The potential energy of a particle varies wit | vedclass

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Question

$[\mathrm{B}]=\left[\mathrm{x}^{2}\right]=\mathrm{L}^{2}$

$[\mathrm{U}]=\frac{[\mathrm{A}][\mathrm{x}]^{1 / 2}}{[\mathrm{x}]^{2}}=\frac{[\mathrm{A}

Vedclass · Accepted Answer

${M^1}{L^{3/2}}{T^{ - 2}}$

Vedclass · Answer

$[\mathrm{B}]=\left[\mathrm{x}^{2}ight]=\mathrm{L}^{2}$

$[\mathrm{U}]=\frac{[\mathrm{A}][\mathrm{x}]^{1 / 2}}{[\mathrm{x}]^{2}}=\frac{[\mathrm{A}]}{[\mathrm{x}]^{3 / 2}} \Rightarrow[\mathrm{A}]=[\mathrm{u}][\mathrm{x}]^{3 / 2}$

$=\mathrm{MI}^{2} \mathrm{T}^{-2} 	imes \mathrm{L}^{3 / 2}$

$[\mathrm{A}]=\mathrm{ML}^{7 / 2} \mathrm{T}^{-2}$

$\left[\frac{\mathrm{A}}{\mathrm{B}}ight]=\frac{\mathrm{ML}^{7 / 2} \mathrm{T}^{-2}}{\mathrm{L}^{2}}$

$[\mathrm{A} / \mathrm{B}]=\mathrm{ML}^{3 / 2} \mathrm{T}^{-2}$

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 $A/B$

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