what? R by root 3 by R. So this is 1 by root 3. So when I write XC by R, I'm assuming that current is ahead of voltage. The value of theta is pi by 6. Tan theta is 1 by root 3. So theta is 30 degrees. Tan of pi is tan of pi by 4. Sorry about that. Tan of pi is 1 by root 2 for the lower one. So pi is tan inverse 1 by root 2. In this case, theta is pi by 6. So in the upper circuit, current is ahead of voltage. In the lower circuit, current is behind of the voltage. So the total phase difference between i1 and i2 will be theta plus pi only. That is tan inverse of 1 by root 2 plus pi by 6. Now there could be printing error over here because we have assumed at the upper one to be R if this value would have been R by root 2, then 5 would have been equal to pi by 4, which is 45 degree and this would have been 30 degree. So it would have got 75 degrees. Fine. So I think there is a printing error. This has to be R by root 2. I hope this does not, this did not, this didn't come in any of your JE exam. Is this thing clear to all of you? Okay, let's see what we have next. Who gave this question and from where? Ramu, you gave, right? All of you, please try this out. You started your preparation for J-Advanced. Loud robot sound is there in the audio. Is audio not clear? Is the audio not clear? Oh, okay. Some buzz sound is there. Is it clear now? Is it clear now? The sound is clearer? Better now? All right, solve this one. B, multiple correct options are there. It's not a single correct option. Fine, let's first visualize what is actually happening over here. Okay, then it'll be clearer. A hollow conducting spherical shell of inner radius R1 outer radius R2 encloses a charge Q inside, which is located at a distance D from the center of the sphere. Okay, and we need to answer whether these options are correct or not. Now because of this charge Q, which is off center, okay, it is not placed at the center. What will happen? Charges will get induced, right? Charge will get induced on the inner surface as well as on the outer surface. Okay, so let's see the charge distribution first. So let's say this is charge plus Q. Okay, there is a plus Q charge over here. Because of that, the density of negative charge will be more over here and the density of negative charge will become lesser as you go away from the positive one. So that is how charge will get induced on the inner surface of the spherical shell. Okay, now the spherical shell was neutral, right? Spherical shell is neutral. So it cannot gain the charge from, it cannot gain the charge from outside. So charges are conserved. So whatever was a charge in an isolated system before has to be after. So charges can be redistributed, but total charge of this spherical shell should be zero. So if negative charge is coming out in the inner surface, at the outer surface positive charge will be there. Okay, and there cannot be any charge inside the metal, fine? Because inside the metal, the electric field must be zero if charges are stationary. Okay, now the charges on the outer surface of this spherical shell will be uniformly distributed. The inner surface charges are you know like this. They are not uniformly distributed because they are under the influence of this charge which is off-center. Okay, but these outer surface, outer surface the charges are getting shielded from the inner whatever is happening. So this charge will not create an effect on these positive ones. Similarly, these negative charge cannot create a force on these ones, right? Okay, till now you have any doubts? You have any doubts till now? No doubts. Now let's this will have total charges minus Q, the outer surface will have total charge of plus Q. Okay, so although conducting spherical shell, are you able to see the screen? Are you able to see? Yes, I have said that already I said that a couple of times. Please attempt this now. My screen is not visible yet. Okay, are you able to hear me clearly or still there is some noise? Okay, so let us try to solve this question. So for a point outside, okay, for a point outside you can see that you can see that the potential will be some of the three charges I mean these two charge distribution potential and plus potential due to this small Q. Okay, let's say the distance from the center is distance from the center is given as R, capital R. Okay, so the potential because of this positive charge will be what K into positive charge this divided by R, then there will be potential due to the inner charge distribution. Okay, the potential due to the inner one will be equal to KQ, the potential due to the inner one will be see it is not uniformly distributed. So we cannot treat this as a point charge located at the center. So similar question you said charges get added up and acts like a normal shell with the added charge. No Ramcharan you're confusing. You are talking about something as you're talking about electric field and here we are discussing potential. So potential due to this negative one and potential due to this one. This kind of expression I doubt whether this kind of expression you will get. Okay, let me check the A part potential at the center of the shell at the center of the