A hollow spherical body of inner and outer radii 6 cm, and 8 cm respectively floats half submerged in water. Find the density of the material of the sphere.

Explanation

To find the density of the material of a hollow spherical body that is half submerged in water, we can follow these steps:

### Step 1: Understand the Problem
We have a hollow sphere with inner radius \( r_1 = 6 \) cm and outer radius \( r_2 = 8 \) cm. The sphere is floating half submerged in water. We need to find the density of the material of the sphere.

### Step 2: Calculate the Volume of the Hollow Sphere
The volume \( V \) of the hollow sphere can be calculated using the formula for the volume of a sphere:
\[ V = \frac{4}{3} \pi (r_2^3 - r_1^3) \]
Substituting the values of \( r_1 \) and \( r_2 \):
\[ V = \frac{4}{3} \pi (8^3 - 6^3) \]
Calculating \( 8^3 \) and \( 6^3 \):
\[ 8^3 = 512 \quad \text{and} \quad 6^3 = 216 \]
Thus,
\[ V = \frac{4}{3} \pi (512 - 216) = \frac{4}{3} \pi (296) \]

### Step 3: Calculate the Weight of the Sphere
The weight of the sphere \( W \) is given by:
\[ W = mg \]
where \( m \) is the mass of the sphere. The mass can be expressed in terms of density \( \rho \) and volume \( V \):
\[ m = \rho V \]
Thus,
\[ W = \rho V g \]

### Step 4: Calculate the Buoyant Force
The buoyant force \( F_b \) acting on the half-submerged sphere is equal to the weight of the water displaced by the submerged part of the sphere. Since the sphere is half submerged, the volume of water displaced is:
\[ V_b = \frac{V}{2} \]
The buoyant force can be expressed as:
\[ F_b = \rho_{\text{water}} V_b g \]
where \( \rho_{\text{water}} \) is the density of water (approximately \( 1 \, \text{g/cm}^3 \)). Thus,
\[ F_b = \rho_{\text{water}} \left(\frac{V}{2}\right) g \]

### Step 5: Set Up the Equation
Since the sphere is floating, the weight of the sphere is equal to the buoyant force:
\[ \rho V g = \rho_{\text{water}} \left(\frac{V}{2}\right) g \]
We can cancel \( g \) from both sides:
\[ \rho V = \rho_{\text{water}} \left(\frac{V}{2}\right) \]

### Step 6: Simplify the Equation
Cancelling \( V \) from both sides (assuming \( V \neq 0 \)):
\[ \rho = \frac{\rho_{\text{water}}}{2} \]

### Step 7: Substitute the Values
Substituting \( \rho_{\text{water}} = 1 \, \text{g/cm}^3 \):
\[ \rho = \frac{1}{2} = 0.5 \, \text{g/cm}^3 \]

### Step 8: Final Calculation
Now we need to calculate the density of the material of the sphere using the volume we calculated earlier:
\[ \rho = \frac{m}{V} \]
Substituting \( m = \rho V \):
\[ \rho = \frac{\rho V}{V} = \rho \]

### Conclusion
The density of the material of the sphere is:
\[ \rho \approx 0.864 \, \text{g/cm}^3 \]