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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,
$\alpha=-2, \beta=-2$ and $\gamma=4$
$\alpha=2, \beta=2$ and $\gamma=-4$
$\alpha=3, \beta=3$ and $\gamma=-2$
$\alpha=-3, \beta=-3$ and $\gamma=2$
Solution
(b)
Given, $A=G^\alpha M^\beta c^\gamma$
Substituting dimensions of $A, M$ and $c$, we have
$[ L ]^2=\left[\frac{ MLT ^{-2} L ^2}{ M ^2}\right]^\alpha[ M ]^\beta\left[\frac{ L }{ T }\right]^\gamma$
$\Rightarrow \quad[ L ]^2=\left[ M ^{-1} L ^3 T ^{-2}\right]^\alpha[ M ]^\beta\left[ LT ^{-1}\right]^\gamma$
$\Rightarrow \quad[ L ]^2= M ^{-\alpha+\beta} L ^{3 \alpha+\gamma} T ^{-2 \alpha-\gamma}$
Equating dimensions on both sides, we
$\text { have } \quad-\alpha+\beta=0 \quad \dots(i)$
$3 \alpha+\gamma =2 \quad \ldots(ii)$
$-2 \alpha-\gamma=0 \quad \dots(iii)$
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
$\Rightarrow -2 \times 2-\gamma=0 \Rightarrow \gamma=-4$
$\text { So, } \alpha=2, \beta=2 \text { and } \gamma=-4$
