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) \[[T]={{[ML{{T}^{-2}}]}^{a}}{{[M{{L}^{2}}{{T}^{-2}}]}^{b}}{{[L{{T}^{-1}}]}^{c}}\] \[[{{T}^{1}}]=[{{M}^{a+b}}][{{L}^{a+2b+c}}][{{T}^{-2a-1b-c}}]\] 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 \[[T]=k[W{{F}^{-1}}{{v}^{-1}}]\]
You need to login to perform this action.
You will be redirected in
3 sec
