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)

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