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NEET-XII-Physics
exam-1 year:2016
- #1From a disc of radius R and mass M, a circular hole
of diameter R, whose rim passes through the centre
is cut. What is the moment of inertia of the
remaining part of the disc about a perpendicular
axis, passing through the centre ?
(1) 15 ``MR^2``/32
(2) 13 ``MR^2``/32
(3) 11 ``MR^2``/32
(4) 9 ``MR^2``/32digAnsr: 2Ans : 2
Sol.
R
R
M
ITotal disc =
2MR
2
MRemoved =
M
4
(Mass  area)
IRemoved (about same Perpendicular axis)
=
2M (R /2)
4 2
+
 
 
 
2
M R
4 2
=
23MR
32
IRemaing disc = ITotal - IRemoved
=
2MR
2
-
3
32
MR2 =
213 MR
32