A) \[[Fv{{T}^{-1}}]\]
B) \[[Fv{{T}^{-2}}]\]
C) \[[F{{v}^{-1}}{{T}^{-1}}]\]
D) \[[F{{v}^{-1}}T]\]
Correct Answer: D
Solution :
Let \[m\propto {{F}^{x}}{{v}^{y}}{{T}^{2}}\] By substituting the following dimensions: \[[F]=[ML{{T}^{-2}}],[v]=[L{{T}^{-1}}],[T]=[T]\] \[[{{M}^{1}}{{L}^{0}}{{T}^{0}}]={{[ML{{T}^{-2}}]}^{x}}{{[L{{T}^{-1}}]}^{y}}{{[T]}^{z}}\] \[\Rightarrow \]\[[{{M}^{1}}{{L}^{0}}{{T}^{0}}]=[{{M}^{x}}{{L}^{x+y}}{{T}^{-2x-y+z}}]\] \[x=1\] \[x+y=0\] \[\Rightarrow y=-1\] \[-2x-y+z=0\] \[\Rightarrow \]\[-2\times 1-(-1)+z=0\] \[\Rightarrow \]\[z=0\] Hence \[[M]=[{{F}^{1}}{{v}^{-1}}{{T}^{1}}]=[F{{v}^{-1}}T]\]
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