Question

If area $\left(A\right)$, time $\left(T\right)$ and momentum $\left(P\right)$ are assumed as fundamental quantities, then dimensional formula for energy will be

• A. $\left[A{T}^{-1/2}{P}^2\right]$
• B. $\left[A{T}^{-2}{P}^1\right]$
• C. $\left[A^{1/2}{T}^{-1}P\right]$
• D. $\left[A^{1/2}{T}^{-1}{P}^2\right]$

Answer 1

The correct option is C $\left[A^{1/2}{T}^{-1}P\right]$

Let the dimensional formula of energy will be,
$\left[E\right]=\left[A^a{T}^b{P}^c\right]$
$\Rightarrow \left[M^1{L}^2{T}^{-2}\right]=\left[L^{2a}{T}^b{M}^c{L}^c{T}^{-c}\right]$
$\because E=F\times d,P=mv$
(Energy is equivalent to work)
$\Rightarrow \left[M^1{L}^2{T}^{-2}\right]=\left[M^c{L}^{2a+c}{T}^{b-c}\right]$
From principle of homogeneity, equating the coefficients on both sides:
$c=1$
$2a+c=2$
$\Rightarrow 2a=1[\because c=1]$
$\Rightarrow a=\frac{1}{2}$
Similarly, $b-c=-2$
$\Rightarrow b=-2-(-1)=-1$
Hence, $\left[E\right]=\left[A^{1/2}{T}^{-1}{P}^1\right]$