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 ?​

Answer 1

Answer:

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Explanation:  

At distance 25 cm from the pivoted end.

As we have one meter rule pivoted at midpoint there will be a perpendicular distance of 50 cm from the contact of force of 50gf.

Here we have moment of force of r×F.

After substituting we get moment of force=50×50gfcm.

Here this must be balanced by 100gf's moment of force.

Let us think that there would be x perpendicular distance.

On equating we get  

50×50gfcm=100×xgfcm

x=50×50/100=25cm.  

Hence 100gf must be applied at a distance of 25cm.