There are two thin spheres A and B of the same material and same thickness. They behave like black bodies, Radius of A is double that of B and Both have same temperature T. When A and B are kept in a room of temperature `T_0 (lt T)`, the ratio of their rates of cooling is (assume negligible heat exchange between A and B).

Options

A. `2 : 1`
B. `1 : 1`
C. `4 : 1`
D. `8 : 1`

Answer

To solve the problem, we need to determine the ratio of the rates of cooling of two thin spheres A and B, given that sphere A has a radius that is double that of sphere B, and both spheres are made of the same material and have the same temperature.

### Step-by-Step Solution:

1. **Understanding the Heat Loss Mechanism**:
The rate of heat loss (or cooling) of a body can be described using the Stefan-Boltzmann law, which states:
\[ \frac{dQ}{dt} = \epsilon \sigma A (T^4 - T_0^4) \]
where:
- \( \epsilon \) is the emissivity of the material (which is the same for both spheres),
- \( \sigma \) is the Stefan-Boltzmann constant,
- \( A \) is the surface area of the sphere,
- \( T \) is the temperature of the sphere,
- \( T_0 \) is the surrounding temperature.

2. **Calculating the Surface Area**:
The surface area \( A \) of a sphere is given by:
\[ A = 4\pi r^2 \]
For sphere A (with radius \( R \)):
\[ A_A = 4\pi R^2 \]
For sphere B (with radius \( r \)):
\[ A_B = 4\pi r^2 \]
Given that the radius of A is double that of B, we have:
\[ R = 2r \]
Thus, the surface area of A becomes:
\[ A_A = 4\pi (2r)^2 = 16\pi r^2 \]

3. **Calculating the Rate of Cooling for Each Sphere**:
Now, we can express the rate of cooling for both spheres:
For sphere A:
\[ \frac{dQ_A}{dt} = \epsilon \sigma (16\pi r^2) (T^4 - T_0^4) \]
For sphere B:
\[ \frac{dQ_B}{dt} = \epsilon \sigma (4\pi r^2) (T^4 - T_0^4) \]

4. **Finding the Ratio of Rates of Cooling**:
Now, we can find the ratio of the rates of cooling:
\[ \frac{\frac{dQ_A}{dt}}{\frac{dQ_B}{dt}} = \frac{\epsilon \sigma (16\pi r^2) (T^4 - T_0^4)}{\epsilon \sigma (4\pi r^2) (T^4 - T_0^4)} \]
The common terms (\(\epsilon\), \(\sigma\), \(\pi\), \(r^2\), and \((T^4 - T_0^4)\)) cancel out:
\[ \frac{dQ_A/dt}{dQ_B/dt} = \frac{16}{4} = 4 \]

5. **Conclusion**:
Therefore, the ratio of the rates of cooling of spheres A and B is:
\[ \frac{dQ_A/dt}{dQ_B/dt} = 4 \]

### Final Answer:
The ratio of their rates of cooling is \( 4:1 \).