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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 : 85
Total: 100
Moment of inertia of a cylinder of mass
M
,
length
L
and radius
R
about an axis passing through its centre and perpendicular to the axis of the cylinder is
I
=
M
(
R
2
4
+
L
2
12
)
.
If such a cylinder is to be made for a given mass of a material, the ratio
L
∕
R
for it to have minimum possible
I
is:
[Sep. 03, 2020 (I)]
2
3
3
2
√
3
2
√
2
3
Validate
Solution:
Let there be a cylinder of mass
m
length
L
and radius
R
. Now, take elementary disc of radius
R
and thickness
d
x
at a distance of
x
from axis
O
O
′
then moment of inertia about
O
O
′
of this element.
d
I
=
d
m
R
2
4
+
d
m
x
2
⇒
I
=
∫
d
I
=
∫
d
m
R
2
4
+
n
=
−
L
∕
2
∫
n
=
L
∕
2
M
L
d
x
×
x
2
Given :
I
=
M
R
2
4
+
M
L
2
12
⇒
I
=
M
4
×
V
π
L
+
M
L
2
12
⇒
I
=
M
V
4
π
L
+
M
L
2
12
d
I
d
L
=
−
m
V
4
π
L
2
+
M
×
2
L
12
=
0
⇒
V
=
2
3
π
L
3
⇒
π
R
2
L
=
2
3
π
L
3
∴
L
R
=
√
3
2
© examsnet.com
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