WAEC Physics 2016 Paper

Concept:A uniform metre rule has its entire weight acting at its midpoint, which is the 50 cm mark.For balance, the clockwise moment must equal the anticlockwise moment about the fulcrum.Explanation:The fulcrum is at the 35 cm mark.A 120 g mass is suspended at the 10 cm mark.Distance of the 120 g mass from the fulcrum: $35 - 10 = 25$ cm.This mass produces an anticlockwise moment about the fulcrum.The weight of the metre rule acts at its centre, the 50 cm mark.Distance of the centre of the rule from the fulcrum: $50 - 35 = 15$ cm.Let the mass of the metre rule be $M$ g.Using the principle of moments:$$120 \times 25 = M \times 15$$$$M = \frac{120 \times 25}{15}$$$$M = \frac{3000}{15} = 200$$Answer:The mass of the metre rule is $200$ g, which is Option D.