A point source of light of power `P_(0)` is placed at a distance of 4 m from the centre of a thin hemispherical shell as shown in the figure, The shell has a radius of 3 m and it behaves like a perfect black body. If the temperature of the hemisphere is related to the power of the sourcec as `T^(4)=(p_(0))/(n pi sigma)` where `sigma` is the stefan's constant, then find the value of `(n)/(10)`.

Answer

To solve the problem, we need to find the value of \( \frac{n}{10} \) based on the relationship between the temperature \( T \) of the hemispherical shell and the power \( P_0 \) of the light source.

### Step-by-Step Solution:

1. **Understanding the Geometry**:
- We have a point source of light with power \( P_0 \) located 4 m from the center of a thin hemispherical shell of radius 3 m.
- The distance from the source to the edge of the hemisphere can be calculated using the Pythagorean theorem. The distance from the source to the center is 4 m, and the radius of the hemisphere is 3 m.

2. **Finding the Angle**:
- Let \( \theta \) be the angle subtended by the radius of the hemisphere at the source. By the Pythagorean theorem:
\[ \text{Distance from the source to the edge of the hemisphere} = \sqrt{(4^2) + (3^2)} = \sqrt{16 + 9} = 5 \text{ m} \]
- Therefore, we can find \( \cos \theta \):
\[ \cos \theta = \frac{3}{5} \]

3. **Calculating the Solid Angle**:
- The solid angle \( \Omega \) subtended by the hemisphere at the source can be calculated using:
\[ \Omega = \int_0^{2\pi} \int_0^{\theta} \sin \theta \, d\theta \, d\phi \]
- This simplifies to:
\[ \Omega = 2\pi \int_0^{\theta} \sin \theta \, d\theta = 2\pi \left[-\cos \theta\right]_0^{\theta} = 2\pi (1 - \cos \theta) \]
- Substituting \( \cos \theta = \frac{4}{5} \):
\[ \Omega = 2\pi \left(1 - \frac{4}{5}\right) = 2\pi \left(\frac{1}{5}\right) = \frac{2\pi}{5} \]

4. **Energy Radiated**:
- The energy radiated by the source into the solid angle is given by:
\[ \frac{P_0 \cdot \Omega}{4\pi} = \sigma T^4 \cdot A \]
- The area \( A \) of the hemisphere is:
\[ A = \pi r^2 = \pi (3^2) = 9\pi \]
- Therefore, we have:
\[ \frac{P_0 \cdot \frac{2\pi}{5}}{4\pi} = \sigma T^4 \cdot 9\pi \]
- Simplifying gives:
\[ \frac{P_0 \cdot 2}{20} = 9\sigma T^4 \]

5. **Relating Temperature to Power**:
- Rearranging gives:
\[ P_0 = 90 \sigma T^4 \]
- According to the problem, we have:
\[ T^4 = \frac{P_0}{n \pi \sigma} \]
- Equating the two expressions for \( P_0 \):
\[ 90 \sigma T^4 = n \pi \sigma T^4 \]
- Dividing both sides by \( \sigma T^4 \) (assuming \( T \neq 0 \)):
\[ n = 90 \]

6. **Finding \( \frac{n}{10} \)**:
- Finally, we calculate:
\[ \frac{n}{10} = \frac{90}{10} = 9 \]

### Final Answer:
\[ \frac{n}{10} = 9 \]