Question

If momentum$\left(P\right)$, area $\left(A\right)$ and time $\left(T\right)$ are taken to be fundamental quantities, then energy has the dimensional formula $\left[P^{\alpha }{A}^{\beta }{T}^{\gamma }\right]$. Find $(\alpha +2\beta +\gamma )$

Answer 1

Given:
$\left[E\right]=$$\left[P^{\alpha }{A}^{\beta }{T}^{\gamma }\right]\cdots \cdot \left(i\right)$
As we know,
Dimension of $\left[E\right]=\left[M{L}^2{T}^{-2}\right]$
Dimension of $\left[P\right]=\left[ML{T}^{-1}\right]$
Dimension of $\left[A\right]=\left[L^2\right]$
Dimension of $\left[T\right]=\left[T\right]$

By putting all dimensions in equation$\left(i\right)$ we get,

$\left[E\right]={\left[ML{T}^{-1}\right]}^{\alpha }{\left[L^2\right]}^{\beta }{\left[T\right]}^{\gamma }$
$\left[M{L}^2{T}^{-2}\right]={\left[ML{T}^{-1}\right]}^{\alpha }{\left[L^2\right]}^{\beta }\left[T^{\gamma }\right]$

By comparing $L.H.S=R.H.S$

$\alpha =1$

$\alpha +2\beta =2\Rightarrow 2\beta =1$

$\Rightarrow \beta =\frac{1}{2}$

$-\alpha +\gamma =-2\Rightarrow \gamma =-1$

Hence, $\alpha =1,\beta =\frac{1}{2},\gamma =-1$

$\left[E\right]=\left[P^1{A}^{\frac{1}{2}}{T}^{-1}\right]$

Therefore, $(\alpha +2\beta +\gamma )=1+2\times \frac{1}{2}+(-1)=1$
Final Answer:(1)