A) \[\text{ }\!\![\!\!\text{ M}{{\text{L}}^{\text{5/2}}}{{\text{T}}^{\text{-2}}}\text{ }\!\!]\!\!\text{ }\]
B) \[\text{ }\!\![\!\!\text{ M}{{\text{L}}^{\text{2}}}{{\text{T}}^{\text{-2}}}\text{ }\!\!]\!\!\text{ }\]
C) \[\text{ }\!\![\!\!\text{ }{{\text{M}}^{3/2}}{{\text{L}}^{\text{3/2}}}{{\text{T}}^{\text{-2}}}\text{ }\!\!]\!\!\text{ }\]
D) \[\text{ }\!\![\!\!\text{ M}{{\text{L}}^{7/2}}{{\text{T}}^{\text{-2}}}\text{ }\!\!]\!\!\text{ }\]
Correct Answer: D
Solution :
Given, \[v=\frac{A\sqrt{x}}{x+B}\] ?(i) Dimensions of \[v=\] dimensions of potential energy \[=[M{{L}^{2}}{{T}^{-2}}]\] Dimensions of\[B=\]dimensions of\[x=[{{M}^{0}}L{{T}^{0}}]\] \[\therefore \,\,\text{Dimensions}\,\text{of}\,\text{A}\] \[=\frac{\text{dimension}\,\text{of}\,v\times \text{dimensions}\,\text{of}\,(x+B)}{\text{dimensions}\,\text{of}\sqrt{x}}\] \[=\frac{[M{{L}^{2}}{{T}^{-2}}]\,[{{M}^{0}}L{{T}^{0}}]}{[{{M}^{0}}{{L}^{1/2}}{{T}^{0}}]}=[M{{L}^{5/2}}{{T}^{-2}}]\] Hence, dimensions of AB \[=[M{{L}^{5/2}}{{T}^{-2}}]\,[{{\text{M}}^{0}}\text{L}{{\text{T}}^{0}}]=[M{{L}^{7/2}}{{T}^{-2}}]\]
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