shell potential will be because of the small charge Q that will be equal to KQ by D only because this charges at a distance of D from the center. Okay, and then there will be potential due to the negative charge that will be equal to minus of KQ by R1. All right, and there will be a potential due to this charge also the posture charge. So that is KQ by R2. All right, so this should be the potential at the center of the shell. Okay, which is happened to be option D. Okay, so D is correct definitely. If D is correct, A is automatically not correct because they're asking the same thing. Ramcharan that that one was a discussion on electric field. If you talk about electric field total charge enclosed will be, you know, will be that potential of the shell. All of you got why option D is correct? Is there any doubt? Okay, potential of the shell. Of course, entire potential will be at the same, sorry, entire shell will be at the same potential because it is a metal. Okay, since it is a metal, it is at the same potential and potential of the shell we need to find out. So let's try to find out. Amogh Goslaw will give you electric field. It will not give you potential, right? Electric field will be zero because if you, if you enclose, if you draw a Gaussian surface like this, total charge enclosed is zero. Okay, so electric field is zero because of that, dv by dr, zero. So it means potential is constant throughout the shell. So whatever is the potential here will be the potential there. But since the charges are asymmetrically distributed, so it will be, can we find potential still because of those charges? I think B is also correct. I have a, I would have marked B. Okay, the thing is that the charge distribution as long as it remains, as long as it remains on the, at the same distance from the center will not affect the potential because potential is a scalar quantity. The charge distribution will affect the direction of electric field but potential at the center will not get affected. So yeah, Naman, I usually get the right answer, but I'm thinking of some better logic to tell you guys why B is correct. Why B is correct is, okay, let me come back to you on the reasoning of this. Okay, give me a couple of hours time. I'll come back with with the proper reasoning for this one. Okay, I'll go to the next question, otherwise we'll just stuck with this one and keep on spending a lot of time. Ramu is gone? Ramu is still there? Okay, Ramu is not gone. Ramu is confused. What is the confusion Ramu quickly? Please tell me, okay, properly. You, you said charge gets added up and acts like a normal shell. Look at that particular question that we discussed. Okay, that one must be for the electric field. Okay, so when you apply Gauss theorem, the internal charges get cancelled off, the charge on the inner surface of the shell. See, let me show you. Look, when you talk about this Gaussian surface, okay, this negative charge and this positive charge, they will add up to zero, so 12 charges in close to zero. Okay, but if you take a Gaussian surface with it, that is outside, that is outside. Okay, whatever the charge over here appears at the outer surface, it will be treated as if charge Q is on the outer surface. Okay, but that is for electric field. That's not for the potential. Understood Ramu? I know exactly why Ramu keeps quiet. Whenever he has doubt, that is not answered, he'll keep quiet then. Okay, this one. I'm not sure whether it is clearly visible to you guys. Is it visible? Is it visible to you guys? Try solving this. This is from that Lakshabhok. Okay, this is from the Lakshabhok. Naman is, others, are you able to see it? Ramu, Amog, Aditya, are you able to see this? Are you able to read it from the screen? I think it is readable. Naman, just increase the resolution from YouTube's channel. Attempt this. These are pretty nice because whatever we are discussing today, these are pretty good questions because a lot of visualization is required. Okay, this is not a mathematical question. All of you, if you are thinking in that direction that will integrate and calculate to find out, no. If you visualize, you'll be able to get it just in two steps, this answer. C, is that correct? Who gave me this question? Aditya, is this correct? C is correct? Okay, let me know should I solve this? Others, should I solve this? Or should I wait? The hint is look for symmetry. Okay, let us see how we can go about solving this one. All right, so now look at this portion right here. This portion is, suppose, okay, another hint. So, I'll solve little bit and then I'll stop. Okay, the hint is, the hint itself is a question. Okay, how many such portions are required to make a hemisphere? This is the hint. Is this hint sufficient? Did you get the answer now? Okay, you need three of these. All right, alpha is equal to beta is equal to gamma, sorry, pi by three, right? So, alpha and beta is what? From this edge and the horizontal, this is the alpha. Okay, and from that edge to this center, right? So, you need two portion, one on this side and one on that side. Okay, now just imagine how the electric field will be because of the individual portions. Okay, because