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The potential energy of a particle varies with distance $x$ from fixed a origin as $v=\left(\frac{A \sqrt{x}}{x+B}\right)$; where, $A$ and $B$ are constants. The dimensions of $A B$ are
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Verified Answer
The correct answer is:
$\left[\mathrm{ML}^{7 / 2} \mathrm{~T}^{-2}\right]$
Given, $v=\frac{A \sqrt{x}}{x+B}$....(i)
Dimensions of $v=$ dimensions of potential energy
$=\left[\mathrm{ML}^2 \mathrm{~T}^{-2}\right]$
From Eq. (i)
Dimensions of $B=$ dimensions of $x=\left[\mathrm{M}^0 \mathrm{LT}^0\right]$
$\therefore$ Dimensions of $A$
$\begin{aligned}
& =\frac{\text { dimension of } v \times \text { dimensions of }(x+B)}{\text { dimensions of } \sqrt{x}} \\
& =\frac{\left[\mathrm{ML}^2 \mathrm{~T}^{-2}\right]\left[\mathrm{M}^0 \mathrm{LT}^0\right]}{\left[\mathrm{M}^0 \mathrm{~L}^{1 / 2} \mathrm{~T}^0\right]}=\left[\mathrm{ML}^{5 / 2} \mathrm{~T}^{-2}\right]
\end{aligned}$
Hence, dimensions of $A B$
$=\left[\mathrm{ML}^{5 / 2} \mathrm{~T}^{-2}\right]\left[\mathrm{M}^0 \mathrm{LT}^0\right]=\left[\mathrm{ML}^{7 / 2} \mathrm{~T}^{-2}\right]$
Dimensions of $v=$ dimensions of potential energy
$=\left[\mathrm{ML}^2 \mathrm{~T}^{-2}\right]$
From Eq. (i)
Dimensions of $B=$ dimensions of $x=\left[\mathrm{M}^0 \mathrm{LT}^0\right]$
$\therefore$ Dimensions of $A$
$\begin{aligned}
& =\frac{\text { dimension of } v \times \text { dimensions of }(x+B)}{\text { dimensions of } \sqrt{x}} \\
& =\frac{\left[\mathrm{ML}^2 \mathrm{~T}^{-2}\right]\left[\mathrm{M}^0 \mathrm{LT}^0\right]}{\left[\mathrm{M}^0 \mathrm{~L}^{1 / 2} \mathrm{~T}^0\right]}=\left[\mathrm{ML}^{5 / 2} \mathrm{~T}^{-2}\right]
\end{aligned}$
Hence, dimensions of $A B$
$=\left[\mathrm{ML}^{5 / 2} \mathrm{~T}^{-2}\right]\left[\mathrm{M}^0 \mathrm{LT}^0\right]=\left[\mathrm{ML}^{7 / 2} \mathrm{~T}^{-2}\right]$
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