 All right, so now let us take up previous year questions which have actually came in J mains exam. Okay, so option C. Others, you're also getting option C. Yes, sir. Yes, sir. Okay, so let me quickly, you know, a disc of mass m and radius r is rolling with angular speed omega on the plane, the magnitude of angular momentum on the disc about the origin into find the angular momentum about the origin. Okay, how, which formula should I use angular momentum is equal to r cross m vcm plus icm into omega. Fine. Now, from this axis, what is the perpendicular distance of velocity center of mass? This is velocity center of mass, this one. Okay, so this is a perpendicular distance or not from this axis, right? So r is a perpendicular distance. So I will be able to write this as perpendicular distance into m into vcm, which is r into m into vcm. Now you can look at the option it is in terms of omega. So I'll try to write in terms of omega only vcm, which is nothing but omega into r. Okay, plus icm into omega. icm of the disc is what? mr square by 2 omega. Okay, so this comes out to be 3 by 2 mr square omega, which is option C. Okay, this came in 1999. Try to solve the next one. Anyone got the answer? Anyone? Option A. Sir, is it A? No, no. Okay, so tell me one thing. After it hits O, okay, after it hits O, what will happen to the block? The block will try to rotate about O? Yes or no? Yes, sir. Okay, so if it tries to rotate about O, then which concept will you use? Sorry, the answer is A only. I was looking at some other answer. Okay, so the correct answer is A. Alright, so what will you try to use? Which concept will you use? Here, how you have arrived the answer A? You got A, right? Sorry. What is the definition of angular momentum? About which axis? About this corner axis coming out of the plane. Okay, so if you conserve angular momentum about this axis, okay, then you will get the value of omega. Okay, so there is a force because of this ridge, because of this, you know, ridge. But then the torque due to this force about this axis when it hits will become zero. Okay, because the force passes through the corner itself. So angular momentum about that corner is zero. Alright, and since there is no external torque due to this force also, hence I can conserve the angular momentum about this axis. Okay, now is there any torque due to the gravity force? Yes or no? The gravity is acting like this. You are saying there is no torque due to gravity about this axis. Is that correct? There is a torque due to gravity, which is mg into a by 2. But we are ignoring that. Why? You can still ignore the external torque if that is very small. You can conserve the angular momentum just before and after collision. Okay, if the torques are acting very small, Purvik mg is not the internal force. The earth is applying mg force on the block. Okay, earth is not the part of your system. So there is an external torque, but still you are able to conserve the angular momentum because external torque is very small. That is not able to change the angular velocity of the mass very quickly. So it is non-impulsive torque. Okay, it is like when you have to just collide in air, you can just before and after you can conserve linear momentum. Similarly here, since torque due to mg is very small, I can conserve angular momentum just before and after. You know, it hits the ridge. That is the reason why. Fine. So initial angular momentum of this block is mv into a by 2. Okay, this block is just going straight. It behaves like a point. Okay, and this point is located at the center of mass. So angular momentum of a point is perpendicular distance into linear momentum. When do you know you can ignore it? So on the other hand, there are these forces which are constantly acting. For example, gravity force is there which cannot suddenly increase their magnitude or they can't suddenly change the angular velocity. So it is non-impulsive. Gravity is always non-impulsive. Normal reaction is not non-impulsive. Then spring force is also not impulsive. Okay. But normal reaction is impulsive because normal reaction can suddenly change the momentum or angular momentum. So at times you need to make a judgment of yours because friction also can be impulsive at times because friction is nothing but mu times normal reaction. So if normal reaction is impulsive, mu times normal reaction also becomes impulsive. So just a matter of couple of questions, practice, you will be able to develop that understanding when to take it impulsive and when you can ignore it and things like that. But whenever you are in doubt, for example here, you don't have any other choice than to use the conservation of angular momentum. Shruti, where MG to be considered, what direction will it rotate the block? MG is trying to rotate it in anti-clockwise direction. So about that axis, about that corner. Fine. So MV into A by 2 is the initial angular momentum. Now when it hits the ridge, immediately the block will come to rest and it will try to rotate about this axis. So the whole thing will just start rotating like that, tries to rotate like that. So this distance is important to know here because I am going to write here angular momentum to be equal to I about fixed axis into omega. Now I about fixed axis is what? This axis. Fine. So if the axis is fixed, torque is equal to I alpha and angular momentum is also equal to fixed axis. Moment of inertia into omega. So since there is a fixed axis, I am trying to use that fixed axis angular, sorry fixed axis. Moment of inertia is I center of mass plus M into this distance square MR square into omega. Now we know that R is root 2 A by 2. So it is A by root 2. Okay, so when you substitute the values, you will get option A to be correct. Fine. So this also came in 1999. Let's move to next question. Doing good. Getting answers quickly. But then you got to minus 1 also. Yes. So okay, a cubical block of length L rest on a rough horizontal surface with coefficient of friction mu. Hirental force is applied if the coefficient of friction is sufficiently high so that the block does not slide before toppling. The minimum force required to topple the block. Okay, now suppose the block is about to topple, then what are changes we will observe here? So the normal reaction will shift. Exactly. The point of contact from the surface will become just this point when it is about to rotate. So normal reaction will be acting from this point force like this. Okay. And the MG force is from the center of the mass because gravity is uniform everywhere. So this is MG. Okay. Now when it is about to topple, the torque about this fixed axis is about to be greater than 0. But right now it is 0 only when it is just about to rotate. Okay. So F into L is trying to rotate this whole block like this. Okay. The whole block F into L tries to rotate like that. But MG torque is trying to rotate like this. So these two torques are equal in opposite. Sorry. These two torques. Yeah. These two torques are equal in opposite right now. Okay. So F into L minus MG into, this is a cubical block. Right. So MG into L by 2, this should be equal to 0. Okay. So if you increase your force beyond