A uniform metre scale is in equilibrium position. Calculate the mass of the ruler.

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Solution

$τ$ = $r \times F$
Let the mass of the ruler be $M$ grams.
Centre of mass of the ruler is at it's center.
Therefore, centre of mass of the ruler is $20$ cm
away from the hinge .
It is given that the ruler is in equilibrium position.
Balancing the torques due to the mass of the rod and due to the 40 gram
mass suspended from the ruler about the hinge :

$40$ * $25$ = $M$ * $20$
$M$ = $50$ grams.
Hence , the mass of the ruler is $50$ grams .

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