Question

A uniform metre rule of weight $10gf$ is pivoted at its $0$ mark. How can it be made horizontal by applying a least force?

• A. By applying a force $10gf$ downwards at the $100cm$ mark.
• B. By applying a force $10gf$ upwards at the $100cm$ mark.
• C. By applying a force $5gf$ upwards at the $100cm$ mark.
• D. By applying a force $5gf$ downwards at the $100cm$ mark.

Answer 1

The correct option is C By applying a force $5gf$ upwards at the $100cm$ mark.

As we have a $10gf$ weight at $0$ mark and it will have a perpendicular distance of $50cm$ from point $O$ that is from the centre of the scale we will have moment of force = $50\times 10gfcm$ = $500gfcm$.

Hence it will have $500gfcm$ to depress the rule.

As we have to apply least force, hence the distance should be maximum, i.e, at the end of meter rule. Equating $500gfcm=100\times x$
$x=500/100gf$. Hence there would be of $5gf$ force at least to make it balanced along horizontal.