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```				   9.     A solid sphere of mass M, radius R and having moment of inertia about an axis passing through the centre of mass as I, is recast into a disc of thickness t, whose moment of inertia about an axis passing through its edge and perpendicular to its plane remains I. Then, radius of the disc will be
(A) 2R/√15
(B) R√(2/15)
(C) 4R/√15
(D) R/4
```

6 years ago

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```										Dear Mainak
moment of inertia of solid sphere I =2/5 MR2
mass will remain conserve
let the new radius is R1
so I = MR12/2 + MR12
=3/2 MR12

so 3/2 MR12 =2/5 MR2
R1 =2R/√15
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
```
6 years ago

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