A hollow spherical shell is made of a metal of density `4.5 g per cm^(3)`. If its internal and external radii are 8 cm and 9 cm, respectively find the weigth of the shell.

Answer

To find the weight of the hollow spherical shell, we will follow these steps:

### Step 1: Identify the given values
- Density of the metal, \( \rho = 4.5 \, \text{g/cm}^3 \)
- Internal radius, \( r_1 = 8 \, \text{cm} \)
- External radius, \( r_2 = 9 \, \text{cm} \)

### Step 2: Calculate the volume of the hollow spherical shell
The volume \( V \) of the hollow spherical shell can be calculated using the formula:
\[ V = V_2 - V_1 \]
where \( V_2 \) is the volume of the outer sphere and \( V_1 \) is the volume of the inner sphere.

The volume of a sphere is given by:
\[ V = \frac{4}{3} \pi r^3 \]

Calculating \( V_2 \):
\[ V_2 = \frac{4}{3} \pi (r_2)^3 = \frac{4}{3} \pi (9)^3 = \frac{4}{3} \pi (729) = \frac{2916}{3} \pi = 972 \pi \, \text{cm}^3 \]

Calculating \( V_1 \):
\[ V_1 = \frac{4}{3} \pi (r_1)^3 = \frac{4}{3} \pi (8)^3 = \frac{4}{3} \pi (512) = \frac{2048}{3} \pi \, \text{cm}^3 \]

Now, substituting \( V_2 \) and \( V_1 \) into the volume formula:
\[ V = V_2 - V_1 = 972 \pi - \frac{2048}{3} \pi \]

To subtract these volumes, we need a common denominator:
\[ V = \left(972 \cdot \frac{3}{3} - \frac{2048}{3}\right) \pi = \left(\frac{2916 - 2048}{3}\right) \pi = \frac{868}{3} \pi \, \text{cm}^3 \]

### Step 3: Calculate the mass of the shell
Using the formula for density:
\[ \text{Density} = \frac{\text{Mass}}{\text{Volume}} \]
We can rearrange this to find the mass:
\[ \text{Mass} = \text{Density} \times \text{Volume} \]

Substituting the values:
\[ \text{Mass} = 4.5 \, \text{g/cm}^3 \times \frac{868}{3} \pi \, \text{cm}^3 \]

Calculating the mass:
\[ \text{Mass} = 4.5 \times \frac{868 \pi}{3} \]
Calculating \( 4.5 \times 868 \):
\[ 4.5 \times 868 = 3906 \]
So,
\[ \text{Mass} = \frac{3906 \pi}{3} \approx 4092 \, \text{g} \quad (\text{using } \pi \approx 3.14) \]

### Step 4: Convert mass to kilograms
To convert grams to kilograms:
\[ \text{Mass in kg} = \frac{4092}{1000} = 4.092 \, \text{kg} \]

### Final Answer
The weight of the hollow spherical shell is approximately \( 4.092 \, \text{kg} \).

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