If the force is given by `F=at+bt^(2)` with `t` is time. The dimensions of `a` and `b` are

Options

A. `[MLT^(-4)],[MLT^(-2)]`
B. `[MLT^(-3)],[MLT^(-4)]`
C. `[ML^2T^(-3)],[ML^2T^(-2)]`
D. `[ML^2T^(-3)],[ML^3 T^(-4)]`

Answer

B

Explanation

Dimension of at = Dimension of F
`[at] = [F]`
`[a] = [F/t]`
`[a] = [(MLT^(-2))/T]`
`[a] = [MLT^(-3)]`
Dimension of `bt^2` = Dimension of `F`
`[bt^2] = [F]`
`[b] = [F/(t^2)]`
`[a] = [(MLT^(-2))/(T^2)]`
`[a] = [MLT^(-4)]`