Two balls of same material and finish have their diameters in the ratio 2:1. Both are heated to the same temperature and allowed to cool by radiation. Rate of cooling of big ball as compared to smaller one will be in the ratio:

Options

A. `1 : 1`
B. `1 : 2`
C. `2 : 1`
D. `4 : 1`

Explanation

To solve the problem of comparing the rate of cooling of two balls of the same material with diameters in the ratio of 2:1, we can follow these steps:

### Step-by-Step Solution:

1. **Understand the Problem**: We have two balls made of the same material, and their diameters are in the ratio of 2:1. We need to find the ratio of their rates of cooling when both are heated to the same temperature and allowed to cool by radiation.

2. **Identify the Relevant Law**: The rate of cooling of an object by radiation is described by the Stefan-Boltzmann law, which states that the power radiated (rate of heat loss) is proportional to the surface area of the object and the fourth power of its absolute temperature. The formula can be expressed as:
\[ \frac{dQ}{dt} \propto A \cdot \sigma \cdot T^4 \]
where \( A \) is the surface area, \( \sigma \) is the Stefan-Boltzmann constant, and \( T \) is the temperature.

3. **Calculate the Surface Area**: The surface area \( A \) of a sphere is given by:
\[ A = 4\pi r^2 \]
Since the diameter \( d \) is twice the radius \( r \) (i.e., \( d = 2r \)), we can express the surface area in terms of the diameter:
\[ A \propto d^2 \]

4. **Relate the Diameters**: Let the diameter of the larger ball be \( d_1 = 2d \) and the diameter of the smaller ball be \( d_2 = d \). The surface areas will then be:
\[ A_1 \propto (2d)^2 = 4d^2 \quad \text{and} \quad A_2 \propto d^2 \]

5. **Determine the Mass**: The mass \( m \) of a sphere is given by:
\[ m = \rho V \]
where \( V \) is the volume. For a sphere, \( V = \frac{4}{3}\pi r^3 \), so:
\[ V \propto d^3 \]
Thus, the mass is:
\[ m \propto d^3 \]

6. **Express the Rate of Cooling**: The rate of cooling can be expressed as:
\[ \frac{dT}{dt} \propto \frac{dQ/dt}{m} \]
Substituting the expressions for \( dQ/dt \) and \( m \):
\[ \frac{dT}{dt} \propto \frac{A \cdot \sigma \cdot T^4}{\rho \cdot V} \]
This simplifies to:
\[ \frac{dT}{dt} \propto \frac{A}{\rho \cdot d^3} \]

7. **Find the Ratio of Rates of Cooling**: Now, we can find the ratio of the rates of cooling for the two balls:
\[ \frac{dT_1/dt}{dT_2/dt} \propto \frac{A_1}{A_2} \cdot \frac{d^3_2}{d^3_1} \]
Substituting the areas and diameters:
\[ \frac{dT_1/dt}{dT_2/dt} \propto \frac{4d^2}{d^2} \cdot \frac{d^3}{(2d)^3} = \frac{4}{1} \cdot \frac{1}{8} = \frac{4}{8} = \frac{1}{2} \]

8. **Final Ratio**: Therefore, the ratio of the rate of cooling of the big ball to the small ball is:
\[ \frac{R_1}{R_2} = \frac{1}{2} \]

### Conclusion:
The final answer is that the rate of cooling of the big ball as compared to the smaller one will be in the ratio of **1:2**.