Question

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

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

Answer 1

Answer: (D)

$[FV^{-1}T]$

Answer 2

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. $(i)$, $[m] = [FV^{-1}T]$