Question

When a meter scale is balanced above a wedge, \(1 \mathrm{~kg}\) mass is hung at \(10 \mathrm{~cm}\) mark and a \(2 \mathrm{~kg}\) mass is hang at the $85 \mathrm{~cm}$ mark. To which mark on the meter scale, the fulcrum be shifted (Neglect mass of meter scale) to balance the scale? (A) \(70 \mathrm{~cm}\) (B) \(50 \mathrm{~cm}\) (C) \(60 \mathrm{~cm}\) (D) \(65 \mathrm{~cm}\)

Short answer

The fulcrum should be shifted to the \(60 \mathrm{~cm}\) mark to balance the scale. The correct answer is (C) \(60 \mathrm{~cm}\).

Step-by-step solution

Identify the forces acting on the system

We have two weights hanging on the meter scale at different positions. The first weight is 1 kg at the 10 cm mark, and the second weight is 2 kg at the 85 cm mark.

Set up an equation for the torques

We will assume that the fulcrum is at position x cm from the left side of the meter scale. So, the distance between 1 kg mass and the fulcrum is (x - 10) cm, and the distance between 2 kg mass and the fulcrum is (85-x) cm. Since the system is balanced, the total torque on the meter scale due to both masses must be zero. Therefore, the torque due to the 1 kg mass must be equal and opposite in direction to the torque due to the 2 kg mass. \[ (1 \mathrm{~kg})(x - 10 \mathrm{~cm}) =( 2 \mathrm{~kg})(85-x \mathrm{~cm}) \]

Solve the equation for the position of the fulcrum

Now, we will solve the equation in step 2 for the position of the fulcrum (x). \[ x-10 = 170 - 2x \] By adding 2x to both sides and adding 10 to both sides, we get, \[ 3x = 180 \] Now, divide both sides by 3 to find the position of the fulcrum (x). \[ x = 60 \mathrm{~cm} \] So, the fulcrum should be shifted to the 60 cm mark to balance the scale. The correct answer is (C) \(60 \mathrm{~cm}\).