If m and M are the masses of two bodies that are tied at two ends of a meter scale that is balanced on a sharp edge of a heavy board wedge. If `M=20`g, its distance from centere = 30 cm and distance of mass m from centre is 25cm when metre scale is balanced, then m is

Options

A. 23g
B. 24g
C. 25g
D. 26g

Answer

To solve the problem, we need to apply the principle of moments (or torque) about the balance point. The condition for equilibrium states that the sum of clockwise moments about a point must equal the sum of anticlockwise moments about that same point.

### Step-by-Step Solution:

1. **Identify the Given Values:**
- Mass \( M = 20 \) g
- Distance of mass \( M \) from the center \( L_1 = 30 \) cm
- Distance of mass \( m \) from the center \( L_2 = 25 \) cm

2. **Write the Torque Equation:**
The torque due to mass \( M \) (anticlockwise) and mass \( m \) (clockwise) must balance out:
\[ M \cdot g \cdot L_1 = m \cdot g \cdot L_2 \]
Here, \( g \) (acceleration due to gravity) cancels out since it is present in both terms.

3. **Rearranging the Equation:**
We can simplify the equation to:
\[ M \cdot L_1 = m \cdot L_2 \]
Rearranging gives:
\[ m = \frac{M \cdot L_1}{L_2} \]

4. **Substituting the Values:**
Now substitute the known values into the equation:
\[ m = \frac{20 \, \text{g} \cdot 30 \, \text{cm}}{25 \, \text{cm}} \]

5. **Calculating the Value of \( m \):**
\[ m = \frac{600 \, \text{g cm}}{25 \, \text{cm}} = 24 \, \text{g} \]

6. **Final Answer:**
The mass \( m \) is \( 24 \) g.