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If $x=a t+b t^{2}$, where $x$ is measured in metre and $t$ in second, then the dimension of $(b / a)$ is
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The correct answer is:
$\left[\mathrm{T}^{-1}\right]$
Given, $x=a t+b t^{2}$
Here, $x$ is measured in metre $(\mathrm{m})$ and $t$ is in second $(\mathrm{s})$.
So, for dimensional consistency of equation,
Dimensions of $a=[a]=\frac{[x]}{[t]}=\frac{[L]}{[T]}=\left[L^{-1}\right]$
Similarly, for $b,[b]=\frac{[x]}{\left[t^{2}\right]}=\frac{[L]}{\left[\mathrm{T}^{2}\right]}=\left[\mathrm{LT}^{-2}\right]$
$\therefore$ Dimensions of $\left[\frac{b}{a}\right]=\frac{\left[L \mathrm{~T}^{-2}\right]}{\left[L \mathrm{~T}^{-1}\right]}=\left[\mathrm{T}^{-1}\right]$
Here, $x$ is measured in metre $(\mathrm{m})$ and $t$ is in second $(\mathrm{s})$.
So, for dimensional consistency of equation,
Dimensions of $a=[a]=\frac{[x]}{[t]}=\frac{[L]}{[T]}=\left[L^{-1}\right]$
Similarly, for $b,[b]=\frac{[x]}{\left[t^{2}\right]}=\frac{[L]}{\left[\mathrm{T}^{2}\right]}=\left[\mathrm{LT}^{-2}\right]$
$\therefore$ Dimensions of $\left[\frac{b}{a}\right]=\frac{\left[L \mathrm{~T}^{-2}\right]}{\left[L \mathrm{~T}^{-1}\right]}=\left[\mathrm{T}^{-1}\right]$
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