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 = Kv^x F^y T^z...(i)

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

[Q] = [LT^-1]^x[MLT^-2]^y[T]^z

i.e.,   [Q] = [M^y][L^x+yT^-x-2y+z]            ...(ii)

​Now

Q = mass i.e.,   [Q] = [M]

So Equation (ii) becomes

[M] = [M^yL^x+yT^-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 = Kv^-1FT