 For two gases, A and B with molar masses ${{M}_{A}}$ and ${{M}_{B}}$, it is observed that at a certain temperature T, the mean velocity of A is equal to the root mean square velocity of B. Thus, the mean velocity of A can be made equal to the mean velocity of B, if:A. A is at temperature T and B at T’, T’>T.B. A is lowered to a temperature ${{T}_{2}}$ C. Both A and B are raised to a higher temperature.D. Both A and B are placed at lower temperature. Verified
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Hint Try to think about the Kinetic theory of gases. According to the data given in the question we need to have an equation that can be used to relate the root mean square speed and the average speed of gas molecules.

Now,${{v}_{avg}}=\sqrt{\dfrac{8RT}{\pi M}}$ and ${{v}_{rms}}=\sqrt{\dfrac{3RT}{M}}$, at the temperature T, both velocities are equal. So,
$\sqrt{\dfrac{8RT}{\pi {{M}_{A}}}}=\sqrt{\dfrac{3RT}{{{M}_{B}}}}$,${{M}_{A}}\,and\,{{M}_{B}}$ are the molar masses.
So,$\dfrac{{{M}_{B}}}{{{M}_{A}}}=3\times \dfrac{\pi }{8};{{M}_{B}}>{{M}_{A}}$. That means, when they have same mean velocities, we have:
\begin{align} & \sqrt{\dfrac{8RT'}{\pi {{M}_{B}}}}=\sqrt{\dfrac{8RT}{\pi {{M}_{A}}}} \\ & Taking\,square,we\,have: \\ & \dfrac{8RT}{\pi {{M}_{B}}}=\dfrac{8RT}{\pi {{M}_{A}}} \\ & \dfrac{T'}{T}=\dfrac{{{M}_{B}}}{{{M}_{A}}}\,\,\,So,\,T'>T \\ \end{align}
\begin{align} & \dfrac{8RT}{\pi {{M}_{B}}}=\dfrac{8R{{T}_{2}}}{\pi {{M}_{A}}} \\ & So,\, \\ & \dfrac{T}{{{T}_{2}}}=\dfrac{{{M}_{B}}}{{{M}_{A}}};\,T>{{T}_{2}} \\ \end{align}
NOTE: The order of the 3 types of velocities can be remembered by recalling the word “RAM” where ‘R’ is r.m.s, ‘A’ means average and ‘M’ means most probable. ${{V}_{rms}}>{{V}_{average}}>{{V}_{most\,probable}}$.