JEE Advanced 2020 · Paper 1, Chemistry question 1

The plotted distribution is symmetric about its peak.

Most probable vs average. For a symmetric distribution the mean lies at the peak, so $v_{mp}=\bar v$.

rms speed. Since $\overline{v^2}-\bar v^2$ is the variance, which is positive, $v_{rms}>\bar v$.

Only $1:1:1.224$ has this pattern. ($1:1.128:1.224$ is the ratio for the skewed Maxwell distribution, not the one drawn.)

(B)