Question

In a new system of units energy $(E)$, density $(d)$ and power $(P)$ are taken as fundamental units, then the dimensional formula of universal gravitational constant $G$ will be

• A. $\left[E^{-1}{d}^{-2}{P}^2\right]$
• B. $\left[E^{-2}{d}^{-1}{P}^2\right]$
• C. $\left[E^2{d}^{-1}{P}^{-1}\right]$
• D. $\left[E^1{d}^{-2}{P}^{-2}\right]$

Answer 1

Answer: (B)

$G=\left[E^a{d}^b{P}^c\right]$
$E=\left[M{L}^2{T}^{-2}\right]$
$d=\left[M{L}^{-3}\right]$
$P=\left[M{L}^2{T}^{-3}\right]$
$G=\left[M^{-1}{L}^3{T}^{-2}\right]$
$\left[M^{-1}{L}^3{T}^{-2}\right]={\left[M{L}^2{T}^{-2}\right]}^9{\left[M{L}^{-3}\right]}^b{\left[M{L}^2{T}^{-3}\right]}^e$
$a+b+c=-1$
$2a-3b+2c=3$
$-2a-3c=-2\Rightarrow 2a+3c=2$
On solving,
$a=-2$
$b=-1$
$c=2$
So, $G=[E^{-2}{d}^{-1}{P}^2\mid$