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)

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