N molecules each of mass m of gas A and 2N molecules each of mass 2m of gas B are contained in the vessel which is maintained at a temperature T. The mean square of velocity of the molecules of B type is denoted by `v^(2)` and the mean square of the x-component of the velocity of A type is denoted by `w^(2).` The ratio of `w^(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)`