Question

The potential energy of a particle varies with distance x from a fixed origin as U=A√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}U=

x

2

+B

A

x

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}][ML

2

7

T

−2

]