 Those who are online, please type in your name on YouTube chat box so that I'm knowing who all are there. Welcome, Simeer. Koshal, let's wait for everybody to join in. Meshnavi, till when your exams are on, like still you have some school exams going on, right? So we are going to basically revisit kinematics and optics, fine? So I hope all of you have come with a basic read of these two chapters and also you have tried your own some questions, fine? So today's class will not discuss theory as such, but then whenever we solve a particular question, I will talk about the concepts involved and some basics of the theory I will touch while solving the numericals, okay? All right, so, okay, practicals till 1st of October, okay, let's wait for maybe one more minute to see if anybody else wants to join. Meanwhile, in case you have any doubts on kinematics and optics, please WhatsApp me your doubts I'll be taking up here first. You can just take a snap of your doubt and send me over WhatsApp, okay? So I guess we can start now and you know, see I was expecting, I was expecting a lot of doubts from the students, but then unfortunately I got only couple of doubts and that too from only one student. So see the thing is these revision sessions are very important. These are like, you can treat them like your crash course, okay? You will never be visiting these topics again and you will not be this comfortable again while revisiting these topics. So make sure you participate in the revision with full energy and come with a good amount of read before attending these sessions. So slightly disappointed, but I can understand there are your semester exams going on, but then still I was expecting a much better response, okay? All right. So fine. So I have, I think we have good enough number of students we can start, right? So Simeer has sent me a lot of doubts, okay? These are from optics, okay Simeer? So what I will do? I will first take up questions on kinematics and then once we go towards the optics part then I will take up your doubts Simeer, okay? So I am going to take questions from kinematics from past year J, all right? And let's see what kind of questions get asked in the exam. Fine. Just one second. We will start with this question. All of you like, when you get a question, you know, if you are able to solve it quickly immediately type in your answer and let's make it more interactive, okay? Question number one, a particle moves in a circle of radius R, all right? It moves in a circle of radius R in half period of revolution. In half the period, in half the period of revolution, you need to find how much is the displacement, okay? And how much is the distance covered by the object? All of you quickly do this and message me. Distance and the displacement covered by the object by a particle moving in a circle of radius R in half the period. Displacement is what? Good loyalty, good to see, immediate answer, 2R is a displacement. So if a particle is moving in a circle, okay? In one time period, it will complete one full circle, right? So the displacement will become 0, okay? But in half the time period, it will cover only half the circle. So if a particle starts from this point, right? So I think almost everybody got it correct. So particle will start from 1 and will go towards 2. So displacement is the shortest distance between the two positions of the object, okay? So this is first position, this is second position. If you connect it with a straight line, that is the shortest distance, all right? And the shortest distance happens to be the diameter of the circle. So that is why it is 2R. But the distance is the amount of, you know, path it has covered. So that is equal to half the circumference. Total circumference is 2 pi R, so pi into R is the distance, okay? So this was asked in 1983, the way back. This is the year when I was born. Okay, let's see what next. See, when I say kinematics, I am covering not only motion in 1D, but I am covering motion in 2D also, okay? So it involves projectiles, circular motion and relative motion in 2D, okay? Let's do this thing. Write down in one second, in one second, particle goes from A to B, okay? Moving in a semicircle of radius 1 meter, okay? So there is a diagram to refer. So this is the semicircle, radius is 1 meter, this is A and that is B, fine? So the radius is 1 meter. You need to find magnitude of the average velocity. What is the average velocity? It goes from A to B like this in one second. Quickly tell me the value of average velocity. Lawyer is saying 2 meter per second. Any other answer? Okay, great. So what is the formula for average velocity? Average velocity is total displacement by time, right? Displacement is how much? From A to B, displacement is 2 and time is 1 second. So average velocity is 2 meter per second, fine? So this came in 1999, okay? All right, let us take another question. A ball is dropped vertically, all right? A ball is dropped vertically from a height D from the ground. The ball is dropped from a height D like this, okay? And then it, after hitting the floor, it bounces up to a height of D by 2, fine? So I hope the question is clear. So considering this motion, like this is the start and after bouncing, it reaches this point, okay? So this is the end. You need to find out how velocity varies with the height above the ground, okay? So these are the four graphs. You need to tell me which of the graph represents variation of velocity with height. On y-axis, we have velocity. On x-axis, we have height from the ground, velocity and height, okay? So let's call it as A, this is B, C and that as D. So type in your answer A, B, C or D. Now you are getting different answers, okay? Fine, tell me one thing. Is the direction of velocity changing or not? In this entire motion, is the direction of velocity remains the same or it changes its direction? It changes, right? So since direction of velocity changes its direction, so for one particular amount of motion, if it is positive for other, it should be negative. So the direction of velocity is changing. So it cannot retain the same sign for entire motion, okay? So you can straight away cancel out B and D, okay? It cannot be positive all the while and it cannot be negative all the while, fine? So it is between A and C, alright? So see, this is velocity and this is height, okay? Now when the height is D, okay? When it is coming down, you are saying that velocity could be negative, okay? That's how these two graphs represent. The velocity was 0 when height is D, okay? And as soon as height becomes 0, immediately velocity changes its direction. So initially it was negative and as soon as it hits the ground, it becomes positive once it leaves the ground, fine? And then it will reach till D by 2, fine? So that is why this point is smaller than that point in magnitude, fine? But then they both represent the same thing, okay? So we need to find out whether velocity has a linear variation with height or it has a curvilinear relation, fine? We know that v square is equal to u square plus 2g into, let's say this distance is, you know, y, okay? So if I am taking distance from the ground, let us say I am taking it as h. So 2g, y, it should come here, so it will be D minus h, I will write, okay? Because on x-axis we have h, we don't have y over here. So you can see that it's not a straight line relation, fine? So v is equal to under root 2g D minus h, okay? So that is the reason why a is correct, fine? So I think many of you got a, alright? So this came in 2000j, hmm, okay, let's take the next question. Aja, how many of you have actually, you know, revised kinematics and optics? You can type in your response in the chat box that you have revised and come for the class. Good, Purvik. Pratik is very honest, understood. So the feedback is, guys, this is not the way you should prepare for JEE, okay? And I don't want you to find out that after you write your JEE exam, fine? So I am