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A force $F$ is given $F=a t+b t^2$, where, $t$ is time. What are the dimensions of $a$ and $b$ ?
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Verified Answer
The correct answer is:
$\left[\mathrm{MLT}^{-3}\right]$ and $\left[\mathrm{MLT}^{-4}\right]$
Force $F=a t+b t^2$
From principle of homogeneity,
Dimension of at $=\frac{[F]}{[t]}=\frac{\left[\mathrm{MLT}^{-2}\right]}{[\mathrm{T}]}$
$=\left[\mathrm{MLT}^{-3}\right]$
Similarly, dimensions of
$b=\frac{[F]}{\left[t^2\right]}=\frac{\left[\mathrm{MLT}^{-2}\right]}{\left[\mathrm{T}^2\right]}\left[\mathrm{MLT}^{-4}\right]$
From principle of homogeneity,
Dimension of at $=\frac{[F]}{[t]}=\frac{\left[\mathrm{MLT}^{-2}\right]}{[\mathrm{T}]}$
$=\left[\mathrm{MLT}^{-3}\right]$
Similarly, dimensions of
$b=\frac{[F]}{\left[t^2\right]}=\frac{\left[\mathrm{MLT}^{-2}\right]}{\left[\mathrm{T}^2\right]}\left[\mathrm{MLT}^{-4}\right]$
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