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N molecules each of mass m of gas A and 2N molecules each of mass 2m of gas B are contained in the same vessel at temperature T. The mean square of the velocity of molecules of gas B is v² and the mean square of x component of the velocity of molecules of gas A is w². The ratio $\large \frac{{{w^2}}}{{{v^2}}}$ is

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$\frac{1}{3}$

$\frac{2}{3}$

answer is D.

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Detailed Solution

{V_A}\, = \,\sqrt {\frac{{3KT}}{m}} \,\,\,

{V_B}\, = \,\sqrt {\frac{{3KT}}{{2m}}} \,\,\,

V_A^2\, = \,V_{Ax}^2 + V_{^{Ay}}^2 + V_{Az}^3\, - - - \left( 1 \right)

$b u t V_{A X} = w$

_{From equation (1) 3W² = V²A}

3{w^2}\, = \,\frac{{3KT}}{m}\,\, \to (2)\, \Rightarrow \,{V^2}\, = \,V_B^2\, = \,\frac{{3KT}}{{2m}} \to \left( 3 \right)

\frac{{3{W^2}}}{{{V^2}}}\, = \,\frac{{3KT}}{m}\frac{{2m}}{{3KT}}\, \Rightarrow \,\frac{{{W^2}}}{{{V^2}}}\, = \,\frac{2}{3}

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