N molecules, each of mass m of gas `A and 2 N` molecules each of mass `2m` of gas `B` are containted in the same vessel which is maintained at temperature T. The mean square velocity of molecules of B type is denoted by `v^(2)` and the mean square velocity of A type is denoted by `(omega)^(2)`. the `omega^(2)//v^(2)` is:

Answer

2

Explanation

For gas `A` `V_(A)^(2)=(3KT)/(m)`
since `V_(X)^(2)=V_(y)^(2)=V_(z)^(2)` for molecule
hence `V_(A)^(2)=V_(X)^(2)+V_(y)^(2)=3V_(X)^(2)`
`V_(A)^(2)=3w^(2)=(3KT)/(m)`
for gas `B` `V_(B)^(2)=V^(2)=(3KT)/(2m)`
divide `=(V_(A)^(2))/(V_(B)^(2))=(w^(2))/(V^(2))=(2)/(3)`