Answer to Question #30812 in Mechanics | Relativity for disha

We posit that
M = C v^a * p^b * g^c
with C a dimensionless constant, and a , b and c (dimensionless) exponents.
Let us determine what values of a, b and c are allowed to make the formula dimensionally correct.

v is s speed, so length/time ( L / T) .
p is a density, so mass/volume (m / L^3)
g is an acceleration, so length / time^2 ( L / T^2)

This means that in terms of dimensions, the equation is
m = (L / T)^a * (m / L^3)^b * ( L / T^2)^c
So, combining powers in each quantity (L, m, T):

m = L^(a - 3b + c) * m^b * T^(-a -2c)

Whence we conclude that

b = 1
a - 3b + c = 0
-a -2c = 0

I.e.

b = 1
a + c = 3
a + 2 c = 0

The latter two equations are solved by
c = -3
a = 6

So

M = C * v^6 * p / g^3