To solve this problem, we need to find the net acceleration of a point on the rim of the disc at the end of 2 seconds.$\\$Given:$\\$• Radius of the disc $r = 50$ cm$= 0.5$ m$\\$• Angular acceleration $\alpha = 2.0$ rad/s$^2 \\$• Time $t = 2.0$ s$\\$First, calculate the angular velocity $\omega$ at the end of 2 seconds using the equation:$\\ \omega = \omega_0 + \alpha t \\$Since the disc starts from rest, $\omega_0 = 0.$ Therefore,$\\ \omega = 0 + 2.0 \times 2.0 = 4.0$ rad/s$\\$Next, calculate the tangential acceleration $a_t$ using the formula:$\\ a_t = \alpha \times r \\ a_t = 2.0 \times 0.5 = 1.0$ m/s$^2 \\$Now, calculate the centripetal acceleration $a_c$ using the formula:$\\ a_c = \omega^2 \times r \\ a_c = (4.0)^2 \times 0.5 = 16.0 \times 0.5 = 8.0$ m/s$^2 \\$The net acceleration $a_{net}$ is the vector sum of $a_t$ and $a_c.$ Since these are perpendicular,$\\ a_{net} = \sqrt{a_t^2 + a_c^2} \\ a_{net} = \sqrt{(1.0)^2 + (8.0)^2} \\ a_{net} = \sqrt{1.0 + 64.0} \\ a_{net} = \sqrt{65.0} \\$Approximating $\sqrt{65.0} \approx 8.06$ m/s$^2 \\$Therefore, the net acceleration at the end of 2 seconds is approximately $8.0$ m/s$^2. \\$This corresponds to Option 3.