Question

If mean square of $x-$ component of the velocity of molecules is denoted by ${\omega }^2$, then R.M.S velocity of molecules will be

• A. $\omega$
• B. $\frac{{\omega }^2}{3}$
• C. $3{\omega }^2$
• D. $\frac{\omega }{3}$

Answer 1

The correct option is C $3{\omega }^2$

Let r.m.s velocity of molecules be $v_{\text{rms}}$.

We know that:
$v_{rms}=\sqrt{\frac{\sum v_i^2}{n}}$
Here, $v_i$ is the velocity of individual particles.

Also,
$v_i^2=v_{ix}^2+v_{iy}^2+v_{iz}^2$

Therefore,

$v_{rms}=\sqrt{\frac{\sum (v_{ix}^2+v_{iy}^2+v_{iz}^2)}{n}}$
$\Rightarrow v_{rms}=\sqrt{\frac{\sum v_{ix}^2}{n}+\frac{\sum v_{iy}^2}{n}+\frac{\sum v_{iz}^2}{n}}$
$\Rightarrow v_{rms}=\sqrt{3{\omega }^2}$
$[\because \frac{\sum v_{ix}^2}{n}=\frac{\sum v_{iy}^2}{n}=\frac{\sum v_{iz}^2}{n}={\omega }^2(\text{Given}\left)\right]$

$\Rightarrow v_{rms}=\sqrt{3{\omega }^2}=\sqrt{3}\omega$