Final answer:
The problem is solved using the principle of levers, balancing the moments about the pivot point. Assuming the scale is uniform, the calculated weight is 22.5 gf, which is closest to Option B, 20 gf, the correct answer.
Step-by-step explanation:
The question is about finding the weight of a half-meter scale balanced at the 5 cm mark when a weight of 10 gf is suspended at one end. To solve this, we can use the principle of moments, which states that for an object in equilibrium, the sum of the clockwise moments about any pivot is equal to the sum of the anticlockwise moments about that pivot (the principle of levers).
In this case, the pivot is at the 5 cm mark, and the weight is hanging at the 50 cm mark, which means that the distance from the pivot to the weight is 45 cm (since the scale is balanced at 5 cm from one end). We can then set up the equation: (weight of the scale at the pivot × distance from pivot to CG of scale) = (weight hanging × distance from pivot to weight).
Assuming the scale is uniform, the center of gravity (CG) would be at its midpoint, which is 25 cm from the pivot. Therefore, we can write the equation as follows: (weight of the scale) × 20 cm = 10 gf × 45 cm. Solving for the weight of the scale gives us: weight of the scale = (10 gf × 45 cm) / 20 cm = 22.5 gf.
However, since we only have options in whole numbers, the closest option the weight could be is Option B, 20 gf, since weights are typically rounded down to the nearest whole number when dealing with moments of force and balance.