Force $F = a t + b t^{2}$ where t is time. The dimensions of $a$ and $b$ are:

{M L T}^{- 3}

and

{M L T}^{- 4}

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{M L T}^{- 3}

and

{M L T}^{- 2}

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{M L T}^{- 1}

and

{M L T}^{0}

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{M L T}^{- 4}

and

{M L T}^{- 1}

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Solution

The correct option is D ${M L T}^{- 3}$ and ${M L T}^{- 4}$
$F = a t + b t^{2}$
$↓$
$(M L T^{- 2})$
$M L T^{- 2} = a t$
$a = M L T^{- 3}$
$M L T^{- 2} = b t^{2}$
$\Rightarrow b = M L T^{- 4}$

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[M L T^{- 1}]

[M L T^{- 1}] + [M L T^{- 1}] =

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