Question

If force $(F)$, velocity $(V)$ and time $(T)$ are taken as fundamental units, then the dimensions of mass are

• A. $[FV{T}^{-1}]$
• B. $[FV{T}^{-2}]$
• C. $[F{V}^{-1}{T}^{-1}]$
• D. $[F{V}^{-1}T]$

Answer 1

Answer: (D)

Let mass $m\propto F^a{V}^b{T}^c$
or $m=k{F}^a{V}^b{T}^c\dots (i)$
where $k$ is a dimensionless constant and $a$, $b$ and $c$ are the exponents.
Writing dimensions on both sides, we get
$[M{L}^0{T}^0]=[ML{T}^{-2}{]}^a[L{T}^{-1}{]}^b[T{]}^c$
$[M{L}^0{T}^0]=[M^a{L}^{a+b}{T}^{-2a-b+c}]$
Applying the principle of homogeneity of dimensions, we get
$a=1\dots (ii)$
$a+b=0\dots (iii)$
$-2a-b+c=0\dots (iv)$
Solving eqns. $(ii)$, $(iii)$ and $(iv)$, we get
$a=1$,
$b=-1$,
$c=1$
From eqn. $(i)$, $[m]=[F{V}^{-1}T]$