If momentum `(p)`, area `(A)` and time`(t) `are taken to be fundamental quantities then energy has the dimensional formula

Options

A. `[pA^(-1)T^(-1)]`
B. `[p^2AT]`
C. `[pA^(-1//2)T]`
D. `[pA^(1//2)T]`

Answer

D

Explanation

Given , fundamenal quantities are momentum (p),
area (A) and time (T) .
We can write energy E as
`E prop p^a A^b T^c`
`E= kp^a A^A T^c`
Where k is dimensionless constant of proportionality.
Dimensions of E `=[R] = [ML^2 T^(-2)]` are `[P] = [MLT^(-1)]`
`[A] = [L^2)]`
`[T] = [T]`
`[E] = [K] [p]^a [A]^b.[T]^c`
Putting all the dimensions , we get
`ML^2 T^(-2) = [MLT^(-1)]^a [L^2]^b [T]^c`
`= M^a L^(2b+a)T^(-a+c)`
By principle of homogeneity of dimensions
`a=1 , 2b+a=2`
`implies 2b+1=2`
`implies b=1//2 -a + c =-2`
`implies c=-2 + a =-2 + 1 =-1`
Hence , `E= pA^(1//2)T^(-1)`