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Physics · PYQ NEET UG 2019(2)

हिन्दी

A 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,

  1. A
    \( \frac{V_0^2}{4\pi n r} \text{ rad/s}^2 \)
  2. B
    \( \frac{V_0^2}{n} \text{ rad/s}^2 \)
  3. C
    \( \frac{V_0^2}{2\pi n r^2} \text{ rad/s}^2 \)
  4. D
    \( \frac{V_0^2}{4\pi n r^2} \text{ rad/s}^2 \)
View Answer & Solution

Correct answer · Option D

\( \frac{V_0^2}{4\pi n r^2} \text{ rad/s}^2 \)

Explanation

Using \( \omega^2 = \omega_0^2 + 2\alpha \theta \), where \( \omega = \frac{V_0}{r} \) and \( \theta = 2\pi n \), we get \( \alpha = \frac{V_0^2}{4\pi n r^2} \).
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