A uniform metre rule is balanced on a fulcrum placed at the 35cm mark by suspending a mass of 120g at the 10cm mark. Calculate the mass of the metre rule.

Analyze the question: A uniform metre rule has its weight acting at its centre, the 50 cm mark. It balances on a pivot at 35 cm, with a 120 g mass hung at 10 cm. We use the principle of moments.

Concept: For equilibrium, the sum of clockwise moments equals the sum of anticlockwise moments about the pivot.

Moment = force × perpendicular distance from the pivot. Since g is the same on both sides, we can use mass directly.

Solution:

• The 120 g mass is at 10 cm, so its distance from the pivot is 35 − 10 = 25 cm. It turns the rule anticlockwise.
• The rule's weight acts at 50 cm, so its distance from the pivot is 50 − 35 = 15 cm. It turns the rule clockwise.

Taking moments about the pivot:

120 × 25 = M × 15

3000 = 15M

M = 3000 ÷ 15 = 200 g

Answer: The mass of the metre rule is 200 g, which is option D.

Tip: In any uniform-rule question, first mark the centre of gravity at 50 cm. Then find each distance from the pivot, not from the end of the rule. Many students lose marks by using 10 cm and 50 cm directly instead of 25 cm and 15 cm.