To solve this problem, we need to find the angular acceleration of a hollow cylinder when a force is applied.$\\$Given:$\\$• Mass of the hollow cylinder $m = 3$ kg$\\$• Radius of the cylinder $R = 40$ cm$= 0.4$ m$\\$• Force applied $F = 30$ N$\\$The moment of inertia $I$ for a hollow cylinder about its central axis is given by:$\\ I = m R^2 \\$Substitute the known values:$\\ I = 3 \times (0.4)^2 \\ I = 3 \times 0.16 \\ I = 0.48$ kg m$^2 \\$The torque $\tau$ exerted by the force is given by:$\\ \tau = F \times R \\$Substitute the known values:$\\ \tau = 30 \times 0.4 \\ \tau = 12$ N m$\\$The angular acceleration $\alpha$ is given by the relation:$\\ \alpha = \frac{\tau}{I} \\$Substitute the values of $\tau$ and $I: \\ \alpha = \frac{12}{0.48} \\$Calculate $\alpha: \\ \alpha = 25$ rad/s$^2 \\$Therefore, the angular acceleration of the cylinder is $25$ rad/s$^2. \\$This corresponds to Option 2.