d potential energy of a particle varies wid distance 'x' from a fixed origin as;-U=A root /x square + B , WHERE A and B are dimensional constants then WHAT WILL be dimensional formula for AB?

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Solution

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$U = \frac{A \sqrt{x}}{x^{2} + B} . . . . . . . . . . . . . . . . . (1) Where U = Potential energy x = distance Dimensional formula of potential energy, U = [M^{1} L^{2} T^{- 2}] Dimensional formula of distance, x = [M^{0} L^{1} T^{0}] From equation (1), we can write x^{2} = - B B = - {[M^{0} L^{1} T^{0}]}^{2} B = - [M^{0} L^{2} T^{0}] Dimensional formula of B = [M^{0} L^{2} T^{0}] We can write equation (1) as, U = \frac{A \sqrt{x}}{x^{2}} U = \frac{A}{x^{\frac{3}{2}}} A = {Ux}^{\frac{3}{2}} A = [M^{1} L^{2} T^{- 2}] \times {[M^{0} L^{1} T^{0}]}^{\frac{3}{2}} A = [M^{1} L^{2} T^{- 2}] \times [M^{0} L^{\frac{3}{2}} T^{0}] A = [M^{1 + 0} L^{2 + \frac{3}{2}} T^{- 2 + 0}] A = [M^{1} L^{\frac{7}{2}} T^{- 2}] A . B = [M^{1} L^{\frac{7}{2}} T^{- 2}] [M^{0} L^{2} T^{0}] A . B = [M^{1} L^{\frac{7}{2} + 2} T^{- 2}] A . B = [M^{1} L^{\frac{11}{2}} T^{- 2}]$

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