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Physics · PYQ NEET UG 2019(2)
हिन्दीA particle starting from rest, moves in a circle of radius \'r\'. It attains a velocity of \( V_0 \) m/s in the \( n^{\text{th}} \) round. Its angular acceleration will be,
Correct answer · Option 4
\( \frac{V_0^2}{4\pi n r^2} \text{ rad/s}^2 \)
Explanation
Using \( \omega^2 = \omega_0^2 + 2\alpha \theta \), where \( \omega = \frac{V_0}{r} \) and \( \theta = 2\pi n \), we get \( \alpha = \frac{V_0^2}{4\pi n r^2} \).
