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JEE Main levelSingle correct optionDifficulty 2 of 3ig133

If the rms speed of $CH X_{4}$ (16) at $T_{1}$ equals the most probable speed of $He$ (4) at $T_{2}$ , then $T_{1} / T_{2}$ is

A $8/3$
B $3/8$
C $2/3$
D $3/2$

Answer: A

\frac{3 R T _{1}}{16} = \frac{2 R T _{2}}{4}

. Squaring:

\frac{3 T _{1}}{16} = \frac{T _{2}}{2}

, so

\frac{T _{1}}{T _{2}} = \frac{16}{6} = \frac{8}{3}

.

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