of this portion, because this symmetric, the electric field will be along the line of symmetry like this. Okay, let us say this is E1. Okay, and because of that portion, because of that portion, electric field will be like this. The magnitude of this electric field also be E1 only, isn't it? Just that directions are different. That's the only difference. Okay, similarly, because of this portion, this one, the electric field would be in this direction, E1. Okay, this angle is pi by three and even that angle is pi by three. Pi by three, pi by three. Okay, so horizontally electric field is getting canceled and vertically electric field is what? 2 E1 cos of pi by three plus E1, right? This should be equal to electric field due to the entire hemisphere, which is E0. Okay, cos pi by three is how much? Cos is 3 half. So, there's two times of E1 is equal to E0. So, from here, you'll get the value of E1 to be equal to E0 by two. Clear, all of you? Okay, let's take up the next doubt. There is this app that comes, okay, there is a cam scanner app that is there for the Android mobile, which can be used to take a pic which looks very similar to scanned image. All right, so next time when you send me doubt, try using that cam scanner app. Is this visible? Try solving this one. The hint is potential difference divided by resistance is current. This is the hint. Should I solve this one or should I wait? Quickly tell me. Should I solve or wait? D, is that correct? Who has given this? I think Aditya, is D correct answer? D for Delhi? Is it C? Aditya, circular guy Aditya, this is spherical, this is spherical shell. You can though use the spherical capacitance formula, okay, you can do that way also, but we are trying to solve it in a more basic manner. Okay, let's try to see how we can solve this one. Potential of the inner sphere, okay. You have solved it using spherical capacitance, is it? If Q is the movement of charge from outer to inner sphere, okay, then potential of the inner sphere is due to the inner sphere that is Q1 plus Q divided by R, okay, plus K times Q2 minus Q divided by 2R, fine. Ultimately, you will arrive at the same expression, okay. So potential of the second sphere is KQ1 plus Q divided by 2R plus K times Q2 minus Q divided by 2R. All of you understood this, how potential at the first and second sphere has come? Any doubts? Any doubts here? My voice is clear, right, that there is no disturbance that was coming earlier, right. I am of the voice is clear now, okay. Just give me a moment, somebody is calling me again and again. Yes, I will call you after the class, right now I am in a class. All right, so the potential at the inner sphere minus potential at the outer sphere, this is the potential difference, okay. So this is equal to what? When you subtract, this term will get cancelled off anyways and you will get K times Q1 plus Q divided by 2R. This is the potential difference. This should be equal to current times resistance R, all right. Current is what? DQ by dt where that times R is equal to K times Q plus Q to R, fine. So this will be equal to K divided by 2RR into dt. But how come this factor of 8 is coming over here? Oh, okay, okay, okay. This K is 1 by 4 pattern not correct. So this, you just integrate this from 0 to t and the value of Q you can get from 0 to Q like this, okay. All right and once you get Q, the current is DQ by dt. Understood? All of you understood now actually current should not grow with time. Current should tend towards this piece tending towards infinity, okay. That is why there is a factor of minus 1 that should come somewhere, all right. So here actually we should write it as minus DQ by dt. There has to be a factor of minus that should come up, all right. So current is minus of DQ by dt, all right. That is how you should write, okay. Because when the current is, you know, you are effectively applying Kirchhoff's law only, all right. So when current is flowing, potential difference is decreasing, okay. So the current is not the cause of potential difference, current is the effect of potential difference. So when you apply Kirchhoff's loop rule, although it is not clearly, you cannot clearly see a loop over here, but actual equation is this, that K Q1 plus Q divided by 2 R, okay, plus R times DQ by dt should be equal to 0, okay. But anyways, if you keep that in your mind that, you know, that current should decrease with time and it should tend towards 0, you can put the minus sign anywhere, I mean towards the end of the solution, okay. Expression, you'll get the same, okay. Let's see what next we have. Okay, that's all. Is there any question I missed today? And anything I did not take apart from that semiconductor question, anything you have sent and I could not take it? Okay, so we have some time. So what I'll do, I'll just pose you some situations, please solve this one. Let's say you have uniformly distributed charge, okay. This is uniformly distributed charge Q distributed over volume, fine, spherical volume of radius R, okay, all right. You need to find potential at a distance small r distance away from the center, where small r is less than capital R. Done. No, Aditya, it's not