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Test Index
JEE Main Physics Class 11 System of Particles and Rotational Motion Part 1 Questions
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© examsnet.com
Question : 20
Total: 100
A thin circular plate of mass
M
and radius
R
has its density varying as
ρ
(
r
)
=
ρ
0
r
with
ρ
0
as constant and
r
is the distance from its center. The moment of Inertia of the circular plate about an axis perpendicular to the plate and passing through its edge is
I
=
a
M
R
2
. The value of the coefficient
a
is:
[8 April 2019 I]
1
∕
2
3
∕
5
8
∕
5
3
∕
2
Validate
Solution:
Taking a circular ring of radius
r
and thickness
d
r
as a mass element, so total mass,
M
=
R
∫
0
ρ
0
r
×
2
π
r
d
r
=
2
π
ρ
0
R
3
3
I
C
=
R
∫
0
ρ
0
r
×
2
π
r
d
r
×
r
2
=
2
π
ρ
0
R
5
5
Using parallel axis theorem
∴
I
=
I
C
+
M
R
2
=
2
π
ρ
0
R
5
(
1
3
+
1
5
)
=
16
π
ρ
0
R
5
15
=
8
5
[
2
3
π
ρ
0
R
3
]
R
2
=
8
5
M
R
2
© examsnet.com
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