Kinetic Theory - Result Question 20
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20. $N$ molecules each of mass $m$ of a gas $A$ and $2 N$ molecules each of mass $2 m$ of gas $B$ are contained in the same vessel which is maintained at temperature $T$. The mean square velocity of molecules of $B$ type is $v^{2}$ and the mean square rectangular component of the velocity of $A$ type is denoted by $\omega^{2}$. Then $\omega^{2} / v^{2}$
======= ####20. $N$ molecules each of mass $m$ of a gas $A$ and $2 N$ molecules each of mass $2 m$ of gas $B$ are contained in the same vessel which is maintained at temperature $T$. The mean square velocity of molecules of $B$ type is $v^{2}$ and the mean square rectangular component of the velocity of $A$ type is denoted by $\omega^{2}$. Then $\omega^{2} / v^{2}$
3e0f7ab6f6a50373c3f2dbda6ca2533482a77bed:content/english/neet-pyq-chapterwise/physics/kinetic-theory/kinetic-theory—result-question-20.md (a) 2
(b) 1
(c) $1 / 3$
(d) $2 / 3$
[1991]
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Answer:
Correct Answer: 20. (d)
Solution:
- (d) The mean square velocity of A type molecules $=\omega^{2}+\omega^{2}+\omega^{2}=3 \omega^{2}$
Therefore, $\frac{1}{2} m(3 \omega^{2})=\frac{1}{2}(2 m) v^{2}$
This gives $\omega^{2} / v^{2}=2 / 3$
Mean kinetic energy of the two types of molecules should be equal.
