If $C$ the velocity of light, $h$ Planck's constant and $G$ gravitational constant are taken as fundamental quantities, then the dimensional formula of mass is

Question

Aicharya · Accepted Answer

Let,
$\begin{aligned}
M & =C^a h^b G^c \\
\mathrm{ML}^0 \mathrm{~T}^0 & =\left[\mathrm{LT}^{-1}\right]^a\left[\mathrm{ML}^2 \mathrm{~T}^{-1}\right]^{\mathrm{b}}\left[\mathrm{M}^{-1} \mathrm{~L}^3 \mathrm{~T}^{-2}\right]^{\mathrm{a}}
\end{aligned}$
where, $h=\frac{\text { Energy }}{\text { Frequency }}$
$\begin{aligned}
& =\frac{\left[\mathrm{ML}^2 \mathrm{~T}^{-2}\right]}{\left[\mathrm{T}^{-1}\right]}=\left[\mathrm{ML}^2 \mathrm{~T}^{-1}\right] \\
C & =\frac{\text { Metre }}{\text { Second }}=\left[\mathrm{LT}^{-1}\right] \\
G & =\frac{\text { Force } \times \text { (distance) }}{\text { (mass) }} \\
& =\frac{\left[\mathrm{MLT}^{-2}\right]\left[\mathrm{L}^2\right]}{\left[\mathrm{M}^2\right]}=\left[\mathrm{M}^{-1} \mathrm{~L}^3 \mathrm{~T}^{-2}\right]
\end{aligned}$
Comparing the coefficients $M, L, T$, of both sides we get
$\begin{array}{l}
b-c=1 \\
a+2 b+3 c=0 \\
-(a+b+2 c)=0
\end{array}$
Solve the Eqs. (ii), (iii) and (iv), we get
$a=\frac{1}{2}, b=\frac{1}{2}, c=-\frac{1}{2}$
So, $M=h^{1 / 2} C^{1 / 2} G^{-1 / 2}$

If $C$ the velocity of light, $h$ Planck's constant and $G$ gravitational constant are taken as fundamental…

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