A solid metallic sphere of radius 8 cm is melted to form 64 equal small solid spheres . The ratio of the surface area of this sphere to that of a small sphere is

Options

A. `4 : 1`
B. `1 : 16`
C. `16 : 1`
D. `1 : 4`

Answer

To solve the problem, we will follow these steps:

### Step 1: Calculate the volume of the large sphere.
The volume \( V \) of a sphere is given by the formula:
\[ V = \frac{4}{3} \pi r^3 \]
where \( r \) is the radius of the sphere.

For the large sphere with a radius of 8 cm:
\[ V_{\text{large}} = \frac{4}{3} \pi (8)^3 \]
Calculating \( (8)^3 \):
\[ (8)^3 = 512 \]
So,
\[ V_{\text{large}} = \frac{4}{3} \pi (512) = \frac{2048}{3} \pi \text{ cm}^3 \]

### Step 2: Calculate the volume of one small sphere.
Since the large sphere is melted to form 64 small spheres, the volume of one small sphere \( V_{\text{small}} \) is:
\[ V_{\text{small}} = \frac{V_{\text{large}}}{64} = \frac{\frac{2048}{3} \pi}{64} \]
Calculating this:
\[ V_{\text{small}} = \frac{2048 \pi}{192} = \frac{64 \pi}{3} \text{ cm}^3 \]

### Step 3: Find the radius of the small spheres.
Using the volume formula for a sphere, we can set the volume of the small sphere equal to \( \frac{64 \pi}{3} \):
\[ \frac{4}{3} \pi r^3 = \frac{64 \pi}{3} \]
Cancelling \( \frac{\pi}{3} \) from both sides:
\[ 4r^3 = 64 \]
Dividing both sides by 4:
\[ r^3 = 16 \]
Taking the cube root:
\[ r = \sqrt[3]{16} = 2 \sqrt[3]{2} \text{ cm} \]

### Step 4: Calculate the surface area of the large sphere.
The surface area \( A \) of a sphere is given by:
\[ A = 4 \pi r^2 \]
For the large sphere:
\[ A_{\text{large}} = 4 \pi (8)^2 = 4 \pi (64) = 256 \pi \text{ cm}^2 \]

### Step 5: Calculate the surface area of one small sphere.
For the small sphere with radius \( r = 2 \sqrt[3]{2} \):
\[ A_{\text{small}} = 4 \pi (2 \sqrt[3]{2})^2 = 4 \pi (4 \cdot \sqrt[3]{4}) = 16 \pi \sqrt[3]{4} \text{ cm}^2 \]

### Step 6: Find the ratio of the surface areas.
Now, we find the ratio of the surface area of the large sphere to that of one small sphere:
\[ \text{Ratio} = \frac{A_{\text{large}}}{A_{\text{small}}} = \frac{256 \pi}{16 \pi \sqrt[3]{4}} = \frac{256}{16 \sqrt[3]{4}} = \frac{16}{\sqrt[3]{4}} \]

### Final Answer:
The ratio of the surface area of the large sphere to that of a small sphere is:
\[ \frac{16}{\sqrt[3]{4}} \]