Question

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

• A. $h^{-\frac{1}{2}}{G}^{\frac{1}{2}}{C}^o$
• B. $h^{\frac{1}{2}}{C}^{\frac{1}{2}}{G}^{-\frac{1}{2}}$
• C. $h^{-\frac{1}{2}}{C}^{\frac{1}{2}}{G}^{-\frac{1}{2}}$
• D. $h^{-\frac{1}{2}}{C}^{-\frac{1}{2}}{G}^{-\frac{1}{2}}$

Answer 1

Answer: (B)

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