To determine which expression is dimensionless, we need to analyze the dimensions of each term.$\\$Given the force $F = \alpha t^2 + \beta t,$ where $\alpha$ and $\beta$ are constants.$\\$The dimension of force $[F]$ is $[M L T^{-2}]. \\$Let's find the dimensions of $\alpha$ and $\beta. \\$Since $\alpha t^2$ is part of the force, its dimension must be the same as force:$\\ [\alpha t^2] = [M L T^{-2}] \\$The dimension of $t^2$ is $[T^2]. \\$Thus, the dimension of $\alpha$ is:$\\ [\alpha] = \frac{[M L T^{-2}]}{[T^2]} = [M L T^{-4}] \\$Similarly, for $\beta t: \\ [\beta t] = [M L T^{-2}] \\$The dimension of $t$ is $[T]. \\$Thus, the dimension of $\beta$ is:$\\ [\beta] = \frac{[M L T^{-2}]}{[T]} = [M L T^{-3}] \\$Now, let's check each option to find the dimensionless expression:$\\$Option 1: $\alpha \beta t \\ [\alpha \beta t] = [\alpha][\beta][T] = [M L T^{-4}][M L T^{-3}][T] = [M^2 L^2 T^{-6}] \\$Option 2: $\frac{\alpha \beta}{t} \\ \left[\frac{\alpha \beta}{t}\right] = \frac{[\alpha][\beta]}{[T]} = \frac{[M L T^{-4}][M L T^{-3}]}{[T]} = [M^2 L^2 T^{-8}] \\$Option 3: $\frac{\beta t}{\alpha} \\ \left[\frac{\beta t}{\alpha}\right] = \frac{[\beta][T]}{[\alpha]} = \frac{[M L T^{-3}][T]}{[M L T^{-4}]} = [T] \\$Option 4: $\frac{\alpha t}{\beta} \\ \left[\frac{\alpha t}{\beta}\right] = \frac{[\alpha][T]}{[\beta]} = \frac{[M L T^{-4}][T]}{[M L T^{-3}]} = [1] \\$The dimensionless expression is $\frac{\alpha t}{\beta}. \\$Therefore, the correct option is 4.