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JEE Main Physics Class 11 System of Particles and Rotational Motion Part 1 Questions
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© examsnet.com
Question : 44
Total: 100
Moment of inertia of an equilateral triangular lamina
A
B
C
, about the axis passing through its centre
O
and perpendicular to its plane is
I
0
as shown in the figure. A cavity DEF is cut out from the lamina, where
D
,
E
,
F
are the mid points of the sides. Moment of inertia of the remaining part of lamina about the same axis is:
[Online April 8, 2017]
7
8
I
C
15
16
I
o
3
I
o
4
31
I
0
32
Validate
Solution:
According to theorem of perpendicular axes, moment of inertia of triangle
(
A
B
C
)
I
0
=
k
m
l
2
........(i)
B
C
=
1
Moment of inertia of a cavity DEF
I
D
E
F
=
K
m
4
(
l
2
)
2
=
k
16
m
l
2
From equation (i),
I
D
E
F
=
I
0
16
Moment of inertia of remaining part
I
remain
=
I
0
−
I
0
16
=
15
I
0
16
© examsnet.com
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