telling you upfront that this is not the way you should study for JEE, okay? Anyways, so let's move to next question, okay? So particle, write down a particle starts sliding from a frictionless inclined plane. It starts sliding from an inclined plane that is frictionless, no friction, okay? Sn is the distance travelled by it from t equal to n minus 1 second, okay, to t equal to n seconds. So basically Sn is the distance travelled in the nth second of its journey, alright? You need to find out what is the ratio of Sn with S of n plus 1. This is how much? All of you quickly do it. Which one of these is the answer? Do you remember the formula for the distance travelled in the nth second by a particle which is uniformly accelerated? Okay, Koshal saying A. Yes, Koshal, that is correct. See guys, kinematics, see kinematics can be broadly divided into two parts, okay? One is the situations dealing with a uniform acceleration, okay? And other is more generic, which need not be a uniform acceleration, fine? There are other ways to divide the entire kinematics, but for uniform acceleration, you can use these equations. v is equal to u plus At, I hope you remember this. v square equal to u square plus 2 As, okay? S equals to ut plus half At square and distance travelled in the nth second of the journey is u plus half A into 2n minus 1, okay? So this equation people forget, fine? So this is the distance travelled in the nth second, fine? Use the initial velocity and n is which second we are talking about. So if you use it, you will get Sn divided by Sn plus 1 to be A, fine? Okay, fine. So let us move to next question, all right? All of you please draw this diagram. This is time, that is your x-axis and this is acceleration, fine? So this is meter per second square and time is in seconds, fine? Now the graph which is given here is like this. This is the graph. This point is 10 and this point is 11, okay? So 10 means the point on the y-axis is 10 meter per second square and the point on the x-axis is 11 second, this point, fine? So this is the graph. We need to find out maximum velocity attained by the object, v max. This is how much? At t equal to 0, assume initial velocity to be 0, okay? So this is not a uniform acceleration. See, acceleration is changing every moment it is changing, all right? So you need to tell me which one of these it is. Maximum velocity attained by the object is what? At what time maximum velocity will be attained? Any of you, can you tell me what will be the time in seconds? In seconds, when maximum velocity will be attained? It will be 11 seconds only because acceleration is all the while positive. So, acceleration will always increase the velocity, fine? So, till acceleration goes to 0, the velocity will keep on increasing. That's correct. Now, find out what will be that maximum possible, sorry, the maximum velocity, what it is in this particular situation? Option B, Urvik is getting, then Rohan is getting, Atmesh, Saimir, okay, so many of you are getting option B. Let me check. Yes, it is B. It is B. How you do this? See, acceleration is dv by dt, okay? dv by dt is acceleration. This is true, no matter, acceleration is constant or not. This is a differential relation that always exists. This is how you define what is acceleration, fine? So, this will never change. So, dv is equal to Adt, fine? So, when you integrate dv, you will get change in velocity. That will be equal to integral of Adt, fine? And integral of Adt is nothing but area of this graph, fine? This is area as a function of time. So, integral Adt is the area of this, which is what? Half times base 11 and altitude 10. So, this is 55 meter per second, okay? So, this will be the change in velocity from t equal to 0 till t equal to 11 seconds, fine? You can say that v11 minus v0 is 55, okay? And v0 is 0. It has started at t equal to 0. So, v at 11 seconds is 55 meter per second. So, that is why option v is correct. Yeah, Purvik, you can do it many ways. But what I have just discussed is a way in which, you know, you can easily get the answer. Or you can probably get acceleration as a function of time. You can get a straight line equation, okay? Then you can integrate to get velocity as a function of time, okay? And then maximize the velocity, fine? All right, do you guys have any doubts or any clarification that you need in this particular question? Please message. You have any doubts? Yes or no? All right. So, we will move to next question. So, I will, you know, slightly increase the level. Probably if you do difficult questions, then we'll learn more. You guys do this. So, now I'm taking up subjective questions, all right? These are the questions which used to come when J was subjective paper. So, just like your school exams where you have to actually show the steps of solving the question, that time also it was like that for J. This is from 1978. This question is 1978, way back when probably our dads were writing the J exam. So, car accelerates from rest at a constant rate, alpha, okay? Car accelerates from rest at a constant rate. Let's say that is alpha, okay? So, initially it accelerates with a uniform acceleration alpha for some time, okay? Then it decelerates at constant rate beta and then it comes to rest, all right? So, if total time in this entire process, total time is T seconds. T seconds is a total time, all right? You need to evaluate or you need to find out maximum velocity that was reached and total distance traveled, total distance that was traveled. All of you try this for next couple of minutes and type in your answers. Max velocity is alpha T by 2, okay? Any other answer? Those who agree can just say, okay, agree with Pratik. Mr. Pratik, Guma Raju, okay? So, we are getting different answers. Visist is getting completely different answer and others, yes, Loitia, Visist, both same answer and that happens to be correct, okay? Both Visist and Loitia are correct and Pratik, you are probably assuming that the time of acceleration and time of deceleration is same. It need not be symmetric. The graph need not be symmetric, right? So, if you plot a graph between velocity and time, the graph will need not be symmetric. It could be like this, could be, you know, it could be like this, all right? So, but then we don't know how much time, till what time it accelerates, fine? So, let us say that is T1, okay? So, this time period will be what? T minus T1, okay? All right? So, this is acceleration, what? Alpha and this is deceleration of beta, okay? And this point represents the maximum velocity. This is Vmax, okay? So, basically, I can, you know, this is a straight line. So, the slope of the straight line in VT graph should be the acceleration, right? So, Vmax divided by T1 will be equal to alpha, right? And Vmax divided by T minus T1, this will be beta, fine? So, we have two equations. See, I have not taken care of signs here. I am just bothered about the magnitudes, okay? And also, you can solve the same question in different, different ways. So, I have chosen to solve it graphically right now. So, if you divide it, if you divide it, you will get the value of T1, okay? Vmax will get cancelled out, all right? Once you get the value of T1, then you can use the first relation to find the value of Vmax, okay? And that will come out to be Vmax will come out to be alpha beta divided by alpha plus beta into T, okay? Now, this car is moving in a straight line. So, area under the graph of VT represents how much distance it has traveled, all right? So, even though there is deceleration going on, this is negative acceleration, but velocity always remained positive. The car was moving forward all the way, okay? So, you need to just add up the areas. This area, you need to add to that area, okay? Or you can find out the entire area in one shot, okay? So, the total area will represent, will represent how much distance the car has traveled, okay? Any doubt, any of you please type in yes or no? You