A body of mass 200 gram is tied to a spring of spring constant 12.5… | MHT CET 2023 (10 May, Shift 1)

Q22MHT CET 2023 · 10 MAY · SHIFT 1Medium

Centripetal Acceleration & Force

A body of mass 200 gram is tied to a spring of spring constant $12.5\text{ N/m}$ , while other end of spring is fixed at point ' $O$ '. If the body moves about ' $O$ ' in a circular path on a smooth horizontal surface with constant angular speed $5\text{ rad/s}$ then the ratio of extension in the spring to its natural length will be

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Answer: (c)

Spring force gives the centripetal force: $kx = m\omega^2(l + x)$ .
$m\omega^2 = 0.2 \times 25 = 5$ , so $12.5x = 5(l + x)$ .
$7.5x = 5l \Rightarrow \frac{x}{l} = \frac{2}{3}$
So the answer is (c) $2 : 3$ .

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Centripetal Acceleration & Force question from MHT CET 2023 (10 May, Shift 1)

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