A uniform half metre rule can be balanced at the 29.0 cm mark when a mass 20g is hung from its one end.
(a) Draw a diagram of the arrangement.
(b) Find the mass of the half metre rule.
(c) In which direction would the balancing point shift if 20g mass is shifted inside from its one end?

(a) Diagram of the arrangement is shown below:

(b) Suppose, if the mass of the meter rule be M.

Given, uniform meter rule, so weight (Mg) will act at 25 cm.

Mg produces anticlockwise moment about point o.

Now, in order to balance the 20g weight is tied at 50cm point and it acts in the clockwise direction.

As we know, the principle of moments states that

Anticlockwise moment = Clockwise moment

Mg (29 – 25) = 0.02g (50 – 29)

⇒ M = 21 × 0.02\over 4 = 105 g

The weight of half the meter rule is 105 g.

(c) The balancing point will shift towards the 25 cm mark.