QJEEAdv15Ph0a78: JEE Advanced 2015 Physics - Class 11 JEEAdv15Ph0a78 Solution: Planck's constant h, speed of light c and gravitational…

The dimensions of Planck's constant is $h = [{M^1}{L^3}{T^{ - 2}}]$

Speed of light is $c = [{L^1}{T^{ - 1}}]$

Gravitational constant is $G = [{M^{ - 1}}{L^3}{T^{ - 2}}]$

Let $L \propto {h^x}{c^y}{G^z}$ .

$[L] = {[{M^1}{L^2}{T^{ - 1}}]^x}{[{L^1}{T^{ - 1}}]^y}{[{M^{ - 1}}{L^3}{T^{ - 2}}]^z}$

Comparing, we get

$$\begin{array}{c}\hfill x-z=0\\ \hfill 2x+y+3z=1\\ \hfill -x-y-2z=0\end{array}\}$$

Solving, we get

$x = z$

$y + 5x = 1$

$- y - 3x = 0$

or, $2x = 1$

$x = {1 \over 2} = z$

$y = - {3 \over 2}$

Therefore,

$L \propto {h^{1/2}}{C^{ - 3/2}}{G^{1/2}}$

$L \propto \sqrt h$

$L \propto \sqrt G$

Let $M \propto {h^a}{C^b}{G^c}$

$[M] = {[{M^1}{L^2}{T^{ - 1}}]^a}{[{L^1}{T^{ - 1}}]^b}{[{M^{ - 1}}{L^3}{T^{ - 2}}]^c}$

Comparing, we get $a - c = 1$ .

$$\begin{array}{c}2a+b+3c=0\hfill \\ -a-b-2c=0\hfill \end{array}\}a+c=0$$

Therefore,

$2a = 1$

$a = {1 \over 2}$

$c = - {1 \over 2}$

$b = {1 \over 2}$

Therefore,

$M \propto {h^{1/2}}{c^{1/2}}{G^{ - 1/2}}$

$M \propto \sqrt c$