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A circular turn of radius 0.5 m has a smooth groove as shown in fig. (3). A ball of mass 90 g is placed inside
the groove along with a spring of spring constant 10² N/cm. The ball is at a distance of 0⋅1 m from the centre when the turn table is at rest. On rotating the turn table with a constant angular frequency of $10^{2} \sec^{- 1}$ , the ball moves away from the centre by a distance nearly equal to

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$10^{- 1} m$

$10^{- 3} m$

$2 \times 10^{- 1} m$

$10^{- 2} m$

answer is B.

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Detailed Solution

When the turn table rotates with angular speed ω, the particle of mass m describes a circle of radius r. The centrifugal force experienced is

${mrω}^{2} = (9 \times 10^{- 2}) (10^{- 1}) {(10^{2})}^{2} = 9 \times 10 N$

As a result of this force, the spring is compressed by a distance x. The restoring force of the spring

$= kx = 10^{4} x$

In equilibrium, $9 \times 10 = 10^{4} x$

or $x ≅ 10^{- 2} m$

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