A large solid sphere is melted and moulded to form identical right circular cones with base radius and height same as the radius of the sphere. One of these cones is melted and moulded to form a smaller solid sphere. Find the ratio of the surface area of the smaller sphere to the surface are of the larger sphere.

Options

A. `1:3^(4//3)`
B. `1:2^(3//2)`
C. `1:3^(2//3)`
D. `1:2^(4//3)`

Answer

To solve the problem, we need to find the ratio of the surface area of a smaller sphere to the surface area of a larger sphere after a series of transformations involving melting and molding.

### Step-by-Step Solution:

1. **Define the Radius of the Larger Sphere**:
Let the radius of the larger sphere be \( R_1 \).

2. **Volume of the Larger Sphere**:
The volume \( V_1 \) of the larger sphere is given by the formula:
\[ V_1 = \frac{4}{3} \pi R_1^3 \]

3. **Volume of the Cone**:
The identical right circular cone has a base radius and height both equal to \( R_1 \). The volume \( V_c \) of one cone is given by:
\[ V_c = \frac{1}{3} \pi r^2 h = \frac{1}{3} \pi R_1^2 R_1 = \frac{1}{3} \pi R_1^3 \]

4. **Number of Cones Formed**:
The number of identical cones formed from the larger sphere can be calculated by dividing the volume of the sphere by the volume of one cone:
\[ \text{Number of cones} = \frac{V_1}{V_c} = \frac{\frac{4}{3} \pi R_1^3}{\frac{1}{3} \pi R_1^3} = 4 \]
Thus, 4 cones can be formed from the larger sphere.

5. **Volume of the Smaller Sphere**:
One of these cones is melted to form a smaller solid sphere. Let the radius of the smaller sphere be \( R_2 \). The volume \( V_2 \) of the smaller sphere is:
\[ V_2 = \frac{4}{3} \pi R_2^3 \]
Since the volume of the cone equals the volume of the smaller sphere:
\[ \frac{1}{3} \pi R_1^3 = \frac{4}{3} \pi R_2^3 \]
Canceling \( \frac{1}{3} \pi \) from both sides gives:
\[ R_1^3 = 4 R_2^3 \]

6. **Finding \( R_2 \) in terms of \( R_1 \)**:
Rearranging the equation gives:
\[ R_2^3 = \frac{R_1^3}{4} \implies R_2 = \left(\frac{R_1^3}{4}\right)^{\frac{1}{3}} = \frac{R_1}{\sqrt[3]{4}} = \frac{R_1}{2^{\frac{2}{3}}} \]

7. **Surface Area of the Larger Sphere**:
The surface area \( S_1 \) of the larger sphere is:
\[ S_1 = 4 \pi R_1^2 \]

8. **Surface Area of the Smaller Sphere**:
The surface area \( S_2 \) of the smaller sphere is:
\[ S_2 = 4 \pi R_2^2 = 4 \pi \left(\frac{R_1}{2^{\frac{2}{3}}}\right)^2 = 4 \pi \frac{R_1^2}{\left(2^{\frac{2}{3}}\right)^2} = 4 \pi \frac{R_1^2}{2^{\frac{4}{3}}} \]

9. **Finding the Ratio of Surface Areas**:
The ratio of the surface area of the smaller sphere to that of the larger sphere is:
\[ \frac{S_2}{S_1} = \frac{4 \pi \frac{R_1^2}{2^{\frac{4}{3}}}}{4 \pi R_1^2} = \frac{1}{2^{\frac{4}{3}}} \]

10. **Final Ratio**:
Thus, the ratio of the surface area of the smaller sphere to the surface area of the larger sphere is:
\[ \frac{S_2}{S_1} = \frac{1}{2^{\frac{4}{3}}} \]