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If the force is given by $\mathrm{F}=\mathrm{at}+\mathrm{bt}^{2}$ with $\mathrm{t}$ as time. The dimensions of a and b are
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The correct answer is:
$\left[\mathrm{MLT}^{-3}\right]$ and $\left[\mathrm{MLT}^{-4}\right]$
Dimension of at = Dimension of F
$[a t]=[F] \Rightarrow[a]=\left[\frac{F}{t}\right]$
$[b]=\left[\frac{M L T^{-2}}{T}\right] \Rightarrow[a]=\left[M L T^{-3}\right]$
Dimension of $\mathrm{bt}^{2}=$ Dimension of $\mathrm{F}$
$\left[\mathrm{bt}^{2}\right]=[\mathrm{F}] \Rightarrow[\mathrm{b}]=\left[\frac{\mathrm{F}}{\mathrm{t}^{2}}\right]$
$[\mathrm{b}]=\left[\frac{\mathrm{ML} \mathrm{T}^{-4}}{\mathrm{~T}^{2}}\right] \Rightarrow[\mathrm{b}]=\left[\mathrm{MLT}^{-4}\right]$
$[a t]=[F] \Rightarrow[a]=\left[\frac{F}{t}\right]$
$[b]=\left[\frac{M L T^{-2}}{T}\right] \Rightarrow[a]=\left[M L T^{-3}\right]$
Dimension of $\mathrm{bt}^{2}=$ Dimension of $\mathrm{F}$
$\left[\mathrm{bt}^{2}\right]=[\mathrm{F}] \Rightarrow[\mathrm{b}]=\left[\frac{\mathrm{F}}{\mathrm{t}^{2}}\right]$
$[\mathrm{b}]=\left[\frac{\mathrm{ML} \mathrm{T}^{-4}}{\mathrm{~T}^{2}}\right] \Rightarrow[\mathrm{b}]=\left[\mathrm{MLT}^{-4}\right]$
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