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JEE Main Physics Class 11 System of Particles and Rotational Motion Part 1 Questions
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© examsnet.com
Question : 27
Total: 100
Two masses
m
and
m
2
are connected at the two ends of a massless rigid rod of length
l
. The rod is suspended by a thin wire of torsional constant
k
at the centre of mass of the rod-mass system (see figure). Because of torsional constant
k
,
the restoring toruque is
τ
=
k
θ
for angular displacement
θ
. If the rod is rotated by
θ
0
and released, the tension in it when it passes through its mean position will be:
[9 Jan. 2019 I]
3
k
θ
0
2
l
2
k
θ
0
2
l
k
θ
0
2
l
k
θ
0
2
2
l
Validate
Solution:
As we know,
ω
=
√
k
I
ω
=
√
3
k
m
l
2
[
∵
I
rod
=
1
3
m
l
2
]
Tension when it passes through the mean position,
=
m
ω
2
θ
0
2
l
3
=
m
3
k
m
l
2
θ
0
2
l
3
=
k
θ
0
2
l
© examsnet.com
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