have any doubts? Fine, so we'll move to next question. I assume nobody have any doubts. Okay, next question guys, write down. The displacement X, okay? Displacement of a particle moving in one dimension. So, this is motion in 1D. The displacement of a particle moving in one dimension under a constant force is related with time like this. T is equal to under root X. Yes, Khushal, that is correct. Okay, T is equal to root X plus 3, where X is in meters and T is in seconds. You have to find out first, displacement of the particle or you need to find the coordinate of the particle when velocity becomes 0. Okay, all of you do it. Type in your answers. I'll have a better background next class onwards. I'll put a scenery at the back. Okay, people are typing in answers. Okay, many of you are getting 0. Say here, you differentiate it with time. See what you'll get. You'll get 1 is equal to 1 by 2 root X dx by dt. Okay, so dx by dt is what? Velocity. So, from here velocity will come out to be 2 root X. So, when velocity is 0, X should be 0. Fine, I think this was a straight forward one. Okay, do one thing. You need to find out work done by the force in first 6 seconds. In 6 seconds, how much work is done by the force? Find out work done by the force which is causing this motion in first 6 seconds is what? 2 root 3 meter from Lohetia. Anybody else? 2 root 3 meter Lohetia. It's a work done. It should be joules, right? Or you retracted your message? Smart, huh? Mass is mass given. No, mass is not given. If, see, mass is not given, you assume it to be m. Okay, if it is independent of mass, it will come out independent of mass. Okay, none of the answers are correct yet. Ratik is getting 1 into mass. No, it should be independent of mass because mass is not given. See, I guess you guys do not remember work energy theorem. Work done by the force that is causing this acceleration is equal to change in kinetic energy. Half m v square. No sine here. That's not correct. Half m q square. Now, we need to find initial velocity and final velocity. Okay, so can you find out initial velocity? How much is u at t equal to 0? At t equal to 0, can you find out initial velocity? How much it is at t equal to 0? What's the initial velocity, guys? 0, no. See, you have to be careful. Initial velocity need not be 0. Guys, it's okay to be wrong. Don't retract your message. It's all right. Okay, you're getting minus 6, 0, minus 6. Okay, let's see. At t equal to 0, when you put t equal to 0 here. When you put t equal to 0 here, you'll get 0 is equal to root x plus 3. So x will come out to be 9 meters. So at t equal to 0, the particle is at x equal to 9 meters. And we have velocity as a function of x, which is this. So at t equal to 0, when x is 9 meters, velocity will be 2 into root 9, which is 2 into 3. Okay, so 6 meters per second. Fine. So I will not be able to determine the direction of the velocity. So I'm just taking it like mod of velocity. Mod of initial velocity is given as 6 meter per second. Okay, can you find out what is the final velocity? At t equal to 6 seconds, what is the final velocity? How much it is? At t equal to 6 seconds, you'll get 6 is equal to root x plus 3. So again, you will get x as 9 meters. Fine. And again, the velocity, you'll get here as 6 meter per second. Fine. So mod of final velocity is also 6 meter per second. So is kinetic energy changing? Kinetic energy is not changing. Okay. So work done will be 0. Okay. This work done will come out to be 0. Fine. So at times, there will be a question in which multiple chapters concepts are involved. So this is one of such questions. So it is very common. Okay. I think more than 50% of the questions will be like that only where multiple chapters will be there, which will be tested in a single question itself. Any doubt from the previous question? So while I'm looking for the next one, you can type in your doubts. Okay. Guys, here is the next question. It came in 1979 when our dads were attempting J. There was time and this is position x. Okay. Now draw this graph. Tell me, is the time variation of the position as shown in the figure observed anywhere in the nature? If yes, give some example. Reason, reason why it is not observed. What is so unnatural about it? Correct. So if, see time axis is always tricky because time has this constraint, multiple constraint time has. Okay. Delta t should always be greater than zero and an object cannot have two locations at the same time. Okay. And once you increase the time, you can't go back on time. You can't decrease the time. Okay. So there's so many problems with this particular graph. All right. So do not look at these graphs as if there's just any mathematical functions graph and anything is possible. You have to also include the knowledge of the physical world on every graph you see. Okay. Then only it is physics. Right. Otherwise what is the difference between physics and maths? Fine guys. So let us take up questions on motion in 2D. Okay. This is, this came in 1996 for five months. Okay. Here is the question. Guns situated on top of a hill of height 10 meters. Two guns are situated at top of hill of height 10 meters. Okay. They fire one shot each with same speed. They fire one shot each with the same speed of five root three meter per second. Okay. At some interval of time, they are not firing the bullet simultaneously. Okay. So one after the other they are firing. So there is a time difference. One gun fires horizontally. So suppose this is the hill. Okay. So one gun fires horizontally with five root three meter per second. Okay. And the other fires upwards at an angle 60 degree with the horizontal. So the other gun fires like this where this is 60 degrees. Okay. Both have five root three meter per second as magnitude of the velocity. Okay. The shots collide in air at a point P. The shots, they collide midair. So this is what happens. They collide these two shots collide midair. This is point P. Okay. You need to find out time intervals between the firing. Okay. So if one gun gets fired at T equal to zero. Suppose this is the first gun. Okay. This is the first gun. And the second gun gets fired at T equal to T seconds. You need to find out what is the value of T. What is the difference in times of the firing. Okay. All of you understood the question. If not, please message. Very fast. Sixth question. No lawyer. That's not correct. It's pretty lengthy. How can you get so quickly? I'll just switch on the lights. No lawyer. There are two projectiles. Should I solve it or you guys are in between? You want more time? This is how it will be in J advanced. It's given lawyer that they are meeting. They are meeting. Yes. Simear. It is one second. Simear got it. Correct. Anybody else? How will answer be independent of T? It is not Vaishnavi. We are finding T only. Okay. So what is T for which both of them will meet? Okay. So you are throwing two projectiles with the same velocity. So there should be some exact time after which you will throw the second projectile. Then only it will meet the first one. That is what we are asking. How much? Okay. Gaurav is also attending the class. Gaurav, did you get the answer? Okay. So should I do it now? Now one thing you should appreciate that both of them have traveled same distance. Horizontally, let's say that is x. Okay. And both of them will travel same distance vertically. Fine. Let's say that is y. Fine. See in projectile motion, we always divide the motion into two parts. One is in x-axis and other is along y-axis. And usually we keep our x-axis horizontal and y-axis vertical. All right. So for projectile one, let's say projectile one takes time T. Okay. Projectile one takes time T from here to reach there. Okay. So x is what? Ajah, what do you think? Which one would have been thrown first? The horizontal projectile is thrown first or this one is thrown first? Which one? What do you think? Exactly. Correct. Because the horizontal projectiles entire