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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 : 18
Total: 100
The moment of inertia of a uniform thin rod about a perpendicular axis passing through one end is
I
1
. The same rod is bent into a ring and its moment of inertia about a diameter is
I
2
. If
I
1
I
2
is
x
π
2
3
, then the value of
x
will be___
[29-Jun-2022-Shift-2]
Your Answer:
Validate
Solution:
I
1
=
M
L
2
3
.
.
.
.
.
(
1
)
For ring:
I
2
=
M
R
2
2
and
2
π
R
=
L
⇒
I
2
=
M
2
(
L
2
4
π
2
)
.
.
.
.
.
⇒
I
1
I
2
=
8
π
2
3
⇒
x
=
8
© examsnet.com
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