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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 : 68
Total: 100
A spring mass system (mass
m
,
spring constant
k
and natural length
l
) rests in equilibrium on a horizontal disc. The free end of the spring is fixed at the centre of the disc. If the disc together with spring mass system, rotates about it's axis with an angular velocity
ω
,
(
k
>
>
m
ω
2
)
the relative change in the length of the spring is best given by the option:
[9 Jan. 2020 II]
√
2
3
(
m
ω
2
k
)
2
m
ω
2
k
m
ω
2
k
m
ω
2
3
k
Validate
Solution:
Free body diagram in the frame of disc
∴
m
ω
2
(
l
0
+
x
)
=
k
x
⇒
x
=
m
l
0
ω
2
k
−
m
ω
2
For
k
>
>
m
ω
2
⇒
x
l
0
=
m
ω
2
k
© examsnet.com
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