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