Q 11-12-106JEE MainJEE Main 2021 (20 Jul, Shift 1)Easy
Consider a mixture of gas molecule of types $A$, $B$ and $C$ having masses $m_A < m_B < m_C$. The ratio of their root mean square speeds at normal temperature and pressure is:
Answer: (D) $\frac{1}{v_A} < \frac{1}{v_B} < \frac{1}{v_C}$
$v_{rms} = \sqrt{\dfrac{3kT}{m}}$. At the same temperature the lightest molecule is fastest: $v_A > v_B > v_C$, so $\dfrac1{v_A} < \dfrac1{v_B} < \dfrac1{v_C}$.
Solution by Sreeraj P, M.Sc Physics