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
`Epropp^(a)A^(b)T^(c)`
`E=kp^(a)A^(A)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`
`Rightarrow2b+1=2`
`Rightarrowb=1//2-a+c=-2`
`Rightarrow c=-2+a=-2+1=-1`
`"Hence, "E=pA^(1//2)T^(-1)`