A hollow metallic sphere of radius 20cm surrounds a concentric metallic sphere of radius 5cm. The space between the two sphere is filled with a nonmetallic material. The inner and outer sphere are maintained at `50^(@)C` and `10^(@)C` respectively and it is found that 100J of heat passes from the inner sphere to the outer sphere per second. Find the thermal conductivity of the material between the sphere.

Explanation

To solve the problem, we need to find the thermal conductivity of the non-metallic material between the two concentric spheres. We will use the formula derived from Fourier's law of heat conduction for spherical coordinates.

### Step-by-Step Solution

1. **Identify the Given Values:**
- Inner radius \( R_1 = 5 \, \text{cm} = 0.05 \, \text{m} \)
- Outer radius \( R_2 = 20 \, \text{cm} = 0.20 \, \text{m} \)
- Temperature of the inner sphere \( T_1 = 50 \, \text{°C} \)
- Temperature of the outer sphere \( T_2 = 10 \, \text{°C} \)
- Heat transfer rate \( \Delta Q = 100 \, \text{J/s} \)

2. **Use the Heat Transfer Formula for Spheres:**
The formula for the heat transfer rate through a spherical shell is given by:
\[ \Delta Q = \frac{4 \pi k R_1 R_2 (T_1 - T_2)}{R_2 - R_1} \]
where \( k \) is the thermal conductivity we need to find.

3. **Substitute the Known Values:**
Substitute the known values into the equation:
\[ 100 = \frac{4 \pi k (0.05)(0.20)(50 - 10)}{0.20 - 0.05} \]

4. **Simplify the Equation:**
Calculate the temperature difference:
\[ T_1 - T_2 = 50 - 10 = 40 \, \text{°C} \]
The denominator \( R_2 - R_1 \) becomes:
\[ 0.20 - 0.05 = 0.15 \, \text{m} \]
Thus, the equation simplifies to:
\[ 100 = \frac{4 \pi k (0.05)(0.20)(40)}{0.15} \]

5. **Rearranging the Equation:**
Rearranging to solve for \( k \):
\[ k = \frac{100 \times 0.15}{4 \pi (0.05)(0.20)(40)} \]

6. **Calculate \( k \):**
Now calculate the value:
\[ k = \frac{15}{4 \pi (0.05)(0.20)(40)} \]
First calculate the denominator:
\[ 4 \pi (0.05)(0.20)(40) = 4 \times 3.14 \times 0.05 \times 0.20 \times 40 \]
\[ = 4 \times 3.14 \times 0.01 \times 40 = 4 \times 3.14 \times 0.4 = 5.024 \]
Now substituting back:
\[ k = \frac{15}{5.024} \approx 2.98 \, \text{W/m°C} \]

7. **Final Result:**
Rounding off, we find:
\[ k \approx 3 \, \text{W/m°C} \]

### Final Answer:
The thermal conductivity of the non-metallic material is approximately \( 3 \, \text{W/m°C} \).