Physics - Unit & Dimension Question with Solution | TestHub

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 $T$,

$Q\mathit{ }\mathit{=}\mathit{ }K{v}^{\mathit{x}}\mathit{ }{F}^{\mathit{y}}\mathit{ }{T}^{\mathit{z}}\mathit{ }\mathit{ }\mathit{.}\mathit{.}\mathit{.}\left(\mathrm{i}\right)$     

where K is a dimensionless constant of proportionality. The above equation dimensionally becomes

    $\left[Q\right] = \left[L{T}^{-1}{]}^x\right[ML{T}^{-2}{]}^y[T{]}^z$

i.e.,   $\left[Q\right] = \left[M^y{L}^{(x+y)}{T}^{(-x-2y+z)}\right]$, ​Now

Q = mass i.e.,   $\left[Q\right]=\left[M\right]$

So Equation (ii) becomes

$\left[M\right] = \left[M^y{L}^{(x+y)}{T}^{(-x-2y+z)}\right]$
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

$Q\mathit{=}K\mathit{ }{v}^{\mathit{-}\mathit{1}}\mathit{ }F\mathit{ }T$