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MHT CET 2026 (13 May, Shift 1) · Centripetal Acceleration & Force

Centripetal Acceleration & Force question from MHT CET 2026 (13 May, Shift 1)

Q1MHT CET 2026 · 13 MAY · SHIFT 1Medium

Centripetal Acceleration & Force

The real force F acting on particle of mass m performing circular motion acts along the radius of circle r and is directed forwards the centre of the circle. The magnitude of square root of such a force is (T = periodic time)

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Answer: (a)

Centripetal force $F = m\omega^2 r = \frac{4\pi^2 m r}{T^2}$ .
$\sqrt{F} = \frac{2\pi\sqrt{mr}}{T}$
So the answer is (a) $\frac{2\pi\sqrt{mr}}{T}$ .

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