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

Options

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

Answer

D

Explanation

Let, energy `E=kp^(a)A^(b)t^(c)` . . .(i)
where k is a dimensionless constant of proportionality.
Equating dimensions on both sides of (i), we get
`[ML^(2)T^(-2)]=[MLT^(-1)]^(a)[M^(0)L^(2)T^(0)]^(b)[M^(0)L^(0)T]^(c)` ltBrgt `=[M^(a)L^(a+2b)T^(-a+c)]`
Applying the principle of homogeity of dimensions, ltBrgt we get
`a=1` . .. (ii)
`a+2b=2` . . .(iii)
`-a+c=-2` . . .(iv)
On solving eqs. (ii), (iii) and (iv), we get
`a=1,b=(1)/(2),c=-1 therefore[E]=[p^(1)A^(1//2)t^(-1)]`