The potential energy 'U' of a particle veries with distance 'x' from a fixed origin as `U=(AX)/(x^(2)+B).` Where A and B are dimensional constants. Find the dimension of length in AB.

Answer

[0005]

Explanation

`U=(Ax)/(x^(2)+B)`
`therefore[B] =L^(2)`
`ML""^(2)T^(-2)=([A]L)/(L^(2))`
`therefore[A] =ML""^(3)T^(-2)`
`[AB]=ML""^(5)T^(-2)" "]`