A solid weights 32 gf I air and 28.8 gf in water. The R.D. of the solid is

Options

A. 8.9
B. 10
C. 29.12
D. 20

Answer

To find the relative density (R.D.) of the solid, we can use the formula:

\[ \text{Relative Density (R.D.)} = \frac{W_1}{W_1 - W_2} \times \text{Relative Density of Water} \]

Where:
- \( W_1 \) is the weight of the solid in air.
- \( W_2 \) is the weight of the solid in water.
- The relative density of water is 1 (since we are using grams-force).

### Step-by-Step Solution:

1. **Identify the weights**:
- Weight of solid in air, \( W_1 = 32 \, \text{gf} \)
- Weight of solid in water, \( W_2 = 28.8 \, \text{gf} \)

2. **Calculate the difference in weights**:
- \( W_1 - W_2 = 32 \, \text{gf} - 28.8 \, \text{gf} = 3.2 \, \text{gf} \)

3. **Substitute the values into the formula**:
\[ \text{R.D.} = \frac{W_1}{W_1 - W_2} = \frac{32 \, \text{gf}}{3.2 \, \text{gf}} \]

4. **Perform the division**:
\[ \text{R.D.} = \frac{32}{3.2} = 10 \]

5. **Conclusion**:
- The relative density of the solid is \( 10 \).

### Final Answer:
The R.D. of the solid is **10**.