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Physics · SYSTEMS OF PARTICLES AND ROTATIONAL MOTION

हिन्दी

An oxygen molecule has mass \( M = 5.30 \times 10^{-26} \text{ kg} \) and a moment of inertia \( I = 1.94 \times 10^{-46} \text{ kg m}^2 \) about an axis perpendicular to the line joining the two atoms. If the mean translational speed of the molecule in a gas is \( 500 \text{ m s}^{-1} \) and its rotational kinetic energy is two-thirds of its translational kinetic energy, its average angular velocity \( \omega \) is approximately:

  1. \( 4.41 \times 10^{11} \text{ rad s}^{-1} \)
  2. \( 1.15 \times 10^{12} \text{ rad s}^{-1} \)
  3. \( 2.14 \times 10^{13} \text{ rad s}^{-1} \)
  4. \( 6.74 \times 10^{12} \text{ rad s}^{-1} \)

Correct answer · Option 4

\( 6.74 \times 10^{12} \text{ rad s}^{-1} \)

Explanation

\( K_{\text{trans}} = \frac{1}{2} M v^2 = \frac{1}{2}(5.30 \times 10^{-26})(2.5 \times 10^5) = 6.625 \times 10^{-21} \text{ J} \). \( K_{\text{rot}} = \frac{2}{3} K_{\text{trans}} = 4.417 \times 10^{-21} \text{ J} \). Since \( \frac{1}{2} I \omega^2 = K_{\text{rot}} \), \( \omega = \sqrt{\frac{2 K_{\text{rot}}}{I}} = \sqrt{\frac{2 \times 4.417 \times 10^{-21}}{1.94 \times 10^{-46}}} = \sqrt{4.553 \times 10^{25}} \approx 6.74 \times 10^{12} \text{ rad s}^{-1} \).
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