correct. That's not correct. No, no, that's not how it is. I mean, use the basics, okay. Use the basics. Yes, Naman is small integration is there. Those who have done, can you send me over WhatsApp your final expression? Okay, let me solve this now. Ramcharan has sent me. That's correct. All right, let me quickly give you hint, all right. So I need to find potential at a distance r from the center, okay. Now, whatever is a charge inside this, inside the radius r, inside r, okay, Naman has sent me, Amov has sent. No, Amov, that's not yes. Naman, I think that is correct. Let me quickly show you how we can get that. So inside r, all the charge can be assumed to be located at center, right. So how much is the charge inside r that is q divided by 4 by 3 pi r cube into 4 by 3 pi small r cube. So the charge inside it is q r cube by r cube, all right. So it will behave like as if it is located at the center. So potential due to the charge inside this v1 will be equal to k times charge q r cube by r cube divided by r, fine. So it will be simply k cube r square by r cube. Now this is potential only because of this r radius charge, okay. Now there will be potential due to the charges which are outside, okay. So let's assume that I have taken up a dr width sphere at a distance r away, okay. So for that dr width sphere, this point will have same potential as if this point is on the surface of this dr width sphere, okay. So small potential due to this dr width will be equal to k dq which is inside this dr divided by r, all right. Now dq is what? dq is charge per unit volume which is q divided by 4 by 3 pi r cube which is volume charge density into the volume of that dr width sphere which is 4 pi r square dr, okay. So this is the charge dq that divided by r, okay. So you can see that 4 pi goes off and 1 r is also gone. So dv is equal to k cube by 3 r cube r dr, fine. So v is an integral of this where small r goes from small r. Actually this I should have taken as x, okay. If I take it as x, it will be x square dx, okay. So this is x dx instead of r, okay. Now x will change from small r to capital R because this is the potential v2 that is because of the charge on the outer part of this distance r, getting it. So my integration will be from small r to capital R, okay. So once you integrate, you get v2 and the total potential will be equal to v1 plus v2, okay. So if in case somebody asks you to find out the potential energy between let us say a charge is kept inside this sphere which has volume surface, volume charge density, then potential energy will be equal to charge into potential at that particular point, fine. 3 will be in the numerator. Fine. So that's it for today. Hope you've learned something new. We will plan out something which we should do next class onwards. May I know your school schedule like how it is, your pre-bords, then your boards. Pre-board 1 is over, right? Pre-board 1 is over. Pre-board 2 from 1 to 1 and tentative date for boards. What it is? Can you quickly tell me? Practicals from, all pre-boards are over. Practicals from 21st to 21st to 28th. Boards from 2nd March. So entire FEB is free. It's free. How much time you're planning to devote for board exam preparation? Devote for the board preparation. See when you prepare for boards, it should be like you're preparing for JMAINs, okay. And also when you prepare for boards, keep in mind that this is the last opportunity for you to improve on inorganic chemistry for J advanced, okay. So 2 hours every day. That should be sufficient, okay. So the requirement for J advanced is that you should be 75% and above, okay. Just recheck whether this year it has changed or it has remained the same. But I think more or less it will be that only. So don't try to go overboard, fine. Suppose you are, you are expecting 95% marks in board exams, but then you are pushing so hard for that extra 1% marks, which doesn't even matter. So 95%, suppose you put double the effort, 95% will become 96%. But is it worth it? Okay, so ask yourself that question. But then I'm not saying that you totally ignore the board exam preparation, okay. That will in fact help you to do well in J as well. If you prepare it the way JMAINs gets prepared, okay. Or you can do one thing that when you prepare for board exam, so you learn about the derivations and all the experiments which are there, don't just stop when you prepare for boards. Don't just prepare for boards. You prepare for boards and then solve numericals which are of JMAINs or J advanced level also from the same chapter, okay. In fact, if you see last year physics paper in board exam, it was mainly application based. A lot of numericals were there and couple of questions were from the previous J advanced questions, okay. Not J advanced, but they were of higher level questions, okay. And mathematics, barring last year, last year and year before that, mathematics came extremely difficult. So the board exam also is orienting itself towards more and more logic based, not memory based, so things like that, okay. So keep that in mind. You can relax for couple of days and then the race again starts, all right. So that's it for today. We'll see you soon. Bye.