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)

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 Eq. (i),
$[T]=k\left[F^{-1} \,W^{1} \,v^{-1}\right]$
Taking $k=1$ in SI system, we have
$[T]=\left[W F^{-1} v^{-1}\right]$