In a new system of units energy (E), density | vedclass

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Question

(b)

$G=\left[E^g d^b P^c\right]$

$E=\left[M^2 T^{-2}\right]$

$d=\left[ ML ^{-3}\right]$

$P=\left[ ML ^2 T ^{-3}\right]$

$G=\left[M^{-1}

Vedclass · Accepted Answer

$\left[E^{-2} d^{-1} P^2\right]$

Vedclass · Answer

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

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 .......

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