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