Let us consider the torques about the wedge placed at the $40 \text{ cm}$ mark. Mass of the rod, $M = 500 \text{ g} = 0.5 \text{ kg}$. Since the rod is uniform, its weight acts at its centre of gravity, which is at the $100 \text{ cm}$ mark. Distance of the $2 \text{ kg}$ mass from the wedge = $40 \text{ cm} - 20 \text{ cm} = 20 \text{ cm} = 0.2 \text{ m}$. Distance of the centre of gravity of the rod from the wedge = $100 \text{ cm} - 40 \text{ cm} = 60 \text{ cm} = 0.6 \text{ m}$. Distance of the unknown mass $m$ from the wedge = $160 \text{ cm} - 40 \text{ cm} = 120 \text{ cm} = 1.2 \text{ m}$. For the rod to be in rotational equilibrium, the net torque about the wedge must be zero. By the principle of moments, the sum of anticlockwise moments equals the sum of clockwise moments. Anticlockwise moment = Moment due to $2 \text{ kg}$ mass = $2 \times g \times 0.2$ Clockwise moment = Moment due to rod's weight + Moment due to mass $m$ = $0.5 \times g \times 0.6 + m \times g \times 1.2$ Equating the moments: $2 \times g \times 0.2 = 0.5 \times g \times 0.6 + m \times g \times 1.2$ $0.4 = 0.3 + 1.2m$ $0.1 = 1.2m$ $m = \frac{0.1}{1.2} = \frac{1}{12} \text{ kg}$.