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 is k a dimensionless constant proportionality equating dimension an 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)`
`[L]= [M^(a)L^(a+2b)T^(a+c)]`
Applying the principle of homogeneity of dimensions 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`
`:. [E] = [p^(1)A^(1//2)c^(-2)]`