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

Options

A. `[M^(1)L^(7//2)T^(-2)]`
B. `[M^(1)L^(11//2)T^(-2)]`
C. `[M^(1)L^(5//2)T^(-2)]`
D. `M^(1)L^(9//2)T^(-2)]`

Answer

B

Explanation

In the expression `U= (Asqrt(x))/(x^(2)+B)`
B must have the dimensions of `A= [(Ux^(2))/(sqrt(x))]= ([ML^(2)T^(-2)][L^(2)])/([L^(1//2)])= [ML^(7//2)T^(-2)]`
`:.` the dimensional formula for `AB = [ML^(7//2)T^(-2)][L^(2)]= [M^(1)L^(11//2)T^(-2)]`