velocity 5 root 3 is horizontal. That will remain unchanged horizontally. Okay. So it is moving faster horizontally. Okay. So if you throw this projectile first, the horizontal one, then they will never meet. So that is the reason why this 5 root 3 projectile is thrown first. Okay. All right. So let us say it takes time T1. It takes time T1 for the horizontal projectile to cover a distance x. So x will be what? 5 root 3 into T1. All right. This is our first equation. And along y axis, s equals to ut plus half at square. This is what we will use. So s is minus y. Initial velocity is 0 and acceleration is g. So this is half g T1 square. All right. This is your second equation. All right. So this is with respect to projectile 1. Now let's talk about projectile 2. For second projectile, you know, although the coordinates are same, but they have taken different times to reach the same coordinate. Okay. That is the reason why you will throw the vertical projectile first and then you throw the horizontal projectile. Fine. Assume that the projectile which is at 60 degree takes T2 amount of time to cover the same distance x. Okay. So you will get x is equal to 5 root 3 cos 60 degree into T2. All right. This is your third equation. And similarly, you will be using this equation, vertical direction. So what you will get here is again minus y is equal to, now it has initial velocity along y axis. That is 5 root 3 sin 60. So this is 5 root 3 sin 60 ut minus half into g T2 square. All right. This is equation 4. Fine. Now you will get, again, you know, these x and y variable you can eliminate. You can equate these two. Right hand sides you can equate. You will get a relation between T1 and T2. And then you can equate the right hand sides of these. So 2 and 4 you can equate. And 1 and 3 you can equate. All right. The answer will be T2 minus T1. Okay. So whatever is the difference in time, however much the second projectile requires the extra time, you have to throw second projectile that much before the first one. Then only they will meet midair. All of you clear about it? Please type in yes or no. Basically the first equation you will get is 5 root 3 T1 is equal to 5 root 3 cos of 60 degree into T2. Okay. This is your fifth equation and more useful one. And the sixth equation will be half G into T1 square is equal to 5 root 3 sin 60 into T2 minus half G T2 square. Okay. So you need to basically solve these two equations to get the value of T1 and T2. And then you just have to subtract T1 from T2 and get the value of delta T. Getting it? Fine. So I hope you have learned many things from this particular question. We'll move to the next question. This came in J2011 as integer type question. See I might be taking slightly different or difficult type of questions that are at a higher level compared to J mains. But then you learn many things quickly if you solve difficult ones. Okay. A train is moving along a straight line. Okay. Write down. A train is moving or train moves along a straight line. Okay. Moves along a straight line with a constant acceleration A. It has a constant acceleration A. All right. A boy standing in a train throws a ball forward with a speed of 10 meter per second. So a boy on a train throws a ball with 10 meter per second. An angle of 60 degree. Okay. So this velocity is actually relative velocity. So boy when he is standing in a train is throwing 10 meter per second related to the train at an angle 60 degree. The boy has to move forward. The boy has to move forward by 1.15 meters inside the train to catch the ball back. To catch the ball back. Okay. So you can ignore the height of the boy or you can say that the ball is thrown from the same height and it is caught also at the same height. So it's a normal projectile. Okay. You need to find the acceleration of the train. What is the value of A in meter per second square? All of you understood this. If you have any doubts, please message. Yes. The boy is moving relative to the train and it goes forward by 1.15 meter to catch the ball back. Yes. You need to find the value of acceleration of the train. So as a hint, I will just draw a diagram over here. 9.2. Loitya is getting others. It's an integer type question. Loitya, you have to say 9. Atma is getting 2. No. That's not correct. Yes. Purvik that is right. How did you do Purvik? Can you summarize? Can you give some inputs on how to solve this? Yes. Okay. Purvik got the range of the projectile. Now is the range 1.15 meters? Is that the range? Is the range 1.15 meters? It's not. Why it is not? Because that's the, that's an apparent one. That's an apparent range. Okay. You're feeling that is range because you are inside a train which is accelerating. Getting it? Okay. So for a boy who is standing on the train, for this boy, you'll feel that it is just 1.5 meters. But a person standing on the ground will not feel so. Okay. So the range is not 1.15. So, but the good thing is that train is accelerating horizontally only. Let's say the acceleration is A in this direction. Vertically there is no acceleration. So vertical direction, you can write down this equation S equal to ut plus half at square. You can apply this along y direction and you can put S is equal to 0. U will be equal to what? 10 sin 60 into t minus half gt square. It is 10 into t square. Fine. So, you will get t equals to root 3 seconds. So, basically the time of flight is root 3 seconds. Time doesn't depend on the frame of your reference unless you are traveling with speed of light. Anyway, we haven't learned so we will not go there. So, t is equal to root 3 seconds is a time of flight. I hope all of you understood this. Now, horizontally the situation is slightly different. So, if I use S equal to ut plus half at square along horizontal direction then u is what? u is 10 cos 60 into t. t is given as t we have found, right? t is fine. So, basically now I am writing the equation with respect to this person. This person is watching what is going on. So, this is equal to 1.15 plus a distance the train has traveled. So, train will be traveling a distance of half into a into t square. So, when you solve these two equations you will get the answer. Any doubt on this question please type in this was J advance 2011. See here, here there are two persons. This person is inside the train. This person will feel that the distance traveled by the ball is just 1.15 meters. But that's not true. This person what will feel that you know not only this distance plus what distance train has traveled. As the distance traveled by the ball. Getting it distance of this point to that point is 1.15. But this point itself shifts here after some time. So, the total distance is 1.15 plus this extra distance that is traveled. So, that is the reason why when you write down this equation. You are writing for this person who is standing on the ground. So, for this person you will see that the velocity is this assuming train initially at rest. So, this into root 3 is should be the total distance traveled by the ball. This total distance which you have got is not equal to 1.15. This total distance which ball has actually traveled is 1.15 plus how much train has traveled in that time. See, you just spend couple of hours learning about kinematics after the class also. Then you will see that your comfort level will exponentially grow. Because you have done everything whatever physics you had to learn you have learned. Now is the time to do lot of problem practice. So, if you start doing problem practice slowly and slowly everything will fall in place. Then you will be able to solve some of the difficult questions also. So, do not go back and start reading the chapter. Do not read the theory. Again and again I am telling you this is for physics and maths. Do not read the theory. Read the theory only when you get stuck. Refer it. Do not read it and you will be like now I am done with the concept and then I start solving the questions. Start