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The position of a particle at time $t$ is given by the equation $x(t)=\frac{v_0}{A}\left(1-e^{A t}\right) v_o=$ constant and $A>0$. Dimensions of $v_o$ and A respectively are
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Verified Answer
The correct answer is:
$\left[\mathrm{M}^0 \mathrm{LT}^{-1}\right]$ and $\left[\mathrm{M}^0 \mathrm{~L}^0 \mathrm{~T}^{-1}\right]$
The dimensions of $x=$ dimensions of $\frac{v_0}{A}$
Therefore, out of the given options $v_0$ has dimensions equal to $\left[\mathrm{M}^{\circ} \mathrm{LT}^{-1}\right]$ and $A$ has dimensions equal to $\left[\mathrm{M}^{\circ} \mathrm{L}^{\circ} \mathrm{T}^{-1}\right]$
So, that
$\begin{aligned}
\frac{\left[v_0\right]}{[A]} & =\frac{\left[\mathrm{M}^{\circ} \mathrm{LT}^{-1}\right]}{\left[\mathrm{M}^{\circ} \mathrm{L}^{\circ} \mathrm{T}^{-1}\right]}=[\mathrm{L}] \\
& =\text { dimension of } x .
\end{aligned}$
Therefore, out of the given options $v_0$ has dimensions equal to $\left[\mathrm{M}^{\circ} \mathrm{LT}^{-1}\right]$ and $A$ has dimensions equal to $\left[\mathrm{M}^{\circ} \mathrm{L}^{\circ} \mathrm{T}^{-1}\right]$
So, that
$\begin{aligned}
\frac{\left[v_0\right]}{[A]} & =\frac{\left[\mathrm{M}^{\circ} \mathrm{LT}^{-1}\right]}{\left[\mathrm{M}^{\circ} \mathrm{L}^{\circ} \mathrm{T}^{-1}\right]}=[\mathrm{L}] \\
& =\text { dimension of } x .
\end{aligned}$
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