Since no external torque acts on the system (man + platform), the total angular momentum is conserved. The initial angular momentum is zero. Let the angular velocity of the platform be $\omega_p$. The velocity of the man relative to the ground is $v = 1\text{ m/s}$. His angular velocity relative to the ground is $\omega_m = \frac{v}{R} = \frac{1}{2} = 0.5\text{ rad/s}$. According to the law of conservation of angular momentum: $L_{\text{man}} + L_{\text{platform}} = 0$ $m v R + I \omega_p = 0$ $(50)(1)(2) + 200 \omega_p = 0 \implies 100 + 200 \omega_p = 0 \implies \omega_p = -0.5\text{ rad/s}$. The negative sign indicates that the platform rotates in the opposite direction to the man's motion. The angular velocity of the man relative to the platform is: $\omega_{\text{rel}} = \omega_m - \omega_p = 0.5 - (-0.5) = 1.0\text{ rad/s}$. The time taken by the man to complete one revolution on the platform is: $T = \frac{2\pi}{\omega_{\text{rel}}} = \frac{2\pi}{1.0} = 2\pi\text{ s}$.