Answer:
The mass of the rule is 56.41 g
Explanation:
Given;
mass of the object suspended at zero mark, m₁ = 200 g
pivot of the uniform meter rule = 22 cm
Total length of meter rule = 100 cm
0 22cm 100cm
-------------------------Δ------------------------------------
↓ ↓
200g m₂
Apply principle of moment
(200 g)(22 cm - 0) = m₂(100 cm - 22 cm)
(200 g)(22 cm) = m₂(78 cm)
m₂ = (200 g)(22 cm) / (78 cm)
m₂ = 56.41 g
Therefore, the mass of the rule is 56.41 g
