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One end of a massless spring of spring constant 100 Nm^–1 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. If the mass rotates at an angular velocity of 2 rad s^–1, the elongation of the spring is approximately

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4 cm

3 cm

1 cm

2 cm

answer is C.

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Detailed Solution

Centrifugal force is balanced by spring force, thus,
kx = mω²(l+x)

⇒x =mω²l/(k−mω²)

=0.01m

=1cm

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