$A$ metre stick is balanced on a knife edge at its centre. When two coins,each of mass $5\; g$ are put one on top of the other at the $12.0 \;cm$ mark,the stick is found to be balanced at $45.0\; cm$. What is the mass of the metre stick?

(66 G) Let $M$ be the mass of the metre stick. The weight of the stick $Mg$ acts at its centre of mass,which is at the $50.0 \;cm$ mark.
The two coins have a total mass of $m = 2 \times 5 \;g = 10 \;g$. Their weight $mg$ acts at the $12.0 \;cm$ mark.
The knife edge (fulcrum) is now at the $45.0 \;cm$ mark.
For rotational equilibrium about the fulcrum,the clockwise torque must equal the counter-clockwise torque.
Clockwise torque due to the weight of the stick: $\tau_{cw} = Mg \times (50.0 - 45.0) = Mg \times 5.0 \;cm$.
Counter-clockwise torque due to the weight of the coins: $\tau_{ccw} = mg \times (45.0 - 12.0) = 10g \times 33.0 \;cm$.
Equating the torques: $Mg \times 5.0 = 10g \times 33.0$.
$M = \frac{10 \times 33.0}{5.0} = 66 \;g$.
Thus,the mass of the metre stick is $66 \;g$.