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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 ?
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 ?
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Solution
Let 'a' is the distance from the centre where 100 gf (gram force) is suspended. From the principle of moments,
50gf × (L/2) = 100gf × a
a = L/4
Length of rule is 1 metre. So, a = 0.25m
Let 'a' is the distance from the centre where 100 gf (gram force) is suspended. From the principle of moments,
50gf × (L/2) = 100gf × a
a = L/4
Length of rule is 1 metre. So, a = 0.25m
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