A) \[{{V}_{1}}+{{V}_{0}}\]
B) \[{{V}_{1}}\]
C) \[{{\{V_{1}^{2}+V_{0}^{2}\}}^{1/2}}\]
D) None
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, \[{{V}_{rms}}=\sqrt{\frac{\sum\limits_{{}}^{{}}{{{\left| \overrightarrow{{{V}_{0}}}+\overrightarrow{V}' \right|}^{2}}}}{n}}\] \[(n\to \] 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). |
You need to login to perform this action.
You will be redirected in
3 sec