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 ?

Solution:

Let W and W ′ represent the weights of the coin and the metre stick, respectively.
The centre of the metre stick's bulk, or the 50 cm point, is where it is concentrated.

The metre stick's mass is m ′.

Each coin weighs 5g.

The centre of mass shifts from point R to end P by 5 cm when the coins are positioned 12 cm from end P. The distance between point P and the centre of mass is 45 cm.

In the case of rotational equilibrium about point S, the net torque will be conserved.

$10\mathrm{g}(45-12)-\mathrm{m}\text{'}\mathrm{g}(50-45)=0$

$\mathrm{m}\text{'}=66\mathrm{g}$

Hence the correct answer is 66.0g.