Torque balance about the midpoint for equilibrium

Steps:

• The rod is horizontal, supported at midpoint (25 cm from left end), with 50 g mass at right end (25 cm from midpoint).
• Torque due to suspended mass about midpoint: 50 × 25 = 1250 g·cm (clockwise).
• Torque due to rod's mass about midpoint: 100 × (25 - d) g·cm (counterclockwise).
• Set torques equal for equilibrium: 100(25 - d) = 1250.
• Solve: 25 - d = 12.5, d = 12.5 cm.

Why A is correct:

• Not enough information (calculated d = 12.5 cm does not match options; diagram "as shown" is missing, making setup ambiguous).

Why the others are wrong:

• B: 15 cm would give left torque 1000 g·cm ≠ 1250 g·cm.
• C: 25 cm would give zero rod torque, unbalanced.
• D: 40 cm would give clockwise rod torque, worsening imbalance.

Final answer: Not enough information.