Question

The oxygen molecule has a mass of $5.30\times 10^{-26}kg$ and a moment of inertia of $1.94\times 10^{-46}kg-m^{-2}$ about an axis through its centre perpendicular to the lines joining the two atoms. Suppose the mean speed of such a molecule in a gas is $500m/s$ and that is $KE$ of rotation is $\frac{2}{3}$ of its KE translation. Find the average angular velocity of the molecule.

• A. $3.75\times 10^{12}rad/s$
• B. $5.75\times 10^{12}rad/s$
• C. $9.75\times 10^{12}rad/s$
• D. $6.75\times 10^{12}rad/s$

Answer 1

Answer: (D)

Mass of oxygen molecule
$(M)=5.30\times 10^{-26}kg$
Moment of inertia $I=1.94\times 10^{-46}kg-m^2$
Mean speed of the molecule $(v)=500m/s$
Given, $KE$ of rotation $=\frac{2}{3}\times KE$ of translation
$\frac{1}{2}I{\omega }^2=\frac{2}{3}\times \frac{1}{2}M{v}^2$
or $\omega =\sqrt{\frac{2M{v}^2}{l}}$
$=\sqrt{\frac{2\times 5.30\times 10^{-26}\times (500{)}^2}{1.94\times 10^{-46}}}$
$=1.35\times 10^{10}\times 500$
$=6.75\times 10^{12}rad/s$