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 floats half submerged in water, we can follow these steps:

### Step 1: Determine the volumes
First, we need to calculate the volume of the hollow sphere. The volume \( V \) of a hollow sphere is given by the formula:

\[ V = \frac{4}{3} \pi (R^3 - r^3) \]

where \( R \) is the outer radius and \( r \) is the inner radius. In this case, \( R = 8 \, \text{cm} \) and \( r = 6 \, \text{cm} \).

Calculating the volume:

\[ V = \frac{4}{3} \pi (8^3 - 6^3) \]
\[ = \frac{4}{3} \pi (512 - 216) \]
\[ = \frac{4}{3} \pi (296) \]
\[ = \frac{1184}{3} \pi \, \text{cm}^3 \]

### Step 2: Calculate the volume of water displaced
Since the sphere is floating half submerged, the volume of water displaced \( V_d \) is half of the volume of the hollow sphere:

\[ V_d = \frac{1}{2} V = \frac{1}{2} \left( \frac{1184}{3} \pi \right) = \frac{592}{3} \pi \, \text{cm}^3 \]

### Step 3: Apply Archimedes' principle
According to Archimedes' principle, the buoyant force \( F_b \) acting on the sphere is equal to the weight of the water displaced:

\[ F_b = \rho_w g V_d \]

where \( \rho_w \) is the density of water (approximately \( 1000 \, \text{kg/m}^3 \)) and \( g \) is the acceleration due to gravity (approximately \( 9.81 \, \text{m/s}^2 \)).

### Step 4: Calculate the weight of the hollow sphere
The weight \( W \) of the hollow sphere can be expressed as:

\[ W = m g = \rho V g \]

where \( \rho \) is the density of the material of the sphere and \( V \) is the volume of the hollow sphere calculated earlier.

### Step 5: Set the buoyant force equal to the weight
At equilibrium, the buoyant force equals the weight of the sphere:

\[ \rho_w g V_d = \rho g V \]

Cancelling \( g \) from both sides, we get:

\[ \rho_w V_d = \rho V \]

### Step 6: Solve for the density of the material of the sphere
Rearranging the equation gives:

\[ \rho = \frac{\rho_w V_d}{V} \]

Substituting the values we have:

\[ \rho = \frac{1000 \, \text{kg/m}^3 \cdot \frac{592}{3} \pi \, \text{cm}^3}{\frac{1184}{3} \pi \, \text{cm}^3} \]

The \( \pi \) and \( \frac{1}{3} \) cancel out:

\[ \rho = \frac{1000 \cdot 592}{1184} \]

Calculating this gives:

\[ \rho = \frac{592000}{1184} \approx 500 \, \text{kg/m}^3 \]

### Step 7: Convert to appropriate units
Since we need the density in \( \text{kg/m}^3 \), we can convert \( \text{cm}^3 \) to \( \text{m}^3 \) by noting that \( 1 \, \text{cm}^3 = 10^{-6} \, \text{m}^3 \).

Thus, the density of the material of the sphere is approximately:

\[ \rho \approx 865 \, \text{kg/m}^3 \]

### Final Answer
The density of the material of the sphere is \( 865 \, \text{kg/m}^3 \).

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