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

Answer 1

Answer: (D)

$[WF^{-1}v^{-1}]$

Answer 2

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