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IIT JEE Advanced 2006 Paper 1
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© examsnet.com
Question : 9
Total: 120
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
2
R
√
15
R
√
2
15
4
R
√
15
R
4
Validate
Solution:
Since the mass remains the same, we get
2
3
M
R
2
=
(
M
r
2
2
+
M
r
2
)
=
3
2
M
r
2
where R is the radius of the solid sphere and r is the radius of the disc. Therefore,
r =
2
R
√
15
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