A) \[[M{{L}^{5/2}}{{T}^{-2}}]\]
B) \[[M{{L}^{2}}{{T}^{-2}}]\]
C) \[[{{M}^{3/2}}{{L}^{3/2}}{{T}^{-2}}]\]
D) \[[M{{L}^{7/2}}{{T}^{-2}}]\]
Correct Answer: D
Solution :
Given, \[v=\frac{A\sqrt{x}}{x+B}\] ?(i) Dimensions of v= dimensions of potentialenergy \[=[M{{L}^{2}}{{T}^{-2}}]\] From Eq. (i), Dimensions ofB= dimensions of\[x=[{{M}^{0}}L{{T}^{o}}]\] \[\therefore \]Dimensions of A \[=\frac{\text{dimensions}\,\text{of}\,v\times \text{dimensitons}\,\text{of}\,(x+B)}{\text{dimensions}\,\text{of}\,\sqrt{x}}\] \[=\frac{[M{{L}^{2}}{{T}^{-2}}][{{M}^{0}}L{{T}^{0}}]}{[{{M}^{0}}{{L}^{1/2}}{{T}^{0}}]}\] \[=[M{{L}^{5/2}}{{T}^{-2}}]\] Hence, dimensions ofAB \[=[M{{L}^{5/2}}{{T}^{-2}}][{{M}^{0}}L{{T}^{0}}]\] \[=[M{{L}^{7/2}}{{T}^{-2}}]\]
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