Answer:
40 gf
Step-by-step explanation:
The balance point of the uniform meter rule with the suspended weights = 40 cm = The pivot point
The location where the 25 gf weight is suspended = 10 cm
The location where the 10 gf weight is suspended = 75 cm
Let W represent the weight of the meter rule.
We have that the location of the application of the weight of the meter rule is at the center, 50 cm mark, point
Given that the meter rule is balanced, and taking moment about the pivot point, we have;
The moment om the left hand side, LHS, of the pivot point = The moment on the right hand side, RHS, of the pivot point
The moment on the LHS = 25 gf × (40 cm - 10 cm) = 750 gf·cm
The moment on the RHS = W × (50 cm - 40 cm) + 10 gf × (75 cm - 40 cm)
The moment on the RHS = W·(10 cm) + 350 gf·cm
∴ 750 gf·cm = W·(10 cm) + 350 gf·cm
W·(10 cm) = 750 gf·cm - 350 gf·cm = 400 gf·cm
W = (400 gf·cm)/(10 cm) = 40 gf
The weight of the meter scale (rule), W = 40 gf.