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

$100g \times (50-40)cm = m \times (40-20)cm \\[0.5em] \Rightarrow 100 \times 10 = m \times 20 \\[0.5em] \Rightarrow m = 50g \\[0.5em]$

(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.

$100gf \times (50-40)cm = 1000gfcm$

Therefore, the resultant moment will be

$1500gfcm - 1000 gfcm = 500gfcm (anticlockwise)$

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

Anticlockwise moment = Clockwise moment

So,

$100gf \times (50-40)cm + 50gf \times d = 50gf \times (40-10)cm \\[0.5em] \Rightarrow 1000gfcm + 50gf \times d = 1500gfcm \\[0.5em] \Rightarrow 50gfd = 500gfcm \\[0.5em] \Rightarrow d = 10 cm$

Hence, we can balance 50gm at 50cm