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```        thanks for the  answer of my post(on 3/9/2010).The question was:
MI  of a  ring is mr^2.
If  we  treat rings as  elements of a disc as well as of a hemisphere,then MI of   a disc and of hemisphere should be  same (becoz mass  as well as perp. distance of  every corresponding ring in both from  the axis will be same , but  actually  not. What is wrong in the concept?
(here  axis  is the axis passing through centre and perp. to plane in both cases)
but if we donot use the concept  : MI= ∫ dm.r^2  but simply see the relative  positions of rings in disc and  hemisphere(as someone has opened the  disc  to  form a  hemisphere),then what is   wrong in that?
MI(of both)=∑m.r^2             m=mass of every ring = same  in both cases
r=perp.  distance of every ring from axis=same in both cases.```
7 years ago

147 Points
```										Dear neeru
In case of hemisphere you can use this concept because mass distribution will not be same throght out its radius.
in case of disk you will get uniform mass distribution  through out its area.
but if you will deal hemisphere as a disk then as you move away from the rotation axis then for same dr distance you will get more mass because hemisphere surface become more and more verticle . (mass distribution is not same in this case).
Please feel free to post as many doubts on our discussion forum as you can.If you find any question Difficult to understand - post it here and we will get you the answer and detailed  solution very  quickly. We are all IITians and here to help you in your IIT JEE preparation.All the best. Regards,Askiitians ExpertsBadiuddin
```
7 years ago
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