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One end of a massless spring of spring 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 table is made to rotate at an angular velocity of 2 rad/s, find the elongation of the spring(approximately).


A

1 cm

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B

51 cm

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C

50 cm

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D

None of these

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Solution

The correct option is A

1 cm


The particle is moving in a horizontal circle, so it is accelerated towards the centre with magnitude v2r . The horizontal force on the particle is due to the spring and equals kl, where l is the elongation and k is the spring constant. Thus, kl=mv2r=mω2r=mω2(l0+l).
kl=12(4)(l+l0)
100l=2(l+12)
l=1 cm

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