A) 1.4
B) 2.0
C) 2.8
D) 4.0
Correct Answer: A
Solution :
\[{{\overrightarrow{v}}_{av}}=\sqrt{\frac{8\,\,RT}{\pi \,M}}\] or \[{{\overrightarrow{v}}_{av}}=\,\,\propto \sqrt{T}\] Given, \[{{T}_{1}}=T,\,\,{{T}_{2}}=2T\] \[\frac{{{({{\overrightarrow{v}}_{av}})}_{1}}}{{{({{\overrightarrow{v}}_{av}})}_{2}}}=\sqrt{\frac{T}{2\,T}}=\frac{1}{\sqrt{2}}\] \[{{({{\overrightarrow{v}}_{av}})}_{2}}=\sqrt{2}\,{{({{\overrightarrow{v}}_{av}})}_{1}}\] \[=1.414\times {{({{\overrightarrow{v}}_{av}})}_{1}}\] Thus, average velocity is increased by a factor 1.4.
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