Question

The potential energy u of a parricles varies with distance x from a fixed origin as u =A√x/x+B, where A and B are constants . The dimensions of A and B are respectively

Answer 1

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

from dimensional analysis,

dimension of x² = dimension of B

so, dimension of B = [L²]

we know, dimension of potential energy = [ML²T-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, [ML²T-²] = dimension of A × [L½]/[L²]

or, [ML²T-²][L²]/[L½] = dimension of A

or, dimension of A = [ML^{7/2}T-²]

hence, dimension of A $\bf{[ML^{\frac{7}{2}}T^{-2}]}$