solving the question right away. Because you have already learned everything that is there. You might have forgotten small things. You will be able to recollect them as you solve questions. Do not spend any time on reading lot of theoretical stuff. Anyways, let's move to next question. This came in J-Advance 2014. Fine. So, let us first draw the diagram which is taken as a reference in this question. So, please draw this diagram first. This is your aeroplane A, second aeroplane 60 degrees. This is the scenario. Airplanes A and B are flying with constant velocity in same vertical plane. They are flying in the same vertical plane at an angle 30 degrees and 60 degrees. With respect to the horizontal as shown in the figure. So, it is stating the obvious thing. The speed of A, the speed of A, velocity of A is given as 100 root 3 meter per second. At t equal to 0, an observer in A, there is an observer sitting inside the train, not the train, flight. There is an observer inside the flight A or airplane A. It finds B at a distance of 500 meter. So, this is B, B to A distance the observer finds at, finds as 500 meters. Okay. Observer sees B moving with constant velocity perpendicular to the line of motion of A. Observer sees velocity of B perpendicular to velocity of A. Getting it? So, observer inside A will observe relative velocity of B. And it observes that B is moving as if it is perpendicular to the motion of A. Okay. This is at t equal to t naught. Okay. Okay. This is not at t equal to t naught. This is, this particular situation, this thing happens at t equal to 0. t equal to 0, this happens and at t equal to t naught, A just escape B. You need to find out value of t naught. How much is t naught? I will repeat the question once again. The velocity of A is given 100 root 3 meter per second. Fine. 100 root 3 meter per second is the velocity of A. Okay. At t equal to 0, an observer in A, there is someone sitting inside A that sees B at a distance of 500 meters. Okay. So, not only he observes that, he also observes that B is moving with a constant velocity perpendicular to the line of motion of A. So, relative velocity of B with respect to A at that time is perpendicular to the line of A. Okay. That is what it sees. And at t equal to t naught, A just escapes B. You have to find out what is t naught. All of you clear about the question? In case you have any doubts, please type in or start solving it if you are clear about it. It's a very simple question if you think simply. Anybody got the answer? Anyone? See, one more hint I'll give you. If A is seeing B moving perpendicular to itself, okay, then the velocity of B, okay, relative velocity of B along the direction of A should be 0. There should not be any relative velocity component of B along the direction of A. So, along the direction of A, velocity of B should be equal to velocity of A. Then only along the direction of A, there is no component of the relative velocity. So, if along A is no component, then all the component of relative velocity will be perpendicular to A. Okay, we have one answer. Vishisht, 5 seconds. Yes, Vishisht. Nice. Impressed. Okay, should I do it? See, this is B, okay, this is velocity of B and this is velocity of A, okay. V A is given, 100 root 3, okay, and V B is unknown, right. So, let us say that V B is V B only, okay. So, this angle is how much that angle is equal to this angle, isn't it? This is 60 degrees, this is 30 degrees, okay. This should be 30 degrees, this is 30, fine. This is 60, this is 30, exterior angle, right, so this is 30 degrees. So, the velocity that is 100 root 3. So, if V B is seen to move perpendicular to V A, then the relative velocity of V B along this line should be 0 or the velocity of B along the motion of A should be equal to the velocity of A only, okay. So, I can say that 100 root 3 is equal to V B cos 30, the component of V B along the direction of A, okay. So, this, the component of V B along A is equal to velocity of A, then along the direction of motion of A, there is no relative velocity, okay. So, only relative velocity left will be perpendicular to motion of A, okay. So, this is 100 root 3 is equal to V B cos 30 root 3 by 2. So, we will get V B is equal to 200 meters per second, okay. Now, V B is 200, now the component of velocity perpendicular direction will be what? V B sin 30, okay. So, this is perpendicular that is equal to 200 sin 30 which is 100. So, this is 100 meter per second, fine. So, B is appearing to move perpendicular to A, fine. And that scenario will not change because they are moving with constant velocity, there is no acceleration. So, if it is appearing to move perpendicular to A, it will always move perpendicular to A only, alright. And it is found that the distance, relative distance it has to cover is 500 meters, fine. So, 500 which is distance should be equal to speed into time. So, T will be equal to 5 seconds, fine. So, this is how you do this particular question. So, I hope you have now a good idea about what type of questions are asked in J advanced, okay. So, this is with respect to kinematics, we have done many questions on kinematics. So, let's revise optics now, although optics is slightly bigger topic, it may not get over by the time class is finished. What we'll do is that we'll finish part of optics and then extend the revision of optics to the next session also, okay. So, now we'll take a 10 minutes break. Right now it is 6.17 pm. Right now it is 6.17. We will meet at 6.27 pm, okay. Alright, in case you have any doubts on this particular question, please type in or let's take a break. Should we wait for everybody to come in? Many people left. Okay guys, so let's quickly write down the formulas or equations that we'll be using to revise ray optics. Who are there yet? Please type in your names, who all are able to hear me? Fine, so optics, optics we have learned in great detail. It's a study of light and light behaves in two ways, two ways. One is ray, another is wave, fine. So ray optics and ray optics are two different topics. So today we'll be focusing on ray optics, fine. Wave optics we will focus probably next class, fine. So we'll quickly list down all the equations that we know of because it has a lot of equations, right. Write down for mirrors, we have mirror formula 1 by v plus 1 by u is equal to 1 by f, okay. Then for spherical interface, we have mu 2 by v minus mu 1 by u is equal to mu 2 minus mu 1 by r. So you can apply this spherical interface formula or the mirror formula not only just for the spherical surface but for the planar surface you can treat as if it has radius of curvature of infinity. So you can put r equal to infinity, you can apply this for planar interface, you can put f equal to infinity, you can apply this mirror formula for plane mirror also. Okay, this is for spherical interface then for lens. You can use the lens formula only when both sides have the same refractive index. This and we have a lens maker formula also, 1 by f is equal to mu 2 by mu 1 minus 1, 1 by r2 minus sorry 1 by r1 is 1 by r2, okay. This is lens maker formula. For mirrors, there is a straightforward relation between focal length and radius of curvature. f is equal to r by 2 but that's not so straightforward for the lens. For the lens, focal length is related to the radius of curvature and also to the refractive index and not only just the refractive index of the lens but also of the surroundings of the lens, okay. So these are the equations that we'll be using for the ray optics, lens, mirror or spherical interface problems, okay. Of course there is one whole sections on optical instruments like microscope, telescope and magnification because of all that, fine. So that also we are not doing today. Today our focus is only on application of these formulas which I have written in front of your screen, right. And of course the basics of optics, I am assuming everybody understands that when the light from the object, the