The potential energy u of a particle varies with distance x from a fixed origin as u=A rootx/x+b find dimension of A and B

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Solution

$P o t e n t i a l e n e r g y i s g i v e n b y u = A \frac{\sqrt{x}}{x} + B u = \frac{A}{\sqrt{x}} + B W e k n o w, t w o p h y s i c a l q u a n t i t e s c a n b e o n l y a d d e d w h e n t h e y h a v e s a m e d i m e n s i o n s . s o d i m e n s i o n o f B i s s a m e a s o f e n e r g y . D i m e n s i o n o f B i s [M^{1} L^{2} T^{- 2}] S i m i l a r l y D i m e n s i o n o f \frac{A}{\sqrt{x}} i s s a m e a s o f e n e r g y \frac{A}{\sqrt{x}} = [M^{1} L^{2} T^{- 2}] [A] = [M^{1} L^{2} T^{- 2}] . [L^{1 / 2}] D i m e n s i o n o f A = [M^{1} L^{5 / 2} T^{- 2}]$

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