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