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JEE Main 11 Jan 2019 Shift 1 Solved Paper
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© examsnet.com
Question : 11
Total: 90
An equilateral triangle ABC 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 perpendicularto the plane of the triangle is
I
0
. It the smaller triangle DEF is removed from ABC, the moment of inertia of the remaining figure about the same axis is I. Then
I =
9
16
I
0
I =
3
4
I
0
I =
1
4
I
0
I =
15
16
I
0
Validate
Solution:
Suppose M is mass and a is side of largertriangle, then
M
4
and
a
2
will be mass and sidelength of smaller triangle.
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
16
I
0
© examsnet.com
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