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JEE Main Physics Class 11 System of Particles and Rotational Motion Part 2 Questions
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© examsnet.com
Question : 79
Total: 100
The linear mass density of a thin rod AB of length L varies from
A
to
B
as
λ
(
x
)
=
λ
0
(
1
+
x
L
)
,
where
x
is the distance from
A
. If
M
is the mass of the rod then its moment of inertia about an axis passing through A and perpendicular to the rod is:
[Sep. 06, 2020 (II)]
5
12
M
L
2
7
18
M
L
2
2
5
M
L
2
3
7
M
L
2
Validate
Solution:
Mass of the small element of the rod
d
m
=
λ
.
d
x
Moment of inertia of small element,
d
I
=
d
m
.
x
2
=
λ
0
(
1
+
x
L
)
.
x
2
d
x
Moment of inertia of the complete rod can be obtained by integration
I
=
λ
0
L
∫
0
(
x
2
+
x
3
L
)
d
x
=
λ
0
|
x
3
3
+
x
4
4
L
|
0
L
=
λ
0
[
L
3
3
+
L
3
4
]
⇒
I
=
7
λ
0
L
3
12
.......(i)
Mass of the thin rod,
M
=
L
∫
0
λ
d
x
=
L
∫
0
λ
0
(
1
+
x
L
)
d
x
=
3
λ
0
L
2
∴
λ
0
=
2
M
3
L
∴
I
=
7
12
(
2
M
3
L
)
L
3
⇒
I
=
7
18
M
L
2
© examsnet.com
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