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JEE Main 2023 (24 Jan, Shift 2) · Centripetal Acceleration & Force

Centripetal Acceleration & Force question from JEE Main 2023 (24 Jan, Shift 2)

Q24JEE Main 2023 · 24 JAN · SHIFT 2Medium

Centripetal Acceleration & Force

A body of mass 200g is tied to a spring of spring constant 12.5 N/m, while the 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 rad/s, then the ratio of extension in the spring to its natural length will be:

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Answer: (b)

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 (b) $2 : 3$ .

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