Question

A uniform meter rule is pivoted at its mid-point. A weight of $50gf$ suspended at one end of it. Where should a weight of $100gf$ be suspended to keep the rule horizontal :

• A. At distance $50cm$ from the pivoted end.
• B. At distance $12.5cm$ from the other end.
• C. At distance $37.5cm$ from the other end.
• D. At distance $25cm$ from the pivoted end.

Answer 1

Answer: (D)

At distance $25cm$ 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 $r\times F$.
After substituting we get $momentofforce=50\times 50gfcm$.
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
$50\times 50gfcm=100\times xgfcm$
$x=50\times 50/100=25cm$.

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