A circular table of radius 0.5 m has a smooth diametrical groove. A ball of mass 90 g is placed inside the groove along with a spring of spring constant 102 Ncm-1. One end of the spring is tied to the edge of the table and the other end to the ball. The ball is at a distance of 0.1 m from the center when the table is at rest. On rotating the table with a constant angular frequency of 102 rad s-1, the ball moves away from the center by a distance nearly equal to
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a
10−1m
b
10−2m
c
10−3m
d
2×10−1m
answer is B.
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Detailed Solution
k=102 Ncm−1=104 Nm−1Let the ball move distance x away from the center as shown in figure. kx=mω2(0.1+x)⇒104x=901000×1022×(0.1+x) Solve to get x≈10−2 m