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. $\left[\right.$ $K$ is a dimensionless constant $\left]\right.$

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

Answer 1

Answer: (A)

Let the quantity be Q, then,
$Q=f\left(\right.v,F,T\left.\right)$
Assuming that the function is the product of power functions of $v,F$ and $T$ ,
$Q=Kv^{x}F^{y}T^{z}...\left(\right.i\left.\right)$
where K is a dimensionless constant of proportionality. The above equation dimensionally becomes
$\left[\right.Q\left]\right.=\left[\right.LT^{- 1}\left]\right.^{x}\left[\right.MLT^{- 2}\left]\right.^{y}\left[\right.T\left]\right.^{z}$
i.e., $\left[\right.Q\left]\right.=\left[\right.M^{y}L^{\left(\right. x + y \left.\right)}T^{\left(\right. - x - 2 y + z \left.\right)}\left]\right.$ , ​Now
Q = mass i.e., $\left[\right.Q\left]\right.=\left[\right.M\left]\right.$
So Equation (ii) becomes
$\left[\right.M\left]\right.=\left[\right.M^{y}L^{\left(\right. x + y \left.\right)}T^{\left(\right. - x - 2 y + z \left.\right)}\left]\right.$
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
$Q=Kv^{- 1}FT$