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 dimensional formula for $A B$ is
  • $\left[ ML ^{11/2} T ^{-2}\right]$
  • B
    $\left[ ML ^{7 / 2} T ^{-2}\right]$
  • C
    $\left[M^2 L^{9 / 2} T^{-2}\right]$
  • D
    $\left[ ML ^{13 / 2} T ^{-3}\right]$

Answer

Correct option: A.
$\left[ ML ^{11/2} T ^{-2}\right]$
a
(a)

$x =$ distance from a fixed origin

$u =\frac{ A \sqrt{ x }}{ x ^2+ B }$

unit of $B$ is same as $x ^2$. Unit of $x ^2=\left[ L ^2\right]$

$B =\left[ L ^2\right]$

$u =\frac{\left[ x ^{1 / 2}\right] A }{ x ^2+ B }$

$A =\frac{ u \left( x ^2+ B \right)}{\sqrt{ x }}=\frac{ kgm }{ s ^2} \times \frac{ m ^2}{ m ^{1 / 2}}$

$A =\frac{ kgm }{ s ^2}$

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

Dimensions of $AB =\left[ ML ^{7 / 2} T ^{-2}\right]\left[ L ^2\right]$

Dimensions of $AB =\left[ ML ^{11 / 2} T ^{-2}\right]$

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