Question

The potential energy of a particle varies with distance `x` from a fixed origin as `U = (A sqrt(x))/( x^(2) + B)`, where `A and B` are dimensional constants , then find the dimensional formula for `AB`.

Answer 1

From the principle of dimensional homogenity ,
`[x^(2)] = [ B]`
:. `[B] = [L^(2)]`
As well as `[U] = ([A] [x^(1//2)])/( [x^(2)] + [B]) rArr [ ML^(2)T^(-2)] = ([A] [L^(1//2)])/([L^(2)])`
:. `[A] = [ ML^(7//2)T^(-2)]`
Now `[AB] = [ ML^(7//2)T^(-2)] xx [L^(2)] = [ML^(11//2)T^(-2)]`