Question

If velocity, force and time are taken as the fundamental quantities, then using dimensional analysis choose the correct dimensional formula for mass among the following. $[$ $K$ is a dimensionless constant $]$

• 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 $T$ ,
$Q=K{v}^x{F}^y{T}^z...(i)$
where K is a dimensionless constant of proportionality. The above equation dimensionally becomes
$[Q]=[L{T}^{-1}{]}^x[ML{T}^{-2}{]}^y[T{]}^z$
i.e., $[Q]=[M^y{L}^{(x+y)}{T}^{(-x-2y+z)}]$ , ​Now
Q = mass i.e., $[Q]=[M]$
So Equation (ii) becomes
$[M]=[M^y{L}^{(x+y)}{T}^{(-x-2y+z)}]$
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
$Q=K{v}^{-1}FT$