A) \[M{{T}^{-2}}\]
B) \[{{M}^{2}}L{{T}^{-\,3}}\]
C) \[M{{L}^{3}}{{T}^{-\,1}}\]
D) \[L{{T}^{-3}}\]
Correct Answer: A
Solution :
\[[a]\,\,=\,\,[{{T}^{2}}]\,\,and\,\,[b]\,\,=\,\,\frac{\left[ a-{{t}^{2}} \right]}{[P]\,[x]}\,\,=\,\,\frac{{{T}^{2}}}{[M{{L}^{-1}}{{T}^{-2}}][L]}\] \[\Rightarrow \,\,\,\,\,[b]\,\,=\,\,[{{M}^{-1}}{{T}^{4}}]\] \[So\,\,\,\,\,\left[ \frac{a}{b} \right]\,\,=\,\frac{[{{T}^{2}}]}{[{{M}^{-1}}{{T}^{4}}]}\,\,=\,\,[M{{T}^{-2}}]\]
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