Question

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

• A. $ h^{-1/2} \, G^{1/2} \, C^{0} $
• B. $ h^{1/2} C^{1/2} \, G ^{-1/2} $
• C. $ h^{-1/2} \, C^{1/2} \, G^{-1/2} $
• D. $ h^{-1/2} \, C^{-1/2} \, G^{-1/2} $

Answer 1

Answer: (B)

Let, $M=C^{a}\, h^{b}\, G^{c}$
$ML ^{0} \,T ^{0}=\left[ LT ^{-1}\right]^{ a }\left[ ML ^{2}\, T ^{-1}\right]^{ b }\left[ M ^{-1}\, L ^{3} \,T ^{-2}\right]^{ a } \ldots$ (i)
where, $h=\frac{\text { Energy }}{\text { Frequency }}$
$=\frac{\left[ ML ^{2} T ^{-2}\right]}{\left[ T ^{-1}\right]}=\left[ ML ^{2} T ^{-1}\right] $
$ C =\frac{\text { Metre }}{\text { Second }}=\left[ LT ^{-1}\right] $
$ G=\frac{\text { Force } \times(\text { distance })^{2}}{( mass )^{2}} $
$=\frac{\left[ MLT ^{-2}\right]\left[ L ^{2}\right]}{\left[ M ^{2}\right]}=\left[ M ^{-1} L ^{3} T ^{-2}\right] $
Comparing the coefficients $M, L, T$, of both sides we get
$b-c=1 \,\,\,...(i)$
$a+2 b+3 c=0 \,\,\,...(ii)$
$-(a+b+2 c)=0\,\,\,...(iv)$
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} $