A metre stick is balanced on a knife edge at its centre. When two coins, each of mass `5 g` are put one on one of the other at the `12 cm` mark, the stick is found to balanced at `45 cm`. The mass of the metre stick is.

Options

A. 56 g
B. 66 g
C. 76 g
D. 36 g

Explanation

To solve the problem, we need to analyze the situation using the principles of torque and equilibrium.

### Step-by-Step Solution:

1. **Identify the Initial Conditions:**
- The meter stick is balanced at its center (50 cm mark).
- Two coins, each of mass 5 g, are placed at the 12 cm mark.

2. **Calculate the Total Mass of the Coins:**
- Total mass of the coins = 5 g + 5 g = 10 g.

3. **Determine the New Balance Point:**
- After placing the coins, the meter stick balances at the 45 cm mark.

4. **Calculate the Distance from the Center:**
- The center of the stick is at 50 cm.
- The distance from the center to the new balance point = 50 cm - 45 cm = 5 cm (to the left).

5. **Set Up the Torque Equation:**
- For the stick to be in equilibrium, the clockwise torque must equal the counterclockwise torque about the new balance point.
- Let the mass of the meter stick be \( M \).
- The weight of the meter stick acts at its center (50 cm).
- The weight of the coins acts at the 12 cm mark.

6. **Calculate the Torques:**
- Torque due to the coins (clockwise):
\[ \text{Torque}_{\text{coins}} = \text{Force} \times \text{Distance} = (10 \, \text{g} \cdot 9.8 \, \text{m/s}^2) \times (50 \, \text{cm} - 12 \, \text{cm}) = 10 \times 9.8 \times 38 \, \text{cm} \]
- Torque due to the meter stick (counterclockwise):
\[ \text{Torque}_{\text{stick}} = M \cdot g \cdot (50 \, \text{cm} - 45 \, \text{cm}) = M \cdot 9.8 \cdot 5 \, \text{cm} \]

7. **Set the Torques Equal:**
\[ 10 \times 9.8 \times 38 = M \cdot 9.8 \cdot 5 \]

8. **Cancel out \( 9.8 \) from both sides:**
\[ 10 \times 38 = M \cdot 5 \]

9. **Solve for \( M \):**
\[ M = \frac{10 \times 38}{5} = 76 \, \text{g} \]

### Final Answer:
The mass of the meter stick is **76 g**.