A) \[[M{{L}^{2}}{{T}^{-2}}]\]
B) \[[{{M}^{3/2}}{{L}^{3/2}}{{T}^{-2}}]\]
C) \[[{{M}^{7/2}}{{T}^{-2}}]\]
D) \[[M{{L}^{3/2}}{{T}^{-2}}]\]
Correct Answer: C
Solution :
Given,\[E=\frac{P\sqrt{h}}{h+Q}\] Dimension of E = Dimension of potential energy \[=[M{{L}^{2}}{{T}^{-2}}]\] From Eq.(i). we get Dimension of \[Q=\]Dimension of\[h=[{{M}^{0}}L{{T}^{0}}]\] \[\therefore \] Dimension of P \[\text{=}\,\frac{\text{Dimension}\,\text{of}\,\text{E}\,\text{ }\!\!\times\!\!\text{ }\,\text{Dimension}\,\text{of}\,\text{(h}\,\text{+}\,\text{Q)}}{\text{Dimension}\,\text{of}\,\sqrt{\text{h}}}\] \[=\frac{[M{{L}^{2}}{{T}^{-2}}][{{M}^{0}}{{L}^{0}}{{T}^{0}}]}{[{{M}^{0}}{{L}^{1/2}}{{T}^{0}}]}=[M{{L}^{5/2}}{{T}^{-2}}]\] Hence, dimensions of PQ \[=[M{{L}^{5/2}}{{T}^{-2}}][{{M}^{0}}L{{T}^{0}}]=[M{{L}^{7/2}}{{T}^{-2}}]\]
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