Question

a uniform half metre rule of mass 100 gm is balanced at fulcrum of 20 cm mark by suspending an unknown mass m at 10 cm mark. find the value of m. to which side the rule will tilt if the mass is moved to 5 cm mark​

Answer 1

Answer:

Explanation:

(i) From the principle of moments, Clockwise moment = Anticlockwise moment 100g × (50 − 40) cm = m × (40 − 20) cm 100g × 10 cm

= m × 20 cm

= m = 50 g

(ii) The rule will tilt on the side of mass m (anticlockwise), if the mass m is moved to the mark 10cm.

(iii) Anticlockwise moment if mass m is moved to the mark 10 cm = 50g × (40−10)cm = 50 × 30 = 1500 g cm Clockwise moment = 100g × (50 − 40) cm = 1000g cm Resultant moment

= 1500g cm − 1000g cm

= 500g cm (anticlockwise)

(iv) From the principle of moments, Clockwise moment = Anticlockwise moment To balance it, 50g weight should be kept on right hand side so as to produce a clockwise moment. Let its distance from fulcrum be d cm. Then, 100g × (50 − 40) cm + 50g × d

= 50g × (40 − 10)cm 1000g cm + 50g × d

= 1500 g cm 50 g × d = 500g cm

Answer 2

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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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