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 ?.
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Answers
Since, uniform metre rule of mass 100 g is balanced on a fulcrum at mark 40 cm.
so, distance of c.o.m. of rule from balanced point is 10cm.
Net moment at balancing point should be zero.
So, m×20 = 100×10 m = 50g
if m is moved to mark of 10cm then rule will tilt to the side where m is suspended.
Now, for balance this
50×30 = 100×10+ 50×(x-40) (x-40) = (1500- 1000)/50 = 10cm X = 50cm.
So, another 50g mass will be suspended at 50cm mark
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.