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IIT JEE Advanced 2012 Paper 1
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© examsnet.com
Question : 19
Total: 60
A lamina is made by removing a small disc of diameter 2R from a bigger disc of uniform mass density and radius 2R, as shown in the figure. The moment of inertia of this lamina about axes passing through O and P is
I
O
and
I
P
respectively. Both these axes are perpendicular to the plane of the lamina. The ratio
I
O
I
P
to the nearest integer is _____.
Your Answer:
Validate
Solution:
We have
I
P
=
[
4
m
R
2
2
+
m
(
4
R
2
)
]
−
[
m
4
R
2
2
+
m
4
5
R
2
]
=
m
R
2
[
(
2
+
4
)
−
(
1
8
+
5
4
)
]
=
m
R
2
[
(
2
+
4
)
−
(
1
8
+
5
4
)
]
I
p
=
m
R
2
(
6
−
11
8
)
=
37
8
m
R
2
(1)
I
O
=
(
4
m
R
2
2
)
−
(
m
4
R
2
2
+
m
4
R
2
)
I
O
=
m
R
2
[
2
−
(
1
8
+
1
4
)
]
=
m
R
2
[
2
−
3
8
]
=
m
R
3
(
13
8
)
(2)
Dividing Eq. (1) by Eq. (2), we get the ratio
I
O
I
P
to the nearest integer as
I
P
I
O
=
37
8
13
8
~ 3
© examsnet.com
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