An object is solid throughout . When the object is completely submerged in ethyl alcohol , its apparent weight is 15.2 N . When completely submerged in water , its apparent weight is 13.7 N . What is the volume of the object ?

Options

A. `1.6 xx 10^(-3) m^(3)`
B. `7.9 xx 10^(-4) m^(3)`
C. `1.0 xx 10^(-3) m^(3)`
D. `8.6 xx 10^(-4) m^(3)`

Answer

To solve the problem, we need to find the volume of the solid object using the information about its apparent weight when submerged in ethyl alcohol and water.

### Step-by-Step Solution:

1. **Understanding Apparent Weight**:
The apparent weight of an object submerged in a fluid is given by:
\[ W' = W - F_b \]
where \( W \) is the actual weight of the object, and \( F_b \) is the buoyant force acting on it.

2. **Buoyant Force Calculation**:
The buoyant force can be calculated using Archimedes' principle:
\[ F_b = \rho_f \cdot V \cdot g \]
where \( \rho_f \) is the density of the fluid, \( V \) is the volume of the object, and \( g \) is the acceleration due to gravity (approximately \( 9.81 \, \text{m/s}^2 \)).

3. **Setting Up Equations**:
For ethyl alcohol (density \( \rho_{eth} = 789 \, \text{kg/m}^3 \)):
\[ W - \rho_{eth} \cdot V \cdot g = 15.2 \, \text{N} \quad \text{(1)} \]
For water (density \( \rho_{water} = 1000 \, \text{kg/m}^3 \)):
\[ W - \rho_{water} \cdot V \cdot g = 13.7 \, \text{N} \quad \text{(2)} \]

4. **Subtracting the Equations**:
By subtracting equation (2) from equation (1), we eliminate \( W \):
\[ (\rho_{water} - \rho_{eth}) \cdot V \cdot g = 15.2 - 13.7 \]
\[ (1000 - 789) \cdot V \cdot g = 1.5 \]
\[ 211 \cdot V \cdot g = 1.5 \]

5. **Solving for Volume \( V \)**:
Rearranging gives:
\[ V = \frac{1.5}{211 \cdot g} \]
Substituting \( g = 9.81 \, \text{m/s}^2 \):
\[ V = \frac{1.5}{211 \cdot 9.81} \]
\[ V = \frac{1.5}{2077.91} \approx 0.000722 \, \text{m}^3 \]

6. **Final Calculation**:
Converting to cubic centimeters (1 m³ = 1,000,000 cm³):
\[ V \approx 0.000722 \, \text{m}^3 \times 10^6 \approx 722 \, \text{cm}^3 \]

### Final Answer:
The volume of the object is approximately \( 0.000722 \, \text{m}^3 \) or \( 722 \, \text{cm}^3 \).