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, Fig.
For equilibrium about C', the `45 cm` mark,
`10 g (45 - 12) = mg (50 - 45)`
`10g xx 33 = mg xx 5`
`m = (10 xx 33)/(5) = 66 gram`