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The problem is about two spherical shells made of different materials, A and B, with the following conditions:

- The first spherical shell has a radius of 1 cm, is made of material A, and is filled with air. 
- The second spherical shell has a radius \( x \) cm, is made of material B, and its interior is evacuated (no air).
- The ratio of the mass per unit area of material A to material B is \( 5:4 \).
- The ratio of the mass of air enclosed in shell A to the mass of shell A itself is \( 1:25 \).
- Both shells have the same total mass (including air in A and the mass of B).

### To solve for \( x \):

Let’s break the problem down step by step:

#### 1. Mass of shell A (with air):
- The surface area of a spherical shell with radius \( r \) is given by \( 4\pi r^2 \).
- The total mass of shell A consists of the mass of the material plus the mass of air inside it.

Let \( m_A \) be the mass of material A per unit area. The mass of the spherical shell made of material A with radius \( 1 \, \text{cm} \) is:
\[
\text{Mass of shell A (material)} = m_A \cdot 4\pi \cdot (1)^2 = 4\pi m_A
\]
Let the mass of air inside the sphere be \( m_{\text{air}} \). According to the problem, the ratio of the mass of air to the mass of the material A is \( \frac{m_{\text{air}}}{4\pi m_A} = \frac{1}{25} \), so:
\[
m_{\text{air}} = \frac{4\pi m_A}{25}
\]
Thus, the total mass of shell A (material + air) is:
\[
\text{Total mass of A} = 4\pi m_A + \frac{4\pi m_A}{25} = 4\pi m_A \left(1 + \frac{1}{25}\right) = 4\pi m_A \cdot \frac{26}{25}
\]

#### 2. Mass of shell B:
The radius of shell B is \( x \) cm, and its interior is evacuated. The mass of shell B is given by the ratio of the mass per unit area of materials A and B. Since the ratio of \( m_A \) to \( m_B \) is \( 5:4 \), we have \( m_B = \frac{4}{5} m_A \).

The surface area of shell B is \( 4\pi x^2 \), so the mass of shell B is:
\[
\text{Mass of shell B} = m_B \cdot 4\pi x^2 = \frac{4}{5} m_A \cdot 4\pi x^2 = \frac{16\pi m_A x^2}{5}
\]

#### 3. Equating the total masses:
The total mass of shell A (including air) and the mass of shell B must be equal, so:
\[
4\pi m_A \cdot \frac{26}{25} = \frac{16\pi m_A x^2}{5}
\]
Simplify by dividing both sides by \( 4\pi m_A \):
\[
\frac{26}{25} = \frac{16x^2}{5}
\]
Multiply both sides by 25:
\[
26 = 80x^2
\]
Solve for \( x^2 \):
\[
x^2 = \frac{26}{80} = \frac{13}{40}
\]
Finally, solve for \( x \):
\[
x = \sqrt{\frac{13}{40}} = \frac{\sqrt{13}}{\sqrt{40}} = \frac{\sqrt{13}}{2\sqrt{10}} = \frac{\sqrt{130}}{20}
\]

So the value of \( x \) is \( \frac{\sqrt{130}}{20} \).

Would you like further clarification on any step, or do you have additional questions?

Here are five related questions:
1. How is the surface area of a sphere derived?
2. What is the formula for the mass of a spherical shell based on its surface area and density?
3. How would the solution change if the ratio of mass per unit area was \( 3:2 \) instead of \( 5:4 \)?
4. What would happen if the second shell (B) was also filled with air?
5. How does the ratio of masses affect the final radius in this type of problem?

**Tip:** When solving ratio problems involving areas or volumes, remember that areas scale with the square of the radius, while volumes scale with the cube of the radius.

A spherical shell of radius 1 cm is formed from a material A and filled with air. A similar shell of radius x is formed from material B with the interior evacuated. The ratio of the mass per unit area of materials A and B is 5:4. The ratio of the mass of air enclosed in A to its shell is 1:25. What is the value of x, given that the spheres have the same total mass?

This solution explains how to solve a spherical shell problem involving the ratio of materials and mass per unit area. The spheres have the same total mass, and the radius of the second shell is determined through surface area formulas and algebraic manipulation.

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