Question

If velocity, force and time are taken to be fundamental quantities, then the dimensional formula for mass is

• A. $Q=K{v}^{-1}FT$
• B. ^{$Q=K{v}^{3}F{T}^2$}
• C. $Q=2K{v}^{-2}FT$
• D. $Q=-3K{v}^{2}FT$

Answer 1

Answer: (A)

Let the quantity be $Q$, then,
$Q=f(v,F,T)$
Assuming that the function is the product of power functions of $V,F$ and $T$,
$Q=K{v}^x{F}^y{T}^z\dots \text{ (i) }$
where $K$ is a dimensionless constant of proportionality.
The above equation dimensionally becomes
$[Q]={\left[L{T}^{-1}\right]}^x{\left[ML{T}^{-2}\right]}^y[T{]}^z$
i.e., $[Q]=\left[M^y\right]\left[L^{x+y}{T}^{-x-2y+z}\right]\dots$ (ii)
Now
Now $Q=$ mass i.e., $[Q]=[M]$
So Equation (ii) becomes
$[M]=\left[M^y{L}^{x+y}{T}^{-x-2y+z}\right]$
its dimensional correctness requires
$y=1,x+y=0$ and $-x-2y+z=0$
which on solving yields
$x=-1,y=1$ and $z=1$
Substituting it in Equation (i), we get
$Q=K{V}^{-1}FT$