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JEE Advanced Model Paper 15 with solutions for online practice
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© examsnet.com
Question : 2
Total: 60
From a solid hemisphere of radius ‘R’ a cone of base radius ‘R’ and height ‘R’ is removed as shown in the figure. The moment of inertia of the remaining body about an axis BB’ in the plane of the base and passing through the centre ‘O’ is
I
0
.
I
1
is the moment of inertia about AA’ which is parallel to BB’ and
I
2
is moment of inertia about an axis perpendicular to BB’, and passing through 'O’, then
I
1
=
I
0
I
2
= 2
I
0
I
1
=
I
0
2
I
2
= 3
I
0
Validate
Solution:
Mass of hemi sphere = M ρ2/3 π
r
3
Mass of cone =
M
1
= ρ × I / 3π
R
3
=
M
2
Y
c
o
m
=
(
M
)
(
3
R
∕
8
)
−
(
M
2
)
(
R
4
)
M
−
M
2
=
R
2
I
B
B
′
=
I
c
m
+ M
(
r
2
)
2
=
I
0
I
A
A
′
=
I
c
m
+ M
(
R
2
)
2
=
I
0
By symmetry
I
2
=
I
0
© examsnet.com
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