The potential energy of a particle varies wit | vedclass

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Question

Dimension of $\mathrm{A}$ is equals to dimension of $\frac{U\left(x+B\right)}{\sqrt{x}}$ or $\left[M L^{2} T^{-2}\right] \cdot \frac{L}{L^{\frac{1}{2}

Vedclass · Accepted Answer

$M^1 L^{7/2} T^{-2}$

Vedclass · Answer

Dimension of $\mathrm{A}$ is equals to dimension of $\frac{U\left(x+B\right)}{\sqrt{x}}$ or $\left[M L^{2} T^{-2}\right] \cdot \frac{L}{L^{\frac{1}{2}}}$ i.e $\left[M L^{\frac{5}{2}} T^{-2}\right]$

And dimension of $\mathrm{B}$ is equals square of $x$ that is $\left[L\right]$ (because two quantities can be added only when their dimension is the same)

So dimension of $A . B$ will be $\left[M L^{\frac{7}{2}} T^{-2}\right]$

The potential energy of a particle varies with distance $x$ from a fixed origin as $V = \frac{{A\sqrt x }}{{x + B}}$,where
$A$ and $B$ are constants. The dimensions of $AB$ are

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