Kinetic Theory Of Gases Question 173
Question: A box has been placed on train moving uniformly with speed $ V _{0}. $ The box contains ideal gas. The value of root mean square speed with respect to an observer present in the trains is $ V _{1}. $ What will be the value of root mean square speed with respect to an observer standing on the platform?
Options:
A) $ V _{1}+V _{0} $
B) $ V _{1} $
C) $ {{{V _{1}^{2}+V _{0}^{2}}}^{1/2}} $
D) None
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Answer:
Correct Answer: C
Solution:
[c] Let velocity of a molecule w.r.t. observer present in the train is $ \overrightarrow{V’} $ . Velocity of the molecule w.r.t. observer on the plate form is $ \overrightarrow{V _{0}}+\overrightarrow{V}’\overrightarrow{V _{0}} $ is the velocity of train w.r.t. plate form Now, $ V _{rms}=\sqrt{\frac{\sum\limits _{{}}^{{}}{{{| \overrightarrow{V _{0}}+\overrightarrow{V}’ |}^{2}}}}{n}} $
$ (n\to $ total number of molecule)
$ =\sqrt{\frac{\sum ({{{\vec{V}}} _{0}}+\vec{V}’)\cdot ({{{\vec{V}}} _{0}}+\vec{V}’)}{n}} $
$ =\sqrt{\frac{\sum (\text{V} _{0}^{2}+\text{V}{{’}^{2}})\cdot (2{{{\vec{V}}} _{0}}\cdot \vec{V}’)}{n}} $
$ =\sqrt{\frac{\sum V _{0}^{2}}{n}+\frac{\sum V’}{n}+\frac{\sum 2(\overrightarrow{V _{0}}.\overrightarrow{V})}{n}} $
$ =\sqrt{V _{0}^{2}+{{(V _{1})}^{2}}+2\overrightarrow{V _{0}}.\frac{(\sum \overrightarrow{V’})}{n}} $
$ \sum \overrightarrow{V’}= $ zero (Relative to the box molecules are moving arbitrarily in all directions).
BETA
