 So here is a question on rigid body motion. Let's see how we can go about solving this. So first, let us read this question. A boy is pushing a ring of mass 2 kg and radius 0.5 meter with a stick as shown. The stick applies a force of 2 Newton on the ring and rolls it without slipping. And acceleration of the ring is given. So this is linear acceleration of center of mass which is 0.3 meter per second square. The coefficient of friction between ground and ring is large enough so that rolling always occurs and the coefficient of friction between stick and the ring is P by 10. Now it talks about coefficient of friction between stick and the ring. So there is a friction between stick and the ring, okay? So that you should take care. All right, so let us see how we can analyze this situation. So like always, first of all, we will try to make a free body diagram of this ring. So this is the ring, okay? So there is a force of 2 Newton being applied, okay? Now when the ring will roll, okay? It will have a sense of rotation like this. Depending on the situation, it will roll like this, okay? And if it is rolling like this, then with respect to stick, which is always vertical, this point will slide in upward direction. So that is the reason why it will feel a friction in this direction. So this is, let us say, FR1, okay? And from here, there will be a normal reaction. Let's say this is N and also there is a force of friction over here, FR2, okay? Now FR2 is a static friction. The magnitude of FR2 is unknown, but FR1 is a kinetic friction. The magnitude of kinetic friction is mu times normal reaction, okay? And normal reaction between the stick and the ring is whatever force the stick is applying on the ring, which is 2 Newton. So mu into two, okay? There is another force, which is gravitational force, which will be acting like this. This is Mg, okay? Now let me write down the Newton's second law equation along the horizontal direction, okay? So net force along the horizontal direction, along the direction of the acceleration is two Newtons, okay? 2 minus FR2 is the net force. This is equal to mass time acceleration A, okay? So let me put the values also here so that we can solve it. 2 minus FR2 is equal to mass is what? 2 and acceleration is 0.3, okay? So FR2 will come out to be equal to 2 minus 0.6. So 1.4 Newton, the value of FR2 should be, okay? So this is something which we have already got. Now let us try to write the torque equation, okay? Torque about center of mass should be equal to I center of mass into alpha, okay? So torque about center of mass is what? Now you can see that the FR2 is trying to rotate in one way and FR1 is looped in other way. FR2, they are in the direction, okay? Subtracted to FR2 minus R into FR1, okay? This will be equal to ICM into alpha. What is ICM? For the ring, the ICM value is MR square, so this into alpha, fine? So I get the value of FR2 minus FR1 to be equal to MR into alpha, fine? And since there is no slipping at the point of contact, R times alpha is also equal to acceleration of center of mass, okay? So this is equal to M into A, which is equal to what? M is two, A is 0.3, so this is 0.6, okay? Now FR2 is 1.4, right? So I'll substitute here, 1.4 is equal to FR1 is equal to 0.6, okay? So the value of FR1 comes out to be 0.8 newtons, all right? This should also be equal to two times coefficient of friction mu, all right? So from here, I will get coefficient of friction to be equal to 0.4, okay? Now it is said that coefficient of friction, if it is P by 10, what is the value of P, okay? So 0.4, if I write that to be equal to P by 10, what I'll get P as P, I'll get it as 4, okay? So like this, you have to solve this particular question.