The total surface area of a solid iron cube and a solid aluminium cuboid are the same. The length, breadth and height of the cuboid are in the ratio `1 : 2 : 4`. Both are melted together in a vessel. What is the ratio of iron and aluminium in the resultant mixture?

Options

A. `(14//3)^(3//2) : 8`
B. `8 : (14//3)^(3//2)`
C. `(3//14)^(3//2):8`
D. `8:(3//14)^(3//2)`

Answer

To solve the problem, we need to find the ratio of the volumes of the solid iron cube and the solid aluminum cuboid, given that their total surface areas are equal and the dimensions of the cuboid are in the ratio 1:2:4.

### Step-by-Step Solution:

1. **Define Variables:**
- Let the side length of the iron cube be \( A \).
- Let the dimensions of the aluminum cuboid be \( L \) (length), \( 2L \) (breadth), and \( 4L \) (height) based on the given ratio.

2. **Calculate the Total Surface Area:**
- The total surface area of the iron cube is given by:
\[ \text{Surface Area of Cube} = 6A^2 \]
- The total surface area of the aluminum cuboid is given by:
\[ \text{Surface Area of Cuboid} = 2(L \cdot 2L + L \cdot 4L + 2L \cdot 4L) = 2(2L^2 + 4L^2 + 8L^2) = 2(14L^2) = 28L^2 \]

3. **Set the Surface Areas Equal:**
- Since the total surface areas are the same:
\[ 6A^2 = 28L^2 \]

4. **Solve for \( A^2 \):**
- Rearranging gives:
\[ A^2 = \frac{28L^2}{6} = \frac{14L^2}{3} \]

5. **Find \( A \):**
- Taking the square root:
\[ A = \sqrt{\frac{14}{3}}L \]

6. **Calculate the Volumes:**
- The volume of the iron cube is:
\[ \text{Volume of Cube} = A^3 = \left(\sqrt{\frac{14}{3}}L\right)^3 = \frac{14\sqrt{14}}{3\sqrt{3}}L^3 \]
- The volume of the aluminum cuboid is:
\[ \text{Volume of Cuboid} = L \cdot 2L \cdot 4L = 8L^3 \]

7. **Find the Ratio of Volumes:**
- The ratio of the volume of the iron cube to the volume of the aluminum cuboid is:
\[ \text{Ratio} = \frac{\text{Volume of Cube}}{\text{Volume of Cuboid}} = \frac{\frac{14\sqrt{14}}{3\sqrt{3}}L^3}{8L^3} \]
- Cancelling \( L^3 \):
\[ \text{Ratio} = \frac{14\sqrt{14}}{3\sqrt{3} \cdot 8} = \frac{14\sqrt{14}}{24\sqrt{3}} = \frac{7\sqrt{14}}{12\sqrt{3}} \]

8. **Simplify the Ratio:**
- To express the ratio in simplest form, we can rationalize the denominator:
\[ \text{Ratio} = \frac{7\sqrt{14} \cdot \sqrt{3}}{12 \cdot 3} = \frac{7\sqrt{42}}{36} \]

### Final Answer:
The ratio of iron to aluminum in the resultant mixture is \( \frac{7\sqrt{42}}{36} \).