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`.

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

Here `x^(2)` has the dimensiions of `L^(2) , B = [L^(2)]`
Also `M L^(2) T^(-2) = (AL^(1//2))/(L^(2))` or `A = ML^(7//2 T^(-2))`
`:. A xx B = M L^(1//2 T^(-2))`