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

Answer: (D)

$[W{{F}^{-1}}{{V}^{-1}}]$

Answer 2

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 $[T]=[W{{F}^{-1}}{{v}^{-1}}]$