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

Options

A. `[p^(1)a^(-1)t^(1)]`
B. `[p^(0)a^(1)t^(1)]`
C. `[p^(1)a^(-1/2)t^(1)]`
D. `[p^(1)a^(1/2)t^(-1)]`

Answer

D

Explanation

Here `p^(1)=MLT^(-1)`
`:.M=(p^(1))/(L^(1)T^(-1))`
Also `a^(1)=L^(2)` or `L=a^(1//2)`
`:.` Energy `=ML^(2)T^(-2)`
`=(P^(1))/(L^(1)T^(-1))xxaxxT^(-2)`
`(P^(1)xxaxxT^(-2))/(a^(1//2)T^(-1))`
`=p^(1)a^(1/2)T^(-1)`
`:.(d)` is correct.