A) \[\left[ {{M}^{2}}L{{T}^{-3}} \right]\]
B) \[\left[ M{{T}^{-2}} \right]\]
C) \[\left[ L{{T}^{-3}} \right]\]
D) \[\left[ M{{L}^{3}}{{T}^{-1}} \right]\]
Correct Answer: B
Solution :
Given \[P=\frac{a-{{t}^{2}}}{bx}\] or \[Pb\,x=a-{{t}^{2}}\] By the law of homogenity of dimensional equation. Dimensions of a = dimensions of \[{{t}^{2}}=[{{T}^{2}}]\] Dimensions of b = dimensions of \[\frac{{{t}^{2}}}{Px}=[{{M}^{-1}}{{T}^{4}}]\] So, dimensions of \[\frac{a}{b}\], is \[[M{{T}^{-2}}]\].
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