Question

the potential energy of a particle varies with the distance x from fixed origin as U = A under root 'x'/ x^2+B , where A and B are dimensional constant then dimensional formula for AB is

Answer 1

given, potential energy of a particle varies with distance from fixed origin as $U=\frac{A\sqrt{x}}{x^2+B}$

here, dimension of x² = dimension of B

so, dimension of B = [L^2]

and we know, dimension of potential energy = [ML^2T^-2 ]

now, dimension of U = dimension of {A√x}/dimension of {x² + B}

or, dimension of U = dimension of A × dimension of √x/dimension of x²

or, [ML^2T^-2] = dimension of A × [L^{1/2}]/[L^2]

or, [ML^2T^-2][L^2]/[L^{1/2}] = dimension of A

or, dimension of A = [ML^{7/2}T^-2]

hence, answer should be $[ML^{\frac{7}{2}}T^{-2}]$

Answer 2

hope it helps you......frnds