A uniform metre rule of mass 100 g is balanced on a fulcrum at mark 40 cm by suspending an unknown mass m at the mark 20 cm.
(i) Find the value of m.
(ii) To which side the rule will tilt if the mass m is moved to the mark 10 cm?
(iii) What is the resultant moment now?
(iv) How can it be balanced by another mass 50 g?

(i) As we know, the principle of moments states that

Anticlockwise moment = Clockwise moment

100 × (50 – 40) = m × (40 – 20)

⇒ 1000 = 20 m

⇒ m = 1000\over 20 = 50 g

(ii) When the mass m is shifted to mark 10cm , it results in rule being shifted on the side of mass m in anticlockwise direction.

(iii) Anticlockwise moment is produced when a mass of m grams is moved towards the mark of 10cm.

100 × (50 – 40) = 1000

Therefore, the resultant moment will be

1500 – 1000 = 500 gf cm (anticlockwise)

(iv) As we know, the principle of moments states that

Anticlockwise moment = Clockwise moment

So,

100 × (50 – 40) + 50 × d = 50 × (40 – 10)

⇒ 1000 + 50d = 1500

⇒ 50d = 1500 – 1000 = 500

⇒ d = 500\over50 = 10 cm

Hence, we can balance 50 gm at 50 cm