The ratio of masses of two metal spheres A and B is 8 : I. If their temperatures are 2000 K and 1000 K respectively, then the ratio of the rates of their energy emission will be

Options

A. `64:1`
B. `128:1`
C. `16:1`
D. `4:1`

Answer

To solve the problem, we need to find the ratio of the rates of energy emission (power) of two metal spheres A and B, given their mass ratio and temperatures.

### Step-by-Step Solution:

1. **Identify the Given Information**:
- Mass ratio of spheres A and B: \( m_A : m_B = 8 : 1 \)
- Temperatures: \( T_A = 2000 \, K \) and \( T_B = 1000 \, K \)

2. **Express Mass in Terms of Volume and Density**:
- For sphere A:
\[ m_A = \rho \cdot \frac{4}{3} \pi r_A^3 \]
- For sphere B:
\[ m_B = \rho \cdot \frac{4}{3} \pi r_B^3 \]
- Here, \( \rho \) is the density, which is the same for both spheres since they are made of the same metal.

3. **Set Up the Mass Ratio Equation**:
- From the mass ratio, we have:
\[ \frac{m_A}{m_B} = \frac{\rho \cdot \frac{4}{3} \pi r_A^3}{\rho \cdot \frac{4}{3} \pi r_B^3} = \frac{r_A^3}{r_B^3} \]
- Given \( \frac{m_A}{m_B} = 8 \), we can write:
\[ \frac{r_A^3}{r_B^3} = 8 \]

4. **Find the Ratio of Radii**:
- Taking the cube root:
\[ \frac{r_A}{r_B} = 2 \]

5. **Calculate the Ratio of Power Emission**:
- The power emitted by a black body is given by the Stefan-Boltzmann law:
\[ P = \sigma \cdot A \cdot T^4 \]
- Where \( A \) is the surface area, \( \sigma \) is the Stefan-Boltzmann constant, and \( T \) is the temperature.
- The surface area \( A \) of a sphere is given by \( A = 4 \pi r^2 \).

6. **Set Up the Power Ratio**:
- The power ratio of spheres A and B is:
\[ \frac{P_A}{P_B} = \frac{\sigma \cdot A_A \cdot T_A^4}{\sigma \cdot A_B \cdot T_B^4} = \frac{A_A \cdot T_A^4}{A_B \cdot T_B^4} \]
- Substituting the surface areas:
\[ \frac{P_A}{P_B} = \frac{4 \pi r_A^2 \cdot T_A^4}{4 \pi r_B^2 \cdot T_B^4} = \frac{r_A^2 \cdot T_A^4}{r_B^2 \cdot T_B^4} \]

7. **Substituting the Values**:
- We already found \( \frac{r_A}{r_B} = 2 \), so:
\[ \frac{r_A^2}{r_B^2} = 2^2 = 4 \]
- Now substituting the temperatures:
\[ \frac{P_A}{P_B} = 4 \cdot \frac{T_A^4}{T_B^4} = 4 \cdot \frac{(2000)^4}{(1000)^4} \]
- Simplifying the temperature ratio:
\[ \frac{(2000)^4}{(1000)^4} = \left(\frac{2000}{1000}\right)^4 = 2^4 = 16 \]
- Therefore:
\[ \frac{P_A}{P_B} = 4 \cdot 16 = 64 \]

8. **Final Ratio of Power Emission**:
- The ratio of the rates of energy emission (power) is:
\[ \frac{P_A}{P_B} = 64 : 1 \]

### Final Answer:
The ratio of the rates of energy emission of spheres A and B is \( 64 : 1 \).