A uniform meter scale weighs 50g. It is pivoted at the 70cm mark. Where should a 40 g mass be placed so that the scale is in equilibrium?

Correct option is A

Weight of the meter scale (50 g) acts at the center of mass (which is at 50 cm). The 40 g mass is placed at an unknown position . Torque equation: Clockwise Torque = Counterclockwise Torque
50 × (70-50) = 40 × (x)
50 × 20 = 40 × (x)
1000 = 40 × x
1000/40 = x
x = 25 cm
So, the 40 g mass will be placed at = 70+25 = 95 cm