light from the object reaches the observer's eyes, then only the image gets formed, okay, fine. So between object and observer, there can be multiple, you know, optical instruments may be there. Could be there is a lens, could be spherical interface, could be mirror, you know, could be anything, alright. So like that between object and observer many things can come and this is how you deal with that. So if a mirror comes, we'll be using mirror formula, spherical interface comes and we're using this, if lens comes then we'll be using lens formula, fine. So our focus is to find the location of object and the nature of the, sorry, location of image and the nature of the image, like most of the time. Okay, okay, everybody, I think many joined in now. We have seen that when revision happens, the attendance drops drastically because you might have your own plans that you'll be like, okay, let me finish first my own, then I'll start solving questions. That time will never come, okay. I am seeing it year after year, you, what I'm telling you is the learning from past 10 or 12 years, okay. So take it as it is, whatever I'm saying, please follow it as it is. Okay, let's do a question. This is a fill in the blanks that came in 1983, okay. Light wave having frequency 5 into 10 raised to the power of 14 hertz, this light enters a medium of refractive index 1.5, okay, fine. If the medium, in the medium, the velocity of wave is what? The velocity of this light in the medium is what? V is what? And what will be its wavelength? All of you quickly find out. Anyone from Rajasinagar class 11th joined in, please type in the name. So we got Sundaria, okay. So that's how the refractive index is defined. That's how the property of any medium is defined, okay. So that's how we quantify a medium that the refractive index is velocity of light in air divided by velocity of light in the medium, fine. So this is given as 1.5, okay. This is equal to 3 into 10 is power 8 divided by velocity of light, which will come out to be equal to 2 into 10 is power 8 meter per second, okay. Now what is the wavelength? How will you find wavelength? When you move from one medium to other medium, the wavelength will not change. Sorry, the frequency will not change, sorry. The frequency in medium 1 should be equal to frequency in medium 2, okay. So frequency in medium 1, how will you get? Velocity divided by wavelength in medium 1, okay. V1 by lambda 1 is equal to V2 by lambda 2, okay. V1 was what? Speed of light, okay. So speed of light divided by V2 comes out to be lambda 1 by lambda 2, okay. And this is nothing but refractive index only, 1.5, alright. So lambda 2 is lambda 1 divided by 1.5, okay. Lambda 1 you can get because this frequency is given and the speed is given as 3 into 10 is power 8, right. So lambda 1 will come out to be C divided by frequency. So this is 3 into 10 is power 8 divided by 5 into 10 is power 14, fine. So this is lambda 1 and once you substitute that lambda 1 over here, you will get the lambda 2, okay. Alright, next question. A thin lens, there is a thin lens of refractive index 1.5. Thin lens of refractive index 1.5 has a focal length of 15 centimeters. It has a focal length of 15 centimeter in air, okay. This is a focal length in air. When the lens is placed in a refractive index of 4 by 3, okay. How much the focal length becomes? F2 is what? All of you please try it. In case you have any doubts, please type in. Anyone got the answer? Please type in 60. Yes, that's correct. Focal length is 60. See we have a formula 1 by F is equal to mu 2 by mu 1 minus 1 times what? 1 by R1 minus 1 by R2, okay. 1 by R1 minus 1 by R2, okay. Now when you change the medium, what will happen? The value of mu 1 will change, okay. R1 and R2 will not change because R1 and R2 are geometrical properties, fine. R1 and R2 depends on from which sphere you are cutting out this particular lens, right. So it doesn't depend on the medium. Hence in air, you will be able to write down the equation like 1 by 15 is equal to mu 2 which is the lens refractive index minus 1 1 by R1 minus 1 by R2, fine. So this is your first equation, fine. And second equation, let us say focal length has become F. So 1 by F is equal to 1.5 divided by... Now medium has changed, it has become 4 by 3 minus 1 1 by R1 minus 1 by R2, fine. So this is equation 2, right. So when you divide equation 1 and 2, what you will get is F by 15 is equal to 1.5 minus 1 divided by 1.5 divided by 4 by 3 minus 1, okay. So from here, you will get the value of F as 60 centimeters, okay. So like this, you have to solve this particular question, okay. Any doubts on this question, please type in. I'm moving to the next one, okay. Write down a diminished image of an object is obtained on a screen that is 1 meter from it. 1 meter from it. I'll read the question again. A diminished image of an object is obtained on a screen 1 meter from it. This is achieved approximately by placing concave mirror, fine. Convex mirror between object and image. Convex lens of focal length less than 0.25 meters or a concave lens. Is it a real image or a virtual image? All of you, is it real or virtual? It is real because you are putting it on the screen. You cannot put a virtual image on the screen, fine. And we know that, we also know that convex mirror and concave lens, they will always form virtual image, right. So B and D, they are out of question. So it is either C or A. This is a diagram. It is a screen, okay. This is your lens or mirror, let's say, and this is the object, so of course. First, let us try out whether lens is possible or not. This is 1 meters, okay. This distance, if it is lens, if it is convex lens, this distance must be equal to 2f plus x. Why I am saying that? Because until this you are beyond 2 times the focal length of the lens, diminished image will not get formed. It will be enlarged image, fine. So that is the reason why you, you can say is it's little twisted question. U is this, so I can use this equation 1 by v minus 1 by u is equal to 1 by f, fine. So I will write down here 1 by v minus 1 by minus of 2f plus x is equal to 1 by f, fine. And we also know that v plus u should be equal to 1, okay. So this we can use over here and x should be greater than 0. So if we use these conditions, okay, it will take time, but ultimately you will arrive at this particular option, option number C, fine. So let us move to next question. Why not opt for A? So if it is a concave mirror, fine. So if it is a concave mirror, first of all the image will be behind, okay, all right. So the screen will be behind. So this distance, the object to image distance should be 1 meter, fine. And this distance should be beyond 2f. Then only a diminished image will get formed, fine. If that distance is less than 2f, then diminished image from this concave mirror will not get formed, okay. And that distance is 1 meter, all right. So you can say that this is u. So v should be equal to what? 1 meter plus 2f plus, this is 2f plus x, let us say, okay. This is 2f plus x, fine. So you can actually check yourself, right. x should come out to be greater than 0 in this particular scenario, but it will not come out. So that is the reason why concave mirror is not possible, okay. So we haven't analyzed for concave mirror, but then a very similar analysis can be done for concave mirror also here. Fine, Vaishnavi. All right, guys, let us move to the next question. Draw this diagram, all of you. This has a refractive index of mu. There is a source s, okay. So incident rays are coming like this. This angle is i and that angle is i, okay. The thickness of the slab is given. This slab has a thickness of t, fine. So the question goes like this, all of you, please listen. A divergent beam of light from a point source s. So this is a point source s, a divergent beam comes like this, fine. Having divergent angle alpha. So divergent angle is alpha, so this angle. This angle is alpha. It falls symmetrically on a