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MHT CET 2025 (22 Apr, Shift 1) · Centripetal Acceleration & Force

Centripetal Acceleration & Force question from MHT CET 2025 (22 Apr, Shift 1)

Q7MHT CET 2025 · 22 APR · SHIFT 1Medium

Centripetal Acceleration & Force

A body of mass 100 gram is tied to a spring of spring constant $8\text{ N/m}$ , while the other end of a spring is fixed. If the body moves in a circular path on smooth horizontal surface with constant angular speed $8\text{ rad/s}$ then the ratio of extension in the spring to its natured length will be

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Answer: (d)

Spring force gives the centripetal force: $kx = m\omega^2(l + x)$ , where $l$ is the natural length.
$8x = 0.1 \times 64\,(l + x) = 6.4(l + x)$
$1.6x = 6.4l \Rightarrow \frac{x}{l} = 4$
So the answer is (d) $4 : 1$ .

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