Question

A metre stick is balanced on a knife edge at its centre. When two coins, each of mass 5 g are put one on top of the other at the 12.0 cm mark, the stick is found to be balanced at 45.0 cm. What is the mass of the metre stick?

Answer 1

Hello friend,

◆ Answer - 66 g

● Explaination-
# Given-
m = 5+5 = 10 g
d = 12 cm
D = 45 cm
M = ?

# Solution-
Without weights stick will balance at 50 cm.

Here, for rotational equilibrium, net torque is conserved, thus
mg(D-d) = Mg(50-D)
10×10(45-12) = M×10×(50-45)
M = 33×10/5
M = 66 g

Thus mass of the metre stick will be 66 g.

Hope that was helpful.

Answer 2

Explanation:-

Consider a meter scale AB. Let it’s mass be concentrated at C at 50 cm. mark. Upon placing coins, balancing point of the scale and the coins system is shifted to C’ at 45 cm mark.

For equilibrium about C’

10(45-12) = m (50-45)

m= 10*33/5

m= 66 g

Therefore the mass of the meter scale is 66 gram.