Problem 14 - Entrance Test

For the meter stick to be in equilibrium, the net torque about any pivot point must be zero. Since the meter stick is uniform, its center of mass is at the 50 cm mark. If we choose the 50 cm mark as our pivot, the weight of the meter stick itself creates no torque about this point. Let the pivot point be P = 50 cm. 1. **Torque due to the 0.5 kg mass (m1):** Mass m1 = 0.5 kg Position of m1 = 20 cm Distance from pivot (r1) = |20 cm - 50 cm| = 30 cm = 0.3 m Torque τ1 = m1 * g * r1 = 0.5 kg * g * 0.3 m = 0.15g Nm. This torque tends to rotate the stick counter-clockwise. 2. **Torque due to the 0.3 kg mass (m2):** Mass m2 = 0.3 kg Let the position of m2 be x cm. To balance the stick, m2 must be placed to the right of the pivot (x > 50 cm) to create a clockwise torque. Distance from pivot (r2) = (x - 50) cm = (x - 50)/100 m Torque τ2 = m2 * g * r2 = 0.3 kg * g * (x - 50)/100 m. 3. **Equilibrium Condition (Στ = 0):** For equilibrium, the counter-clockwise torque must equal the clockwise torque: τ1 = τ2 0.15g = 0.3g * (x - 50)/100 Cancel 'g' from both sides: 0.15 = 0.3 * (x - 50)/100 Divide by 0.3: 0.15 / 0.3 = (x - 50)/100 0.5 = (x - 50)/100 Multiply by 100: 50 = x - 50 Solve for x: x = 50 + 50 x = 100 cm Therefore, the 0.3 kg mass should be placed at the 100 cm mark. The correct choice is D.