A physical balance has its arms of unequal length. A body weight 18 kg if kept in one pan and weight 8 kg. If kept in the other pan. The true weight of the body is

Options

A. 13 kg
B. 12 kg
C. 10 kg
D. 16 kg

Answer

To find the true weight of the body using the information given about the physical balance with unequal arms, we can follow these steps:

### Step-by-Step Solution:

1. **Understanding the Problem**:
- We have a physical balance with two pans of unequal lengths.
- When an 18 kg weight is placed in one pan, it balances with the unknown weight (W) in the other pan.
- When an 8 kg weight is placed in the other pan, it also balances with the same unknown weight (W).

2. **Convert Weights to Forces**:
- The force due to the weights can be calculated using the formula:
\[ \text{Force} = \text{mass} \times g \]
- Assuming \( g = 10 \, \text{m/s}^2 \):
- Force when 18 kg is used:
\[ F_1 = 18 \, \text{kg} \times 10 \, \text{m/s}^2 = 180 \, \text{N} \]
- Force when 8 kg is used:
\[ F_2 = 8 \, \text{kg} \times 10 \, \text{m/s}^2 = 80 \, \text{N} \]

3. **Setting Up Torque Equations**:
- For the first case (18 kg):
\[ 180 \, \text{N} \times L_1 = W \times L_2 \]
- For the second case (8 kg):
\[ W \times L_1 = 80 \, \text{N} \times L_2 \]

4. **Expressing Ratios**:
- From the first equation:
\[ \frac{L_1}{L_2} = \frac{W}{180} \]
- From the second equation:
\[ \frac{L_1}{L_2} = \frac{80}{W} \]

5. **Equating the Two Ratios**:
- Since both expressions are equal to \( \frac{L_1}{L_2} \):
\[ \frac{W}{180} = \frac{80}{W} \]

6. **Cross-Multiplying**:
- Cross-multiplying gives:
\[ W^2 = 80 \times 180 \]

7. **Calculating W**:
- Calculate \( 80 \times 180 \):
\[ 80 \times 180 = 14400 \]
- Taking the square root:
\[ W = \sqrt{14400} = 120 \, \text{N} \]

8. **Finding the True Weight in kg**:
- To convert the weight back to mass (in kg):
\[ \text{True Weight (mass)} = \frac{120 \, \text{N}}{10 \, \text{m/s}^2} = 12 \, \text{kg} \]

### Final Answer:
The true weight of the body is **12 kg**.

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