An uniform metre scale of weight `50 g` is balanced at `60 cm` mark, when a weight of `15 g` is suspended at `10 cm` mark. Where must a weight `100 g` be suspended to balance metre scale.

Options

A. `72.5 cm`
B. `70 cm`
C. `71.5 cm`
D. `74.5 cm`

Explanation

To solve the problem, we need to balance the uniform meter scale with the given weights. Here’s a step-by-step solution:

### Step 1: Understand the System
We have a uniform meter scale weighing 50 g, balanced at the 60 cm mark with a 15 g weight suspended at the 10 cm mark. We need to find where to suspend a 100 g weight to balance the meter scale.

### Step 2: Calculate the Torque
The torque (moment of force) is calculated as the product of the force (weight) and the distance from the pivot point (60 cm mark).

1. **Torque due to the 15 g weight:**
- Distance from the pivot (60 cm) to the 10 cm mark = 60 cm - 10 cm = 50 cm
- Torque = Weight × Distance = 15 g × 50 cm = 750 g·cm

2. **Torque due to the weight of the meter scale:**
- The center of mass of the uniform scale is at the 50 cm mark.
- Distance from the pivot (60 cm) to the 50 cm mark = 60 cm - 50 cm = 10 cm
- Torque = Weight × Distance = 50 g × 10 cm = 500 g·cm

### Step 3: Total Torque on One Side
The total torque on the left side (due to the 15 g weight and the weight of the scale) is:
- Total Torque = Torque from 15 g + Torque from scale = 750 g·cm + 500 g·cm = 1250 g·cm

### Step 4: Set Up the Equation for the 100 g Weight
Let the distance from the pivot (60 cm) to where we suspend the 100 g weight be \( x \) cm. The torque due to the 100 g weight is:
- Torque = Weight × Distance = 100 g × x

### Step 5: Balance the Torques
For the system to be balanced, the total torque on both sides must be equal:
\[ 100 g \cdot x = 1250 g \cdot cm \]

### Step 6: Solve for x
Now, we can solve for \( x \):
\[ x = \frac{1250 g \cdot cm}{100 g} \]
\[ x = 12.5 cm \]

### Step 7: Determine the Position of the 100 g Weight
Since the 100 g weight is to be suspended at a distance of 12.5 cm from the 60 cm mark:
- Position = 60 cm + 12.5 cm = 72.5 cm

### Final Answer
The 100 g weight must be suspended at the **72.5 cm mark** to balance the meter scale.

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