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

Giben, fundamental quantities are momentum (`p`), area (`A`) and time (`T`).
We can write energy `E` as
`" "Epropp^(a)A^(b)T^(C)"rArrE=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)]rArr["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`
`rArr" "2b+1=2rArrb=1//2`
`" "-a+c=-2`
`rArr" "c=-2+a=-2+1=-1`
Hence, `" "E=pA^(1//2)T^(-1)`