Question

The potential energy of a particle varies with distance x from a fixed origin as $U = \frac{A\sqrt{x}}{x^{2} + B},$ where A and B are dimensional constants then dimensional formula for AB is

• A. ML^7/2T$- 2$
• B. $ML^{11/2}T^{- 2}$
• C. $M^{2}L^{9/2}T^{- 2}$
• D. $ML^{13/2}T^{- 3}$

Answer 1

Answer: (B)

$ML^{11/2}T^{- 2}$

Answer 2

From the dimensional homogeneity $\lbrack x^{2}\rbrack = \lbrack B\rbrack$

∴ [B] = [L²]

As well as $\lbrack U\rbrack = \frac{\lbrack A\rbrack ⥂ \lbrack x^{1/2}\rbrack}{\lbrack x^{2}\rbrack + \lbrack B\rbrack}$ ⇒ $\lbrack ML^{2}T^{- 2}\rbrack = \frac{\lbrack A\rbrack\lbrack L^{1/2}\rbrack}{\lbrack L^{2}\rbrack}$ $\therefore\lbrack A\rbrack = \lbrack ML^{7/2}T^{- 2}\rbrack$

Now $\lbrack AB\rbrack = \lbrack ML^{7/2}T^{- 2}\rbrack \times \lbrack L^{2}\rbrack = \lbrack ML^{11/2}T^{- 2}\rbrack$