Q 11-12-140JEE MainJEE Main 2019 (9 Jan, Shift 1)Easy
A mixture of 2 moles of helium gas (atomic mass $= 4\ \text{u}$), and 1 mole of argon gas (atomic mass $= 40\ \text{u}$) is kept at $300\ \text{K}$ in a container. The ratio of their rms speeds $\dfrac{V_{rms}(\text{helium})}{V_{rms}(\text{argon})}$ is close to
Answer: (D) $3.16$
Both gases are at the same temperature and $v_{rms} = \sqrt{3RT/M}$, so the number of moles does not matter:
$$\frac{v_{He}}{v_{Ar}} = \sqrt{\frac{M_{Ar}}{M_{He}}} = \sqrt{\frac{40}{4}} = \sqrt{10} \approx 3.16$$
Solution by Sreeraj P, M.Sc Physics