QJEEAdv23Ph60f1: JEE Advanced 2023 Physics - Class 11 JEEAdv23Ph60f1 Solution: Young's modulus of elasticity is expressed in terms of three…

Given the equation, $Y = c^\alpha h^\beta G^\gamma$

where Y represents Young's modulus, c the speed of light, h is Planck's constant, G is the gravitational constant, and α, β, and γ are powers to be determined.

We start by expressing the quantities involved in terms of their fundamental dimensions: Mass (M), Length (L), and Time (T). The equation in terms of dimensions is :

$[M^1 L^{-1} T^{-2}] = [M^0 L^1 T^{-1}]^\alpha [M^1 L^2 T^{-1}]^\beta [M^{-1} L^3 T^{-2}]^\gamma$

Next, by equating the powers of the dimensions M, L, and T on both sides of the equation, we form three separate equations:

1. Equating the powers of M, we get : $\beta - \gamma = 1$

2. Equating the powers of L, we get : $\alpha + 2\beta + 3\gamma = -1$

3. Equating the powers of T, we get : $-\alpha - \beta - 2\gamma = -2$

Now we can solve this system of equations.

From the first equation, we get $\beta = \gamma + 1$ .

Substituting $\beta$ from the first equation into the second and third equations, we get :

For the second equation, substituting $\beta$ we get : $\alpha + 2(\gamma + 1) + 3\gamma = -1$ , which simplifies to $\alpha = -5\gamma - 3$ .

For the third equation, substituting $\beta$ we get : $-\alpha - (\gamma + 1) - 2\gamma = -2$ , which simplifies to $\alpha = -3\gamma - 1$ .

Setting the two equations for $\alpha$ equal to each other gives : $-5\gamma - 3 = -3\gamma - 1$ , which simplifies to $\gamma = -2$ .

Substituting $\gamma = -2$ back into the equations for $\alpha$ and $\beta$ we find $\alpha = 7$ and $\beta = -1$ .

So, the solution to the system is $\alpha = 7, \beta = -1, \gamma = -2$ , which corresponds to Option A.