A metre stick is balanced on a knife edge at its centre. When two coins, each of mass 5g 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?

Explanation

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

For equilibrium about C', i.e. at the 45 cm mark,
10g (45 - 12) = mg (50 - 45)
`10g xx 33 = mg xx 5`
`m=(10xx33)/(5) = 66 gm`