Question

A metre scale is supported on a wedge at its centre of gravity. A body of weight $w$ is suspended from the $20 \text{ cm}$ mark and another weight of $25 \text{ gram}$ is suspended from the $74 \text{ cm}$ mark to balance it and the metre scale remains perfectly horizontal. Neglecting the weight of the metre scale, the weight of the body is

• A. $20 \text{ gram-wt}$
• B. $15 \text{ gram-wt}$
• C. $33 \text{ gram-wt}$
• D. $30 \text{ gram-wt}$

Hint:

Always measure lever arm distances relative to the pivot point, not from the zero end of the scale! A common mistake is using the raw mark values ($20 \text{ cm}$ and $74 \text{ cm}$) instead of calculating the displacement from the balancing wedge at $50 \text{ cm}$.

Answer 1

The Correct Option is A

Step 1: Understanding the Question:
A uniform metre scale is balanced at its center of gravity ($50 \text{ cm}$ mark). Two weights are placed at opposite sides of the support wedge. For the system to remain horizontal in rotational equilibrium, the total clockwise moment must equal the total counterclockwise moment about the pivot point.

Step 2: Key Formula or Approach:
The principle of moments states:
$$\text{Counterclockwise Torque} = \text{Clockwise Torque}$$ $$\sum \tau_{\text{ccw}} = \sum \tau_{\text{cw}}$$ $$\tau = \text{Force} \times \text{Perpendicular Distance}$$ The distance of each mass is determined relative to the central support point located at $50 \text{ cm}$.

Step 3: Detailed Explanation:
Let's find the lever arm distances from the central pivot at $50 \text{ cm}$:
1. The weight $w$ is attached at the $20 \text{ cm}$ mark. Its lever arm distance is:
$$d_1 = 50 \text{ cm} - 20 \text{ cm} = 30 \text{ cm}$$ This weight exerts a counterclockwise rotational torque. 2. The weight of $25 \text{ g-wt}$ is suspended at the $74 \text{ cm}$ mark. Its lever arm distance is:
$$d_2 = 74 \text{ cm} - 50 \text{ cm} = 24 \text{ cm}$$ This weight exerts a clockwise rotational torque. Applying the principle of moments about the $50 \text{ cm}$ mark:
$$w \times d_1 = 25 \text{ g-wt} \times d_2$$ Substitute the calculated distances into the formula:
$$w \times 30 = 25 \times 24$$ Isolating $w$:
$$w = \frac{25 \times 24}{30}$$ Simplify the fraction by dividing the numerator and denominator by 6:
$$w = \frac{25 \times 4}{5} = 5 \times 4 = 20 \text{ gram-wt}$$

Step 4: Final Answer:
The weight of the body is $20 \text{ gram-wt}$, which corresponds to option (A).