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?

Let m be the mass of the scale. The centre of mass of scale is at 50 cm. thus scale is balanced at this point.

When 2 coins of mass 5 g are placed at 12 cm, the equilibrium has shifted to 45 cm mark.

Centre of mass of coins are at 12 cm mark and centre of mass of meter stick is at 50 cm mark.

For equilibrium about 45 cm mark,

W_coin × d_coin = W_{meter stick} × d_{meter stick}

Where

W_coin is weight of coin which is 10×g

W_{meter scale} is weight of meter stick which is mg

d_coin is distance of centre of mass of coin from equilibrium

d_{meter scale} is distance of centre of mass of meter scale from equilibrium

Thus,

10 g (45 - 12) = m g (50 - 45)

10 g × 33 = mg × 5

m = 10 × 33 /5

m = 66 grams

I.e. The mass of meter stick is 66 grams