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N molecules each of mass m of gas A and ...

N molecules each of mass m of gas A and 2 N molecules each of mass 2m of gas B are contained in the same vessel which is maintined at a temperature T. The mean square of the 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 tye is denoted by `omega^(2)`. What is the ratio of `omega^(2)//v^(2) = ?`

A

2

B

1

C

`(1)/(3)`

D

`(2)/(3)`

Text Solution

Verified by Experts

The correct Answer is:
D
Mean kinetic energy of two types of molecules should be equal.
so, `(1)/(2)m(3omega^(2))=(1)/(2)(2m)v^(2)rArr(omega^(2))/(v^(2))=(2)/(3)`
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