Explanation:
Here, \({v=540}\) r.p.m., \({t=6 {~s}}\)
Angular acceleration \({(\alpha)}\) of the particle is\(\alpha = \frac{\omega }{t}\;\;\;{\mkern 1mu} {\kern 1pt} (\omega = {\text{ angular velocity }})\)
\(\,\,\,\,\,\,\,\, = \frac{{2\pi v}}{t} = \frac{{2\pi \times 540}}{{60 \times 6}}rad/{s^2}\)
\(\therefore \;\;\;\alpha = 3\pi \) \(rad/{s^2}\).
So correct option is (1)