To solve this problem, we will use the principle of moments (torque) for equilibrium.$\\$Given:$\\$• Length of the rod $= 200$ cm$\\$• Mass of the rod $= 500$ g$= 0.5$ kg$\\$• Mass suspended at 20 cm $= 2$ kg$\\$• Unknown mass $m$ suspended at 160 cm$\\$• Wedge (pivot) is at 40 cm mark$\\$• Gravitational acceleration $g = 10$ m/s$^2 \\$The rod is uniform, so its center of mass is at the midpoint, 100 cm.$\\$Let's calculate the torques about the pivot point at 40 cm.$\\$Torque due to the 2 kg mass:$\\$Distance from pivot $= 40 - 20 = 20$ cm$\\$Torque $= 2 \cdot 10 \cdot 20 = 400$ N cm$\\$Torque due to the rod's weight:$\\$Distance from pivot to center of mass $= 100 - 40 = 60$ cm$\\$Torque $= 0.5 \cdot 10 \cdot 60 = 300$ N cm$\\$(This torque acts in the opposite direction)$\\$Torque due to the unknown mass $m: \\$Distance from pivot $= 160 - 40 = 120$ cm$\\$Torque $= m \cdot 10 \cdot 120 = 1200m$ N cm$\\$For equilibrium, the sum of clockwise torques must equal the sum of counterclockwise torques:$\\ 400 + 1200m = 300 \\$Solve for $m: \\ 1200m = 300 - 400 \\ 1200m = -100 \\ m = \frac{-100}{1200} \\ m = \frac{-1}{12}$ kg$\\$Since mass cannot be negative, we must have made an error in direction.$\\$Re-evaluate the direction:$\\$The correct equation should be:$\\ 400 = 300 + 1200m \\$Solve for $m$ again:$\\ 1200m = 400 - 300 \\ 1200m = 100 \\ m = \frac{100}{1200} \\ m = \frac{1}{12}$ kg$\\$Therefore, the value of $m$ is $\frac{1}{12}$ kg.$\\$This corresponds to Option 4.