The scale of uniform mass of 1 kg as shown below is in equilibrium. The mass m suspeneded at 90cm mark has a mass of
The weight of the meter scale will be acting at its mid-point, i.e., at 50 cm mark.
We calculate the torque about the fulcrum, i.e. the 60 cm mark.
By the principle of moments,
Torque of the weight of the rod = Torque of the uniform mass at 90 cm mark
1000×g×(60−50) cm =m×g×(90−60) cm
or, 10000 = 30 m
or, m = 333.33 grams.