Question

A uniform meter rule of mass 100 g is a 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?

Answer 1

Answer:

Therefore, mass 50g will be placed at mark 50cm to balance the rule.

Step-by-step explanation:

(i) Finding the value of m,

The clockwise moment is given by,

⇒M×d1 ….(i)

Where, M is mass of rule i.e. 100g and d1 is the distance of fulcrum from the center of mass of rule.

d1=50cm−40cm=10cm ….(ii)

The anticlockwise moment is given by,

⇒m×d2 ….(iii)

Where, m is an unknown mass and d2is the distance of mass m from the fulcrum.

d2=40cm−20cm=20cm ….(iv)

Now equating the clockwise moment equal to the anticlockwise moment, we get,

From (i) and (iii),

M×d1=m×d2 ….(v)

Putting the values from (ii) and (iv) in (v), we get,

⇒100g×10cm=m×20cm

⇒m=50g

(ii) If the mass m is moved to the mark 10cm, then the balance will shift and the rule will tilt towards mass m.

(iii) The resultant moment when the mass m is moved to the mark 10cm is given by,

⇒m×d=50g×30cm=1500g⋅cm (anticlockwise moment)

Where, d=40cm−10cm=30cm

(iv) To balance the rule by another mass of 50g we must place the new mass to the right of the fulcrum so as to counter the anticlockwise moment with clockwise moment,

Let the mark at which mass 50g will be placed be x,

Hence, clockwise moment is,

⇒1000g⋅cm+50g×(x−40) ….(vi)

And, anticlockwise moment is,

⇒1500g⋅cm ….(vii)

Thus, equating (vi) and (vii) and solving for x, we get,

⇒1000g⋅cm+50g×(x−40)=1500g⋅cm

⇒x=50cm