glass slab as shown, right. The angle of incidence of the two extreme rays are equal. So angle of incidence is i for this and for that also it is i, okay. If the thickness of glass slab is t and refractive index is mu, then the divergent angle of the emergent beam, when the beam will come out from here, okay, then what will be the angle between these two lines? These two rays, when they will come out from here, what will be the angle between them? This angle is what, fine. So I hope you have understood the question. Please try solving it. Will it be alpha itself? Let me check, okay. See when this particular ray hits the glass slab, this ray will be parallel to that ray, right. The rays will be parallel but just laterally shifted, fine. Similarly, this ray will be parallel to that ray, okay. So this ray, since it is parallel to that ray and this particular ray is parallel to this ray, the angle will be alpha only, that's correct. All right, so I hope you have understood this. So through a parallel slab, if there is a ray that is incident, then when it get transmitted from the slab, the ray that comes out will be parallel to the incident ray, okay. So that is the reason why the angle of divergence remains alpha only. Let's move to next question, guys. All of you please draw the diagram with me. This is N1, this angle is alpha, okay. And outside the refractive index is N2. Outside of this slab, the refractive index is N2. Now let me read the question. The question goes like this. A rectangular glass slab ABCD of refractive index N1 is immersed in water of refractive index N2, fine. There is water outside. N1 is greater than N2. This is given. A ray of light is incident at surface AP of the slab as shown. You have to find out what is the maximum value of alpha, what is the maximum value of this angle alpha, fine. What is the maximum value of alpha? Such that ray comes out only from the other surface. As in, this is ABCT. You have to find out what is the maximum value of alpha and this ray only comes out from CD, fine. So it doesn't come out from AD. If it hits AD, it should be total internal reflection, okay. I'll write down the options here. You guys can start solving the question and in case you have any doubts, please type in. So whenever you solve objective questions, you can immediately reject the most obvious wrong answers, you know, since N1 is more than N2, sine inverse N1 by N2 doesn't make any sense because N1 by N2 will be more than one. Not sure, sir, raise it A. Not sure, Kushal, is it correct? See, you have to make sure that the ray, see, N1 is more so it would bend towards the normal, so it would bend like this. So this should hit the upper surface at critical angle. This should be critical angle. Then only it will reflect. Critical angle was slightly more than critical angle. Then only it will reflect, fine. And you can say it will come out from CD or even if it grazes the surface at critical angle, it will still come out from CD only. If you increase alpha, this angle will increase and this angle will decrease. So there's a limit up to which you can increase alpha because if you increase this, this angle will decrease and then it will no longer be a critical angle, fine. So that is why maximum angle alpha is asked. Alright, so let me do this question now. Let's say that this angle is theta, fine. So I can write here that N2 sin alpha max should be equal to N1 sin theta, okay. So sin theta comes out to be N2 by N1. So this will be N1 sin theta, okay. So this is your first equation, alright. And theta should be equal to what? Theta should be equal to 90 degree minus critical angle, right. Because this will be a 90 degree and this is a triangle. So C plus theta should be 90 degrees, okay. So at this point, at this point of refraction, I can write down S Purvik. N1 sin C, both of you are correct, Koshal and Purvik. N1 sin C is equal to N2 sin of 90 degrees because at critical angle, refraction is 90 degrees. So C is equal to sin inverse N2 by N1, okay. So theta is what? 90 minus sin inverse of N2 minus N1. Theta will come out to be what? Theta will be equal to cos inverse of, theta will come out to be 90 minus sin inverse N2 by N1, okay. So when you substitute this theta over here, you get the value of alpha, alright. So I hope you have done inverse signometric function in mathematics. So you can, from here on you can do it yourself. Yes. Who is this? Saptarishi Das. I'm hearing this name for the first time. Okay, let's do this question. All of you, again this refers to figure. So draw this, all of you. Okay, this is the question. As in this is the diagram, let me read the question. This came for five marks in 1999 when J used to be subjective paper. So a quarter cylinder of radius R and refractive index 1.5. The cylinder has a refractive index of 1.5, okay. So this is a quarter of a cylinder, fine. It is placed on a table. A point object P is kept at a distance of MR from it. Find the value of M, what is the value of M? Okay, you have to find what is M. For which a ray from P will emerge parallel to the table as shown in the figure. For which the ray from this point P will emerge parallel to the table, fine. All of you understood this question? Please attempt it. Okay, so Saptarishi Das has answered M is equal to 2. Okay, so there is a quarter of a cylinder, fine. There's a quarter cylinder of radius R and refractive index 1.5. It is placed on a table, okay. A point object kept at a distance of MR from it, find the value of M for which a ray from P will emerge parallel to the table as shown in the figure. Okay, let me check the answer. No Saptarishi, that's not correct. Yes, Sai Meer, you're correct. M is equal to 4 by 3. Anybody else got it? Sai Meer is impressive today, huh? Quickly answering all the questions. Keep working hard. Should I do it now? This is not a lens, okay? This is not a lens. This is interface, okay? First, you have a flat interface. This is a flat interface, this one. This is a flat interface. And then you have a curved interface, fine? So you have to use the interface formula twice, okay? When you are using interface formula for the flat interface, you have to measure the distance from this point, okay? But when you are using it for this curved surface, you should measure the distance from this point because your optical center changes, okay? Yes, Koshal. Alright, so let's first write down the spherical interface formula mu2 by V minus mu1 by U is equal to mu2 minus mu1 by R, okay? So mu1 is what? Mu1 is the refractive index of the object and mu2 is the other refractive index, fine? So when you are talking about the flat interface, okay? When I am saying for the flat interface, mu2 is 1.5, okay? Let's say V is this, mu1 is what? 1 and U is what? U is minus MR. I am using it with sine convention, okay? This is equal to mu2 1.5 minus 1 divided by R. R is what? Radius of curvature for the interface, which is what? Infinity, alright? So this term goes to 0. So you will get here 1.5 by V is equal to 1 by MR, okay? So V will be equal to 1.5 times MR, so that is 3 by 2 times MR, okay? This is your first equation, alright? So basically we have got the image due to the flat surface, okay? And it is coming out to be positive, alright? So we know that the image should be that side somewhere, okay? Now I am using spherical interface formula for the curved surface. The optical center for the curved surface is this, alright? So if I use this formula, spherical interface formula for the curved surface, now what is mu2? Mu2 is the AS effective index, which is 1, okay? Minus final image. Final image is where? At infinity. This ray is coming out to be parallel. So all the rays should come out parallel as the question suggests. So this divided by infinity minus mu1 is 1.5, okay? Minus what? Minus the object distance. The object for the curved surface is the image of the flat surface, okay? So this is r, 3 by 2 r will be somewhere here, let us say. This