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MHT CET 2019 (2 May, Shift 2) · Centripetal Acceleration & Force

Centripetal Acceleration & Force question from MHT CET 2019 (2 May, Shift 2)

Q43MHT CET 2019 · 2 MAY · SHIFT 2Easy

Centripetal Acceleration & Force

A particle is performing U.C.M. along the circumference of a circle of diameter 50 cm with frequency 2 Hz. The acceleration of the particle in $\text{m/s}^2$ is

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Answer: (d)

Radius $r = 50/2 = 25\text{ cm} = 0.25\text{ m}$ , $n = 2\text{ Hz}$ .
$a = 4\pi^2 n^2 r = 4\pi^2 (4)(0.25)$
$a = 4\pi^2\text{ m/s}^2$
So the answer is (d) $4\pi^2$ .

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