A metre scale of weight 50 gf can be balanced at 40 cm mark without any weight suspended on it.

(a) If this ruler is cut at its centre then state which part [0 to 50 cm or 50 to 100 cm] of the ruler will weigh more than 25 gf.

(b) What minimum weight placed on this metre ruler can balance this ruler when it is pivoted at its centre?

(a) Given, wt of scale = 50 gf

Let wt. from [0 to 40] cm be W₁ and [40 to 100] cm be W₂

W₁ + W₂ = 50

⇒ W₁ = 50 - W₂ …..(1)

In equilibrium,

Anti-clockwise moment = Clockwise moment

(40 - 0) × W₁ = (100 - 40)W₂

40W₁ = 60W₂

40 [50 - W₂] = 60W₂ […..From (1)]

40 x 50 - 40W₂ = 60W₂

2000 = 60W₂ + 40W₂

2000 = 100W₂

W₂ = $\dfrac{2000}{100}$

W₂ = 20 gf

Now, W₁ = 50 - 20 = 30 gf

∴ [0 to 40 cm] part of ruler weighs 30 gf.

Hence, the part [ 0 to 50 cm] of the ruler will weigh more 25 gf.

(b) Given the metre scale balances at 40 cm mark, so its weight of 50 gf acts at 40 cm mark.

Its weight of 50 gf produces anticlockwise moment about the centre.

Let the minimum weight needed be m.

In equilibrium,

Anti-clockwise moment = Clockwise moment

50 x (50 - 40) = m x (100 - 50)

⇒ 50 x 10 = m x 50

⇒ m = $\dfrac{50 \times 10}{50}$

⇒ m = 10 gf

Hence, minimum weight needed to balance the metre ruler when it is pivoted at its centre = 10 gf.