# Question 7de72

Jun 3, 2017

Dimension of the velocity of light, c $= \left[L\right] {\left[T\right]}^{-} 1$

Dimension of the Gravitational constant, G $= {\left[M\right]}^{-} 1 {\left[L\right]}^{3} {\left[T\right]}^{-} 2$

Dimension of Planck's constant, h $= \left[M\right] {\left[L\right]}^{2} {\left[T\right]}^{-} 1$

Let the dimension of mass in new system be

$\left[M\right] = {\left[c\right]}^{x} {\left[G\right]}^{y} {\left[h\right]}^{z}$

Hence

[M]=([L][T]^-1)^x ([M]^-1[L]^3[T]^-2)^y] ([M][L]^2[T]^-1)^z#

$\implies \left[M\right] {\left[L\right]}^{0} {\left[T\right]}^{0} = {\left[M\right]}^{z - y} {\left[L\right]}^{x + 3 y + 2 z} {\left[T\right]}^{- x - 2 y - z}$

so comparing the power of both sides we have

$z - y = 1. \ldots . . \left[1\right]$

$x + 3 y + 2 z = 0. \ldots . . \left[2\right]$

$- x - 2 y - z = 0. \ldots . \left[3\right]$

$y + z = 0. \ldots . \left[4\right]$

$2 z = 1 \implies z = \frac{1}{2}$

so $\frac{1}{2} - y = 1 \implies y = - \frac{1}{2}$

Inserting y and z in 

$x + 3 \left(- \frac{1}{2}\right) + 2 \times \frac{1}{2} = 0$

$\implies x - \frac{3}{2} + 1 = 0$

$\implies x = \frac{1}{2}$

Hence

$\left[M\right] = {\left[c\right]}^{\frac{1}{2}} {\left[G\right]}^{- \frac{1}{2}} {\left[h\right]}^{\frac{1}{2}}$