Question

17. 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 of
50 g ?
Ans. (i) m = 50 g, (ii) on the side of mass m,
(iii) 500 gf x cm (anticlockwise),
(iv) by suspending the mass 50 g at the mark 50 cm.​

Answer 1

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From the principle of moments,

Clockwise moment = Anticlockwise moment

100 g × (50 – 40) cm = m × (40 – 20) cm

100 g × 10 cm = m × 20 cm

m = 50 g

If the mass m is moved to the mark 10 cm, the rule will tilt on the side of mass m (anticlockwise)

Anticlockwise moment if mass m is moved to the mark 10 cm

= 50 g × (40 – 10) cm

= 50 g × 30 cm

= 1500 g cm

Clockwise moment = 100 g × (50 – 40) cm

= 100 g × 10 cm

= 1000 g cm

Resultant moment = 1500 g cm – 1000 g cm

= 500 g cm (anticlockwise)

According to the principle of moments.

Clockwise moment = Anticlockwise moment

To balance it, 50 g weight should be kept on right-hand side so as to produce a clockwise moment. Let d cm be the distance from the fulcrum. Then,

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

100 g × 10 cm + 50 g × d = 50 g × 30 cm

1000 g cm + 50 g × d = 1500 g cm

50 g × d = 500 g cm

Then, d = 10 cm

It can be balanced by suspending the mass 50 g at the mark 50 cm.

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