Question

If force (F), work (W) and velocity (v) are taken as fundamental quantities, then the dimensional formula of time (T) is

• A. $ [WFv] $
• B. $ [WF{{v}^{-1}}] $
• C. $ [{{W}^{-1}}{{F}^{-1}}v] $
• D. $ [W{{F}^{-1}}{{V}^{-1}}] $

Answer 1

The Correct Option is D

Let $ T\propto {{F}^{a}}{{W}^{b}}{{V}^{c}} $ $ [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 $ ?(i) $ a+2b+c=0 $ ?(ii) $ -2a-2b-c=1 $ ?(iv) Solving Eqs. (ii),(iii) and (iv),we get $ a=-1,\,b=1,c=-1 $ Therefore, from E(i), $ [T]=k[{{F}^{-1}}{{W}^{1}}{{v}^{-1}}] $ Taking $ k=1 $ in SI system, we have $ [T]=[W{{F}^{-1}}{{v}^{-1}}] $