Torque Equilibrium on a Seesaw

Steps:

• Total bar length is 30 cm; pivot is 20 cm from unknown mass, so 10 cm from 1.0 kg mass.
• For balance, clockwise torque equals counterclockwise torque: $m \times 20 = 1.0 \times 10$ (g and cm cancel).
• Solve for $m$: $m = \frac{10}{20} = 0.5$ kg.
• Convert to grams: 0.5 kg = 500 g.

Why B is correct:

• 500 g satisfies the torque balance equation $m \times d_1 = m_2 \times d_2$, from the principle of moments for rotational equilibrium.

Why the others are wrong:

• A: 333 g gives torque of 6.66 kg·cm, less than 10 kg·cm, so unbalanced.
• C: 667 g gives torque of 13.34 kg·cm, more than 10 kg·cm, so unbalanced.
• D: 1000 g gives torque of 20 kg·cm, more than 10 kg·cm, so unbalanced.

Final answer: B