Gujarati
1.Units, Dimensions and Measurement
hard

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,

A

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

B

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

C

$\alpha=3, \beta=3$ and $\gamma=-2$

D

$\alpha=-3, \beta=-3$ and $\gamma=2$

(KVPY-2015)

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

$\Rightarrow -2 \times 2-\gamma=0 \Rightarrow \gamma=-4$

$\text { So, } \alpha=2, \beta=2 \text { and } \gamma=-4$

Standard 11
Physics

Similar Questions

Start a Free Trial Now

Confusing about what to choose? Our team will schedule a demo shortly.