Question

A uniform metre rule of mass $100g$ is balanced on a fulcrum at mark $40cm$ by suspending an unknown mass $m$ at the mark $20cm$ . To which side the rule will tilt if the mass $m$ is moved to the mark $10cm$?

• A. on the side of mass $m$
• B. on the other side of mass $m$
• C. can't say
• D. it won't tilt

Answer 1

The correct option is A on the side of mass $m$

As 100 g is balanced at fulcrum it will have a perpendicular distance of 40 cm from itself and mass m will have $100-20=80cm$.

Here moment of forces are equal that is balance each other.Hence $m\times 80=100\times 40\Rightarrow$

$m=100/2=50g$.

If there is mark at 10 cm then perpendicular distance would be 90 cm and moment of force would be $50\times 90=4500\times gdyne-cm$

where as first moment of force is $100\times 40=400\times gdyne-cm$.

Hence it tilts on mass m side as it has more moment of force.