The potential energy of a particle varies with distance x from a … - Vidyadip

MCQ

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

✓

Answer

Correct option: B.

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

b
$[\mathrm{B}]=\left[\mathrm{x}^{2}\right]=\mathrm{L}^{2}$

$[\mathrm{U}]=\frac{[\mathrm{A}][\mathrm{x}]^{1 / 2}}{[\mathrm{x}]^{2}}=\frac{[\mathrm{A}]}{[\mathrm{x}]^{3 / 2}} \Rightarrow[\mathrm{A}]=[\mathrm{u}][\mathrm{x}]^{3 / 2}$

$=\mathrm{MI}^{2} \mathrm{T}^{-2} \times \mathrm{L}^{3 / 2}$

$[\mathrm{A}]=\mathrm{ML}^{7 / 2} \mathrm{T}^{-2}$

$\left[\frac{\mathrm{A}}{\mathrm{B}}\right]=\frac{\mathrm{ML}^{7 / 2} \mathrm{T}^{-2}}{\mathrm{L}^{2}}$

$[\mathrm{A} / \mathrm{B}]=\mathrm{ML}^{3 / 2} \mathrm{T}^{-2}$

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