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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 : 23
Total: 100
An equilateral triangle
A
B
C
is cut from a thin solid sheet of wood. (See figure)
D
,
E
and
F
are the mid-points of its sides as shown and
G
is the centre of the triangle. The moment of inertia of the triangle about an axis passing through
G
and perpendicular to the plane of the triangle is
I
0
. If the smaller triangle DEF is removed from
A
B
C
, the moment of inertia of the remaining figure about the same axis is I. Then:
[11 Jan. 2019 I]
I
=
15
16
I
0
I
=
3
4
I
0
I
=
9
16
I
0
I
=
I
0
4
Validate
Solution:
Let mass of the larger triangle
=
M
Side of larger triangle
=
l
Moment of inertia of larger triangle
=
m
a
2
Mass of smaller triangle
=
M
4
Length of smaller triangle
=
l
2
Moment of inertia of removed triangle
=
M
4
(
a
2
)
2
∴
I
removed
I
original
=
M
4
M
.
(
a
2
)
2
(
a
)
2
I
removed
=
I
0
16
So,
I
=
I
0
−
I
0
16
=
15
I
0
16
© examsnet.com
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