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

Options

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

Answer

B

Explanation

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