Given that F =at +bt², where F denotes force and f time, then the dimensions of a and b are respectively

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LT^-1 andT^-2

MLT^-3 and MLT^-4

T and T^-2

L T^-2 and T^-2

answer is D.

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Detailed Solution

$\begin{array}{l} F = at + {bt}^{2} \\ ∴ [a] = \frac{[F]}{[t]} = \frac{[{MLT}^{- 2}]}{[T]} = [{MLT}^{- 3}] \\ [b] = \frac{[F]}{[t^{2}]} = \frac{[{MLT}^{- 2}]}{[T^{2}]} = [{MLT}^{- 4}] \end{array}$

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