A spherical solid blakc body of radius 'r' radiates power 'H' and its rate of cooling is 'C'. If density is constant then which of the following is/are true.

Options

A. `H prop r` and `c prop r^(2)`
B. `H prop r^(2)` and `c prop (1)/(r )`
C. `H prop r` and `c prop (1)/(r^(2))`
D. `H prop r^(2)` and `c prop r^(2)`

Answer

To solve the problem, we need to analyze the relationships between the power radiated by a spherical solid black body, its radius, and the rate of cooling. Let's break it down step by step.

### Step 1: Understand the Power Radiated
The power radiated by a black body can be expressed using the Stefan-Boltzmann law:
\[ H = \sigma e A T^4 \]
where:
- \( H \) is the power radiated,
- \( \sigma \) is the Stefan-Boltzmann constant,
- \( e \) is the emissivity (which is 1 for a black body),
- \( A \) is the surface area,
- \( T \) is the absolute temperature.

### Step 2: Calculate the Surface Area of the Sphere
For a sphere, the surface area \( A \) is given by:
\[ A = 4\pi r^2 \]
where \( r \) is the radius of the sphere.

### Step 3: Substitute the Surface Area into the Power Equation
Substituting the surface area into the power equation, we get:
\[ H = \sigma (1) (4\pi r^2) T^4 \]
This simplifies to:
\[ H = 4\pi \sigma r^2 T^4 \]
From this equation, we can see that the power \( H \) is proportional to the square of the radius \( r \):
\[ H \propto r^2 \]

### Step 4: Relate Power to Rate of Cooling
The rate of cooling \( C \) can be expressed as:
\[ C = \frac{dT}{dt} \]
Using the relationship between power and cooling, we can express power in terms of mass and specific heat:
\[ H = m s \frac{dT}{dt} \]
where:
- \( m \) is the mass,
- \( s \) is the specific heat capacity.

### Step 5: Express Mass in Terms of Density and Volume
The mass \( m \) can be expressed as:
\[ m = \rho V \]
where \( \rho \) is the density and \( V \) is the volume of the sphere. The volume of a sphere is given by:
\[ V = \frac{4}{3} \pi r^3 \]
Thus,
\[ m = \rho \left(\frac{4}{3} \pi r^3\right) \]

### Step 6: Substitute Mass into the Cooling Rate Equation
Substituting for mass in the cooling rate equation gives:
\[ H = \rho \left(\frac{4}{3} \pi r^3\right) s \frac{dT}{dt} \]
Rearranging this, we find:
\[ \frac{dT}{dt} = \frac{H}{\rho \left(\frac{4}{3} \pi r^3\right) s} \]

### Step 7: Analyze the Rate of Cooling
From the above equation, we can see that the rate of cooling \( C \) is inversely proportional to the cube of the radius \( r \):
\[ C \propto \frac{1}{r^3} \]

### Conclusion
From our analysis, we have established two key relationships:
1. The power \( H \) is proportional to \( r^2 \).
2. The rate of cooling \( C \) is inversely proportional to \( r^3 \).

### Final Answer
- **True Statements**:
- \( H \) is proportional to \( r^2 \).
- \( C \) is inversely proportional to \( r^3 \).