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

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(B) Given,$A = G^\alpha M^\beta c^\gamma$.The dimensions of the quantities are:$[A] = [L]^2$$[G] = [M^{-1} L^3 T^{-2}]$$[M] = [M]$$[c] = [L T^{-1}]$Su

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

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(B) Given,$A = G^\alpha M^\beta c^\gamma$.The dimensions of the quantities are:$[A] = [L]^2$$[G] = [M^{-1} L^3 T^{-2}]$$[M] = [M]$$[c] = [L T^{-1}]$Substituting these into the equation:$[L]^2 = [M^{-1} L^3 T^{-2}]^\alpha [M]^\beta [L T^{-1}]^\gamma$$[L]^2 = M^{-\alpha + \beta} L^{3\alpha + \gamma} T^{-2\alpha - \gamma}$Comparing the powers of $M, L,$ and $T$ on both sides:For $M: -\alpha + \beta = 0 \implies \beta = \alpha \quad (i)$For $L: 3\alpha + \gamma = 2 \quad (ii)$For $T: -2\alpha - \gamma = 0 \implies \gamma = -2\alpha \quad (iii)$Substituting $(iii)$ into $(ii)$:$3\alpha - 2\alpha = 2 \implies \alpha = 2$Since $\beta = \alpha$,$\beta = 2$.Since $\gamma = -2\alpha$,$\gamma = -2(2) = -4$.Thus,$\alpha = 2, \beta = 2$ 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,

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