Force F is given by F= at +bt², where t is time. What are the dimensions of a and b ?

The correct option is [MLT⁻³] and [MLT⁻⁴].

To find the dimensions of the constants $a$ and $b$ in the given equation:
$F=at+b{t}^2$
we use the principle of homogeneity of dimensions. According to this principle, terms that are added or subtracted must have the same dimensions, and the dimensions on both sides of a physical equation must be equal.

Therefore, the dimensions of each term on the right-hand side must be equal to the dimensions of the force ($F$) on the left-hand side:
1. Dimension of $F$ = Dimension of $at$
2. Dimension of $F$ = Dimension of $b{t}^2$

Let us write down the dimensions of the known quantities:
- The dimensional formula of force ($F$) is:
$\left[F\right]=\left[ML{T}^{-2}\right]$
- The dimensional formula of time ($t$) is:
$\left[t\right]=\left[T\right]$

Step 1: Find the dimensions of $a$
From the relation $\left[F\right]=\left[a\right]\left[t\right]$, we get:
$\left[a\right]=\frac{\left[F\right]}{\left[t\right]}$
Substituting the dimensions:
$\left[a\right]=\frac{\left[ML{T}^{-2}\right]}{\left[T\right]}=\left[ML{T}^{-3}\right]$

Step 2: Find the dimensions of $b$
From the relation $\left[F\right]=\left[b\right]\left[t^2\right]$, we get:
$\left[b\right]=\frac{\left[F\right]}{\left[t^2\right]}$
Substituting the dimensions:
$\left[b\right]=\frac{\left[ML{T}^{-2}\right]}{\left[T^2\right]}=\left[ML{T}^{-4}\right]$

Thus, the dimensions of $a$ and $b$ are [MLT⁻³] and [MLT⁻⁴] respectively.