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