 All right, friends, so here is a question on rigid-bodied dynamics. So this is with respect to the movement of inertia concept, OK? So during the past few years, many questions were asked from this small topic. So this is a small disjoint set. Or let us say this particular concept does not mix up with many other concepts in this chapter. So there can be questions which can be only on movement of inertia. So this is one of it. So at times, if you find, for example, the entire chapter to be very difficult, you can probably pick a few topics like movement of inertia and be good at it. Because if a question on such topic comes, then you should definitely be able to get it, OK? So let's see how we can solve this particular question. So we have four solid spheres. Each of diameter root 5 centimeter and mass 0.5 kg, they are placed with the centers at the corner of a square of side 4 centimeter, OK? So let me draw a square over here. So let's say this is the square. This question is talking about. The square has the side length of 4 centimeter. And you have four solid spheres. 1, 2, 3, 4, OK? So we have four solid spheres at the corner of the four, of the corners of the square. We need to find moment of inertia of a system about the diagonal of the square, OK? So let me draw that axis itself. So basically, my interest is to find moment of inertia about this line, OK? So let us see how we can proceed with this particular question, OK? So the mass and the radius of all the sphere is same, isn't it? So let me call that as M and R itself. You can see that there is this axis, this axis, which is passing through the center of mass of the sphere. And also it is parallel to this axis, OK? So what I'll do, I'll move this axis using parallel axis theorem from here till there, OK? Where this distance, where this distance from this point to this, this is the perpendicular distance, right? This distance is what? If side length is A, let us say A is a side length, this entire diagonal length will be A root 2 and this will be A root 2 by 2. So that will be A by root 2, OK? So moment of inertia of this particular sphere, I'm just talking about this one sphere. So I center of mass is 2 by 5 M R square, OK? Now I'm translating this axis from here till there, so it will be 2 by 5 M R square, which is I plus M into D square, D is what? A by root 2, so this becomes A square by 2, OK? Now similarly, there is one more sphere this side, OK? So I'll just multiply with 2, so I get moment of inertia of these two spheres about this line, OK? Now this is moment of inertia of these two. What about the other two ones? These two, the moment of inertia about this line is 2 by 5 M R square only, OK? So the total moment of inertia is 2 by 5 M R square for, let's say, sphere A as well as for sphere B about the specified axis, OK? So this into 2 will count both of them and plus this particular expression, OK? Which is 4 by 5 M R square plus M into A square, OK? So total moment of inertia will come out to be equal to 8 by 5 M R square plus M into A square, OK? Now let's substitute the values, so it will be 8 by 5. Mass is 0.5 kgs, so this is 0.5 R square. Diameter is root 5, so radius will be what? Root 5 by 2, OK? So R square will be 5 by 4, OK? Plus mass is again 0.5 kgs into A square. Now side length is 4 centimeter, right? So A is 4, so it will be 16, all right? And everything is given in terms of centimeter, so this should be multiplied with 10 raised to the power minus 4, OK? Now let us simplify this further, so 5 and 5 will go away and this is 4 to the 8, OK? So what I'll get here is 2 into 0.5 is 1 plus this is 0.5 into 16, 0.5 is essentially 1 by 2 only. So this is 8 into 10 raised to the power minus 4, so I'm getting here as 9 into 10 raised to the power minus 4, all right? Now if I read the question again, it says that moment of inertia is N into 10 raised to the power minus 4 and then asking what is N, OK? So clearly N is equal to 9, OK? So that's how you have to solve this particular question.