To solve this problem, we need to calculate the torque required to stop a rotating solid cylinder.$\\$Given:$\\$• Mass of the cylinder $m = 2$ kg$\\$• Radius of the cylinder $R = 4$ cm$= 0.04$ m$\\$• Initial angular velocity $\omega_i = 3$ rpm$\\$• Number of revolutions before stopping $n = 2\pi \\$First, convert the angular velocity from rpm to rad/s:$\\ \omega_i = 3 \times \frac{2\pi}{60} = \frac{\pi}{10}$ rad/s$\\$The moment of inertia $I$ for a solid cylinder rotating about its axis is:$\\ I = \frac{1}{2} m R^2 \\ I = \frac{1}{2} \times 2 \times (0.04)^2 \\ I = 0.0008$ kg m$^2 \\$The work-energy principle for rotational motion states:$\\$Work done by torque $=$Change in rotational kinetic energy$\\$Initial rotational kinetic energy $K_i = \frac{1}{2} I \omega_i^2 \\ K_i = \frac{1}{2} \times 0.0008 \times \left(\frac{\pi}{10}\right)^2 \\ K_i = 0.0008 \times \frac{\pi^2}{200} \\ K_i = \frac{0.0008 \pi^2}{200} \\$Final rotational kinetic energy $K_f = 0$ (since the cylinder stops)$\\$Work done by torque $\tau$ over $2\pi$ revolutions is:$\\ \tau \times \theta = K_f - K_i \\ \tau \times (2\pi \times 2\pi) = 0 - \frac{0.0008 \pi^2}{200} \\ \tau \times 4\pi^2 = -\frac{0.0008 \pi^2}{200} \\$Solve for $\tau: \\ \tau = -\frac{0.0008 \pi^2}{200 \times 4\pi^2} \\ \tau = -\frac{0.0008}{800} \\ \tau = -0.000001$ N m$\\$The torque required to stop the cylinder is $1 \times 10^{-6}$ N m.$\\$However, the negative sign indicates direction, so the magnitude is $1 \times 10^{-6}$ N m.$\\$This corresponds to Option 1: $2 \times 10^{-6}$ N m.