To find the density of the material of the sphere, we can follow these steps:
### Step 1: Understand the problem
We have a sphere floating in two different liquids: oil and mercury. The sphere is half immersed in oil and half in mercury. We need to find the density of the sphere.
### Step 2: Write the equation for the weight of the sphere
The weight of the sphere (W) can be expressed as:
\[ W = V \cdot \rho \cdot g \]
where:
- \( V \) is the volume of the sphere,
- \( \rho \) is the density of the sphere,
- \( g \) is the acceleration due to gravity.
Since the sphere is homogeneous, its volume \( V \) can be represented as:
\[ V = \frac{4}{3} \pi R^3 \]
Thus, the weight of the sphere becomes:
\[ W = \frac{4}{3} \pi R^3 \cdot \rho \cdot g \]
### Step 3: Write the equation for the buoyant force
The buoyant force (upthrust) acting on the sphere is equal to the weight of the liquid displaced by the sphere. Since half of the sphere is immersed in oil and the other half in mercury, we can express the buoyant force (F_b) as:
\[ F_b = F_{b, \text{oil}} + F_{b, \text{mercury}} \]
where:
- \( F_{b, \text{oil}} = \text{Volume}_{\text{oil}} \cdot \text{Density}_{\text{oil}} \cdot g \)
- \( F_{b, \text{mercury}} = \text{Volume}_{\text{mercury}} \cdot \text{Density}_{\text{mercury}} \cdot g \)
Since half of the sphere is in each liquid, we have:
\[ F_{b, \text{oil}} = \frac{1}{2} V \cdot \rho_{\text{oil}} \cdot g \]
\[ F_{b, \text{mercury}} = \frac{1}{2} V \cdot \rho_{\text{mercury}} \cdot g \]
Thus, the total buoyant force is:
\[ F_b = \frac{1}{2} V \cdot \rho_{\text{oil}} \cdot g + \frac{1}{2} V \cdot \rho_{\text{mercury}} \cdot g \]
### Step 4: Substitute the values
Substituting the values of the densities:
- \( \rho_{\text{oil}} = 0.8 \, \text{g/cm}^3 \)
- \( \rho_{\text{mercury}} = 13.6 \, \text{g/cm}^3 \)
We get:
\[ F_b = \frac{1}{2} V \cdot 0.8 \cdot g + \frac{1}{2} V \cdot 13.6 \cdot g \]
### Step 5: Set the weight equal to the buoyant force
Since the sphere is floating, the weight of the sphere equals the buoyant force:
\[ \frac{4}{3} \pi R^3 \cdot \rho \cdot g = \frac{1}{2} V \cdot 0.8 \cdot g + \frac{1}{2} V \cdot 13.6 \cdot g \]
### Step 6: Simplify the equation
We can cancel \( g \) and \( V \) (since \( V = \frac{4}{3} \pi R^3 \)):
\[ \rho = \frac{1}{2} \left( 0.8 + 13.6 \right) \]
### Step 7: Calculate the density
Calculating the right side:
\[ \rho = \frac{1}{2} (0.8 + 13.6) = \frac{1}{2} (14.4) = 7.2 \, \text{g/cm}^3 \]
### Final Answer
The density of the material of the sphere is:
\[ \rho = 7.2 \, \text{g/cm}^3 \]
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