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Question
A uniform metre rule is pivoted at its mid-point. A weight of 50 gf is suspended at one end of it. Where should a weight of 100 gf be suspended to keep the rule horizontal?
Solution
Length of meter scale = 1 m
Weight W1 = 50 gf
Weight W2 = 100 gf
distance = ?
W1 × d = W2 × d
50 × 50 cm = 100 × d
d = `(50 ×50)/100 = 25` cm from other end
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