A system consists of two concentric spherical shells of radii a and `b (b>a)` maintained at tempareture `T_1` and `T_2` respectively.The radial rate of flow of heat through subtance between the two concentric spherical shells is proportional to

Options

A. b - a
B. `(ab)/(b - a)`
C. `log_e(b/a)`
D. `(b - a)/(ab)`

Answer

To find the radial rate of flow of heat through the substance between two concentric spherical shells of radii \( a \) and \( b \) (where \( b > a \)) maintained at temperatures \( T_1 \) and \( T_2 \) respectively, we can follow these steps:

### Step-by-Step Solution:

1. **Understand the Concept**:
The heat transfer through the spherical shells can be analyzed using Fourier's law of heat conduction. The law states that the rate of heat transfer \( Q \) is proportional to the negative gradient of temperature and the area through which heat is being transferred.

2. **Set Up the Equation**:
According to Fourier's law:
\[ Q = -k \cdot A \cdot \frac{dT}{dr} \]
where \( k \) is the thermal conductivity, \( A \) is the surface area, and \( \frac{dT}{dr} \) is the temperature gradient.

3. **Determine the Area**:
For a spherical shell at a distance \( r \) from the center, the area \( A \) is given by:
\[ A = 4\pi r^2 \]

4. **Substitute the Area into the Equation**:
Substitute \( A \) into the heat transfer equation:
\[ Q = -k \cdot 4\pi r^2 \cdot \frac{dT}{dr} \]

5. **Rearranging the Equation**:
Rearranging gives:
\[ \frac{dT}{dr} = -\frac{Q}{4\pi k r^2} \]

6. **Integrate the Equation**:
We need to integrate this equation from \( r = a \) to \( r = b \) and from \( T = T_1 \) to \( T = T_2 \):
\[ \int_{T_1}^{T_2} dT = -\frac{Q}{4\pi k} \int_{a}^{b} \frac{1}{r^2} dr \]

7. **Calculate the Integral**:
The integral \( \int \frac{1}{r^2} dr = -\frac{1}{r} \). Evaluating this from \( a \) to \( b \):
\[ -\left[-\frac{1}{b} + \frac{1}{a}\right] = \frac{1}{a} - \frac{1}{b} = \frac{b-a}{ab} \]

8. **Substituting Back**:
Substituting this back into the equation gives:
\[ T_2 - T_1 = -\frac{Q}{4\pi k} \cdot \frac{b-a}{ab} \]
Rearranging gives:
\[ Q = \frac{4\pi k (T_1 - T_2) ab}{b - a} \]

9. **Conclusion**:
From this expression, we can conclude that the radial rate of flow of heat \( Q \) is directly proportional to \( \frac{ab}{b-a} \).

### Final Answer:
The radial rate of flow of heat through the substance between the two concentric spherical shells is proportional to \( \frac{ab}{b-a} \).