Center of mass at pivot for rotational equilibrium Steps:

• Mass of box: $m_b = \frac{0.43}{9.8} \approx 0.044$ kg; its CG at 6 cm from supported end.
• Sphere touches supported end; its CG at 1.2 cm from supported end (radius).
• Supported length = sphere diameter = 2.4 cm, so system CG at 2.4 cm from supported end.
• Set $\frac{m_b \cdot 6 + m_s \cdot 1.2}{m_b + m_s} = 2.4$ (in cm); solve: $m_s = m_b \frac{6 - 2.4}{2.4 - 1.2} = 0.044 \cdot 3 = 0.13$ kg. Why B is correct:
• Matches calculation using center of mass formula for equilibrium, where net torque is zero about pivot. Why the others are wrong:
• A: Too low; underestimates shift needed to place CG at 2.4 cm.
• C: Too high; overestimates mass, shifting CG past 2.4 cm toward supported end.
• D: Excessively high; would place CG well inside supported region, not at edge.

Final answer: B