Question

A metre scale is balanced on a knife edge at its centre. When two coins, each of mass 9 g, are kept one above the other at the 10 cm mark, the scale is found to be balanced at 35 cm. The mass of the metre scale is:

• A. \( 15 \) g
• B. \( 30 \) g
• C. \( 45 \) g
• D. \( 60 \) g

Hint:

For problems involving the principle of moments, set up the equation using \( \text{Force} \times \text{Distance} \) for all forces about the pivot and solve for the unknown mass.

Answer 1

The Correct Option is B

Step 1: Understanding the Principle of Moments The principle of moments states that for rotational equilibrium: \[ \text{Sum of clockwise moments} = \text{Sum of anticlockwise moments} \] Given: - A metre scale is balanced at its centre (50 cm) initially. - Two coins of mass \( 9 g \) each are placed at the 10 cm mark. - The new balance point shifts to 35 cm. - Let the mass of the metre scale be \( M \). 

Step 2: Calculate the Moments The moment of the two coins about the new balance point: \[ \text{Moment} = \text{Force} \times \text{Perpendicular Distance} \] \[ (9 + 9) \times (35 - 10) = 18 \times 25 = 450 \] The moment of the metre scale's weight about the new balance point: \[ M \times (50 - 35) = M \times 15 \]

Step 3: Solve for \( M \) Using the equilibrium condition: \[ 18 \times 25 = M \times 15 \] \[ 450 = 15M \] \[ M = \frac{450}{15} = 30 \text{ g} \] Thus, the mass of the metre scale is: \[ \mathbf{30} \text{ g} \]