 All right friends. So here is a question on rigid body motion. Let's see how we can analyze this particular situation. Okay. Let's first read the question. Initial angular velocity of a server disk of mass capital M is omega one. Okay. So there is a server disk. Okay. That is already rotating with omega one angular velocity. The mass is given as capital M. Okay. Then two small spheres of mass small m are attached gently to the diametrically opposite points on the edge of the disk. So what is happening is suppose you take a diameter. Okay. So this is a diameter. All right. So you have two mass which are gently placed over here. Okay. The masses are given as small m. You need to find final angular velocity of the disk. Okay. Now my system here is my system is m and the other two masses. They both contain my total system. Okay. So what is the force between small m and capital M? That is an internal force. Okay. So now since we are trying to find angular velocity, there might be some formula that might be coming in your mind. One could be, you know, can I apply conservation of energy? Okay. You may not be able to apply conservation of energy because there might be some friction between small m and capital M. Okay. So, you know, initially it may slide and then it will stop at the edge by some mechanism. So there might be some energy loss over there. Okay. Then one thing that should come in our mind is that you can conserve angular momentum about the axis of rotation. How we can do that? See there is no external torque about the axis of rotation. You can say that, you know, the mass of the capital M will be pulled by gravity with a mass of, with a force of mg, right? But then that force is applied through the center of the disk and that passes through the axis. So there is no external torque due to the weight of capital M. Okay. Similarly, if you take small m, this small m, if you take alone like this, there might be, you know, you can say that there is external force on this small m and that has some torque about this axis. But then there is an equal and opposite torque due to the weight of this small m. Fine. So, the torque get cancelled out and you can say that net net, still there is no external torque. Okay. So in this entire process before you are putting small m and after you're putting the small m, the net external torque about this axis of rotation is zero. Okay. So what we'll do? We will conserve the angular momentum. Fine. So angular momentum initially will be equal to angular momentum finally. Okay. Now this rigid body is spinning. Okay. Since it is spinning about this axis, I can say angular momentum to be i times omega. The initial angular momentum is i of the disk into omega 1. Okay. And final angular momentum, when you write final angular momentum, you have to take care of the fact that moment of inertia of this disk plus small m is changed. Okay. Now that is moment of inertia of the disk plus m into r square plus m into r square. So you have 2m into r square that times omega. Okay. Now, I know that moment of inertia of the disk is mass of the disk by 2. So I'll use that fact and put the values. Okay. So I can put it here also, mr square by 2 plus 2mr square. Okay. Now you can see that radius is not given in the question. And when you solve it, you'll see that it is not even required. The radius square is getting cancelled from everywhere. Okay. So at times, you want some data, but then it is not given in the question. Then you just assume it to be something like, for example, here, I've assumed it to be r. Okay. So if r values needed automatically, the expression will come out to be in terms of r. Right. If it is not needed, r will get cancelled out like what is happening here. Okay. So omega will come out to be m by 2 divided by m by 2 plus 2m times omega 1. Fine. So angular velocity will come out to be m divided by m plus 4m times omega 1. Right. So you can see that option number c is correct. All right. So like this, you can solve this particular question.