Question

A balance is made using a uniform metre scale of mass 100 g and two plates each of mass 200 g fixed at the two ends of the scale and the balance is pivoted at 45 cm mark of the scale. The error when 300 g weight is placed in the plate at 0 cm to weigh vegetables placed in the plate at 100 cm is

• A. 36.4 g
• B. 63.6 g
• C. 200 g
• D. 100 g

Hint:

For problems involving balances and levers, the principle of moments is key: the sum of clockwise torques about the pivot must equal the sum of counter-clockwise torques. Torque is calculated as Force $\times$ Lever Arm. Remember to account for the weight of the lever itself, acting at its center of mass.

Answer 1

The Correct Option is D

Let's analyze the torques acting on the metre scale when it is balanced. The pivot is at the 45 cm mark.
Forces creating clockwise torque (tending to rotate right): 1. Weight of the right plate: Mass $m_p = 200$ g at the 100 cm mark. Lever arm = $100 - 45 = 55$ cm. 2. Weight of the vegetables: Mass $M_v$ (unknown) at the 100 cm mark. Lever arm = 55 cm. 3. Weight of the right part of the scale: The scale is uniform, so its center of mass is at 50 cm. The right part (from 45 cm to 100 cm) has length 55 cm and mass $55$ g. Its center of mass is at $(45+100)/2 = 72.5$ cm. Lever arm = $72.5 - 45 = 27.5$ cm.
Forces creating counter-clockwise torque (tending to rotate left): 1. Weight of the left plate: Mass $m_p = 200$ g at the 0 cm mark. Lever arm = $45 - 0 = 45$ cm. 2. The added weight: Mass $M_w = 300$ g at the 0 cm mark. Lever arm = 45 cm. 3. Weight of the left part of the scale: The left part (from 0 cm to 45 cm) has length 45 cm and mass $45$ g. Its center of mass is at $45/2 = 22.5$ cm. Lever arm = $45 - 22.5 = 22.5$ cm.
Alternatively, the center of mass of the entire 100g scale is at 50 cm. Its lever arm is $50-45=5$ cm, creating a clockwise torque.
Let's use the second method for the scale's torque. Principle of moments: Clockwise torque = Counter-clockwise torque.
Torque\textsubscript{cw} = (Torque from right plate) + (Torque from vegetables) + (Torque from scale's CoM) $= (200 \times 55) + (M_v \times 55) + (100 \times (50-45))$. $= 11000 + 55M_v + 500 = 11500 + 55M_v$.
Torque\textsubscript{ccw} = (Torque from left plate) + (Torque from added weight) $= (200 \times 45) + (300 \times 45) = (500 \times 45) = 22500$.
Equating the torques: $11500 + 55M_v = 22500$.
$55M_v = 22500 - 11500 = 11000$.
$M_v = \frac{11000}{55} = \frac{1000}{5} = 200$ g.
The measured weight of the vegetables is 200 g. The weight used for balancing is 300 g.
The error is the difference between the true weight of the vegetables and the weight used to measure them.
Error = $|M_v - M_w| = |200 \text{ g} - 300 \text{ g}|$. However, error is typically defined as (Measured Value - True Value) or the discrepancy in what the scale reads. The scale reads 300 g because that's the weight used. The true weight is 200 g. Error = Reading - True Value = $300 - 200 = 100$ g.