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On a seesaw, a 40 kgf child sits 1.5 m from the pivot. At what distance must a 60 kgf child sit on the other side to balance it?
A2.25 m
B1.0 m
C0.9 m
D3.6 m
Answer & Solution
Correct answer: B. 1.0 m
1. Balance requires F₁ x d₁ = F₂ x d₂ (principle of moments).
2. Substitute: 40 x 1.5 = 60 x d₂, so 60 = 60 x d₂.
3. d₂ = 60 / 60 = 1.0 m.
4. Rule out 2.25 m: the heavier child must sit closer, not farther, than the lighter one.
_Source: Samacheer Kalvi (TN SCERT) Class 10 Science, Unit 1 "Laws of Motion", p.12_
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