A metre stick is balanced on a knife edge at its centre. When two coins, each of mass `5 g` are put one on of the other at the `12 cm` mark, the stick is found to balanced at `45 cm`. The mass of the metre stick is.

Explanation

Let m be the mass of the stick concentrated at C, the 50 cm mark, see fig.

For equilibrium about C, the 45 cm mark, 10 g (45-12)=mg (50-45)
`10g xx33=mgxx5`
`= gt m=10xx33//5`
or m=66 grams.