Final answer:
The mass of the uniform half-meter rule that balances at the 5cm mark from end A with a 0.1kg mass suspended is 0.15 kg, calculated by equating the torques on both sides of the fulcrum.
Step-by-step explanation:
The student is asking about how to calculate the mass of a rule using the principles of torque balance in physics. The rule is balanced horizontally on a fulcrum and a mass is suspended at a given point from one end.
To solve this, we consider that the torque due to the mass of the rule must be equal and opposite to the torque due to the hanging mass to maintain equilibrium. Mathematically, the torque (τ) is given by the product of the force (F) and the distance (d) from the pivot point, so τ = F × d. For the rule, the force is its weight (mass × gravitational acceleration), and the distance is half its length since it is uniform. For the hanging mass, the force is its weight, and the distance is the distance from the fulcrum.
Let the mass of the rule be m. Then the torque due to the rule: m × 9.8 m/s² × (50/2) cm and the torque due to the 0.1kg mass: 0.1 kg × 9.8 m/s² × 5 cm. Since these torques are equal, we set up the equation: m × 9.8 m/s² × 25 cm = 0.1 kg × 9.8 m/s² × 5 cm. Solving for m gives us the mass of the rule.
After solving the equation, we find that the correct answer is Option D: 0.15 kg.