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

Answer

D

Explanation

Given fundamental quantities are momentum (p), area (A) and time (T)
We can write energy E as
`Eprop p^(a)A^(b) T^cimplies E=kp^(a)A^b T^c`
where k is dimensionless constant of proportionality Dimensions of `E=[E] = [ML^2 T^(-2) ] and [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`
`implies -a+c=-2`
`implies c =-2 + a =-2 +1 =-1`
Hene , `E=pA^(1//2) T^(-1)`