is the distance of the image from the flat surface. This is 1.5 times MR, fine? So from this point, the image is at what distance? It will be 3 by 2 MR minus r, fine? And it is on the right hand side of the curved surface, fine? So you have to take it as positive. So 3 by 2 MR minus r, fine? This is equal to mu2 minus mu1. That is 1 minus 1.5 divided by r, okay? Now r, I have to substitute here with sine convention. You have to write it minus r over here, all right? So once you solve it, you will get m equals to 4 by 3, okay? Any doubt on this question, anything? Type in yes or no, no doubts? Okay, so I assume Saim here is speaking on behalf of everyone. All right. So let us move to next question. So this is the next one. The focal length of thin bi-convex lens. The focal length of thin bi-convex lens. The next online class, when it happens, what I will do, I will put all of you on Skype so you can speak also, okay? So that way it will be two-way interaction. You can always, you know, I'll share the screen on Skype as well on the YouTube also, okay? So focal length of thin bi-convex lens is 20 centimeter. Fine, this is bi-convex lens. When an object is moved, when the object is moved from a distance of 25 centimeter in front of it to 50 centimeter, okay? Magnification of this image changes from M25 to M50, okay? You need to find out the ratio between the magnifications. M25 divided by M50. This is how much? This is integer type question, so it must be an integer. So I'll repeat the question again just to avoid any confusion. The focal length of a thin bi-convex lens is 20 centimeter. When an object is moved from a distance of 25 centimeter in front of it to 50 centimeter, the magnification of this image changes from M25 to M50. So what is the ratio between M25 and M50? Okay, let me check. The answer is 6. The answer is 6. So it's a bi-convex lens, alright? It's a bi-convex lens. So we have 1 by V minus 1 by U is equal to 1 by F, fine? And magnification is V by U, alright? So when the object is at 25 centimeter and focal length is 20 centimeter, you'll get the, you know, image distance, right? So you'll get M25, the image distance when object is at 25 centimeter, okay? divided by the object distance. So you'll get M25. And then similarly, you'll get M50 also, okay? And then simply you have to take the ratio. This came in J advanced, 2010, okay? So, you know, a J paper will have at least 40 to 50 percent questions that are very easy. So our task when you, when we take any exam is to find out the easy questions and solve it quickly. First, then only we should move to the difficult ones because whether you solve a difficult question or easy question, ultimately you're getting the same number of marks, right? Okay, so fine. See, we are now at a very important phase of the preparation, okay? So we have already started the revision, okay? And we plan to put more effort in the revision. When I say we, I mean, you have to do it, okay? So now the plan is, we have two plans. You have to tell me which of them. Plan A, okay? Is to teach the same thing every batch, okay? To teach the same thing every batch. So we'll be meeting, physics will be done once in a week, okay? That is plan A. So till November 30th, we'll be able to meet at least or at most eight or nine times only now, okay? If that is a plan, okay? So we'll be, if it is physical class, okay? Eight or nine times we'll be meeting, like weekly once physics class, okay? Similarly maths eight or nine times and chemistry eight or nine times, okay? Now in plan B, if we do these problem practices, there is no teaching as such involved, right? So there is more of problem practice. So if we do problem practice together, okay? Maybe three or four batches, is it together do the problem practice? Then we can meet nine into four times only for physics. So we'll be able to meet 36 times. Of course that will not be physically possible. So it will have to be virtual sessions. So it will be either this or eight or nine times, okay? So we plan to do revision where we'll be having problem solving only. So you have to pick which one do you think is better for you? Do you think that physical classes has a lot of advantage so that even if you get 36 classes, almost every alternate day will be your physics class. Or maybe at times in a single day you'll have a physics as well as maths class if it is online. So it will be very rigorous sort of thing, okay? So you have to tell me A or B. Can you just, I mean it's just an initial poll, okay? I'm not going to decide anything based on this. I just wanted to take a feedback. Which one you want, A or B? When it comes to revision only, okay? I'm talking about these number till number. This will be physical class, okay? Three times a week, okay? Three times a week, okay? This will be online, fine? This will be multiple times a week. This could be, this could go up to ten times a week. Yes, Vishis, I understand that, you know, that's correct. It is little time consuming, but this is a desperate time so desperate measure has to be taken and I want you to, you know, I want you to work hard because this is a time where it is very important, okay? You have to work hard now. It is hardly a matter of, you know, two months. Now, if we are not willing to work hard, you know, even for two months, then let's not expect anything. But then, yeah, three hours into 12 times is a little difficult. So that's what we have made ten times. At least once in a day, if we meet, okay? So Andhra is saying meeting in person is required. So we can have a hybrid thing. See, right now what will happen is that, see, I have revised kinematics and optics for you, okay? Now, yes, I'm here. I'm talking something else right now. All three subject Vishis, I mean, it's not a single subject ten times, okay? Fine. So, see, I'm going to do the same thing now for others, all right? So those sessions will not be useful to you guys, fine? But then if I, if suppose I have, you know, if I have more number of classes, then probably same kinematics which we have revised only for two hours, okay? Same kinematics, we would be revising for, let us say, seven, eight hours. We'll be solving many, many different kinds of questions. But if it is once in a week type of affair, then it will be like, you know, max eight to 10 questions and it will be just an introduction sort of thing. But then, yeah, physically, it is better, no doubt about it. But then you have to see, physically, we'll be meeting three times a week compared to online where it is more than 30 times, fine? So, Khushal, I'm not asking suggestion on what I should do on, let's say, test, okay? That I will, I'll let you know what you have to do right now what I'm asking, let's focus on that. Okay, so, all right, guys, quick, okay? Just a final thing, without any explanation, I don't want any explanation. Just type in A or B, that's it, okay? Type in A or B, all of you quickly, then we can end this class, okay? Fine, all right, so that's it for today, okay? And guys, you have online test logins with you, okay? You have online test logins with you, so you have to make sure that tomorrow, without fail, you should take test on kinematics, all the tests which are available on kinematics, tomorrow you guys have to take online test on kinematics as per your convenient time, okay? I'm not saying that this is the time you should take the test. As per your convenient time, please take all kinematics tests which are there available, okay? And if there are ray optics test available, please take level one of ray optics test, okay? Is it fine, all of you? Tomorrow, all of you are taking all the tests on kinematics and level one on ray optics, fine? Okay, so that's it from my side. In case you have any doubts, quickly type in or else we will end this session now. Okay, thank you very much, I hope you have learned many things today and do well, work hard, hardly a matter of couple of months, thank you.