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Q.
If force $(F)$, velocity $(V)$ and time $(T)$ are taken as fundamental units, then the dimensions of mass are
AIPMTAIPMT 2014Physical World, Units and Measurements
Solution:
Let mass $m \propto F^aV^bT^c$
or $m=kF^aV^bT^c \ldots (i)$
where $k$ is a dimensionless constant and $a$, $b$ and $c$ are the exponents.
Writing dimensions on both sides, we get
$[ML^0T^0]=[MLT^{-2}]^a[LT^{-1}]^b[T]^c$
$[ML^0T^0]=[M^aL^{a+b}T^{-2a-b+c}]$
Applying the principle of homogeneity of dimensions, we get
$a=1 \ldots (ii)$
$a+b=0 \ldots (iii)$
$-2a-b+c=0 \ldots (iv)$
Solving eqns. $(ii)$, $(iii)$ and $(iv)$, we get
$a= 1$,
$b = -1$,
$c = 1$
From eqn. $(i)$, $[m] = [FV^{-1}T]$