this point when they are equal, then this block will start to rotate because then this thing will be greater than 0. And this will be equal to I alpha. Are you getting it all of you? So this will give you minimum force. Okay. If you increase beyond this, it will start rotating. So minimum force will come out to be just MG by 2. Okay. So the torque due to friction is not coming in picture here. It acts on the axis. The friction is acting like this. The friction force is passing through the point. So torque due to friction is 0. It will not come. Any doubts on this question? Anyone of you? Please type in. All right. This is, I'm putting here, this is one of the questions that's on calculating moment of inertia, but since, you know, nowadays questions on moment of inertia, they are more regular. Okay. Just on that small concept. So let's take one of such questions. Try to do this quickly. If F was not applied on top, but somewhere say L by 2 on the block, then then you take F into L by 2 because you have to take the perpendicular distance of the line of force. So this is the line of force from this point, which is L. So you just have to see. But if this force passes through this line, then the block will never topple because the torque due to that force becomes 0. So D. Okay. Sime is Purvik saying option D. Yes. Yes. Option number D. See, it is what it is end of the day. It's a loop. Okay. And moment of inertia of a loop should be MR square. Okay. So just find out M in terms of row and R in terms of L and just substitute there. So moment of inertia is MR square. Okay. So linear mass density is row. So the mass is row into L and the length is L. So L should be equal to 2 pi R. Okay. So R is equal to L by 2 pi. So just substitute the value of M and the value of R from here and you get the answer. Okay. So you can see that simple straight forward question is also coming in J mains. In fact, you know, you can, you can see for a fact that every mains paper you write or you see 60 to 70% paper is straight forward. So if you don't panic, if you keep your head cool, you'll definitely score more than 60% marks. Okay. So many times the, you know, keeping that calm head is more important than knowing a concept. Hmm. Do this one. It's a theoretical question. See. Conserved about the center. Angular momentum is what? Mv into R. So if, you know, velocity is changing as in velocity is decreasing and radius is also decreasing, the angular momentum that magnitude is changing. Yes or no? So angular momentum is not conserved. All right. Only direction of angular momentum is conserved. Direction of angular momentum is always perpendicular to the plane. Okay. So you can just curl your hand along the direction of rotation. Your thumb will show the direction of angular momentum. So no matter how slow it is moving, angular momentum direction remains the same. Okay. If it is moving anti clockwise, it is coming out of the paper. Okay. So by angular momentum direction will be always conserved. Okay. And there's no reason why it will spiral towards the center. Okay. And its acceleration is towards the center or not. Suppose there is an object whose velocity is decreasing. So this velocity is decreasing. So it means that it has tangential acceleration and then it has a radial acceleration also. Okay. So net acceleration will be a vector sum of this and that. So it will not be towards the center. It will be somewhere like this. So if there is a tangential acceleration which decreases the mind-reader velocity and this is a radial acceleration, the net acceleration will have a direction like that. So even D is not correct. Okay. So option B is correct. So you can see that the theoretical questions are at times more difficult to answer than the numerical ones. Because numerical you will have a very definitive way of moving forward and arriving at the answer. But theoretically if somebody asks some concept, then you can, you start to wonder in all sorts of directions. Okay. So it becomes slightly difficult. So don't take theoretical questions for granted. Okay. So don't just mark any option which you feel correct and move ahead. So you may not be getting the theoretical question correct. If you don't take it on your ego that you are not getting a theoretical question, it is okay that you are not able to answer it, move ahead. That's what my point is. Okay. Do this. This came in 2005. So is it C? All of you getting C? Yes sir. Yes sir. We showed that spiral has, I mean it has a very regular shape. So anything that radius decreases and velocity also decreases need not be a spiral path. So the speed might be decreasing in some other way. So it might not be a linear uniform way. The speed is decreasing. Okay. So let's see in this question. A block of mass, small m is rest under the action of capital F against a wall. Which of the following statement is incorrect? All right. So now this block is at rest. So it is not rotating. Okay. So this block since it is not rotating the torque about every axis should be equal to 0. Okay. So if there is a normal reaction, let us say let's take about the center. Okay. Let's take the torque about the center. So there's a friction also. Okay. So friction force will be acting upwards. Let us say, okay. And there is a normal reaction. So let's say the normal reaction is passing through the center so that there is no torque due to normal reaction. Then if normal reaction passes through the center, then about center of mass, normal reaction produces zero torque. So will be this external force. Even the external force will not produce any torque because external force is also passing through the center. Okay. So even this external force torque is zero and then even the mg torque is zero. So only torque remains is talked due to this friction. Okay. So there is no torque about center of mass and there is no other force. Right. Hence the normal reaction will act in such a way that it will shift slightly down. Okay. It shifts slightly down so as to counterbalance the torque due to the friction. All right. And hence normal reaction will produce a torque. And that is the reason why option D is the answer for this particular question. Okay. Any doubt with respect to this? Anything? Sir, so why not C? Why not? C. Could you explain that once more? F will not produce torque. See here. The question says which one of these is incorrect. So these incorrect because N doesn't produce. N will produce the torque. Okay. F will also produce a torque. Fine. And F will be equal to normal reaction and friction force should be equal to mg. If you balance the forces. Any other doubt with respect to this? Sir, if both will produce a torque then so both N and F will produce a torque. See about center of mass you have to see about center of mass this capital M is passing through the center of mass. Okay. But normal reaction is not passing through the center of mass. It is shifted down. Okay. And hence the torque due to the capital F about the center of mass is zero. And that is the reason why C is no C is this question is asking which of the following statement is incorrect. So C is correct. All right. It's not that C is not correct.