Young’s modulus of elasticity Y is expressed in terms of...

Dimensions of Young's modulus: $\left[{M}^{1}{L}^{– 1}T{}^{– 2}\right]$

Dimensions of velocity of light: \left[{M}^{0}{L}^{1}{T}^{– 1}\right]$</p><p>Dimensions of Planck's constant: \left[{M}^{1}{L}^{2}{T}^{– 1}\right]$

Dimensions of Gravitational constant: \left[{M}^{– 1}{L}^{3}{T}^{– 2}\right]$</p><p>Now, as given in the question$

$Y = {c}^{\alpha}{h}^{\beta}{G}^{\gamma}$

$⇒ \left[{M}^{1}{L}^{– 1}T{}^{– 2}\right]$ = {\left[{M}^{0}{L}^{1}{T}^{– 1}\right]}^{\alpha}{\left[{M}^{1}{L}^{2}{T}^{– 1}\right]}^{\beta}{\left[{M}^{– 1}{L}^{3}{T}^{– 2}\right]}^{\gamma}

Comparing exponent of the M, L & T both sides, we get$</p><p>$1 = \beta - \gamma ⇒ \beta = 1 + \gamma . . . \left(1\right)$</p><p>$- 1 = \alpha + 2\beta + 3\gamma . . . \left(2\right)

and $- 2 = - \alpha - \beta - 2\gamma . . .\left(3\right)$

From equation(1) and (2), we get$</p><p>$- 1 = \alpha + 2\left(1 + \gamma\right) + 3\gamma ⇒ \alpha + 5\gamma = - 3 . . . \left(4\right)

From equation(1)$and (3), we get</p><p>$- 2 = - \alpha - \left(1 + \gamma\right) - 2\gamma ⇒ \alpha + 3\gamma = 1 . . . \left(5\right)

From equation(4)$and equation(5), we get</p><p>$\alpha = 7, \gamma = – 2$and then$\beta = 1 + \gamma = - 1.