17. A uniform metre rule of mass 100 g is balanced on afulcrum at mark

1.	17. A uniform metre rule of mass 100 g is balanced on afulcrum at mark 40 cm by suspending an unknownmass m at the mark 20 cm.(i) Find the value of m.(ii) To which side the rule will tilt if the mass m ismoved to the mark 10 cm ?(iii) What is the resultant moment now?(iv) How can it be balanced by another mass of50 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» $\LARGE{ \underline{\underline{ \pink{ \bf{Required \: answer:}}}}}$ 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. $\pink{}$ $\red{Mark}$ $\green{As}$ $\blue{Brainliest}$ $\orange{}$ $\pink{}$ $\red{Please}$ $\green{Like}$ $\blue{}$ $\orange{}$ ${\fcolorbox{blue}{black}{\orange{\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\: DecentMortal\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:}}}$

17. A uniform metre rule of mass 100 g is balanced on afulcrum at mark 40 cm by suspending an unknownmass m at the mark 20 cm.(i) Find the value of m.(ii) To which side the rule will tilt if the mass m ismoved to the mark 10 cm ?(iii) What is the resultant moment now?(iv) How can it be balanced by another mass of50 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»

$\LARGE{ \underline{\underline{ \pink{ \bf{Required \: answer:}}}}}$

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.

$\pink{}$ $\red{Mark}$ $\green{As}$ $\blue{Brainliest}$ $\orange{}$

$\pink{}$ $\red{Please}$ $\green{Like}$ $\blue{}$ $\orange{}$

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