 Alright so here is a question on again moment of inertia concept. Let's see what is this about. A lamina is made by removing a small disc of diameter 2r from a bigger disc of uniform mass density and radius 2r. Okay so you can refer the figure also while reading the question. The moment of inertia of this lamina about the axis passing through O and P is I0 and IP respectively. So moment of inertia about the axis that pass through O and perpendicular to this plane. Okay that is I0 and about IP is the moment of inertia which passes through the P. Okay both this axis are perpendicular to plane. The ratio of IP by I0 is what? Okay now let us you know understand first thing that you know this particular figure you can add a small disc to it. You can add this figure on to this smaller disc and what you will get? You will get the full disc. Okay so using principle of superposition if I have to find moment of inertia about O of this figure. Okay then I can say that is equal to moment of inertia about O for this particular figure. Let's say that is I1 and this is I2 then I0 will be equal to I1 minus I2. Okay but both the moment of inertia should be about the same point. So O is here. Okay this is O and here this is O because when you add this smaller disc you are adding it here isn't it? So point O will come out to be at the tip here. Now one more thing you need to take care of here is that it is a uniform mass density. So you can say that the mass density is sigma per unit area. Okay so just multiply sigma with area you will get the mass. Okay now let us first try to find out what is I1. I1 about this O should be what? MR square by 2. Okay now let me replace this capital R with a small r so that we don't get confused. This is radius of this particular disc which happens to be 2r. Okay so the radius is given as 2r here. Now capital M is what? Sigma into pi into 2r. So pi 2r square is the area of the bigger disc so sigma into that is mass. This into 2r square by 2. Alright so what you get here is 4 into 416 divided by 2 it is 8 sigma pi r to the power 4. Okay so this is I1 and let's see what is I2. Now I2 is I'll first find a moment of inertia about this which is center of mass of I2 and then shift that axis till O. Okay so what I'll get is mass of the smaller disc which is sigma into pi r square which is the area of the smaller disc. Okay this is M MR square by 2. Okay plus this is M into r square. Fine so when you simplify what you get here is 3 by 2 sigma pi r to the power 4. Okay so I0 will come out to be equal to 8 minus 3 by 2 sigma pi r to the power 4. Okay this happens to be 13 by 2 sigma pi r to the power 4. Okay so let me see let me say that this is you know equation number one. Okay now let's try to find out moment of inertia about this point. Okay now moment of inertia about P is again using super principle of superposition will be I1 dash minus I2 dash. Okay now I1 dash will be what I1 dash is moment of inertia of this bigger disc about that point this is P. Okay so that is equal to I center of mass which we have already found which is 8 sigma pi r to the power 4 so that is this plus M into r square. M is what sigma into pi 2r square this is M into r square. Okay now r you have to constantly remind yourself is 2r over here. Okay so this is what this will come out to be 4 into 4 16 16 plus 8 so this will come out to be 24 sigma pi r to the power 4. Okay so this is I1 dash now what is I2 dash? I2 dash I need to first find out moment of inertia about this point. I'll find about this point moment of inertia which is center of mass and then I'll translate my axis from that point. Okay I will transmit my axis from here till point P. Okay so this is what distance this distance is the distance between center of mass axis and the axis of P. Okay these are the this is the distance between two parallel axis this happens to be under root 2r square plus r square that is equal to root 5 times r. Okay so you can see that there is a right angle triangle over here so this is under root 2r square plus r square that is why it is root 5r. Okay so this will be equal to M r square by 2 so this is M sigma into area r square by 2 M r square plus this is M into distance square distance square is what 5 into r square fine so this comes out to be equal to this is to have the 10 and this will be 11 by 2 sigma pi r to the power 4. Okay so let me write here it will be equal to Ip I1 dash minus I2 dash that will be 24 minus 11 by 2 times sigma pi r to the power 4. Okay so this is 48 minus 11 so that is 37 by 2 sigma pi r to the power 4. Okay now Ip by Io we need to find out right so Ip divided by Io will come out to be 37 divided by 3 37 divided by 13 okay so this is what you will get and this is closest to integer 3 right so I'll take option which says integer 3 as my answer. Okay