The temperature of an ideal gas in increased from 100 K to 400 K . If ' x ' is the r.m.s. velocity of its molecules at 100 K , then at 400 K , the r.m.s. velocity becomes

The root mean square (RMS) velocity of gas molecules is defined by $v_{\text{rms}} = \sqrt{\frac{3RT}{M}}$, where $R$ is the universal gas constant, $T$ is the absolute temperature, and $M$ is the molar mass.

This shows that $v_{\text{rms}} \propto \sqrt{T}$.

For initial temperature $T_1 = 100$ K and initial velocity $v_1 = x$, and final temperature $T_2 = 400$ K with final velocity $v_2$, the ratio becomes:

$\frac{v_2}{x} = \sqrt{\frac{400}{100}} = \sqrt{4} = 2$

So $v_2 = 2x$, which corresponds to option C.