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JEE Main 2021 (26 Feb, Shift 1) · Centripetal Acceleration & Force

Centripetal Acceleration & Force question from JEE Main 2021 (26 Feb, Shift 1)

Q37JEE Main 2021 · 26 FEB · SHIFT 1Medium

Centripetal Acceleration & Force

A particle is moving with uniform speed along the circumference of a circle of radius R under the action of a central fictitious force F which is inversely proportional to R3. Its time period of revolution will be given by:

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Answer: (d)

$m\omega^2 R = F \propto \frac{1}{R^3}$
$\omega^2 \propto R^{-4} \Rightarrow \omega \propto R^{-2}$
$T = \frac{2\pi}{\omega} \propto R^2$
So the answer is (d) $T \propto R^2$ .

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