Question

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

• A. $Q=Kv^{- 1}FT$
• B. ^{$Q=Kv^{3}FT^{2}$}
• C. $Q=2Kv^{- 2}FT$
• D. $Q=-3Kv^{2}FT$

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 $T$,
$Q=K v^{x} F^{y} T^{z} \ldots \text { (i) }$
where $K$ is a dimensionless constant of proportionality.
The above equation dimensionally becomes
${[Q]=\left[L T^{-1}\right]^{x}\left[M L T^{-2}\right]^{y}[T]^{z}}$
i.e., $[Q]=\left[M^{y}\right]\left[L^{x+y} T^{-x-2 y+z}\right] \ldots$ (ii)
Now
Now $Q=$ mass i.e., $[Q]=[M]$
So Equation (ii) becomes
$[ M ]=\left[ M ^{ y } L ^{ x + y } T ^{- x -2 y + z }\right]$
its dimensional correctness requires
$y=1, x+y=0$ and $-x-2 y+z=0$
which on solving yields
$x=-1, y=1$ and $z=1$
Substituting it in Equation (i), we get
$Q = KV ^{-1} FT$