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The potential energy of a particle moving along $x$-direction varies as $V=\frac{A x^2}{\sqrt{x}+B}$. The dimensions of $\frac{A^2}{B}$ are:
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The correct answer is:
$\left[\mathrm{M}^2 \mathrm{~L}^{1 / 2} \mathrm{~T}^{-4}\right]$
$V=\frac{A x^2}{\sqrt{x}+B}$
As per homogeneous rule $B=\sqrt{L}$
$\mathrm{ML}^2 \mathrm{~T}^{-2}=\frac{A \mathrm{~L}^2}{\mathrm{~L}^{1 / 2}}$
$A=\mathrm{ML}^{1 / 2} \mathrm{~T}^{-2}$
$\frac{A^2}{B}=\frac{\mathrm{M}^2 L T^{-4}}{\mathrm{~L}^{1 / 2}}=\mathrm{M}^2 \mathrm{~L}^{1 / 2} \mathrm{~T}^{-4}$
As per homogeneous rule $B=\sqrt{L}$
$\mathrm{ML}^2 \mathrm{~T}^{-2}=\frac{A \mathrm{~L}^2}{\mathrm{~L}^{1 / 2}}$
$A=\mathrm{ML}^{1 / 2} \mathrm{~T}^{-2}$
$\frac{A^2}{B}=\frac{\mathrm{M}^2 L T^{-4}}{\mathrm{~L}^{1 / 2}}=\mathrm{M}^2 \mathrm{~L}^{1 / 2} \mathrm{~T}^{-4}$
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