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JEE Main 31-Aug-2021 Shift 2 Solved Paper
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© examsnet.com
Question : 3
Total: 90
A system consists of two identical spheres each of mass
1.5
kg
and radius
50
cm
at the end of light rod. The distance between the centres of the two spheres is
5
m
. What will be the moment of inertia of the system about an axis perpendicular to the rod passing through its mid-point?
[31 Aug 2021 Shift 2]
18.75
kgm
2
1.905
×
10
5
kgm
2
19.05
kgm
2
1.875
×
10
5
kgm
2
Validate
Solution:
Given, mass of each sphere, M= 1.5 kg
Radius of each sphere,
R
=
50
cm
=
50
×
10
−
2
m
Distance between centre of spheres, l = 5 m
By using parallel axis theorem, moment of inertia of system is
I
=
2
[
2
5
MR
2
+
Ml
2
4
]
I
=
2
M
[
2
5
R
2
+
l
2
4
]
Substituting the values, we get
I
=
2
×
1.5
[
2
5
×
(
50
×
10
−
2
)
2
+
(
5
)
2
4
]
=
19.05
kg
−
m
2
© examsnet.com
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