A meter scale weighs 50 gms and carries 20 gm at one end , the scale balances when it is suspende at a distance x cm from other end . Then x I equal to

Options

A. `35.7 ` cm
B. `64.3` cm
C. `14.3 ` cm
D. 50 cm

Explanation

To solve the problem, we need to analyze the forces and torques acting on the meter scale. Let's break down the steps:

### Step-by-Step Solution:

1. **Identify the Given Information:**
- Weight of the meter scale, \( W_s = 50 \, \text{g} = 0.05 \, \text{kg} \) (since 1 g = 0.001 kg)
- Weight of the mass at one end, \( W_m = 20 \, \text{g} = 0.02 \, \text{kg} \)
- Length of the scale, \( L = 100 \, \text{cm} \)

2. **Determine the Center of Mass:**
- The center of mass of the meter scale is located at its midpoint, which is at \( 50 \, \text{cm} \) from either end.

3. **Set Up the Torque Equation:**
- The scale is balanced when suspended at a distance \( x \) from the end opposite to the mass.
- The torque due to the weight of the scale about the pivot point (suspension point) is calculated as follows:
- Torque due to the weight of the scale (anticlockwise):
\[ \tau_s = W_s \cdot d_s = 0.05 \cdot g \cdot (x - 50) \]
- Torque due to the weight of the mass (clockwise):
\[ \tau_m = W_m \cdot d_m = 0.02 \cdot g \cdot (100 - x) \]

4. **Set the Net Torque to Zero:**
- For the scale to be in equilibrium, the net torque must be zero:
\[ \tau_s - \tau_m = 0 \]
- Substituting the torque equations:
\[ 0.05 \cdot g \cdot (x - 50) - 0.02 \cdot g \cdot (100 - x) = 0 \]
- Cancel \( g \) from both sides:
\[ 0.05(x - 50) = 0.02(100 - x) \]

5. **Simplify the Equation:**
- Expanding both sides:
\[ 0.05x - 2.5 = 2 - 0.02x \]
- Rearranging the equation:
\[ 0.05x + 0.02x = 2 + 2.5 \]
\[ 0.07x = 4.5 \]

6. **Solve for \( x \):**
- Dividing both sides by \( 0.07 \):
\[ x = \frac{4.5}{0.07} \approx 64.2857 \, \text{cm} \]
- Rounding to two decimal places gives:
\[ x \approx 64.29 \, \text{cm} \]

### Final Answer:
The value of \( x \) is approximately \( 64.3 \, \text{cm} \).

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