A force defined by $F=\mathrm{\alpha <!--\; \alpha \; -->}{t}^2+\mathrm{\beta <!--\; \beta \; -->}t$ acts on a particle at a given time t. The factor which is dimensionless, if 𝛼 and β are constants, is:

The correct option is:
$\frac{\alpha t}{\beta }$

To find the dimensionless factor, we can apply the principle of dimensional homogeneity. According to this principle, terms that are added or subtracted must have the same dimensions, and the dimensions on both sides of an equation must be equal.

The given force equation is:
$F=\mathrm{��}{t}^2+\beta t$

Here, $F$ represents force and $t$ represents time. Let $\left[X\right]$ denote the dimensions of any quantity $X$. The dimension of time $t$ is $\left[T\right]$.

Applying the principle of dimensional homogeneity, each term on the right-hand side must have the dimensions of force:
$\left[\alpha t^2\right]=\left[F\right]$
and
$\left[\beta t\right]=\left[F\right]$

Since both terms are equal in dimension to $\left[F\right]$, we can equate their dimensions directly:
$\left[\alpha t^2\right]=\left[\beta t\right]$

This allows us to find the relationship between the dimensions of the constants $\alpha$ and $\beta$:
$\left[\alpha \right]{\left[T\right]}^2=\left[\beta \right]\left[T\right]$
Dividing both sides by $\left[\beta \right]\left[T\right]$ gives:
$\frac{\left[\alpha \right]\left[T\right]}{\left[\beta \right]}=1$

Replacing the dimension of time $\left[T\right]$ back with $\left[t\right]$, we see that:
$\left[\frac{\alpha t}{\beta }\right]=\left[M^0{L}^0{T}^0\right]$

Thus, the quantity $\frac{\alpha t}{\beta }$ is dimensionless.