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JEE Advanced 2023 Paper 1 Solutions
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© examsnet.com
Question : 13
Total: 51
Two point-like objects of masses
20
gm
and
30
gm
are fixed at the two ends of a rigid massless rod of length
10
cm
. This system is suspended vertically from a rigid ceiling using a thin wire attached to its center of mass, as shown in the figure. The resulting torsional pendulum undergoes small oscillations. The torsional constant of the wire is
1.2
×
10
−
8
N
m
rad
−
1
. The angular frequency of the oscillations in
n
×
10
−
3
rad
s
−
1
. The value of
n
is________.
Your Answer:
Validate
Solution:
m
1
=
30
gm
m
2
=
20
gm
Moment of inertia about the axis of rotation is
I
=
m
1
r
1
2
+
m
2
r
2
2
Clearly
r
1
=
4
cm
And
r
2
=
6
cm
∴
I
=
(
30
×
10
−
3
×
16
×
10
−
4
)
+
(
20
×
10
−
3
×
36
×
10
−
4
)
⇒
I
=
1200
×
10
−
7
kg
m
2
If the system is rotated by small angle '
θ
', the restoring torque is
τ
(
R
)
=
−
k
θ
And
d
2
θ
d
t
2
=
−
k
l
⋅
θ
=
−
ω
2
θ
=
−
1.2
×
10
−
8
1200
×
10
−
7
⋅
θ
∴
ω
2
=
10
−
4
So,
ω
=
1
100
rad
∕
s
⇒
ω
=
10
×
10
−
3
rad
∕
s
© examsnet.com
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