The potential energy of a particle varies with distance x from a fixed origin as U, = ,frac{{Asqrt x }}{{{x^2} + B}} Where A and

Q: 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 $A/B$

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

SECTION - A [PHYSICS MCQ]

GSEB - English Medium

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 $A/B$

A${M^2}{L^1}{T^{ - 2}}$
B${M^1}{L^{3/2}}{T^{ - 2}}$
C${M^0}{L^{1/5}}{T^{ - 3}}$
D${M^2}{L^{2/2}}{T^{ - 3}}$

Medium

STD 11 Science
JEE
STD 11 - 1. units,dimensions and measurement
Physics

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