A metre scale is balanced on a knife edge at its centre. When a coin of mass 15 g is kept at the 12 cm mark, the scale is found to be balanced at 45 cm. What is the mass of the metre scale?

Options

A. 33 gr
B. 99 gr
C. 66 gr
D. 45 gr

Explanation

To solve the problem, we need to analyze the situation using the principles of torque and balance. Here's a step-by-step solution:

### Step 1: Understand the Setup
The meter scale is balanced on a knife edge at its center (50 cm mark). When a coin of mass 15 g is placed at the 12 cm mark, the scale balances at the 45 cm mark.

### Step 2: Identify Forces and Distances
- The weight of the coin (W_coin) = mass × gravity = 15 g (since we are considering g as a constant).
- The distance of the coin from the knife edge (45 cm) = 45 cm - 12 cm = 33 cm.
- The distance of the center of mass of the scale from the knife edge (50 cm) = 50 cm - 45 cm = 5 cm.

### Step 3: Set Up the Torque Equation
For the scale to be balanced, the clockwise torque must equal the counterclockwise torque about the knife edge.

Let M be the mass of the meter scale. The torque due to the coin is:
\[ \text{Torque}_{\text{coin}} = W_{\text{coin}} \times \text{distance}_{\text{coin}} = 15 \, \text{g} \times 33 \, \text{cm} \]

The torque due to the scale's weight is:
\[ \text{Torque}_{\text{scale}} = M \times g \times \text{distance}_{\text{scale}} = M \times 5 \, \text{cm} \]

### Step 4: Write the Torque Balance Equation
Setting the torques equal gives:
\[ 15 \, \text{g} \times 33 \, \text{cm} = M \times 5 \, \text{cm} \]

### Step 5: Solve for M
Rearranging the equation to solve for M:
\[ M = \frac{15 \, \text{g} \times 33 \, \text{cm}}{5 \, \text{cm}} \]

Calculating the right side:
\[ M = \frac{495 \, \text{g cm}}{5 \, \text{cm}} = 99 \, \text{g} \]

### Conclusion
The mass of the meter scale is **99 g**.

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