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Question

# 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 mass is made to rotate at an angular velocity of 2 rad/s. Find the elongation of the spring.

A
1 cm
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B
4 cm
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C
7 cm
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D
10 cm
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Solution

## The correct option is A 1 cmWe know that,kl=mv2/r=mω2r=mω2(l0+l)Here ω is the angular velocity l0 is the natural length (0.5)m and (l0+l)Thus (k−mω2)l=mω2l0l=mω2l0k−mω2Putting valuel=0.5×4×0.5100−0.5×4m=1100m=1cm

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