Physics - Unit & Dimension Question with Solution | TestHub

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

Dimensions of velocity of light: $\left[M^0{L}^1{T}^{–1}\right]$

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

Dimensions of Gravitational constant: $\left[M^{–1}{L}^3{T}^{–2}\right]$

Now, as given in the question

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

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

Comparing exponent of the M, L & T both sides, we get

$1=\beta -\gamma \Rightarrow \beta =1+\gamma ...\left(1\right)$

$-1=\alpha +2\beta +3\gamma ...\left(2\right)$

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

From equation(1) and (2), we get

$$\begin{gathered} -1=\alpha +2\left(1+\gamma \right)+3\gamma \\ \Rightarrow \alpha +5\gamma =-3 ...\left(4\right) \end{gathered}$$

From equation(1) and (3), we get

$$\begin{gathered} -2=-\alpha -\left(1+\gamma \right)-2\gamma \\ \Rightarrow \alpha +3\gamma =1 ...\left(5\right) \end{gathered}$$

From equation(4) and equation(5), we get

$\alpha =7, \gamma =–2$ and then $\beta =1+\gamma =-1$.