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. `[p A^(-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^(c)impliesE=kp^(a)A^(b)T^(c)`where, K is dimensionless contact of proportionality . Dimensions of `E=[E]=[ML^(2)T^(-2)]and [p]=[MLT^(-1)]`
`[A]=[L^(2)]implies [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 principal of homogeity of dimensions,`a=1,2b+a=2`
`implies2b+1=2implies b=1//2`
`-a+c=-2`
`implies c=-2+a=-2+1=-1`
Hence `E=pA^(1//2)T^(-1)`