. 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


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?

 

Best Answer

(i) From the principle of moments,

Clockwise moment= Anticlockwise moment

100 g x (50-40) cm= mx (40-20) cm

100 g x 10 cm = m x 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 10 cm.

(iii)Anticlockwise moment if mass m is moved to the mark 10 cm= 50g x (40-10)cm =50 x 30=1500 g cm

Clockwise moment=100 g x (50-40) cm= 1000 g cm

Resultant moment= 1500 g cm -1000 g cm= 500 g cm (anticlockwise)

(iv) From 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 its distance from fulcrum bed cm. Then,

100g x (50-40) cm + 50g x d =50g x (40-10)cm

1000g cm + 50g x d =1500 g cm

50 g x d= 500g cm

So, d =10 cm

By suspending the mass 50 g at the mark 50 cm, it can be balanced.

 

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