The dimensions of the area A of a black hole | vedclass

Name: PDF Generator App | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(b)

Given, $A=G^\alpha M^\beta c^\gamma$

Substituting dimensions of $A, M$ and $c$, we have

$[ L ]^2=\left[\frac{ MLT ^{-2} L ^2}{ M ^2}\righ

Vedclass · Accepted Answer

$\alpha=2, \beta=2$ and $\gamma=-4$

Vedclass · Answer

(b)

Given, $A=G^\alpha M^\beta c^\gamma$

Substituting dimensions of $A, M$ and $c$, we have

$[ L ]^2=\left[\frac{ MLT ^{-2} L ^2}{ M ^2}ight]^\alpha[ M ]^\beta\left[\frac{ L }{ T }ight]^\gamma$

$\Rightarrow \quad[ L ]^2=\left[ M ^{-1} L ^3 T ^{-2}ight]^\alpha[ M ]^\beta\left[ LT ^{-1}ight]^\gamma$

$\Rightarrow \quad[ L ]^2= M ^{-\alpha+\beta} L ^{3 \alpha+\gamma} T ^{-2 \alpha-\gamma}$

Equating dimensions on both sides, we

$	ext { have } \quad-\alpha+\beta=0 \quad \dots(i)$

$3 \alpha+\gamma =2 \quad \ldots(ii)$

$-2 \alpha-\gamma=0 \quad \dots(iii)$

Adding Eqs. $(ii)$ and $(iii)$, we get

$\Rightarrow \quad 3 \alpha+\gamma-2 \alpha-\gamma=2+0 \Rightarrow \alpha=2 \quad \ldots$ $(iv)$

Now, putting the value of $\alpha$ in Eq. $(i)$, we get

$\Rightarrow \quad-2+\beta=0 \Rightarrow \beta=2$

Again, putting the value of $\alpha$ in Eq. $(iii)$, we get

$\Rightarrow -2 	imes 2-\gamma=0 \Rightarrow \gamma=-4$

$	ext { So, } \alpha=2, \beta=2 	ext { and } \gamma=-4$

The dimensions of the area $A$ of a black hole can be written in terms of the universal gravitational constant $G$, its mass $M$ and the speed of light $c$ as $A=G^\alpha M^\beta c^\gamma$. Here,

Similar Questions

If speed $V,$ area $A$ and force $F$ are chosen as fundamental units, then the dimension of Young's modulus will be :

The characteristic distance at which quantum gravitational effects are significant, the Planck length, can be determined from a suitable combination of the fundamental physical constants $G, h$ and $c$ . Which of the following correctly gives the Planck length?

Using dimensional analysis, the resistivity in terms of fundamental constants $h, m_{e}, c, e, \varepsilon_{0}$ can be expressed as

If force $({F})$, length $({L})$ and time $({T})$ are taken as the fundamental quantities. Then what will be the dimension of density