Two solid spheres are heated to the same temperature allowed to cool under identical conditions Compare (i) initial rates of fall of temperatre, and (ii) initial rates of loss of heat Assume that all the surfaces have the same emissivity and ratios of their radii specific heats and densities are respectively `1: alpha 1: beta, 1 : gamma` .

Options

A. `alpha beta gamma :1`
B. `1 :alpha^(2)`
C. `beta = alpha gamma`
D. `1 : alpha^(3)`

Explanation

To solve the problem, we will analyze the cooling of two solid spheres under identical conditions, focusing on the initial rates of fall of temperature and the initial rates of loss of heat.

### Step 1: Understanding the Cooling Process
Both spheres are initially at the same temperature and are cooling down. According to Newton's law of cooling, the rate of change of temperature of an object is proportional to the difference between its temperature and the surrounding temperature.

### Step 2: Initial Rate of Fall of Temperature
The initial rate of fall of temperature can be expressed as:
\[ \frac{dT}{dt} = -\frac{dQ}{ms} \]
where \(dQ\) is the heat lost, \(m\) is the mass of the sphere, and \(s\) is the specific heat capacity.

### Step 3: Heat Loss by Radiation
The rate of heat loss by radiation can be given by:
\[ \frac{dQ}{dt} = E \sigma A (T^4 - T_0^4) \]
where \(E\) is the emissivity, \(\sigma\) is the Stefan-Boltzmann constant, \(A\) is the surface area, \(T\) is the temperature of the sphere, and \(T_0\) is the surrounding temperature.

### Step 4: Surface Area and Mass
For spheres A and B:
- Let the radius of sphere A be \(r_A\) and that of sphere B be \(r_B\).
- The surface area \(A\) of a sphere is given by \(A = 4\pi r^2\).
- The mass \(m\) of a sphere can be expressed in terms of its density \(\rho\) and volume \(V\):
\[ m = \rho V = \rho \left(\frac{4}{3} \pi r^3\right) \]

### Step 5: Comparing the Initial Rates of Fall of Temperature
For sphere A:
\[ \frac{dQ_A}{dt} = E \sigma (4\pi r_A^2) (T^4 - T_0^4) \]
For sphere B:
\[ \frac{dQ_B}{dt} = E \sigma (4\pi r_B^2) (T^4 - T_0^4) \]

Dividing the two:
\[ \frac{\frac{dT_A}{dt}}{\frac{dT_B}{dt}} = \frac{m_B s_B}{m_A s_A} \cdot \frac{A_B}{A_A} \]
Substituting the expressions for mass and area:
\[ \frac{\frac{dT_A}{dt}}{\frac{dT_B}{dt}} = \frac{\rho_B \left(\frac{4}{3} \pi r_B^3\right) s_B}{\rho_A \left(\frac{4}{3} \pi r_A^3\right) s_A} \cdot \frac{4\pi r_B^2}{4\pi r_A^2} \]
This simplifies to:
\[ \frac{\frac{dT_A}{dt}}{\frac{dT_B}{dt}} = \frac{\rho_B r_B^2 s_B}{\rho_A r_A^2 s_A} \]
Given the ratios of densities and specific heats as \(1 : \alpha\) and \(1 : \beta\), and the radii as \(1 : \gamma\), we can express:
\[ \frac{dT_A}{dt} : \frac{dT_B}{dt} = \frac{\beta \gamma^2}{\alpha} \]

### Step 6: Initial Rates of Loss of Heat
Using the same heat loss equations:
\[ \frac{dQ_A}{dt} = E \sigma (4\pi r_A^2)(T^4 - T_0^4) \]
\[ \frac{dQ_B}{dt} = E \sigma (4\pi r_B^2)(T^4 - T_0^4) \]
Thus,
\[ \frac{dQ_A/dt}{dQ_B/dt} = \frac{A_A}{A_B} = \frac{r_A^2}{r_B^2} = \frac{1}{\alpha^2} \]

### Final Results
1. **Initial rates of fall of temperature**:
\[ \frac{dT_A}{dt} : \frac{dT_B}{dt} = \frac{\beta \gamma^2}{\alpha} \]
2. **Initial rates of loss of heat**:
\[ \frac{dQ_A/dt}{dQ_B/dt} = \frac{1}{\alpha^2} \]