Question

If $C$, the velocity of light, $g$ the acceleration due to gravity and $P$ the atmospheric pressure be the fundamental quantities in MKS system, then the dimensions of length will be same as that of

• A. $\frac{C}{g}$
• B. $\frac{C}{P}$
• C. PCg
• D. $\frac{C^2}{g}$

Answer 1

Answer: (D)

Let $C^x{g}^y{P}^z$ be dimensions of length.
$\Rightarrow {\left[M^{0}L{T}^{-1}\right]}^x{\left[M^{0}L{T}^{-2}\right]}^y{\left[M{L}^{-1}{T}^{-2}\right]}^2=M^{0}L{T}^0$
$\Rightarrow$ Comparing powers of $M,L$ and $T$.
We get, $z=0;x+y=1;-x-2y=0$
$-2y+y=1\Rightarrow x=-2y$
$\Rightarrow y=-1\Rightarrow x=2$
$\therefore dim($ length $)=c^x{g}^y{p}^z$
$=c^2{g}^{-1}{p}^0=\frac{c^2}{g}$.