A) \[[WFv]\]
B) \[[WF{{v}^{-1}}]\]
C) \[[{{W}^{-1}}{{F}^{-1}}v]\]
D) \[[W{{F}^{-1}}{{v}^{-1}}]\]
Correct Answer: D
Solution :
Let\[T\propto {{F}^{a}}{{W}^{b}}{{v}^{c}}\] ... (i) \[[T]={{[ML{{T}^{-2}}]}^{a}}{{[M{{L}^{2}}{{T}^{-2}}]}^{b}}{{[L{{T}^{-1}}]}^{c}}\] \[[{{T}^{1}}]=[{{M}^{a+b}}][{{L}^{a+2b+c}}][{{T}^{-2a-2b-c}}]\] Comparing the powers, we get \[a+b=0\] ... (ii) \[a+2b+c=0\] ... (iii) \[-2a-2b-c=1\] ... (iv) Solving Eqs. (ii), (iii) and (iv), we get \[a=-1,\,\,b=-1,\,\,c=-1\] Therefore, from Eq. (i), \[[T]=k[{{F}^{-1}}{{W}^{1}}{{v}^{-1}}]\] Taking\[k=1\]in SI system, we have \[[T]=[W{{F}^{-1}}{{v}^{-1}}]\]
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