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, $h=\frac{E n e r g y}{F r e q u e n c y}$
$=\frac{\left[M L^{2} \, T^{- 2}\right]}{\left[T^{- 1}\right]}=\left[M L^{2} T^{- 1}\right]$
$C=\frac{M e t r e}{S e c o n d}=\left[L T^{- 1}\right]$
$G=\frac{F o r c e \times \left(d i s t a n c e\right)^{2}}{\left(m a s s\right)^{2}}$
$=\frac{\left[M L T^{- 2}\right] \, \left[L^{2}\right]}{\left[M^{2}\right]}=\left[M^{- 1} L^{3} T^{- 2}\right]$
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, $M=h^{\frac{1}{2}} \, C^{\frac{1}{2}} \, G^{- \frac{1}{2}}$