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Question

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 moleccules of gas A is w2. The ratio w2v2 is

A
1
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B
2
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C
13
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D
23
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Solution

The correct option is D 23
Total KE of A type of molecules = 32mw2
Total KE of A type of molecules is
K.E.A=12[(Vr.m.s)2x+(Vr.m.s)2y+(Vr.m.s.)2z)]
But (Vrms)x=w
So, (Vrms)y=(Vrms)z=w
Total KE of B type of molecules is
=12×2mv2=m.v2
Now as KE is dependent on temperature T
32×mw2=mv2(w2/v2)=23

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