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

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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$

Column $-I$	Column $-II$
$(A)$ Electrical resistance	$(p)$ $M{L^3}{T^{ - 3}}{A^{ - 2}}$
$(B)$ Electrical potential	$(q)$ $M{L^2}{T^{ - 3}}{A^{ - 2}}$
$(C)$ Specific resistance	$(r)$ $M{L^2}{T^{ - 3}}{A^{ - 1}}$
$(D)$ Specific conductance	$(s)$ None of these

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,

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