The oxygen molecule has a mass of $5.30 \times 10^{-26} \; kg$ and a moment of inertia of $1.94 \times 10^{-46} \; kg \cdot m^{2}$ about an axis through its centre perpendicular to the line joining the two atoms. Suppose the mean speed of such a molecule in a gas is $500 \; m/s$ and that its kinetic energy of rotation is two-thirds of its kinetic energy of translation. Find the average angular velocity of the molecule.

(N/A) Given: Mass of oxygen molecule $m = 5.30 \times 10^{-26} \; kg$,Moment of inertia $I = 1.94 \times 10^{-46} \; kg \cdot m^{2}$,Mean speed $v = 500 \; m/s$.
The kinetic energy of translation is $KE_{trans} = \frac{1}{2} m v^{2}$.
The kinetic energy of rotation is $KE_{rot} = \frac{1}{2} I \omega^{2}$.
According to the problem,$KE_{rot} = \frac{2}{3} KE_{trans}$.
Substituting the expressions: $\frac{1}{2} I \omega^{2} = \frac{2}{3} (\frac{1}{2} m v^{2})$.
$I \omega^{2} = \frac{2}{3} m v^{2}$.
$\omega^{2} = \frac{2 m v^{2}}{3 I}$.
$\omega = \sqrt{\frac{2 m v^{2}}{3 I}} = v \sqrt{\frac{2 m}{3 I}}$.
Substituting the values: $\omega = 500 \times \sqrt{\frac{2 \times 5.30 \times 10^{-26}}{3 \times 1.94 \times 10^{-46}}}$.
$\omega = 500 \times \sqrt{\frac{10.60 \times 10^{-26}}{5.82 \times 10^{-46}}} = 500 \times \sqrt{1.821 \times 10^{20}}$.
$\omega = 500 \times 1.349 \times 10^{10} \approx 6.75 \times 10^{12} \; rad/s$.