A) \[{{M}^{2}}{{L}^{2}}{{T}^{-2}}\]
B) \[{{M}^{0}}{{L}^{2}}{{T}^{-4}}\]
C) \[ML{{T}^{-2}}\]
D) \[{{M}^{2}}L{{T}^{-4}}\]
Correct Answer: D
Solution :
The given expression is \[F=\alpha \beta \exp \left( -\frac{{{x}^{2}}}{\alpha kT} \right)\] \[\frac{{{x}^{2}}}{\alpha KT}[{{M}^{0}}{{L}^{0}}{{T}^{0}}]\] \[\Rightarrow \]\[\frac{[{{L}^{2}}]}{[\alpha ][M{{L}^{2}}{{T}^{-2}}{{K}^{-1}}][K]}=[{{M}^{0}}{{L}^{0}}{{T}^{0}}]\] \[[\alpha ]=[{{M}^{-1}}{{T}^{2}}]\] Now,\[[\alpha ][\beta ]=[F]\] \[\Rightarrow \]\[[{{M}^{-1}}{{T}^{2}}][\beta ]=[M{{L}^{2}}{{T}^{-2}}]\] \[\Rightarrow \]\[[\beta ]=[{{M}^{2}}L{{T}^{-4}}]\]
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