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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 find the dimensional formula for $A/B$
${M^2}{L^1}{T^{ - 2}}$
${M^1}{L^{3/2}}{T^{ - 2}}$
${M^0}{L^{1/5}}{T^{ - 3}}$
${M^2}{L^{2/2}}{T^{ - 3}}$
Solution
$[\mathrm{B}]=\left[\mathrm{x}^{2}\right]=\mathrm{L}^{2}$
$[\mathrm{U}]=\frac{[\mathrm{A}][\mathrm{x}]^{1 / 2}}{[\mathrm{x}]^{2}}=\frac{[\mathrm{A}]}{[\mathrm{x}]^{3 / 2}} \Rightarrow[\mathrm{A}]=[\mathrm{u}][\mathrm{x}]^{3 / 2}$
$=\mathrm{MI}^{2} \mathrm{T}^{-2} \times \mathrm{L}^{3 / 2}$
$[\mathrm{A}]=\mathrm{ML}^{7 / 2} \mathrm{T}^{-2}$
$\left[\frac{\mathrm{A}}{\mathrm{B}}\right]=\frac{\mathrm{ML}^{7 / 2} \mathrm{T}^{-2}}{\mathrm{L}^{2}}$
$[\mathrm{A} / \mathrm{B}]=\mathrm{ML}^{3 / 2} \mathrm{T}^{-2}$
