To solve this problem, we need to find the tension in the string that produces a given angular acceleration in the cylinder.$\\$Given:$\\$• Mass of the cylinder $m = 50 \, \mathrm{kg} \\$• Radius of the cylinder $r = 0.5 \, \mathrm{m} \\$• Angular acceleration $\alpha = 2 \, \mathrm{rev/s}^2 \\$First, convert the angular acceleration from revolutions per second squared to radians per second squared:$\\ 1 \, \mathrm{rev} = 2\pi \, \mathrm{rad} \\ \alpha = 2 \, \mathrm{rev/s}^2 \times 2\pi \, \mathrm{rad/rev} = 4\pi \, \mathrm{rad/s}^2 \\$The moment of inertia $I$ of a solid cylinder about its axis is given by:$\\ I = \frac{1}{2} m r^2 \\$Substitute the known values:$\\ I = \frac{1}{2} \times 50 \, \mathrm{kg} \times (0.5 \, \mathrm{m})^2 \\ I = \frac{1}{2} \times 50 \times 0.25 \\ I = 6.25 \, \mathrm{kg \cdot m^2} \\$The torque $\tau$ required to produce the angular acceleration is given by:$\\ \tau = I \alpha \\$Substitute the known values:$\\ \tau = 6.25 \, \mathrm{kg \cdot m^2} \times 4\pi \, \mathrm{rad/s}^2 \\ \tau = 25\pi \, \mathrm{N \cdot m} \\$The tension $T$ in the string is related to the torque by:$\\ \tau = T \cdot r \\$Rearrange to solve for $T: \\ T = \frac{\tau}{r} \\$Substitute the known values:$\\ T = \frac{25\pi \, \mathrm{N \cdot m}}{0.5 \, \mathrm{m}} \\ T = 50\pi \, \mathrm{N} \\$Calculate the numerical value:$\\ T \approx 50 \times 3.14 \\ T \approx 157 \, \mathrm{N} \\$Therefore, the tension in the string required to produce the angular acceleration is $157 \, \mathrm{N}. \\$This corresponds to Option 4.