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Question

One end of a massless spring of constant 100 N/m and natural length 0.5 m is fixed and the other end is connected to a particle of mass 0.5 kg lying on a frictionless horizontal table. The spring remains horizontal. If the mass is made to rotate at angular velocity of 2 rad/s, then elongation of spring is


A
0.1 m
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B
10 cm
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C
1 cm
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D
0.01 cm
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Solution

The correct option is C 1 cm
Centrifugal force is balanced by spring force, thus,
$$kx=m{ \omega  }^{ 2 }\left( l+x \right) \\ \Rightarrow x=\dfrac { m{ \omega  }^{ 2 }l }{ k-m{ \omega  }^{ 2 } } =0.01m=1cm$$

Physics

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