1. Equilibrium Equation (Torque Balance) To find the condition for the minimum mass \( P \), we check when the beam is about to tip. If \( P \) is very light, the heavy 75kg mass on the far left will cause the beam to rotate counter-clockwise around the left rope (Rope 1).

Taking torque about Rope 1: \[ \Sigma \tau = 0 \] \[ (75g \times 0.5) + (T_2 \times 2.25) = (P g \times 3.0) \] This gives an equation relating \( T_2 \) and \( P \).

2. Minimum Mass Condition As \( P \) decreases, the clockwise torque \( P g \times 3.0 \) decreases. To maintain balance, \( T_2 \) must decrease. The limiting case (tipping point) occurs when the rope goes slack: \[ T_2 = 0 \] Substituting \( T_2 = 0 \) into the torque equation: \[ 75g(0.5) = P_{min} g (3.0) \] \[ 37.5 = 3 P_{min} \implies P_{min} = 12.5 \text{ kg} \]

3. Analyzing Options

• (a) Minimum mass is 12.5 kg: This is a correct numerical value.
• (c) When P is minimum, tension vanishes: This is the physical condition derived above ($T_2 = 0$).

In multiple-choice questions focusing on conditions, statement (c) is the fundamental principle used to determine the stability limit. The numerical result (a) depends on specific values, but (c) is the general condition for static equilibrium failure in this mode.