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

Options

A. `[P^(1//2)"AT"^(-1)] `
B. `[PA^(1//2)T^(-1)] `
C. `[PA^(1//2)T^(-1)] `
D. `[P^(2)AT^(-2)] `

Answer

B

Explanation

`[P] = "MLT"^(-1) larr ` momentum
`[A] =M^(0)L^(2)T^(0) larr ` Area
`[T] = M^(0)L^(0)T^(1) larr ` Time
Let [E] = `P^(x)A^(y)T^(z)`
` ML^(2)T^(-2) = [MLT^(-1)]^(x) [L^(2)]^(y) [T]^(z)`
` = M^(x) L^(x+2y_ T^(z-x)`
Comparing both sides :-
` x = 1 " "......(i)`
`x+2y = 2 rArr 1 + 2y = 2 or , y = 1/2 " ".....(ii)`
`z - x = - 2 rArr z - 1 = - 2 or z= - 1 " ".......(iii)`
` :. [E] =[P^(1) A^(1//2) T^(-1)] " " .....(iii)`