The potential energy of a particle varies with distance x as $U = \frac{A x^{1 / 2}}{x^{2} + B},$ where A and B are constants. The dimensional formula for A x B is

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$M^{1} L^{7 / 2} T^{- 2}$

$M^{1} L^{11 / 2} T^{- 2}$

$M^{1} L^{5 / 2} T^{- 2}$

$M^{1} L^{9 / 2} T^{- 2}$

answer is B.

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Detailed Solution

$\begin{array}{l} Here x^{2} has the dimensions of L^{2}, B = [L^{2}] \\ Also M L^{2} T^{- 2} = \frac{A L^{1 / 2}}{L^{2}} or A = M L^{7 / 2} T^{- 2} \\ ∴ A \times B = M L^{11 / 2} T^{- 2} \end{array}$

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